Mathematical mysteries by Lubina

J. A.

L U B I N A

M A T H E M A T I C A L M Y S T E R I E S 2 65 97

3 35 67

131 161 163 193 227 257 259 289 323 353 355 385 419 449 451 481 515 545 547 577 611 641 643 673 707 737 739 769 803 833 835 865 899 929 931 961 995

5 37

11 43

71 101 103 107 133 137 139 167 169 197 199 203 229 233 235 263 265 293 295 299 325 329 331 359 361 389 391 395 421 425 427 455 457 485 487 491 517 521 523 551 553 581 583 587 613 617 619 647 649 677 679 683 709 713 715 743 745 773 775 779 805 809 811 839 841 869 871 875 901 905 907 935 937 965 967 971 997 1001 1003

77 109

41 73

173 205 269 301 365 397 461 493 557 589 653 685 749 781 845 877 941 973

47 79 143 175 239 271 335 367 431 463 527 559 623 655 719 751 815 847 911 943 1007

17 49

23 55

83 113 115 119 145 149 151 179 181 209 211 215 241 245 247 275 277 305 307 311 337 341 343 371 373 401 403 407 433 437 439 467 469 497 499 503 529 533 535 563 565 593 595 599 625 629 631 659 661 689 691 695 721 725 727 755 757 785 787 791 817 821 823 851 853 881 883 887 913 917 919 947 949 977 979 983 1009 1013 1015

89 121

53 85

185 217 281 313 377 409 473 505 569 601 665 697 761 793 857 889 953 985

59 91 155 187 251 283 347 379 443 475 539 571 635 667 731 763 827 859 923 955

29 61 125 157 221 253 317 349 413 445 509 541 605 637 701 733 797 829 893 925

989 1019 1021

31 95 127 191 223 287 319 383 415 479 511 575 607 671 703 767 799 863 895 959 991

2 p ± 1 = 2n/4 p(p') ± 1 = 2n/3 289 - 1 = 288/3 317 - 1 = 316/4 319 287

317 253 221

289 257

223 191

157 125

127 95

193 161 97

259 227 163 131

293 229 197

133 295 263 67 101 199 187 35 31 61 167 155 2 3 37 29 103 91 71 5 59 7 313 281 217 185 121 89 25 41 73 137 169 233 265 11 23 43 55 13 107 119 139 151 53 19 17 77 203 47 215 85 235 49 247 109 79 299 83 311 149 115 173 181 113 205 143 145 245 175 179 269 277 211 301 209 239 241 271 275 307 305 283 251

SEARCHING TO SOLVE THE GREATEST MYSTERY IN MATHEMATICS. ―There are some mysteries that the human mind will never penetrate. To convince ourselves we have only to cast a glance at tables of primes and we should perceive that there reigns neither order nor rule.‖ Leonhard Euler „The scientist dose not study Nature because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful. If Nature were not beautiful, it would not be worth knowing, and if Nature were not worth knowing, life would not be worth living.‖ Jules Henri Poincaré. "Sed omnia in mensura, et numero, et pondere diposuisti.‖

Sapientia 11, 21.

―Mathematicians do not study objects, but relations between objects. Thus, they are free to replace some objects by others so long as the relations remain unchanged. Content to them is irrelevant: they are interested in form only‖. Jules Henri Poincaré. For the memory of Katharina Lubina June 18, 1983

Preface World and all what oneself it on him finds it carries mathematical structures. So God created him with mathematical point of sight. Given the man the strength to him from God of reason, plan of his building can the discoveries. It is the mathematics so the key to understanding of world. In peer with her development, she went the change of aims what her was placed. It does not serve the mathematics the endeavor to better perception only and the understanding of nature, but it has to permit her to master. Both points of sight, chief place of mathematics, strength of granted her certainty and the incontestability, which are useful different disciplines scientific character, as also change of this, what oneself it under this notion understands and what it the thanks were wanted was to reach her, they gave the beginning my trials they would solve problem of primes. So began my adventure with primes. Human spirit and human culture they unrolled such formal system of thinking, to can formulae recognize, to classify and to use. We call him mathematics, because it is mathematician the science of formulae. The only right of existence for mathematician, the desire of discovery of new formulae is and the inherent in rights of nature regularities, as and announcing this, what it will happen. Though looking for formulae and structures it is mathematician's activity, then proper his task is formulation there in irrefutable proofs. Numbers are the simplest mathematical object, and the simplest formulae of nature are numerical, because perfect relations between numbers reign. The primes are used as building blocks of natural numbers in every textbook of number theory, more or less dealt with in detail, and to get along but with relatively few sentence. My work "Mathematical mysteries" presents easy-to number theory, which is based on prime numbers 2 and 3. They are also the answer to until now unreadable classical problems such as Proof of Goldbach's and Fermat conjecture irregularities in the distribution of primes, twin primes, Riemann's Hypothesis and distribution of prime numbers, etc. The "Mathematical mysteries" brings the problems into a form that requires little previous knowledge, first and second, the essential ideas can emerge. It will be important in choice areas of mathematics, their basic idea as in carefully edited, accessible and overviews presented in moderate volume. These depictions not only at professional mathematicians, but also to those who value the work of the mathematization in itself a source of pure joy. It should be emphasized that this work will not be more than an introduction to the fascinating field of prime numbers, which dominate the whole world of numbers. Some problems have her remain. With a mischievous smile, she says: "If you replaced me, will you have nothing to do." Not knowing that mathematicians invent more problems than solve themselves. I wish you, young student, teacher or mathematician, engineer, and all of you that you are the delight of the numbers that you may be tempted to think about the wonderful "Mathematical mysteries" with all its deep-rooted secrets, and not be bored. I invite you to small excursions into the mysteries of the world figure

4 The one - the number of origin and the original ground The basic theorem of algebraically theory of numbers sounds: All numbers descend from one. "O M N I A E X U N O‖ Theorem this be leaning on system of certainties, what Italian mathematician Giuseppe Peano in 1889 r. submitted on unquestionable truth the undemanding proof "parental power‖ of number one, giving the same bearing foundation theorem taking out from one all natural numbers. The forcible model of principle "all of one‖/ OMNIA EX UNO / is the draught of natural numbers 1, 2, 3, 4,.. in which number one, it is for all numbers the "point of exit‖. One is really only corner stone the whole draught of numbers on which is bases here. In gathering of natural numbers the number one is the class alone for me, the "Unity‖ is called also from here. One is only number, which does not change when oneself it divide her by her, or it increases. Geometric he be introduced as point, by what his elusiveness be expressed. Point's the lack of length, width and height, upper or bottom side, any color, and even the position. It was cannot say even, points are round, because taking at all closely they do not widen. This borders on with miracle directly, that attributes number these essential and necessary features without which the whole draught of natural numbers would not can exist. Then she is the "Point‖ of reference, what to which all natural numbers graphic be co-ordinated, introduced on two co-ordinates a, and b. She is the "source‖ even and odd units also from which it comes into being whole row of prime and folded numbers. It exist such "Unity‖, from which the whole wealth of world results, as one axiom will suffice as foundation the fine edifice of arithmetic. "It exist such number 1 possessing property, which treats to every number - n‖: n · 1= n = n + 0 1 · p = p Really comparative size with 2 enters in life, in support about which , all different measurable can pit .She beyond this is with nature the number of "unification‖ from two unit's make one number. 2 1 1= 3 2 1 2 1 2= + 3 3 p 1 p= a + b b= p- a a= 3 p=

p 1 p 1 (p  ) 3 3

p= a + (p - a)

Prime numbers this "building blocks‖, from which be built all different natural numbers. Not we will find them however in multiplication table, because number first cannot be the result "sensible‖ operation of multiplication, but only addition. Every prime numbers is the sum two components defining her place in draught of natural numbers p = a + b. p 1 Component a = then they came into being with divisible numbers even quotient by 3. 3 Component b = p - a then difference among prime number, and even quotient. It number 2 is only even prime number and across her principle "larger about one‖ it will become transferred on next natural numbers, guaranteeing contact and progress in draught. 2 1 1= 3

5 2 =

1 1 2

3 = 1

1 4 = 2

2 1

5 = 2

1 6 = 3

7 = 2

+ (3 - 1)

5 1 3

+ (5 -

5 1 ) 3

7 1 3

+ (7 -

7 1 ) 3

1 8 = 3 + 5 All prime numbers precede or they follow after divisible even number by 4 e.g. 2, 3, 4 5, 7, 8 11, 12 13, 16 17, 19, 20 23, 24, for except 2. Eureka! ―I‘ve found it!‖ p ± 1 = 2n/4 3 + 1 = 4 5 – 1 = 4 7 + 1 = 8/4 11 + 1 = 12/4 13 – 1 = 12/4 p = 2n/4  1 e.g. 1999 = 2000 – 1 p = n + (n ± 1) 1999 = 2000/2 + (1000 – 1) 5 13 17 29 37 41 53 61 73 89 97 101 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100 3 7 11 19 23 31 43 47 59 67 71 79 83

p = n + (n + 1) 3 = 1 + (1 + 1) 5 = 2 + (2 + 1) 7 = 3 + (3 + 1) 11 = 5 + (5 + 1) 100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Serie3 3

7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Serie2 2

9 10 12 15 16 19 21 22 24 27 30 31 34 36 37 40 42 45 49

Serie1 1

9 11 14 15 18 20 21 23 26 29 30 33 35 36 39 41 44 48

6 The fact that a prime number is a natural number greater than one, the only one and itself is divided can be described as characteristic of the prime. It is the property of which defines the prime. In addition, the prime has a lot of other properties, such as those that break down on sum p = a + (2 ± 1), those odd summand is twice greater of plus minus 1 as the even summand a = 2n / 3 equal 3 quotient of the even number. Some of these properties are trivial, but have an impact on numbers that are composed of these primes. Odd numbers, as this results from every multiplication table, are product of prime numbers, and almost prime. n =b n = 3 b = (2 b + b) 3 9 = 2 (3) + 3 15 = 2(5) + 5 21 = 2(7) + 7 27 = 2(9) + 9 In odd numbers the relation of even components to odd is always 2 : 1, we can from here write n = 2b + b e.g. 15 = 2(5) + 5 21 = 2(7) + 7 If decomposes the sum of units of number on the components the being in relation expressed in equation n = 2b + b, then it is surely then the odd number. Triangle of numbers. "Number is collection of units‖, Euclid defines her in VII book "Elements‖ definition 2 so. "Tria juncta in uno" / Three join in one / In triangle of numbers the Principle "larger about one" the links units in integers. If decomposes the sum of individuals of number on the components the being in relation expressed in equation p = a + ( p - a) this is surely then prime number. If every number natural larger from one, can be written in aspect of the sum of unity or the sum primes, and ―unity ‖ is quotient of the sum of prime and ―unity‖ by next number prime, then the infinite sum of natural numbers is equals infinite sum ―unity‖, e. g. 4= 1+1+1+1 N   1 k+k=n

1 + 1 = 2 + 1 = 3, 4 = 2 + 2, 5 = 2 + 3, 6 = 4 + 2, 7 = 4 + 3,

8 = 6 + 2, 9 = 6 + 3,...

0 1

53 52 51 50 50 49 48 48 48 50 47 46 46 46 45 44 44 44 43 42 42 42 41 40 40 40 39 38 38 38 37 36 36 36 35 34 34 34 33 32 32 32 31 30 30 30 29 28 28 28 27 26 26 26 25 24 24 24 23 22 22 22 21 20 20 20 19 18 18 18 17 16 16 16 15 14 14 14 13 12 12 12 11 10 10 10 9 8 8 8 7 6 6 6 5 4 4 3 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3

7 Since natural numbers is infinitely many, then and primes is infinitely many, because all different with them consist, and what with this goes also pair of twin primes. This is yet completely comprehensible! And simultaneously not natural in natural numbers. All natural numbers which carry in me principle ―larger about one‖, can be written as the sum of ones, or primes 2 and 3. 2 1 2k = p + p… 2k = (2) 1= n = p‘ + p‘ n = (3) 3 1 + 1 = 2 = 1(2) 1 + 1 + 1 = 3 = 1(3) 1+1+1+1=4=2+2 1+1+1+1+1=5=2+3 1+1+1+1+1+1=6=2+2+2 1+1+1+1+1+1+1=7=2+2+3 1 + 1 + 1 + 1 + 1 + 1+ 1 + 1 = 8 = 2 + 2 + 2 + 2 1+1+1+1+1+1+1+1+1=9=3+3+3 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =10= 2 + 2 + 2 + 2 + 2 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =11= 2 + 2 + 2 + 2 + 3 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =12= 2 + 2 + 2 + 2 + 2 + 2 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =13= 2 + 2 + 2 + 2 + 2 + 3 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =14= 2 + 2 + 2 + 2 + 2 + 2 + 2 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =15= 3 + 3 + 3 + 3 + 3 15 + 1 =16= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 16 + 1 =17= 7(2) + 1(3) p = n(p) + p‘ 17 + 1 =18= 9(2) 18 + 1 =19= 8(2) + 1(3) 19 + 1 =20= 10(2) 24 + 1 =25=5(2) + 5(3) „p‖= n(p) + n(p‘) 34 + 1 =35=7(2) + 7(3) + 1 =  Every number prime 3 consists from 3 and number even, not divided by three. p = 3 + (p – 3) p – 3 = 2n p = n(2) + 3 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+ \ 3 / + \ 2/ +\2 / +\ 2 /+\ 2/ +\ 2 /+ \ 2/ +\2/ + \2/ + \2/ + \2/ 1 2 3

5 6

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 p = n(2) + 3 23 = 10(2) + 1(3) 37 = 17(2) + 1(3) 29 = 13(2) + 1(3) 41 = 19(2) + 1(3) 31 = 14(2) + 1(3) 43 = 20(2) + 1(3) 47 = 22(2) + 1(3) 67 = 32(2) + 1(3) 53 = 25(2) + 1(3) 71 = 34(2) + 1(3) 59 = 28(2) + 1(3) 73 = 35(2) + 1(3) 61 = 34(2) + 1(3) 79 = 38(2) + 1(3) 83 = 40(2) + 1(3) 89 = 43(2) + 1(3) 97 = 47(2) + 1(3) 107 = 52(2) + 1(3) 101 = 49(2) + 1(3) 109 = 53(2) + 1(3) 103 = 50(2) + 1(3) 113 = 55(2) + 1(3)

The whole infinite file of natural numbers consists from infinite quantity 2 and 3, which are ―units‖ all numbers. N = (2) + (3) = (1)

8  N =   1 =  

2 1 3

2 1 1(2), 1(3). 3 In this way was proved mathematically indirectly that all numbers descend from one, because they consist from ―units‖.

Alone meanwhile ―units‖ they are even and odd multiplicity ―unity‖ 1 =

p  1  p  1 179  1 179  1 e. g. 179 =  2 1  2 1  3 3  3   3  p = [ 2(k) – 2] + 3 727 = [ 2(363) – 2] + 3 = (726 – 2) + 3 As to that indivisibility, Euler announced, that possesses algebraically proof on existence a  bn God. His form looked so:  x , hence God exists. If in place of algebraically signs to n substitute three first numbers, then for mathematician equation this can to be proof on 2  13 indivisibility number 3  1 . Philosopher can tell, that only plurality can to get unite. 3 Theologian meanwhile it will say: Father and Son with triple only Holy Spirit it is indivisible Holy trinity, hence exists one God in three persons. And all are right, because plurality is the form of unity. See this on example primes, which despite that they consist from many individuals, they exist as individual indivisible numbers. On beginning of plan of building cosmos were primes, in which God‘s similarity be reflects. Decoding code of primes is decoding God‘s secret formula and the riddle of universe. In glitter apple pie order reigning in world of numbers, we recognize me and different the creation‘s wonders. p=

2 + 1/3 =1 100% 1

1 1

1 80%

60% 4

40%

28 27

20% 2 2

0% 3

How the slide-fastener closes the principle "larger about one‖ the even numbers and odd in one draught of numbers.

9 They three numbers create base. Three is first number folded 2 + 1 = 3 how Pythagoras qualified this which has beginning, centre and end, and only which only one of these modules possesses as united plurality.  2 1 1=    3  6 7  2 1  2 1 2 3 4 5 1=  1=     3   3   2 1  2 1  2 1  2 1 1=  1=   1=   1=     3   3   3   3   2 1  2 1  2 1  2 1  2 1  2 1 1=  1=  1=   1 =   1=   1=     3   3   3   3   3   3   2 1  2 1  2 1 0-----------1=   ------------- 1 =   --------------------- 1 =    3   3   3   2 1  2 1  2 1  2 1  2 1  2 1 1=  1=  1=    1=   1=   1=     3   3   3   3   3   3   2 1  2 1  2 1  2 1 1=  1=   1=   1=     3   3   3   3   2 1  2 1 1=  1=     3   3  Three, as all odd numbers possesses symmetry creator "centre‖. Centre of three is two, quintuple the three, seven the five etc, hence with 2 and 3 consist all natural numbers and the three the state "centre‖ of all odd numbers, and all natural numbers are the quotient of three. Every plurality is the plurality of „units‖.  2 1  N =  1   3  From those "units" be folded the whole cosmos, world of minerals, plants, animals and human existences. "Man was created on range and similarity of only God's‖, it tells us so many belief, "all consists from the smallest and indivisible particles‖ - natural sciences teach so. Both on their way struggle about formulating one and the same truth: Such Unity is, from which whole plurality descends. 0 0:1= 0 3 2 1 3  1 3 3 2 1 2 1 6 = 2  3 3 3 2 1 2 1 2 1 9 +   3 3 3 3 3 2  1 2  1 2  1 2  1 12 + =   4 3 3 3 3 3 2  1 2  1 2  1 2  1 2  1 15    5 + + 3 3 3 3 3 3

10 2  1 2  1 2  1 2  1 2  1 2  1 18 + + =    6 3 3 3 3 3 3 3

Elusive twins. Prime numbers are a rich and ancient source of mathematical mystery. It has been known for over 2000 years that there are an infinite number of them. If sum two following prime numbers form n and n + 2, it is divisible by 12, that there are surely twin primes. p  ( p  2) p + (p + 2) = 12 12 / \ 5+ 7

24 36 60 / \ / \ / \ 11 + 13 17 + 19 29 + 31

84 / \ 41 + 43

"Twins‖ call a pair of prime numbers if they differ by 2 are, e.g. 5-7, 11-13, 17-19, 29-31, 4143, 59-61, but not pair 131-133, or 10 000 037 - 10 000 039, it because 2, 3 and 4 number can was take apart on prime factors 133 = 7(19), 10 000 037 = 43(232 559) 10 000 039 = 7(1 428 577). Divide the sum of twin pair by 12, we will find out near which following even number 30137  30139 divisible by 3, came into being numbers prime.  5023 because 5023 · 6 = 12 30138/3 During when sequence of the reciprocal of primes is divergent / with reason of growing space 1 (n) 6 /    , sequence of the reciprocal of all twin numbers is convergent / because they pprim p

1 1     < ∞, and his exact value be well-known! In p  2  p  2 prim  p 1919 Viggo Brun proved that the sum of the reciprocals of all the twin primes converged. The sum of the reciprocal of primes grows with subtle slowness. Though so as quantity of primes be infinite, one need to add above 300 000 members, but that the sum there would get number 3, on million 3,068. near mutually on distance 2/



The six- wide array further helps to demonstrate the otherwise still unproven conjecture that there must be infinitely many twin primes. In the six- wide rectangular array, the consecutive multiples of each number higher than three lay on a straight line from zero to that number and beyond, and on periodic parallels to that line further ―down‖ if we begin writing the numbers from the ―top‖ of the array. Soon after this ―factor line‖ leaves the array rectangle on one side, a parallel to it re- enters it on the other side, farther down in the array at the next such multiple. Each so broken factor line thus cascades in evenly spaced stripes down the layers of the array. Whenever the factor lines from all the primes above a given layer in the six- wide array happen to miss the two spaces before and after the 6n column in that layer, the entries there are not multiples of any among those prior primes. They are therefore primes themselves and from a pair of twin primes, as illustrated in following table. This approach to the way Euclid suggested multiplying all the primes, up to a supposedly ―largest‖ one, with each other. He imagined this equally unfeasible multiplication to show that the result plus or minus one is either a prime, or else the product of two or more primes larger than the previously ―largest‖. By this method, he proved that

11 there always exists a prime larger than any allegedly ―largest‖ one, and that there must thus be an infinite quantity of them. It is from in pair‘s twin numbers similarly. Always the foundling oneself the larger pair of twin primes from allegedly "largest‖, and by then sequence there has not the end.

4 10 16 22 28 34 40 46 52 58 64 70 76 82 88 94 100 106 112 118 124 130 136 142 148 154 160 166 172 178 184 190 196 202 208 214

5 11 17 23 29 35 41 47 53 59 65 71 77 83 89 95 101 107 113 119 125 131 137 143 149 155 161 167 173 179 185 191 197 203 209 215

0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102 108 114 120 126 132 138 144 150 156 162 168 174 180 186 192 198 204 210 216

1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 103 109 115 121 127 133 139 145 151 157 163 169 175 181 187 193 199 205 211 217

2 8 14 20 26 32 38 44 50 56 62 68 74 80 86 92 98 104 110 116 122 128 134 140 146 152 158 164 170 176 182 188 194 200 206 212 218

3 9 15 21 27 33 39 45 51 57 63 69 75 81 87 93 99 105 111 117 123 129 135 141 147 153 159 165 171 177 183 189 195 201 207 213 219

12 Mathematicians have known since the 18th century that prime numbers are more common among the smaller numbers (π10 = 4, π100 = 25, π1000 = 168), and get increasingly rare as you look at larger and larger numbers (π10000 = 1229, π100000 = 9592). Twin primes are even rarer than ordinary primes and finding them is no easy feat. The sum the pair of twin primes equals sum of first three successive the pair as the triangular multiplicities number 12 and the next different multiplicities in dependence from this, which they in turn are the pair with infinite set of numbers. p  ( p  2) 12(1,3,6)  12(n)  N n  12 p ( p  2) If you look at the distribution list of the first 4 primes you‘ll see that it contains two twin primes (5 7), in 25 prime numbers are already 14 twin primes (5 7, 11 13, 17 19, 29 31, 41 43, 59 61, 71 73) in 168 prime numbers there are 68 twin primes and in 1229 prime numbers even 408th. The question is not whether there are infinitely many twin primes, how special are the twin primes distributed in the primes. Response you see in prime number distribution. As the asymptotically decreasing number of primes in hidden geometric sequence 3 (q) the sum of the differences between primes and almost primes grows, so too grows the sum of the differences between almost primes nd(pp‘), twin primes d(p, p‘) and remaining primes dr(p). nd(pp ') + d (p, p') + dr (p) = 3(q) 9 + 12 + 9 = 3(10) 157 + 54 + 89 = 30(10) 2 + 12 = 14 + 54 = 68 + 340 = 408 2 + 9 = 11 + 89 = 100 + 721 = 821 2 + 2 = 4 14 + 11 = 25 68 + 100 = 168 408 + 821 = 1229 nd(pp') 0 9 157 1 939

3(q) dr(p) d(p, p') 3 2 2 30 9 12 300 89 54 3 000 721 340

p 4 25 168 1 229

n(p, p') 2 14 68 408

π10 = 4(p) = 2(p, p+2) π100 = 25(p) = 14(p, p+2) π1000 = 168(p) = 68(p, p+2)

2 2

29 31

41 37

59 61

101 107 113

71 67

131 137

97 103 109

127

3 149

167 173 179

191 197

181

193 199

211

251 257 263 269

281

3 151 157 163 4

227 233

239

4 223 229

241

5 293

311 317

307

6 6 367 373 379 7

443 449

139

271 277 283 347 353

359

313

331 337

349

383 389

401

419

431

397

409

421

433

461 467

479

491

503

13 7 439

457 463

8 509

521

523

587 593

9 661

499

557 563 569 541 547

571

577

617

641

647

599

601 607 613 619

631

677 683

701

10 653 659 10

487

673

691

743

761

643 719 709

773

11 727 733 739

751 757

769

787

12 797

809

821 827

839

857

811

823 829

881

877 883

853 859

887

911 907

14 941 947 953

929 919

937

971 977 983

967

p, (p +2),

863

991 997

1009

5, (5 + 2), 11, (11 + 2), 29, (29 + 2), 107, (107 + 2) 181 151 131 101 71 61 41 31 11 5 2

179 149 139

109 89 79 59

3 713 23 43 53 73 83103113

29 19

163 173

17 37 47 67 97 107 127 137 157 167

Theorem: They twin numbers prime, place oneself before and after even number divisible by 3, when sum of their ciphers of units equal 4, 10 or 16. 11 + 12/3 + 13 1+ 3=4

17 + 18/3 + 19 7 + 9 = 16

29 + 30/3 + 31 9 + 1 = 10

2087 + 2088/3 + 2089 7 + 9 = 16

69 67 68 66 67 66 65 64 63 62 61 61 60 60 59 59

73 1 71 72 1 73 72100% 71 70

5 4 5

6 6

7 7

80%

9 10 11 11 12 12 13 13 14 15

60%

40%

16 17 17 18 18

20%

58 57 56

19 19

54 54 53 53 52 51 50 49 48 48 47 47 46

21 22 23 23 24 24 25 26 27 28 45

41 40 39 38 41

34 37 36 35

29 30 29 31 30 32 33 31

37 36

Only primes, which even components are even, create the not only that is to say, of twin number e.g. 5 and 7, 11 and 13, form n and n + 2, but once even number ‖triplets‖: 3, 5, 7, form n and n + 2 and n + 4, in which this even components are even : -1 -3 -5 = 2 It exist also one peer of successive prime 2 and 3 which are not "twins‖ yet only "successive‖. 12

/ \

5+ 7

11 + 13 p, (p +2),

17 + 19 29 +31

pd =

41 + 43

5, (5 + 2), 11, (11 + 2), 29, (29 + 2), 107, (107 + 2) 181 151 131 101 71 61 41 31 11 5 2

179 149 139

109 89 79 59

3 713 23 43 53 73 83103113

29 19

17 37 47 67 97 107 127 137 157 167

163 173

n  (n  2) 12

_1 + _3 = 4

_7 + _9 = 16 71

_9 + _1 = 10

72 73

67 66

61 60 59 5 67 123

11 12 13 17 18 19

54 53

23 24

29 30 31

48 47 43 4241

3736

48 44 40 36 32 28 24 20 16 12 8 4 0 59

4743 3531 2319 11 7 3

1 5 2 6 10 14 18 22 26 30 34 38 42 46 50

1317 2529 3741 49

16 All natural numbers congruent to me according to module n‘ – n  0 mod 6. n' - n = 0 mod 6 250

37 200 31 150

100

18 13 12 11 10 9 8 4

7 6 5 4 3 2 1

33 27

15 14 4

2 4

2 6

2 8

2 10

Serie8

Serie7

Serie6

Serie5

Serie4

Serie3

Serie2 Serie1

4 2

The one light appears diverse, colorful, if it is refracted in the prism. Again, spreading from unit all natural numbers to accept form six waves, with the length 6. Congruent to me according to module 6 numbers they divide on three groups of even numbers and odd / 2, 4, 6 / 3 -2- 5 -2- 7 / keeping among me solid space 2 and 6 in every group 2/8, 3/9, 4/10, 5/11, 6/12, 7/13. It's a funny thing with the prime. For more than two thousand years, it is already known, this indivisible Every natural number prime factorization is unique, but nevertheless it is difficult from a certain size, say of a natural number, whether it is prime or not. Above a certain size, it is even not possible to determine within a human life, whether it is a number is prime. Human beings are very resourceful in order to uncover the secrets of the primes, but how far it may also occur, most of it will probably always have on From first ten numbers prime rise for them four characteristic the number of unity n k+1 11

n+2 k + 3 13

n+6 k + 7 17

n+8 k + 9 19

All the "almost primes" are predecessor or successor is an even number that is divisible by three and therefore decomposes the model: " p"1 " p"1 " p"1 b= )  (" p" 2 )  2b  (" p"2b) 3 3 3 55  1 55  1 55 = 2( ) + (55 - 2 ) = 2(18) + (55 –36) = 36 + 19 3 3

„p― = 2(

17 If before or after an even number divisible by 3, a smaller or larger by 1 odd number occurs, which is divisible by primes greater than 3, and then this is surely an "almost prime". „p‖ = 2n/3 ± 1

341 = 342 – 1

e.g.

25 49 55 85 91 115 121 133 145 169 175 181 187 24 36 48 54 66 78 84 90 96 114 120 126 132 144 156 162 168 174 180 186 204 35

65 77

119 125

143 155 161

If the sum of the following two "almost primes" of the form n and n + 4 / 91, 95 / two times larger than the odd number, that is divisible by 3 and between them lies, then surely that is almost twin primes. 91 \ 115 \ 121 \ 93 x 2 = 186 117 x 2 = 234 123 x 2 = 246 95 / 119 / 125 / Closer to each other they cannot be - they cannot differ by 2, because then an odd of them would have to be divisible by 3, and thus would not almost prime. If they have the distance 2 then there are pairs "following almost primes" as 119 – 121, 143 – 145, 185 – 187. p(p') = 2n/3 ± 1 p(p') = 2(2n/3)+[p(p') - 2(2n/3)] 25=2(24/3) + [25 - 2(24/3)] = 2(8) + [25 - 2(8) = 16 + 9 100%

90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

Serie3 25

115 119 121 125 133 143 145 155

Serie2 16

104

Serie1

Every almost prime numbers we can introduce as sum of 2 and 3 keeping definite proportions. In almost prime numbers, which are multiplicity of number 5 the relation of 2 to 3 amount 11 because 5 = 3 + 2 25 = 3 + 3 + 3 + 3 + 3 + 2 + 2 + 2 + 2 + 2 25 = 5(2) + 5(3) 5(5) „p― = n(2)  n(3) 55 = 11(2) + 11(3) 5(11) 65 = 13(2) + 13(3) 5(13) 85 = 17(2) + 17(3) 5(17) 95 = 19(2) + 19(3) 5(19) 115 = 23(2) + 23(3) 5(23)

18 125 = 25(2) + 25(3) 5(25) 155 = 31(2) + 31(3) 5(31) 185 = 37(2) + 37(3) 5(37) 625 = 125(2) + 125(3) 5(5)(25)

145 = 29(2) + 29(3) 5(29) 175 = 35(2) + 35(3) 5(35) 205 = 41(2) + 41(3) 5(41) 875 = 175(2) + 175(3) 5(7)(25)

In almost prime numbers, which are multiplicity of number 7 the relation of 2 to 3 amount 21 because 7 = 2(2) +3 35 = 3 + 3 + 3 + 3 + 3 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 35 = 10(2) + 5(3) 7(5) „p― = 2n(2)  n(3) 77 = 22(2) + 11(3) 7(11) 119 = 34(2) + 17(3) 7(17) 161 = 46(2) + 23(3) 7(23) 2401 = 686(2) + 343(3) 7(7)(49)

49 = 14(2) + 7(3) 7(7) 91 = 26(2) + 13(3) 7(13) 133 = 38(2) + 19(3) 7(19) 203 = 58(2) + 29(3) 7(29) 2695 = 770(2) + 385(3) 7(7)(55)

In almost prime numbers, which are multiplicity of number 11 the relation of 2 to 3 amount 4 1 because 11 = 4(2) + 3 „p― = 4n(2)  n(3)

121 = 44(2) + 11(3) 11(11) 275 = 100(2) + 25(3) 11(25)

143 = 52(2) + 13(3) 11(13) 385 =140(2) + 35(3) 11(35)

In almost prime numbers, which are multiplicity of number 13 the relation of 2 to 3 amount 5 1 because 13 = 5(2) + 3 „p― = 5n(2)  n(3)

169 = 65(2) + 13(3) 13(13) 637 = 245(2) + 49(3) 13(49)

221 = 85(2) + 17(3) 13(17) 715 = 275(2) + 55(3) 13(55)

In almost prime numbers, which are multiplicity of number 17 the relation of 2 to 3 amount 7 1 because 17 = 7(2) + 3 289 = 119(2) + 17(3) 17(17) 1105 = 455(2) + 65(3) 17(65)

„p― = 7n(2)  n(3)

323 = 133(2) + 19(3) 17(19) 1309 = 539(2) + 77(3) 17(77)

In almost prime numbers, which are multiplicity of number 19 the relation of 2 to 3 amount 8 1 because 19 = 8(2) + 3 361 = 152(2) + 19(3) 19(19) „p― = 8n(2)  n(3) 437 = 184(2) + 23(3) 19(23) In almost prime numbers, which are multiplicity of number 23 the relation of 2 to 3 amount 10 1 because 23 = 10(2) + 3 529 = 230(2) + 23(3) 23(23)

„p‖ = 10n(2) + n(3)

575 = 250(2) + 25(3) 23(25)

In almost prime numbers, which are multiplicity of number 29 the relation of 2 to 3 amount 13 1 because 29 = 13(2) + 3 841 = 377(2) + 29(3) 29(29) 841 = 754 + 87

„p‖ = 13n(2) + n(3)

899 = 403(2) + 31(3) 29(31) 899 = 806 + 93

19 In almost prime numbers, which are multiplicity of number 31 the relation of 2 to 3 amount 14 1 because 31 = 14(2) + 3 In almost prime numbers, which are multiplicity of number 7 the relation of 2 to 3 amount 21 because 7 = 2(2) +3 961 = 434(2) + 31(3) 31(31) 961 = 868 + 93

„p‖ = 14n(2) + n(3)

1147 = 518(2) + 37(3) 31(37) 1147 = 1036 + 111

In almost prime numbers, which are multiplicity of number 37 the relation of 2 to 3 amount 17 1 because 37 = 17(2) + 3 1369 = 629(2) + 37(3) 37(37) „p‖= 17n(2) + n(3) 25271 = 11611(2) + 683(3) 37(683) 1369 = 1258 + 111 25271 = 23222 + 2049 It the whole infinite set of natural numbers consists with infinite quantity of 2 and 3, which are "units‖ of all numbers. N = (2) + (3) = (1) 2 1  N =   1 =   3 2 1 Alone meanwhile "units‖ they are even and odd multiplicity "unity‖ 1(2), 1(3), 1= 3 In this way was proved indirectly the basic theorem of algebraically theory of numbers, that all numbers descend from one, because they consist from "units‖. Only plurality can to get unite, and primes as only they possess this ability, because they are indivisible. Why a number is prime? Because could be written as two smaller numbers multiplied together. That is, it is not possible to represent a prime as the product of two integers a x b with a, b > 1. Let q and r be the quotient and remainder of the division of n by d. That is, for each n and d, let n = d q + r, where r and q are positive integers and 0 ≤ r < d. Because all prime numbers contain in me one 3, it was not possible divide here by two. Superiority meanwhile 2 it causes, that they don´t divide by three also. So they are indivisible by all different numbers, and on this depends the complete primality certificate! p = n(2) + 3 2 = 1(2) + 0 17 = 7(2) + 3 p=

3 = 0(2) + 3

5=2+3

19 = 8(2) + 3

p  1  p  1  2 1 3  3  p = [ 2(k) – 2] + 3

7 = 2(2) + 3

23 = 10(2) + 3 e.g.

11 = 4(2) + 3

233 = 115(2) + 3

13 = 5(2) + 3 251 = 124(2) + 3

179  1 179  1  2 1 3  3  727 = [ 2(363) – 2] + 3 = (726 – 2) + 3

179 =

2 127 1 = 170 141 183 460 469 231 731 687 303 715 884 105 727 3 170 141 183 460 469 231 731 687 303 715 884 105 724 34 279 974 696 877 740 253 374 607 431 768 211 457 3 34 279 974 696 877 740 253 374 607 431 768 211 454

20 The natural numbers in scheme of 2 and 3 If p ≥ 2 and p‘ ≠ 0, are whole numbers not having common divisor, than such arithmetical sequence contains in me all natural numbers. 2, 3, n(2), 2 + 3, n(3), 3 + n(2), n(2), n(3), n(2), ... n(2) + n(3) 2, 3, 4, 5, 6, 7, 8, 9, 10, .... 10 + 15 = 25 P(n) = p, p‘, n(p), p + p‘, n(p‘), p‘+ n(p), .... n(p) + n(p‘),

n(2)

n(3) =

2 3 2(2) 2

3 2(3)

2(2) 4(2)

3 3(3)

5(2) 4(2)

3 4(3)

5(2) 7(2)

3 5(3)

8(2) 7(2)

3 6(3)

8(2) 10(2)

3 7(3)

11(2) 10(2) 5(2) 13(2)

3 8(3) 5(3) 9(3)

14(2) 13(20

3 10(3)

14(2) 16(2)

3 11(3)

17(2) 7(2) 17(2) 19(2)

7(3) 12(3) 3

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

21 13(3) 20(2) 19(2)

39 40 41 42 43 44 45 46

3 14(3)

20(2) 22(2)

3 15(3)

23(2)

And here how with two primes 2 and 3 come into being all natural numbers. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

2 3 2 3

2 2 2 2 2 2 2 2 2 2 2 2 2 2

22 33 34 35

p = 3 + n(2) n (2) =n= n (3)

"p" = n (2) + n (3)

36 35 34 33 32 31 30 3 3 45 3 2 212 2 3 6 7 23 8 32 9 23 32 2 2 10 32 3 2 11 32 3 2 12 32 2 2 23 3 3 3 13

29 28 27 26

14 15

25 16

24 23

17 22

20 19

The scheme of natural numbers And so harmoniously develop natural numbers in support about principle "larger about one‖ on the base of 2 and 3 in 360 ° the circle.

19 18 17

16 15 32

14 13

33 10

6 5 4 321 0

34 36

23 The proprieties of natural numbers repeat oneself periodically, what six numbers according to pattern of primes. Proof: 1 + 2 + 3 = 6 p + 6 = p‘ n + 6 = n‘ „p‘― – „p‖= 6 p + 6 = p'

2 3 5 7 11

29 31 37

43 73

47 53

71 67

With discovery of regularity in sequence of primes, that what 6 numbers repeat oneself the same proprieties, was decoded together pattern how be distributed primes and the basing on him periodicity of natural numbers. 2 1 1= 3 Two first numbers / 1 + 2 / added to me and divided by third next number / 3 /, it equals / 1 / that is to say, again the same first number from three taking part in this working. Three first next numbers added to me give perfect and triangular number 6, defining length of period in what will repeat oneself the same proprieties in whole sequence of natural numbers. Tres faciunt collegium, then it means three numbers they decide about whole scheme of natural numbers. It 2 (3) = 6, was can introduce all numbers from here, as the sum of the ones (+ 1), the twos (+ 2) and the threes (+ 3). The periodical scheme of natural numbers is so perfect, as perfect is first perfect number 6, him untouched basis. 1+2+3=6=2·3 Ranked according to propriety natural numbers create 6 groups. Propriety of numbers in four central groups repeat oneself in turn periodically, what 6 figure. Primes create here two the rows the complementary to two rows of group sixth the almost prime numbers.

24 Periodical scheme of natural numbers. n1

n 2

n 3

n 4

n 5

2 3 5

n 6

4 6

7 11

20 26

12 13

18 19

30 31 36 37

43 48

44 50

49 51

52 55

80 86

66 67 71

72 73 78 79

77 85

90 91

92 96

97 101 103 107 109 113

95 98

100

104

105

106

110 116

111

112

117

118

123

124

102 108 114 120 126

122

115 119 121 125

25 Sieve of Eratosthenes In the six- wide rectangular array, the consecutive multiples of each number higher than three lay on a straight line from zero to that number and beyond, and on periodic parallels to that line further ―down‖ if we begin writing the numbers from the ―top‖ of the array. In six groups of numbers we have 3 group of even numbers (II, IV, VI), and 3 odd (I, III, V). Her multiplicities for prime number 5 on left have lain cascade, until after number almost prime 25 = 5(5). Next multiplicities for prime number 7 on right have lain her cascade, among which we have second almost prime number 35 = 7(5). Parallel line by her runs factor 5 falling on left in pit, until to fourth almost prime number 55 = 5(11). The parallel line factor 7 falls from the multiplicity number 7(7) = 49 in right, until to lying in V group of almost prime number 77 = 7(11). Parallel line factor 5 falling on left in pit it crosses out their 13(5) = 65 and 15(5) = 85 multiplicity. In this way they the parallel lines factors 5 and 7 cross out all almost prime numbers in I and V the group of numbers. So of the sieve Eratosthenes is situated less than 100 numbers 25 primes. 2 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99

Propriety of all natural numbers repeat oneself in turn periodically, what 6 figure. n‘ – n  0 mod. 6

2n = p – 3

(2n - 1) =p(n)

2n = p – 3

p = 2n + 3 "p"= n(2)+n(3)

2n = p(n) = p±1

p = 2n + 3 "p"= n(2)+n(3)

5-3=2

7-3=4

2 +3 = 5

3(2) = 6

2(2) + 3 = 7

11 - 3 = 8

3(3) = 9

13 - 3 = 10

2(4) + 3 = 11

3(4) = 12

2(5) + 3 = 13

17 -3 = 14

3(5) = 15

19 - 3 = 16

2(7) + 3 = 17

3(6) = 18

2(8) + 3 = 19

23 - 3 = 23

3(7) = 21

2(10)+3 = 23

3(8) = 24

5(2 + 3) = 25

29 - 3 = 26

25 + 2 = 27

31 - 3 = 28

2(13)+3 = 29

3(10) = 30

2(14)+3 = 31

35 - 2 = 33

7(2 + 3) = 35

3(12) = 36

41 - 3 =38

3(13) = 39

43 - 3 = 40

3(14) = 42

47 - 3 = 44

3(15) = 45

14(2)+7(3)=49

53 - 3 = 50

49 + 2 = 51

11(2 + 3) = 55

59 - 3 = 56

55 + 2 = 57

61 - 3 = 58

65 - 2 = 63

13(2 + 3) = 65

3(23) = 69

75 + 2 = 77

22(2)+11(3)=77

3(27) = 81

17(2 + 3) = 85

85 + 2 = 87

26(2)+13(3)=91

95-2=91+2

19(2 + 3) = 95

3(33) = 99

100

101

102

103

103 9791 8579 7367 6155 4943 3731 2519 137

126 2418 30 4236 5448 60 7266 8478 90 96 102

98 92 86 80 74 68 62 56 50 44 38 32 26 20 14 8 2

5 11 17 23 29 35 41 47 53 59 65 71 77 83 89 95 101

99 8793 81 6975 5763 51 3945 2733 21 15 39 410 1622 2834 4046 5258 6470 7682 8894 100

2n - 1 = 6n - 3 9 = 6(2) - 3 15 = 6(3) - 3 21 = 6(4) - 3 27 = 6(5) - 3 33 = 6(6) 3 39 = 6(7) - 3 45 = 6(8) - 3 2n = 6n - 4 2 = 6(1) - 4 8 = 6(2) - 4 14 = 6(3) - 4 20 = 6(4) - 4 26 = 6(5) - 4 32 = 6( 6)- 4 38 = 6(7) - 4 p = 6n - 7 5 = 6(2) - 7 11 = 6(3) - 7 17 = 6(4) - 7 23 = 6(5) - 7 29 = 6(6) - 7 p = 6n - 5 7 = 6(2) - 5 13 = 6(3) - 5 2n = 6n - 6 6 = 6(2) - 6 12 = 6(3) - 6 18 = 6(4) - 6 24 = 6(5) - 6 30 = 6(6) - 6 36 = 6(7) - 6 42 = 6(8) - 6

27 2n = 6n - 8 4 = 6(2) - 8 10 = 6(3) - 8 16 = 6(4) - 8 22 = 6(5) - 8 28 = 6(6) - 8 34 = 6(7) - 8 40 = 6(8) - 8 And all runs according to pattern of prime numbers which seems, that they be scattered how savagely growing weeds among natural numbers, but only there where they create fertile soil giving the infinite quantity of natural numbers. From first ten prime numbers can possess four characteristic ends: - 1, - 3, - 7, - 9, resulting with rhythm 2/4 in what 11 +(2), 13 +(4), 17 + (2), 19 step out and they repeat oneself what 21 places in tens divisible by 3. Table of tens in which prime numbers step out. I II III IV V VI VII VIII IX X XI XII XIII XIV .. |E – 1 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 | - 1 : 3 | : 3 | -1 : 3 | : 3 | - 1 : 3 |: 3 | -1 : 3 | : 3 | - 1 : 3 | | | 1 | 3 | 4 | 6 | 7 | 10 | | 13 |15 | | 18 | 19 | 21 | | | | | 24 | 25 | 27 | 28 | 31 | 33 | | | | | 40 | 42 | 43 | | 21x2 | | | 46 | | 49 | 52 | 54 | | 57 | | 60 | | 63 | 64 | | | | 66 | 67 | 69 | 70 | 73 | 75 | 76 | | | 81 | 82 | | | | | | | 88 | | 91 | 94 | | 97 | 99 | |102| 103 |105 | 106 | | E- 7 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | : 3 |- 1 : 3 | : 3 |- 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 | | | 1 | 3 | 4 | 6 | 9 | 10 | 12 | 13 | 15 | 16 | | 19 | | 22 | | | | | 25 | 27 | 30 | 31 | 33 | 34 | 36 | | 39 | | | | | 21x2| | 45 | 46 | 48 | | | 54 | 55 | 57 | 58 | 60 | 61 | 63 | 64 | | | | | 67 | | 72 | | 75 | | 78 | 79 | | 82 | | . | | | 87 | 88 | 90 | 93 | 94 | 96 | 97 | | | | | | | . I II III IV V |E- 3 | - 1 : 3 |+1: 3|-1 : 3|+1: 3|-1 : 3 | | 1 | 2 | 4 | 5 | 7 | | | 23 | 25 | 26 | 28 |21x3 | | 44 | 46 | | | | | 65 | 67 | 68 | | | | 86 | 88 | | 91 |E- 9 | - 1 : 3 |+1: 3 |-1 : 3|+1:3 |-1 : 3 | | 1 | 2 | | 5 | 7 | | | 23 | | 26 | |21x3| | 44 | | 47 | 49 | | | 65 | | | 70 | | | | | | 91

VI VII VIII IX X XI XII XIII XIV XV . |+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1:3 |- 1: 3 | | 8 | 10 | 11 | | | 16 | 17 | 19| | 22 | | 29 | 31 | | | 35 | 37 | 38 | | | 43 | | 50 | 52 | | | 56 | | 59 | 61| | 64 | | | 73 | 74 | | 77 | | | 82| | 85 | | | | 95 | | 98 | | 101| 103| | 106 | |+1:3 |-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1:3 |- 1: 3 | | 8 | 10 | | 13 | 14 | | 17 | 19 | 20 | 22 | | | | | 34 | 35 | 37 | 38 | 40 | 41 | 43 | | 50 | | | | 56 | | 59 | 61 | | | | 71 | 73 | | 76 | | | 80 | 82 | 83 | | | 92 | | | | | 100 | 101| 103| 104 | 106 |

The table of tens in which step out prime numbers betrays us sure regularity what it reigns in sequence of prime numbers. Not accidentally has written down in this table of ending of prime numbers in this way 1 - 7 = 6 = 3 - 9. This shows that the regularity what 6 numbers from what can step out prime numbers, crosses over on the whole sequence the natural numbers, which of propriety what they 6 numbers repeat oneself in six groups. Primes with ending 1 - 7 create XIV ranks, in which their endings repeat oneself what 21 and 42 place, and with ending 3 - 9 create XV ranks, in which their endings repeat oneself what 21, 42 or 63 places and they in both cases are then divisible numbers by 7, which will be further great meaning. Prime, even and odd numbers they create "twelve segmental cycles‖. 5 + 7 = 12 = 2 + 4 + 6 = 12 = 3 + 9 Periodical scheme of prime numbers results with principle the "twelve of segmental cycles‖ in 360 numbers which be comprises 30. Multiply thirty by unitary length of period (7) primes in what step out 30 · 7 = 210 - receive decimal length of period of prime numbers.

28 p'1 p  1 p' '1 p'1   70   3 3 3 3 11 - 3931 and 17 - 4217

+ -1 + 1/3 /3 1/3 11 31 41 241 251 461 661 671 881 1091 1291 1301 1511 1721 1931 2131 2141 2341 2351 2551 2971 3181 3391

37 457 877 1087 1297

2137 2347 2557 2767

3191

3821 47 257 467 677 887 1097 1307

2357 2777

3187 3607 4027

3407 3617

+ + 1/3 1/3 1/3 61 71 101 271 281 311 491 521 691 701 731 911 941 1151 1321 1361 1531 1571 1741 1951 2161 2371 2381 2411 2591 2621 2791 2801 3001 3011 3041 3221 3251 3461 3631 3671 3851 3881 67 97 107 277 307 317 487 727 907 937 947 1117 1327 1367 1567 1747 1777 1787 1987 1997 2207 2377 2417 2617 2797 2837 3037 3217 3257 3457 3467 3637 3677 3847 3877 4057

- 1/3 331 541 751

+ + 1/3 1/3 1/3 131 151 571 761 971 991 1181 1201

1381 1801 2011 2221

1601 1621 1811 1831

+ + 1/3 1/3 1/3 1/3 181 191 211 401 421 431 601 631 641 811 821 1021 1031 1051 1061 1231 1451 1471 1481

2251

1861 1871 2081 2281

2671

2711

2441 2851 3061 3271 3691 127 337 547 757 967

1597 2017 2437 2647 2857 3067

3697 3907

13 - 3793

1901 2111 2311 2521 2731

2531 2741

3361 3571

3371 3581

2861 3301 3491 3511 3701 3911 3931 137 157 167 347 367 557 577 587 787 797 977 997 1187 1217 1427 1607 1627 1637 1847 2027 2237 2267 2447 2467 2477 2657 2677 2687 2887 2897

3917 4127

and

3121 3331 3541 3761 197 397 607

617 827

637

1237 1447 1657 1667 1867 1877 2087 2287 2297 2707 2917

3407 3517 3527

3547

3947 4157

3967 4177

19 - 4409

227 647

1277 1487 1697 1907

2927 3137 3347 3557

2957 3167

3797 4007 4217

29 +1 +1/3 -1/3 +1/3 -1 : 3 53 73 83 103 113 263 283 293 313 503 523 683 733 743 913 953 1103 1123 1153 1163 1373 1523 1543 1553 1583 1733 1753 1783 1973 1993 2003 2203 2213 2383 2393 2423 2593 2633 2803 2833 2843 3023 3253 3413 3433 3463 3593 3613 3623 3643 3673 19 29 59 79 89 109 239 269 449 479 499 509 659 709 719 739 919 929 1109 1129 1289 1319 1499 1549 1559 1579 1709 1759 1789 1949 1979 1999 2129 2179 2339 2389 2399 2549 2579 2609 2789 2819 2969 2999 3019 3049 3209 3229 3259 3389 3469 3659 3889 4019 4049 4079 4099 4229 4259 4289 +1/3 -1/3 13 23 43 233 253 443 463 653 673 863 883 1093 1283 1303 1493 1723 1913 1933 2143 2333 2543 2753 2963

-1 :3

139 349 769

1399 1609 2029 2239 2659 3079 3499 3709 3919 4129 4339

+1 :3

-1 +1 :3 :3 163 173 353 373 383 563 593 773 983 1013 1193 1213 1223 1423 1433 1613 1823 2053 2063 2243 2273 2473 2663 2683 2693 2903 3083 3313 3323 3533 3733 149 179 359 379 389 569 599 809 1009 1019 1229 1429 1439 1619

11 23

1453 1663 1873 2083 2293 2503 2713

3343

-1 :3 223 433 643 853 1063 1483 1693 2113

2953 3163 3373 3583 3793 209 229 419 439

839 1049 1069 1259 1279 1489 1699 1889 2039 2099 2269 2309 2459 2539 2689 2699 2719 2729 2749 2879 2909 2939 3089 3109 3119 3169 3299 3319 3329 3359 3529 3539 3559 3719 3739 3769 3779 3929 3989 4139 4159 4219 4349 4409

2 + 3 = 5 + 7 = 12 + 11 + 13 = 3(12) 3 7 19

613 823 1033

199 409 619 829 1039 1249 1459 1669 1879 2069 2089

Spiral twelve segmental cycles of primes.

2 5 17

-1 +1 :3 :3 193

30 29 41 53

31 43

37 47 59 71 83

67 79

61 73

89 101 113

103

107 131

137 149

127 139 151 163

157 167 179 191

173 197

199 211 223

181 193

227 239 251 263

233 257 269 281 293

97 109

229 241

271 283

277

307

311

313

2 + 3 = 5 + 7 = 12 71

1009

983

149 59

997

73 137

977

919

53 971

907

929

883 887

853

953

967 41

139

131

991 47

911

127 61 43

197

113 107

109

191

211 163 101 103 199 193 157 89 863 859941 823 181 1719 97 179 811 1113 83 151 857 57 173 839 293 23 787 827 269 281 167 821 251 257 263 809 797 227 233 239 709 223 229 733 727 241 307 271 751 739 701 277 313 769 757 283 311 331 631 719 397 337 641547 389 317 743 349 563 409 643 367 467487 761 647 571 347 401 373 421 773 379 661 569 353 653 577 419 499 433 479 673 359 659 587 431 601 439 691 383 503 491 677 457 443 607 593 683 463 449 613 509 523 599 619 461 617 541 521 829

881

877 947

29 37 937 23 31

31 Spiral scheme of natural numbers With spiral arrangement of primes and almost prime result spiral arrangement of all natural numbers, what we see in following table. (- 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6, 7) + n(7) = n (- 6) + 7 = 1 (- 5) + 7 = 2 (- 4) + 7 = 3 (- 3) + 7 = 4 (- 2) + 7 = 5 (- 1) + 7 = 6 0 + 7 = 7

-2

-1

-6

-5

-4

-3

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188

189

190

191

192

193

The spiral sequence of natural numbers / primes and almost prime / a = b mod 17

101 85

84 100

16 17 18

01 2

14 30

13 12 11

20 3 4 5 6

61 59

91 75

79 96

9 8

54 37

88 71

31 47

35 36

48 64

87 70

Spiral of primes.

p' - p= 0 mod 17 = 19 - 2

101

53 71

17 19

31 47

37 89

3 5

97 41 79

43 61 59

Spiral of almost prime "p" - p = mod 23 = 25 - 2 481 437

413

485

391 415

343

365

299

275

323

253 341 295

455

205

161

185

115 203

385 0

133

395

119

25 2

143 121

155

407 361

85 55 77

245

451

259 305

125

175

289 335

235

145 169

221

475

371 325

187

91 247

301

209

215

329

377

445

217 265 287

427 473 497

355

403 425

493

It is true in spirals primes and almost prime congruent according to different modules, however difference between them is common module all natural numbers 23 -17 = mod 6, what show above mentioned graphs. Module 40 = 17 + 23 arranges natural numbers in infinite spiral. 2 + 3 + 5 + 11 + 19 = mod 40 199

198 197 196

158

193

156 155

154

160

161

162 163

157

195 194

159

117 116 115

119 120 121 122 118

164 165

123 124

80 81 82 78 79

125 83

166

77 126 84 167 76 85 75 127 86 39404142 43 113 168 74 44 37 38 152 87 45 36 128 73 112 35 46 88 34 191 47 169 72 151 33 012 345 129 48 111 67 89 32 71 49 89 31 190 150 110 70 30 10 50 90 130 170 11 29 12 51 69 28 13 91 109 14 27 52 131 149 15 68 26 171 16 189 2524 53 92 17 108 67 23 2221201918 54 132 148 66 93 55 107 172 65 56 188 94 133 64 57 106 147 63 62 58 95 61 60 59 173 105 134 187 96 146 104 97 135 103 174 145 102 101 100 99 98 186 136 144 175 137 185 143 138 142 141 139 176 140 184 177 183 182 181 180 179 178 153

192

114

34 Congruence according to module 6 shine numbers with all colors of rainbow. a = b mod 6

5 117 23 2935

179

53 59 1 4 0 16 22 283440 178 65 5273339 4652 177 39121 172 71 8120 2 4263238 4551 58 176 167 44 171 57 719 325 31 175 1 3743 5056 6364 170 49 12 6 0 18 24 0642 55 62 77 174 33 169 4854 61 6970 166 168 165 60 6768 76 164 75 66 74 163 83 162 7273 82 161 79 8081 78 160 159 158157 156 8485 868788 89 9091 92 9394 150 152151 154 153 9697 95 155 98 99 102 144 100 145 103 146 108 104 147 148 138 101 114 109 105 139 132 106 126 120 115 110 140 111 133 149 141 127 121 116 112 142 134 107 128 122 117 135 123 118 129 136 124 130 143 113 173

119

137 125

131

3+3=6=2x3

7 3 2 1

6 12 18 24

5 11 17 23

The cross of numbers is bases on number 6, appointed by primes 2 and 3, which squeeze out his brand on whole scheme of natural numbers.

+ 143 121

119

47 139

137

115

113

13 11

101

103

125

127

107

109

131

133

It comes from structures of cross of natural numbers from congruence of primes and almost prime according to module 2(2)2. +

127 119

121 113

103 95

97 89 79 71

73 65 55 47

49 41 31 23

25 17 7

0 2

13 29 37 53 61 77 85 101 109 125

11 19 35 43 59 67 83 91 107 115

3 + 2(2)2 = 11 + 2(2)2 = 19 + 2[2(2)2] = 35 + 2(2)2 = 43 + 2[2(2)2] = 59 + 2(2)2 = 67 100% 90% 80%

p' + p(p)p = p"

70%

103

119

127

115

125

113

121

109

60% 50%

101

40% 30% 20%

107

10% 0% Serie8

2 0 1

103

119

127

Serie7 5

Serie4

Serie3

Serie6

101

109

125

Serie5 35

Serie2 Serie1

107

115 113

121

Sequence of primes and almost prime in five groups about the same endings are possible the thanks their congruence according to module 5.

-9

-3

-7

-1

-5

95 85

65 55

35 25 89

79 59

49 29

5 0

3 13 23

2 7 17 37

43 53 73 83

47 67 77 97

37 So looks clocks of primes measuring it in rhythm 2/4 (6). (p-1) + (p+1) / 2 = p 137 139 2 3

131

11 13

127 17 19 23

113

29 31

109 107 37 103 101

41 43

97 47

91 89

83 79

73 71

According to this rhythm of primes flow away us days in four times year.

2 + 3 = 5 x 73 = 365

73 71 72

27 48

28 47

29 46

30 45

31 44

42 41 40 39 38 37 36 35 34

2 + 3 = 5 x 73 = 365

72 73

68 67 66 65 64 63 62 61 60 59 11 56 78910 12 13 234 14 58 15 16 17 57 18 19 56 20 21 22 55 23 54 24 25 53 26 52 27 28 51 29 50 30 31 49 32 48 33 47 34 46 4544 3635 43 424140393837

Binary and Ternary Goldbach's Conjecture, equation Pythagoras and great Fermat's theorem. Creative process in mathematics begins from conjecture. Mathematical conjecture really then it becomes theorem, when we have on his truth irrefutable proof. Theorem: Even numbers are "larger about 1‖ from one's odd, prime or almost prime predecessor, and so they are duplication different natural number. Proof:

(2n – 1)  1 \ „p‖  1 = 2n p 1 /

(2n – 1) + 1 = 2n 1 + (2n – 1) = 2n = 3p - p = p + p‘  p‘ + p = 2n 7 – 5  mod 2  5 – 3 p‘- p = n/2 6 – 4  mod 2  10 – 8 3(2) – 2 = 4 3(3) – 3 = 6 3 + 5 = 8 3(5) – 5 = 10 5 + 7 = 12 3(7) – 7 = 14 11 + 5 = 16 This theorem proves the just truth Binary Goldbach's Conjecture, that every even larger number than 2 is the sum two primes. Both even numbers how and primes congruent to me according to modules 2, that is to say differences between them divisible they are by 2. From here simple conclusion, if differences this and sum two primes divisible they are by 2, as even numbers.

2p = 2n = p + p', 2 + 2 = 4, 3 + 3 = 6, 3 + 5 = 8, 5 + 5 = 10, 5 + 7 = 12, 7 + 7 = 14, 5 + 11 = 16, 7 + 11 = 18, 7 + 13 = 20, 5 + 17 = 22, 7 + 17 = 24, 13 + 13 = 26, 11 + 17 = 28, 13 + 17 = 30, 13 + 19 = 32, 17 + 17 = 34 140

120

100 23 80

19 17

60 13 11

2 1 1

Serie3

3 6 3

2 4 2 2

5 10

4 4

7 14

9 9

29 29

Serie2

Serie1

It will permit then us on formulating polynomial describing the solution of Binary Goldbach‘s Conjecture. P(2n) = (p + p‘) (2+2) (3 + 3) (3 + 5) (5 + 5) (5 + 7) (7 + 7) (5 + 11) (5 + 13) (7 + 13) (5 + 17)

p + p' = 2n 2 + 2 = 4 3 + 3 = 6 3 + 5 = 8 17 + 19 = 36 37 + 41 =78 59 + 61 = 120 67 + 71 = 138 101 + 103 = 204 109 + 113 = 222 163 + 167 = 330 193 + 197 = 390 227 + 229 = 456 450 400 349; 359 350 311; 313 300 269; 257 307; 317

250

227; 229 197; 193

200 167; 163

150

113; 109 101; 103

100 71; 67 59; 61 41; 37

50 17; 19 0 -5

0 -50 -100

3; 5

2n = p + p'

4=2+2

6=3+3

8=3+5

10 = 5 + 5

12 = 5 + 7

14 = 7 + 7

16 = 5 + 11

80 38 36

34 32

30 28 26

24 22

20 18 16

14 12

10 8 6

3 3

2 2

17 11

Serie3

Serie2

Serie1

p' + (p + p) = 2n - 1

2n = p + p

(2n - 1) - (p + p) = p

3 + (2 + 2) = 7

4=2+2

7 - (2 + 2) = 3

120

100

49 47 45 43 41

39 37 35 33 31

29 27 25 23 21

19 17 15 13 11

9 7

4 3

Serie3

Serie2

Serie1

Theorem: Every odd number larger than 5, is sum three primes, because difference among odd and even number is always 3. Proof:

2n = p + p 4=2+2

(2n – 1) – (p + p) = p 7 - (2 + 2) = 3



2n – 1 = p + p + p‖ 7 =2+2+3

p + p' = p"

2+3=5

p + p = 2n

5+3=8

p'' + p' + p = 2n - 1

7 + 5 + 3 = 15

30 29 28 27 26 25

25 24 23 22 21

20 19 18 17 16

15 14 13 12 11

10 9 8 7 6

5 4 3 2 1

3 2 1

0 1

Fermat‘s last theorem. a(a) - b(b) = (a - b)(a + b)

5(5) - 3(3) = (5 - 3)(5 + 3)

25 - 9 = 2(8)

700 625 600

576 529 484

500 441 400 400 361 324 289

300 256 225 196

200 169 144 121 100

100

49 47 45 43 41 39 37 35 33 16 31 29 27 25 9 24 25 23 23 22 4 21 20 19 18 17 16 15 14 13 7 15 8 17 9 19 10 21 11 13 12 11 9 7 6 5 5 4 3 3 2 1 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 25

Theorem: Difference among two successive square numbers always odd number, and square number too is. a(a) – b(b) = (a – b)(a + b) = c(c) 25 – 16 = 5(5) – 4(4) = (5 - 4)(5 + 4) = 3(3) 3(3) + 4(4) = 5(5)

42 Next numbers from infinite file of odd numbers added to square minuend create always square subtrahend. That is to say, that equation x n  y n  z n from great theorem Fermat‘s, only near n = 2 has solution, because when add only odd number to square, we receive next square number. 1 and 3 are not square numbers, and above mentioned theorem despite this and on them checks because primes are the multiplicity of number one and only me also. p = 1(p) 2² - 1² = (2 – 1)(2 + 1) = 1(3) 1(1) + 1(3) = 2(2) In other words, equation for n > 2 in infinite file of natural numbers does not possess no solution, because only the square numbers create the ternary Pythagoras. z 2 x 2  y 2

25 – 9 = 16 ( y 2 )  0 mod 2 16 : 2 = 8 Rest 0 x x‘ 2  x 2 x2 1 + 1 1 + 3 2 4 + 5 3 9 + 7 4 16 + 9 5 = 24 25 2 2 ! = 24

22 ! + 12 = 52

3+5+7+9 + 1 = 25

First ternary Pythagoras comes into being, when the sum of differences among successive square numbers reaches value of faculty 2 2 !. It number 2 modulates so formation so square numbers how and ternary Pythagoras, that is to say that differences among odd numbers and squares in ternary Pythagoras they are divisible by 2, therefore squares how and ternary Pythagoras are product 2 factors. 3(3) + 4(4) = 5(5)

[2(5) – 1] + 2(8) = 2(13) – 1

3[2(2) – 1] + 2(2)2(2) = 5[2(3) – 1]

1 + 3 = 4 + 5 = 9 + 7 = 16 + 9 = 25 + 11 = 36 + 13 = 49 + 15 = 64 + 17 = 81 + 19 = 100 + 21.. \ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 / 25 – 9 = 2(8)

169 – 25 = 2(72)

289 – 225 = 2(32) 625 – 49 = 2(288) 841 – 441 = 2(200)

It because 2 is the solid value of differences among two the successive odd numbers becomes she the modules of differences with them square numbers in ternary Pythagoras, where with difference among horizontal the and vertical length the side of triangle square creates, being simultaneously the proof on truth of equation the Pythagoras and the Fermat's conjecture.  xy  2 z 2  4   x  y   2

z 2 = 2xy + x 2 - 2xy + y 2

z2 = x2 + y2

X +

4xy/2+(x - y)^= z^ z^- 4xy/2= (x - y)^ 2xy+(x-y)(x-y)=z(z) 2xy+x(x)-2xy+y(y)=z(z) x(x) = x^ y(y) = y^ z(z) = z^ ^=2 If the product of hypotenuse to be equal to sum two products of legs and product of their difference, that is to say, that the square of hypotenuse is the sum of squares of legs. Theorem: Difference among every odd square in triples Pythagoras even square is, which congruent according to modules y 2  mod 2 Proof: z 2  x 2  y 2 25 – 9 = 16 ( y 2 )  0 mod 2 16 : 2 = 8 Rest 0 [2(5) – 1] + 2(8) = 2(13) – 1 The congruence the even square y 2  0 mod 2 signifies, that in quadrate of hypotenuse(z 2 ) 4 triangles of the same hypotenuse be comprise (z), replenished about quadrate came into being with differences between horizontal and vertical length the sides 4 triangles. e.g. yx y2 + x2 = z2 y–x=n 4 + n2 = z2 2 e.g. 4 2 + 3 2 = 5 2 4–3=1 4(6) + 1 2 = 25 12 2 + 5 2 = 13 2 12 – 5 = 7 4(30) + 7 2 = 169 8 2 + 15 2 = 17 2 15 – 8 = 7 4(60) + 7 2 = 289 24 2 + 7 2 = 25 2 24 – 7 = 17 4(84) + 17 2 = 625 20 2 + 21 2 = 29 2 21 – 20 = 1 4(210) + 1 2 = 841 12 2 + 35 2 = 37 2 35 – 12 = 23 4(210) + 23 2 = 1369 2 2 2 40 + 9 = 41 40 – 9 = 31 4(180) + 31 2 = 1681 28 2 + 45 2 = 53 2 45 – 28 = 17 4(630) + 17 2 = 2809 60 2 + 11 2 = 61 2 60 – 11 = 49 4(330) + 49 2 = 3721 56 2 + 33 2 = 65 2 56 – 33 = 23 4(928) + 23 2 = 4225 84 2 + 13 2 = 85 2 84 – 13 = 71 4(546) + 71 2 = 7225 72 2 + 65 2 = 97 2 72 – 65 = 7 4(2340) + 7 2 = 9409 144 2 + 17 2 = 145 2 144 – 17 = 127 4(1224) + 127 2 =21025 180 2 + 19 2 = 181 2 180 – 19 = 161 4(1710) + 161 2 = 37261

44 Ternary Pythagoras this square equation, and how there are all quadratic functions as graph of function is a parabola. The running by vertex axis of symmetry be shifted in them about 2 in direction on line - x, and about (y – x) 2 in direction on line - y. y = (y - x)^ = 1

x^ + y^ = z^

9 + 16 = 25

(1 + 3 + 5) + (7 + 9) = 25 - 1 = 24 - 3 = 21 - 5 = 16 - 7 = 9 9=0

1 0 0

0,5

1,5

2,5

3,5

y = (y - x)^ =(12 - 5)^ = 49 5^ +12^ =13^ 25+144=169 (1+3+5+7+9) + (11+13+15+17+19+21+23+25) =169 -1=168 - 3 =165 - 5=160 -7=153 - 9 =144 -11=133 -13=120 -15 = 105 -17 = 88 -19 = 69 -21= 48 23=25 - 25 = 0 180 169

169

160

140

120

100

60 49 40

0 0

0,5

1,5

2,5

Number 2 in every semi stabile elliptic curve over rational numbers modular is. (1+3+5+7+9)25 + (11+13+15+17+19+21+23+25)144 = 169 = 49(1+3+5+7+9+11+13) + \2/\2/\2/\2/ \2/ \2/ \2/ \2/ \2/ \2/ \2/ 120(15 +17+19+21+23+25)

3,5

x(2n - 1) + y(2n - 1) = z^ - z(2n -1) = 0 (1 + 3 + 5) + (7 + 9) = 25 - 1 = 24 - 3 = 21 - 5 = 16 - 7 = 9 - 9 = 0 3^ + 4^ = 5^ x^ - [2xy + (y-x)^] + y^ = 0 9 - (24 + 1) + 16 = 0 30 25

20 15 10

5 3

3 1

-5 -10 -15

-11

Serie1 Serie2 Serie3

2xy +(y - x)^ = z^

2(3)4+(4 -3)^ = 25

2(5)12+(12 -5)^ =169 2(15)8+(15 -8)^ =289 2(7)24+(24 -7)^ =625 2(21)20 + (21 - 20)^ =841

5000 2809

2809

841

625

4500

4000

3500

3000

2500

2000

1500

1000

500

289

169 25

49 1

169 25

Theorem: The square of hypotenuse is equal the sum of squares of legs, when it is sum of such quantity of odd numbers how degree of square hypotenuse. Proof: x(2n – 1) = x 2

y(2n ) = y 2

 x2 + y2 = z2

z(2n – 1) = z 2

 z 2 = z(2n – 1)

x+y=z

y=z–x

z 2 - z(2n – 1) = 0

x +(z – x) = z

46 3[2(2) – 1] + 2(2)2(2) = 5[2(3) – 1] 1 + 3 + 5 = 9 = 3(3)

1 + 3 + 5 + 7 = 16 = 4(4)

1 + 3 + 5 + 7 + 9 = 25 = 5(5)

(1 + 3 + 5) + (1 + 3 + 5 + 7) = (1 + 3 + 5 + 7 +9)

Σ(2n - 1) = n(a + z)/2 = n(n)

3(1 +5)/2 = 3(3)

3(3) + 4(4) = 5(5)

4(1 + 7)/2 = 4(4)

1 + 3 + 5 = 9

5(1 + 9)/2 = 5(5)

1 + 3 + 5 + 7 = 16

1 + 5 + 7 + 9

9 + 16 = 25

3 +

= 25

Theorem: If the square of every natural number n, is the sum of consecutive odd numbers, it cannot be decomposed into the sum of powers greater than 2, because in this case the difference of the squares must be a number square. Proof: n² = Σ n (2n – 1) 2² = 1 + 3 3² = 1 + 3 + 5 4² = 1 + 3 + 5 + 7 5² = 1 + 3 + 5 + 7 + 9 then 5² = (1 + 3 + 5) + (7 + 9) = 25 = 9 + 16 j^ + a^ = n^

j^ = n^ - a^ = (a + n)(n – a) = j(j) → j² = 1(a + n) j² = n² - a² →

j² + a² = n²

If the product of sum and difference it is the square number, then she is divisible by number which is square, as square difference two squares. 3= 3² = 4 + 5 → 3² + 4² = 5² = 25 = 9 + 16 5= 5² = 12 + 13 → 5² + 12² = 13² = 169 = 25 + 144 7= 7² = 24 + 25 → 7² + 24² = 25² = 625 = 49 + 576 9= 9² = 40 + 41 → 9² + 40² = 41² = 1681 = 81 + 1600 11 = 11² = 60 + 61 → 11² + 60² = 61² = 3721 = 121 + 3600 13 = 13² = 84 + 85 → 13² + 84² = 85² = 7225 = 169 + 7056 Can be to decompose on the sum squares these odd numbers, which as last term of a sum have squares odd numbers e.g. 9, 25, 49, 81, as well as all the squares of tenth power. e.g. 10², 20², 30², 40²,.. 6² + 8² = 10²

12² + 16² = 20²

18² + 24² = 30²

24² + 32² = 40²

13² = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23) + (25) = 169 = 25 + 144 25² = 625 = 49 + 576 (1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37+39+41+43+45+47)+(49) 1 + 3 = 4 + 5 = 9 + 7 = 16 + 9 = 25 + 11 = (36) + 13 = 49 + 15 = (64) + 17 = 81 + 19 = 100 + 21 = 121 + 23 = (144) + 25 = (169) + (27 + 29 + 31) = (256) + (33 + 35) = 324 + (37 + 39) = (400) + (41 + 43 + 45 + 47) = 576 + (49) = (625) + (51 + 53 + 55 + 57 + 59) = 900

n² = j² + a² 25 = 9 + 16 169 = 25 + 144 625 = 49 + 576 1681 = 81 + 1600 25

169

625

1681

3721

7225

12769

144

576

1600

3600

7056

12544

121

169

225

9 25 1

n² = j² + a² 10² = 36 + 64 20² = 144 + 256 30² = 324 + 576 40² = 576 + 1024

100

400

900

1600

2500

3600

4900

256

576

1024

1600

2304

3136

144

324

576

900

1296

1764

Σ(2n-1) = n(a + z)/2 = n(n) 256 = 16(1 + 31)/2 = 16(16)

1 1 1 1 1

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

5 5 5 5 5 5 5 5 5 5 5 5 5 5 4

7 7 7 7 7 7 7 7 7 7 7 7 7 9 0

9 9 9 9 9 9 9 9 9 9 9 9 16 0

11 11 11 11 11 11 11 11 11 11 11 25

13 13 13 13 13 13 13 13 13 13 36

15 15 15 15 15 15 15 15 15 49

19 19 19 19 19 19 19 81

17 17 17 17 17 17 17 17 64

225

196

169

144

121

21 21

256

100

Because the square of hypotenuse is sum of such quantity of successive odd numbers, as degree of square of hypotenuse, equation Pythagoras was can write as fraction: x 2   y 2  z 2 z 2 z 2 z 2 The common square denominator confirmed that the square of hypotenuse is the sum of the squares of legs. (1 + 3 + 5) + (1 + 3 + 5 + 7) = (1 + 3 + 5 + 7 +9) Σ(2n - 1)= n(a + z)/2 = n(n) 3(1 +5)/2 = 3(3) 4(1 + 7)/2 = 4(4) 3(3) + 4(4) = 5(5) 9 + 16 = 25

0 0

-10

0 4

-10

49 (z – y)(z + y) = x n = (z n  y n ) z

25 169 289 625 841 1369 1681 2809 3721 4225 4225 5329 7225 7225 7921 9409 10201 11881 12769 15625 18769 21025 21025 22201

n=2

(z – y)(z + y) = x (5 – 4)(4 + 5) = 9 (13 – 12)(12 + 13) = (17 – 15)(17 + 15) = (25 – 24)(25 + 24) = (29 – 21)(29 + 21) = (37 – 35)(37 + 35) = (41 – 40)(41 + 40) = (53 – 45)(53 + 45) = (61 – 60)(61 + 60) = (65 – 33)(65 + 33) = (65 – 63)(65 + 63) = (73 – 55)(73 + 55) = (85 – 77)(85 + 77) = (85 – 84)(85 + 84) = (89 – 39)(89 + 39) = (97 – 65)(97 + 65) = (101–99)(101 +99)= (109-91)(109+91) = (113-112)(113+112) = (125-117)(125+117) = (137-105)(137+105) = (145-143)(145+143) = (145-144)(145+144) = (149-51)(149+51) =

9 25 64 49 400 144 81 784 121 3136 256 2304 1296 169 6400 5184 400 3600 225 1936 7744 576 289 19600

30² - 24² = (30 – 24)(30 + 24) = 18² n(n) - a(a) = (n - a)(n + a) = j(j) 5(5) - 4(4) = (5 - 4)(5 + 4) = 3(3)

900

625 576

400 324 256 169 144 100 64 49 36 25 16 9

0 0

0,2

0,4

0 0,6

0,8

1,2

50 If Pythagorean threes come into being, when shorter leg is the square root of sum of hypotenuse and longer leg and such threes have not common divisor then they come into being also, when equation this we multiply about any total number L. x= → x² = y + z → x² + y² = z² 3= → 3² = 4 + 5 → 3² + 4² = 5² L(x) = → (Lx)² = L(Ly + Lz) → (Lx)²/L + (Ly)²/L = (Lz)²/L 2(3) = → [2(3)]² = 2[2(4) + 2(5)] → 6²/2 + 8²/2 = 10²/2 3(3) = → [3(3)]² = 3[3(4) + 3(5)] → 9²/3 + 12²/3 = 15²/3 Triple (j, a, n) is a Pythagorean triple if and only if it is a [Lj, La, Ln], for any positive integer L. Primitive Pythagorean triple is called, if j, a and n have no common divisor. e.g. (3, 4, 5), (5, 12, 13). Thus, each Pythagorean triples can be obtained by dividing the original by the greatest common divisor; and any Pythagorean three can be obtained from the original by multiplying all three elements of the corresponding same number of positive integer. From here the conclusion, that exist infinitely many primitive Pythagorean threes, so much how many natural numbers, and any Pythagorean threes, so much how many products infinitely many primitive Pythagorean threes of multiplied by infinitely many integer.

j(j) + a(a) = n(n) n(n) 10(10 17(17) 26(26) 20(20)

[L(j)]² + [L(a)]² = [L(n)]²

(n - a)(n +a) (10 - 8)(10 + 8) (17 - 15)(17 + 15) (26 - 24)(26 + 24) (20 - 16)(20 + 16)

j(j) 6(6) 8(8) 10(10) 12(12)

51 50(50) 34(34) 30(30) 52(52) 122(122)

(50 - 48)(50 + 48) (34 - 30)(34 + 30) (30 - 24)(30 + 24) (52 - 48)(52 + 48) (122 - 120)(122 + 120)

14(14) 16(16) 18(18) 20(20) 22(22)

Also prime numbers except 2 can introduce as product of difference and sum two natural numbers and they are then prime difference two square numbers. (a - n)(a + n) = p = a(a) - n(n) 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0

0 0,2

0,4

0,6

0,8

2² - 1² = (2 – 1)(2 + 1) = 1(3) 3 = (2 – 1)(2 + 1) 13 = (7 – 6)(7 + 6) p= [

5 = (3 – 2)(3 + 2)

5 1 2 5 1 2 ]  [5  ]  (3  2)(3  2) 2 5= 2 [

1,2

1(1) + 1(3) = 2(2)

7 = (4 – 3)(4 + 3)

17 = (9 – 8)(9 + 8) 19 = (10 – 9)(19 + 9)

p 1 2 p 1 2 ] [ p  ]  (a  b)(a  b) 2 2

3 =[

11 = (6 – 5)(6 + 5)

23 = (12 – 11)(12 + 11).

3 1 2 3 1 2 ]  [3  ]  (2  1)(2  1) 2 2

7 1 2 7 1 2 ]  [7  ]  (4  3)(4  3) 2 7= 2 [

This is Great Fermat's theorem for all values of n proved, because he is for all odd prime values of n valid.

52 Natural numbers, divisibility and prime Central notion within of natural numbers concerns divisibility, and more far order of primes natural number larger than 1, which has not natural divisor, that is to say, no different divisor except 1 or me alone. Sequence of primes has begun since 2, 3, 5, 7, 11, 13, 17, 19, 23,…Already Euclid proved before over 2 000 years, that this sequence does not end, and so there is no the largest prime. Beyond 2 all primes are odd with characteristic endings - 1 - 7 3 - 9. From second side is in force the main theorem of arithmetic: every natural number will give oneself unambiguously to introduce as product of primes. Primes gain by this on meaning for mathematics, as contribution to construction of all different numbers. The every number, which is not prime, will give oneself with these indivisible factors to to put together. Prime numbers 2 and 3 are components of all natural numbers really. Why, for example, can‘t all numbers be built simply by multiplying and adding together different combinations of the primes 2 and 3. e.g. 4 = 2(2), 5 = 2 + 3, 6 = 3 + 3, 7 = 2(2) + 3, 8 = 2(2)2, 9 = 3(3), 10 = 2(2) + 3 + 3, 11 = 2(2)2 + 3, 12 = 3(3)+ 3, 13 = 2(2)+ 3(3), 14 = 2(2)2 + 3 + 3, 15 = 3(3)+ 3 + 3, 16 = 2(2)2(2), 17 =2(2)2+3(3), 18 =2(2)2(2)+2, 19 =2(2)2(2)+3, 20 =2(2)2(2)+2(2), 21 =2(2)2(2)+2+3 22 = 2(2)2(2)+ 3 + 3, 23 = 2(2)2(2)+ 2(2)+ 3, 24 = 2(2)2(2)+ 2(2)2, 25 = 2(2)2(2)+ 3(3), 26 =2(2)2(2)+2(2)2+2, 27 =3(3)3, 28 =2(2)2(2)+3+3(3), 29 =2+3(3)3, 30 =3+3(3)3. All prime be built according to simple formula: p = n(2)+3, 5 =2+3, 7 =2(2)+3 11 =4(2)+3, 13 = 5(2) + 3, 17 = 7(2) + 3, 19 = 8(2) + 3, 23 = 10(2) + 3, 29 = 13(2) + 3. Formula this permits us to divide primes on two classes: they class of basic primes (2, 3, 5, 7), which alone for me are the building material and these, which are already the multiplicity of number 7. e.g.11 = 7+(4) 13 = 7+(6) 17 = 2(7)+(3) 19 = 2(7)+(5) 23 = 3(7) + (2) 29 = 4(7) + (1) And so we write new formula: p = n(7) + The rest (1,2,3,4,5,6)

p = 3 + n(2)

13 11 10 8

19 17 16 14

25 23 22 20

31 29 28 26

37 35 34 32

43 41 40 38

49 47 46 44

53 50

7 5 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 12 222 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1

9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53

53 It was can sequence of primes and write so: 2, 3, 5,-2- 7,-4- 11,-2- 13,-4- 17,-2- 19,-4- 23. In spaces among numbers notice hidden formula: - 2 - 4 - 2 - 4. These two last formulae, they will play further decisive part. Are there formulas that produce some of the prime? Here you are! p = n(2) + 3 2 = 1(2) + 0 3 = 0(2) + 3 5 = 1(2) + 3 7 = 2(2) + 3 11 = 4(2) + 3 13 = 5(2) + 3 17 = 7(2) + 3 19 = 8(2) + 3 23 = 10(2) + 3 13(2) + 3 = 29 14(2) + 3 =31 233 = 115(2) + 3 251 = 124(2) + 3. Irrefutable proof Mathematicians knew, however, that proving the Riemann Hypothesis would be of far greater significance for the future of mathematics than knowing that Fermat‘s equation has no solutions when n is bigger than 2. The Riemann Hypothesis seeks to understand the most fundamental objects in mathematics – prime numbers. The primes are those indivisible numbers that cannot be written as two smaller numbers multiplied together. The primes are the jewels studded throughout the vast expanse of the infinite universe of numbers that mathematicians have explored down the centuries. Their importance to mathematics comes from their power to build all other numbers. Every number that is not a prime can be constructed by multiplying together these prime building blocks /2 and 3/. Mastering these building blocks offers the mathematician the hope of discovering new ways of charting a course through the vast complexities of the mathematical world. Yet despite their apparent simplicity and principal character, prime numbers remain the most mysterious objects studied by mathematicians. They question about distribution of primes belonged to the most difficult. They were the long time then the question of plain theoretical nature, however today found primes in different realms the use. Suddenly the economic interest appears also the question, or proof the Riemann's hypothesis cans you something say about distribution of primes in world of numbers. If centuries of searching had failed to unearth some unknowing formula which would generate the list of prime numbers, perhaps it was time to adopt a different strategy. Look through a list of prime numbers, and you‘ll find that it‘s impossible to predict when the next prime will appear. The list seems chaotic, random, and offers no clues as to how to determine the next number. Can you find a formula that generates the numbers in this list, some unknowing rule that will tell you what the 10 000 000th prime number is? Not the question about quantity of primes in given interval of numbers, but the observation of spaces between two primes, she directed me on sure regularity from what they appear. 2, 3, 5,-2- 7,-4- 11,-2- 13,-4- 17,-2- 19,-4- 23 and so 2, 4, 2, 4, then the smallest space is among two primes and the decisive structure, recognizable in whole does not end sequence of primes. It after 23 number first 29 comes however in space 6 (23,- 2 -25,-4-29), because place between them is for first product of primes, number almost prime 25 = 5(5). Since then all almost prime numbers, as product of primes will take free place in sequence of primes, keeping spaces - 2 - 4 - 2 - 4. Generations have sat listening to the rhythm of the prime number drum as it beats out its sequence of numbers: two beats, followed by three beats, five, seven, eleven. As the beat goes on, it becomes easy to believe that random white noise, without any inner logic, is responsible. At the centre of mathematics, the pursuit of order, mathematicians could only hear the sound of chaos.

54 I do realize, that prime and almost prime numbers appear in interval two and fourth. If it walks about finding formulae and order, then primes are not more unequalled challenge. Knowing in what space sequent prime or relatively prime will appear, we can easily whole their list take down. And when we to this have yet the hand, as to qualify in sequence the sequent number, or it is prime or almost prime numbers, then and list of primes does not appear us as chaotic and accidental. The List of primes is, the heartbeat of mathematics, but a pulse wired regular in rhythm by multiplicity of seven in two – by – four steps. Fractions are the numbers whose decimal expansions have a repeating pattern. For example 1/7 = 0,142 857 142 857 At bases of distribution of primes in sequence of numbers, lies decomposition of their products on prime factors. According to Fermat small theorem numbers to power ( p - 1) minus one, they are divisible without the rest by prime. e.g. - 1 = 999 999/7 = ↓ - 142 857 857 142 Proof: if a ≠ p

p≥3

a≥2

= 64 – 1 = 63/7

= 729 – 1 = 728/7

1/7 = 0,142 857 142 857 1 … 2/7 = 0,2857 142 857 14 … 3/7 = 0,42857 142 857 1 … 4/7 = 0,57 142857 142857 1.. 5/7 = 0,7 142587 142587 1.. 6/7 = 0,857 142587 142587.. 8/7 = 1,142 857 142857 9/7 = 1,2857 142 857 14.. 10/7 = 1,42857 142 857 … 11/7 = 1,57 142 857 1428 … 12/7 = 1,7 142 857 14285 … 13/7 = 1,857 142 857 142 …

Similarly by fractions:

where the quotient in decimal expansion from some place after comma begins repeating in infinity six - digits numbers since 1, and finishing on 7. In practice this marks that every the the six - digit combination of numbers e.g. (x x x x x x)/ 7, (x y x y x y)/ 7, (y x y x y x)/ 7, (xyz xyz)/ 7, (zxy zxy)/ 7, (yzx yzx)/ 7, (zyx zyx)/ 7, (yxz yxz)/ 7, (xzy xzy)/ 7, and their multiplicities divide without the rest by 7. 111 111 111 111 111 111 / 7 = 15 873 015 873 015 873 0 1 2 3 4 5 6 7

1 2 3 4 5 6 7 8

2 3 4 5 6 7 8 9

3 4 5 6 7 8 9 0

4 5 6 7 8 9 0 1

5 6 7 8 9 0 1 2

6 7 8 9 0 1 2 3

7 8 9 0 1 2 3 4

This gets from here, that all numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,.. they are congruent to me according to module 7, as this shows following radar graph

a - 7/7 9

7 4

9 8 7 6

3 2 1 0

7 6 5 4 3 2 1 0

2 3 4 5 6 7 8 9 0

0 1

5 4 5 6 7 8 9

0 3

If difference among given number a, and prime is divisible by prime, then given number be a p 187  17  Vp complex. e.g.  10 p 17 Only difference among two primes divisible it is by 7, because p - (2,3,5,11,13,29) = n(7), and all primes be distributed according to multiplicity of number 7 (23 - 2)/ 7 = 3 (17 - 3)/ 7 = 2 (19 - 5)/ 7 = 2

(53 - 11)/ 7 = 6

(41 - 13)/ 7 = 4

(43 - 29)/ 7 = 2

Primes and almost prime follow after me in rhythm on two fourth. 1. 3. + 2 = 5.- 2 – 7 – 4 – 11 – 2 – 13 – 4 – 17- 2 – 19 – 4 – 23 - 2 – 25 – 4 – 29 – 2 – 31 -

56 Theorem: If difference between odd numbers, the cannot be written as product of two smaller numbers a and b with a, b> 1, divide without rest by 7, then a and b are prim and congruent modulo 7. Proof: a – b = n(p) then a ≡ b (mod p) p  p`mod p 11  53 mod 7 when 11 =1(7)+ 4 and 53 = 7(7)+ 4 then p and p` congruent according to mod p, and difference p`- p is multiplicity p. 53 - 11 = 42/7 = 6 This proof gives mathematics to instruction very quick procedure on qualification of primes about any quantity of places. p - (2,3,5,11,13,29) = n7 p = n(7) + R(1,2,3,4,5,6)

0 7

4 11

17 23 29 35

25 31

41 47

53 59 65 71

55 61

67 73

95 101 107 113

119

97 103

109 115

121

125

127

131

133

137 143 149 155

161

139 145

151 157

163 169 176

6 13

175

177

167 178

179

173 180

181

p'- p = n(7) 19 - 5 = 2(7) 47 - 19 = 4(7) 61 - 47 = 2(7) 89 - 61 = 4(7)

100

90 80 70 60 50 40 30 20 10 0

79 55 61 67 73 77 41 47 53 59 65 71

97 95

13 19 25 31 37 43 49 0 17 23 290 35 11 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 3 0 4 0 5 0 6 7 0 8 9 10 11 12 13 14

Discovery of rhythm beating the heart of mathematics storm the safety of system the RSA and so it turns out, that it 64 numerical factor with 129 numerical code is not number prime, because divisible it is by 7. 3 490 529 510 847 650 949 147 844 619 903 898 133 417 764 638 493 387 843 990820577:7 = 498 647 072 978 235 849 878 263 517 129 128 304 773 966 376 927 626 834 855 831 511 But second 65 numerical factors is prime (32769132993 266 709 549 961 988 190 834 461413 177 642 967 992 942 539 798 288 533 – 5):7 = 4 681304713323815649994569741547780201882 520 423 998 991 791 399 755 504 Prime numbers are numbers that are divisible only by one and themselves. They are the atoms of arithmetic, for any number is either a prime or a product of primes. The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, but despite their simple definition the prime numbers appear to be scattered randomly amid the integers. There is simple way to tell if a number is prime – than they cannot be written as product of two smaller numbers a and b, with a, b>1, and that is the basis for most modern encryption schemes. Solving the Riemann Hypothesis could lead to new encryption schemes and possibly provide tools that would make existing schemes, which depend on the properties of prime numbers, more vulnerable.

58 91 77 97

85 71

35 55 43

41 7 0

13 6

1 2

61 47

23 37

4 11

65 79

95 p, p', p" = 2n + p' 2, 3, 5 = 2 + 3, 7 = 4 + 3, p'" = n(7) + Rest(1,2,3,4,5,6), 11 = 7 + 4, 13 = 7 + 6, 17 = 2(7) + 3, 19 = 2(7) + 5, 23 = 3(7) + 2, 29 = 4(7) + 1, 2, 3, 5, -2- 7, -4- 11, -2- 13, -4- 17, -2- 19, -4- 23, 400 181

350 167 300

151 139

250 103

200 83

150

131

61 100 41 50 0

19 17

13 11 7

61 53

31 23

139 103

173 179

167

181 173

137

151

101 79

163

131 109

157

127

67 59

113 83

101

137

149

107

41 19

109

179 157

107

149

163

Primes do not possess except 1 and only number no factors, but number almost prime are almost so good, because they have at most two factors prime, or one prime and almost prime, or two almost prime e.g. 187 = 11(17) 343 = 7(49) 78 337 = 133(589). 23 is prime, but 25 = 5(5) it is almost prime. So alone numbers 35(5·7), 49(7·7), 55(5·11), 65(5·13), 77(7· 11), 85(5 · 17)

"p" = n(3) + (1,2)

25 = 8(3) + 1 35 = 11(3) + 1 49 =16(3) + 1 55 = 18(3) +1 77 = 25(3) + 2 85 = 28(3) + 1 91 = 30(3) + 1

65 = 21(3) + 2

95 85 335

125

329

121

325

305 295 301

289

287

143 115

319

299

293

145

119 323

133

275

155

161

175 259

235 185

203 215 253 265

169 187

245 205 247

217

251

209 221

Number almost prime built with prime numbers larger than three, they develop how splendid fan in infinity. 5,7,11,13,17,19,23,29,31,37,41,43,(p) = "p" 600 559 500

473

400 301

451

300 215 287

200

205 481

49 77121 43 100 25 35 55 5 7 185 169 11 91 143 37 13 65 0 17 35 85119 175 19 187 31 221 289 255 23 29 25 95 455 385 133 155 115 209 217 145 125 247 161 203 175 323 341 361 253 403 299 275 319 325 377 391 437 425 493 407

259

475

Sequences of numbers almost prime. (10)(20) 25 35 (14)(28) 49

60 55 65 77 85 95 115 125 145 155

91 119 (22)(44) 121 133 143 (52)(26) 169

161 175 185 205 215

187 203 217

209 221

235 245

247 259

253

265 275

(34)(68) 289

287 295 305

299 301 319

325 335

329 343

323 341

355 365

(76)(38) 361 371

377

385 395

391 407

415 425

403

413 427 437

445 455

451 469

475 485

473 481 497

505 515

511

493 517 527

(46)(92) 529

61 In interval what 30 numbers (10-40) on three numbers with ending 5 (15, 25, 35), two of them are almost prime. Primes and almost prime follow after me in interval what 2(p) and 4(p). 25 – 2(5) – 35 – 4(5) – 55 49 – 4(7) – 77 – 2(7) – 91 121 – 2(11) - 143 – 4(11) – 187 17 - 4(5) – 37 – 2(5) – 47 – 4(5) – 67- 2(5) – 77 – 4(5) – 97 – 2(5) – 107 – 4(5) – 127 – 2(5). Triangle of almost prime . 5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59

5 25 -22 -52 -34 -76 -46 100 -58 124 -70 148 -82 172 -94 196 106 220 118

7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59 35 55 65 85 95 115 125 145 155 175 185 205 215 235 245 265 275 295 49 77 91 119 133 161 175 203 217 245 259 287 301 329 343 371 385 413 -44 121 143 187 209 253 275 319 341 385 407 451 473 517 539 583 605 649 -26 169 221 247 299 325 377 403 455 481 533 559 611 637 689 715 767 -68 289 323 391 425 493 527 595 629 697 731 799 833 901 935 1003 -38 361 437 475 551 589 665 703 779 817 893 931 1007 1045 1121 -92 529 575 667 713 805 851 943 989 1081 1127 1219 1265 1357 -50 625 725 775 875 925 1025 1075 1175 1225 1325 1375 1475 116 841 899 1015 1073 1189 1247 2363 1421 1537 1595 1711 -62 961 1085 1147 1271 1333 1457 1519 1643 1705 1829 140 1225 1295 1435 1505 1645 1715 1855 1925 2065 -74 1369 1517 1591 1739 1813 1961 2035 2183 164 1681 1763 1927 2009 2173 2255 2419 -86 1849 2021 2107 2279 2365 2537 188 2209 2303 2491 2585 2773 -98 2401 2597 2695 2891 212 2809 2915 3127 110 3025 3245 236 3481

25 + 2(5) = 35 + 4(5) =55 + 2(5) = 65,… 49 + 4(7) = 77 + 2(7) = 91 + 4(7) = 119,.. 25 55 85 115 145 175 205 235 265 295 325 355 385 415 445 475 505 535 565

35 65 49 95 125 155 185 169 215 245 275 289 305 319 335 365 395 425 455 485 515 545 575

91 121

119 133

143 161 203 221

253 287

187 217 247

301

299 329

323 377 407 437 497 527

361 391

343 403

451 481 511

209

341 371 413 427 473

493

517 533

553

539 551

62 595 625 655 685 715 745 775 805 835 865 895 925 955 985 1015 1045 1075 1105 1135 1165 1195 1225 1255 1285 1315 1345 1375 1405 1435 1465 1495 1525 1555 1585 1615 1645 1675 1705 1735 1765 1795

605 635 665 695 725 755 785 815 845 875 905 935 965 995 1025 1055 1085 1115 1145 1175 1205 1235 1265 1295 1325 1355 1385 1415 1445 1475 1505 1535 1565 1595 1625 1655 1685 1715 1745 1775 1805

601

583

581 611

623 649 679

671 707 737 767

721

703

781

763 793

917 949 979

841 871 901 931 961

803 833

5 29

43 53

79 89

73 83

791 851

913 943 973 1003

37 47

31 41

61 71

689 749 779

817 847

893 923

869 899 959 989

1007 1037 1043 1067 1081 1073 1099 1111 1127 1141 1133 1159 1157 1189 1183 1219 1247 1261 1243 1253 1273 1309 1313 1339 1337 1351 1333 1343 1369 1363 1397 1411 1393 1403 1441 1457 1463 1501 1493 1519 1517 1513 1547 1561 1577 1591 1573 1603 1639 1651 1633 1643 1681 1673 1711 1703 1729 1727 1757 1771 1793

637 667 697

731

799

889

713

629

1001 1027 1057 1079 1121

1139 1147 1169 1177 1199 1211 1207 1241 1271 1267 1331

1349 1357 1379 1391 1387 1421 1417 1469 1477 1507 1529 1541 1537 1589 1631 1661 1691 1687 1717 1751 1781

25 35 49 55 65 77 85

13 17

1649 1679 1739 1769 1799

63 13

97 95 91 109 103 107 101 17 113 115 119 11 127 125 121 19 139 137 131 133 13 149 145 143 157 151 155 13 23 163 167 161 169 179 173 175 17 181 185 187 199 193 197 191 19 29 205 203 209 31 211 215 217 17 229 223 227 221 239 233 235 19 241 245 247 23 37 257 251 259 253 269 263 265 277 271 275 41 283 281 287 23 293 295 299 43 307 305 301 29 313 317 311 319 47 325 329 337 331 335 31 49 349 347 343 341 359 353 355 367 365 29 53 379 373 371 377 389 383 385 397 395 31 37 409 401 407 403 59 419 415 413 61 421 425 427 439 433 431 449 443 445 41 457 455 451 67 463 467 461 469 43 479 475 473 37 487 485 481 71 499 491 497 509 503 505

289

323

361

391

437

493

Theorem: If difference between odd numbers, the can be written as product of two primes, divide without rest by 3, then the number are almost prime and congruent modulo 3. Proof: a – b = n(3) then a ≡ b (mod 3) "p"  "p`" mod p 49  85 mod 3 when 49 =16(3)+1 and 85=28(3)+1, then "p" and "p`" congruent according mod p, and difference "p`"-"p" is multiplicity p. 85 - 49 = 36/3 = 12 „p‖- (5,7,11,13) = n(3)

"p" ≡ "p" mod 3, 91 - 49 = 14(3), 119 - 77 = 14(3), 161 - 143 = 6(3), 169 133 = 12(3), 299 - 287 = 4(3), 301 - 289 = 4(3),

305 299301 287295 275289 259 265 245 215221235247253 203209217 185 205

161 175 169 143155 125 119 133145 121 115 95 77 65 85 91 49 55 250 35 0 0 00 0 0 00 0 0 1 2 3 4 0 0 00 5 6 7 8 0 9 10 11 12 0 0 0 13 14 15 16 0 17 18 19 20 21 22 23 24 25 26 27 28 29 30

350 357 364 371 378 385 392 399 406 413 420 427 434 441 448 455 462 469 476 483 490 497

351 358 365 372 379 386 393 400 407 414 421 428 435 442 449 456 463 470 477 484 491 498

352 359 366 373 380 387 394 401 408 415 422 429 436 443 450 457 464 471 478 485 492 499

353 360 367 374 381 388 395 402 409 416 423 430 437 444 451 458 465 472 479 486 493 500

354 361 368 375 382 389 396 403 410 417 424 431 438 445 452 459 466 473 480 487 494 501

355 362 369 376 383 390 397 404 411 418 425 432 439 446 453 460 467 474 481 488 495 502

356 363 370 377 384 391 398 405 412 419 426 433 440 447 454 461 468 475 482 489 496 503

355

(76)(38) 361

365 377 391 395 403 415 425

407 413 427 437

445 451 455 475

473

469 481

485 493

65 504 511 518 525 532 539 546 553

505 512 519 526 533 540 547 554

506 513 520 527 534 541 548 555

507 514 521 528 535 542 549 556

508 515 522 529 536 543 550 557

509 516 523 530 537 544 551 558

510 505 517 515 524 531 538 535 545 552 559

517 (46)(92)

527

529

533 539

"p"+n(7)="p' " 35+ 2(7)=49+ 4(7)= 77 + 2(7) = 91 + 4(7) = 119 + 2(7) = 133 + 4(7) = 161 + 2(7) = 175 + 4(7) = 203 + 2(7) = 217 + 4(7) = 245 + 2(7) = 259 + 4(7) = 287 287 259 245 217 203 175 161 133 119

299 215 187

235 221

91 77

145

49 35 7

95 25

185

115 143

125 209

265

65 85 121 155 169 205 247

253

275 289

295

It will suffice only to look on table of primes and almost prime, to oneself about this to convince, what order and rules reign here. This would border about absurdity, these just fundamental elements of well ordered world of mathematics, if would they behave savagely and unforeseeable. Prime gaps What is a prime number gap? That's easy. A prime gap is the number of non-prime numbers between two primes. The most common form of prime gaps are the twin primes, two prime numbers p and p‘ = p + 2, In between which is a non-prime. 2617

2615 2611 2621

2633 2647 2657

2659 2663

2635 2645 2641 2651 2665

2627 2623 2629 2639 2653 2669

Above mentioned table shows, that primes and almost prime ranked according to 4 basic

66 numbers unity 9 – 3 – 7 – 1, divide number almost prime about number unity 5 on two complementary parts. Fact this is the denial so far general opinion, about irregularities of occurrence of primes and them the alleged decreasing on further of up growth of sequence numbers. Smaller or larger gaps among primes 2-4-6-8-14-18-20-22-24-26-30-34-36-44-52-60-72-8696-112-114-118-132-148-154-180-210-220-222-234-248-250-282-288-292-320-336-354382-384-394-456-464-468-474-486-490-500-514-516-532-534-540-582-588-602-652-674716-766-778-804-806-906,… be full with numbers almost prime, so as appearing primes in interval 2 and 4. Among primes 1327 and 1361 his place occupies 10 successive numbers divisible by primes, that is almost primes. 1327 + 4 = 1331/11 + 2 = 1333/31 + 4 = 1337/7 + 2 = 1339/13 + 4 = 1343/17 + 2 = 1345/5 + 4 = 1349/19 + 2 = 1351/7 + 4 = 1355/5 + 2 = 1357/23 + 4 = 1361 – 1327 = 34 Similarly is among primes 8467 and 8501. Among 370261 and 370373 we have gap about length 112. For p < N the largest at present well-known maximal gap equal m =1442, p = 804 212 830 686 677 669. 10 000 019+2 = 10 000 021/97+4 = 10 000 025/5+2 = 10 000 027/37+4 = 10 000 031/227 + 2 =10 000 033/397+4 = 10 000 037/43+2 = 10 000039/7+4 = 10 000 43/2089+2 = 10 000 045/5 + 4=10000049/47+2 =10 000 051/73+4 =10 000 055/5+2=10 000 57/79+4= 10 000 061/19+2 =10 000 063/17+4 = 10 000 067/7+2 = 10 000 069/181+4 = 10 000 073/31+ 2 = 10 000 075/5 + 4 = 10 000 079 – 10 000 019 = 60

2689 2699

2683 2693

2677 2687

2671

2681 2695 2705

2707 2719 2729

2713

2731 2741 2753

2 4 6 8 14 18

p 3 7 23 89 113 523

2717 2725 2735 2755 2765

2767

2701

2711

2749

2675

5 11 29 97 127 541

n 292 320 336 354 382 384

2723 2737 2747

2743 2759

2761

P 1453168141 2300942549 3842610773 4302407359 10726904659 20678048297

P 1453168433 2300942869 3842611109 4302407713 10726905041 20678048681

67 20 887 907 22 1129 1151 30 13063 13093 34 1327 1361 36 9551 9587 44 11633 11677 52 19609 19661 60 100000019 100000079 72 31397 31469 86 155921 156007 96 360653 360749 112 370261 370373 114 492113 492227 118 1349533 1349651 132 1357201 1357333 148 2010733 2010881 154 4652353 4652507 180 17051707 17051887 210 20831323 20831533 220 47326693 47326913 222 122164747 122164969 234 189695659 1899695893 248 191912783 191913031 250 387096133 387096383 282 436273009 436273291 288 1294268491 1294268779

394 456 464 468 474 486 490 500 514 516 532 534 540 582 588 602 652 674 716 766 778 804 806 906 1132 1308

22367084959 25056082087 42652618343 127976334671 182226896239 241160624143 297501075799 303371455241 304599508537 416608695821 461690510011 614487453523 738832927927 1346294310749 1408695493609 1968188556461 2614941710599 7177162611713 13829048559701 19581334192423 42842283925351 90874329411493 171231342420521 218209405436543 1693182318746371 749565457554371299

22367085353 25056082543 42652618807 127976335139 182226896713 141160624629 297501076289 303371455741 304599509051 416608696337 461690510543 614487454057 738832928467 1346294311331 1408695494197 1968188557063 2614941711251 7177162612387 13829048560417 19581334193189 42842283926129 90874329412297 171231342421327 218209405437449 1693182318747503 749565457554372607

Theorem about congruence odd number permits faultlessly to distinguish primes from another divisible numbers, that is almost prime. e. g. prime confirms the legitimacy of formula: p = 1 + n(7) p = 2 + n(7) p = 3 + n(7) p = 4 + n(7) p = 5 + n(7) p = 6 + n(7) 2 89 - 1 = 618 970 019 642 690 137 449 562 111 -3 618 970 019 642 690 137 449 562 108/7 = 88 424 288 520 384 305 344 937 444 (3 203 000 719 597 029 781 – 3) : 7 = 457 571 531 371 004 254 (810 433 818 265 726 529 159 – 5) : 7 = 115 776 259 752 246 647 022 and almost prime with numerous iterations inside, as and in quotient of formula ―p‖= 2 + n(3) 7 · 20408163265306122449 = 142 857 142 857 142 857 143 - 2 142 857 142 857 142 857 141/3 = 476 190 476 190 476 190 47 We happen in second factor of following expression sure unusual prime: 10 31 + 1 = 11· 909 090 909 090 909 090 909 090 909 091 = 10 000 000 000 000 000 000 000 000 000 001 Decomposition on primes her product, it lies at bases of iteration in this number. From this, that 1001 = 7 · 143 = 11 · 91 = 13 · 77 and 10 001 = 73 · 137 create following iterations. Products:

68 7 · 1001 = 7007 11 · 1001 = 11011 13 · 1001 = 13013 77 · 1001 = 77077 91 · 1001 = 91091 143 · 1001 = 143143 73 · 1001 = 73073 137 · 1001 = 137137 and 999 multiplicity 1001 e.g. 323 · 1001 = 323 323 and number 10 001, 43 ·10001= 430043 29 · 430 043 = 124 7 124 7 3 · 12 471 247 = 37 41 37 41 We see noteworthy iterations in prime 9 090 909 091 and her square, and so number almost prime 826 644 628 100 826 446 281 and prime 82 644 628 099 173 553 719, in which except iteration see two peers of numbers in reflection mirror. 355

343

361

365

335 329

49 353

323 331

319

293 281

283 277241 269 257

199 205

259

223 233

247

211 203

163

149 157

179 167

193 185

235

227

103 115

137

107

101

125

119

131

145 151

173

215

113 139 127

181 239 229

253 245 263

71 73

109

251

265

59 67

307 313

271

275

317

295

47 37

337 349

299

287

359

301

289

347 325

311 305

367

341

121 133

161 197

221 217

209

191

143 187

175

169

155

On radar graph number almost prime are visible on black background. 9999907 9999901

9999905 9999913

9999929 9999931

9999911

9999925

9999973

9999935 9999949

9999947

9999941

9999955

9999959

9999953

9999961

9999965

9999967

9999977

9999979

9999971 9999991

9999919

9999923

9999937 9999943

9999917

9999985

9999983

9999997

9999995

9999989

10000003 10000001 10000007 10000009 10000019

10000015 10000013 10000021 10000025

10000027

10000039 10000037 10000031 10000033 10000045 10000043 10000049 10000051

10000055 10000057

69 10000069 10000067 10000061 10000063 10000079

10000073 10000075 10000085 10000081 10000087 10000097 10000091 10000099 10000093

Let‘s apply so well-known us a formula to constructing successive primes and almost prime, that could generate this kind of pattern. (1,2,3,4,5,6) + n(7) = p 2 + 3(7) = 23 (1,2) + n(3) = „p― 1 + 8(3) = 25 9 999 901 = 1 428 557(7) + 2 9 999 905 = 3 333 301(3) + 2 9 999 907 = 1 428 558(7) + 1 9 999 913 = 3 333 304(3) + 1 9 999 929 = 1 428 561(7) + 2 9 999 923 = 3 333 307(3) + 2 9 999 931 = 1 428 561(7) + 4 9 999 925 = 3 333 308(3) + 1 9 999 937 = 1 428 562(7) + 3 9 999 935 = 3 333 311(3) + 2 9 999 943 = 1 428 563(7) + 2 9 999 949 = 3 333 316(3) + 1 9 999 971 = 1 428 567(7) + 2 9 999 977 = 3 333 325(3) + 2 9 999 973 = 1 428 567(7) + 4 9 999 985 = 3 333 328(3) + 1 9 999 991 = 1 428 570(7) + 1 9 999 997 = 3 333 332(3) + 1 10 000 019 = 1 428 574(7) + 1 10 000 015 = 3 333 338(3) + 1 10 000 079 = 1 428 582(7) + 5 10 000 085 = 3 333 361(3) + 2 9999907 9999901

9999905 9999911

9999929

9999917

9999925 9999937 9999931

9999923 9999941

9999947

9999955 9999965 9999971

9999949 9999953

9999961

9999977 9999995

9999959

9999967

9999985 9999991

9999919

9999935

9999943

9999973

9999913

9999979 9999983

9999989

9999997 10000001 10000007 10000003 10000009

10000019

10000015 10000025

10000013 10000021 10000027 10000031 10000037 10000033 10000039

10000045 10000055

10000043 10000049 10000051 10000057 10000061 10000067 10000063 10000069

10000079

10000075 10000085

10000073 10000081 10000087 10000091 10000097 10000093 10000099

Here are the primes amongst the 100 numbers either side of 10 000 000. For example in the 100 numbers immediately before 10 000 000 since 9 999 901 to 9 999 991 there are 9 primes, but look now at how few there are in the 100 numbers above 10 000 000: only 2 primes since10 000 001 to 10 000 099.

f(p"p") = p, p', p' + d, p" + 2d, d = 2,

f(p"p") = 2, 3, 3 + 2,

5 + 2,

7 + 4, 11 + 2, 13 + 4, 17 + 2,

53 50

49 47 43 41

40 37 35 31

29 25 23

19 17 13 11

10 7

0 1

5 3

2 11

2 21

2 25

2 27

2 33

2 35

2 41

2 43

2 47

2 49

Arithmetical sequence of primes and almost prime are sequence line and helical growing. P(p) = (2,3,5,11,13,29) + n(7)

59 = 3 + 8(7)

P("p") = (5,7,11,13) + n(3) 55 = 7 + 16(3)

140

2,3, -2- 5, -2- 7, - 4- 11, -2- 13,- 4 - 17,..

120 55 100

49 43

80 37 31

59 53

47 41

35 29

17 7

5 2

3 2 1

4 2

17 2

11 2

5 2

2 1

Serie3

Serie4 Serie2

Serie1

2 20 61 59

4 2

71 Helical sequence of primes and almost prime 3 7

5 11

13 19

23 31

47 53 61

49 55 59

65 71

73 79

83 89 97

77 85 91 95

101 103 107 109 113 115 121 127 125 131 133 137 139 145 151 149 157 155 163 161 167 169 173 175 181 179 187 191 193 197 199 205 211 209 217 223 221 227 229

119

143

185

203 215

Theorem: Primes and almost primes create twin sequences, growing evenly about absolute number 6. The Helical structure of arithmetical twin sequence of primes and almost primes show among two following terms constant difference, i.e. it exists such number d  R possessing property, which treat to all n  N: a n1 an  d 11 – 5 = 6 = 13 – 7 2 + 3 = 5 (2) 7 11 13 17 19

72 23

Theorem: It sequence of primes and relatively primes is the twin sequence from initial term p = 2 and constantly difference d = 6 (- 2 – 4). 2 + 3 = 5 -2- 7 -4- 11 -2- 13 -4- 17 -2- 19 -4- 23 -2- 25

2 55

31 25 19 13 7

3 5 11 17 23 29 35 41 47 53

Therefore though in Riemann‘s conjecture function of location of primes π (x), is the gradual function about high irregularity, as helical arithmetical twin sequence of primes and almost primes, which the difference the d = 6 is constant, it shows the amazing smoothness. 60

7 8 9 4 5 6 1 2 3 1 2 3 4 5 6 7 8 9

13 14 10 11 12

15 16

17 18 19

20 21

26 27 23 24 25

25 4

Serie5

Serie4

8 7

41 42 43

52 53 54

14 13

20 19

30 26

Serie3 Serie1 1

31 32 33

39 40

49 50 51

1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5

Serie6

Serie2

28 29 30

37 38 34 35 36

48 45 46 47

32 31

38 37

46 48

44 43

54 50

73 From here sequence of prime numbers is not similar to accidental sequence of numbers, but to well ordered structure. So basic numbers does not be definite per nature the method of accidental throw with coin. Accident and chaos they are for mathematician simply cruelty. P(p) = (2, 3 + R2 + R4), 2, 3+ 2 = 5, 3 + 4 = 7= 5 + 2, 7 + 4 = 11 + 2 = 13 + 4 = 17 + 2 = 19 + 4 = 23 + 2 = 25 + 4 = 29 + 2 = 31 + 4 = 35 + 2 = 37 + 4 = 41 + 2 = 43 + 4 = 47 + 2 = 49 + 4 = 53 + 2 = 55 140 131 125127 119121 113115 107109 101103

120

100

95 97 89 91 83

77 79 71 73 65 67 59 61

60 53 55 47 49 41 43

35 37 29 31

23 25 17 19 11 13 7 5 4 2 3 2 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Twin sequences of prime and almost prime numbers a congruent to me according to algebraic module 72. p'="p" mod p

65 = 13(2) + 13(3)

137 = 67(2) + 3

p' - "p" = n/p

137 - 65 = 72/2

65 387 381 375 369 363 397 385 391 379 373 385 367 361 379 395 373 389 383 367 377 371 361 365 359 355 349 343 337 331325319313307301295289 287 293 217 299 305 223 311 317 229 323 235 329 335 241 341 247 347 353 253 259 265 271 277 283

59 61

53 55

47 49

137 131

125 119 37 113 29 31 107 23 101 25 95 17 19 89 83 11 13 77 5 7 71 2 3 73 79 85 91 97 103109 115 121 127 133 139 35

215 221 227 233 239 245 251 257 263 269 275 281

143 149 145 155 161 151 167 173 157 179 163 185 191 169 197 175 203 209 181 187 193 199 205 211

Distribution of primes and almost prime according to rules of congruence of modules 7 and 3 is the reason, why these folded with 2 and 3 numbers be place on straight line of line, which confirms the legitimacy of the Riemann hypothesis. The uniformity from what rises the graph

74 of primes e.g.: by 100 000, he owes not quantity of primes to number N what can express with logarithmic function, but proportionate distributing, resulting from congruence of according to modules 7. 61

335

59 331

79 55

329

313

325

311 307

271

283

319

265

269

277

67 91 65 89

275

131

133

215

139

151

173

191

119

115

137

149

233 235

113

109

107

257

259

263

299 281

317 287

101

49 301

289

103

323 305

293

121

143

125

145

155

239 193

217

241 245

175

197

221

157 161

179

163

247 199

223

251

181

203

227

167

185

205 209

229

187

299

335

319

257

293

277 215

251

235

209

173

167

313 271 187

109

145

35 77

227

119

113 155 197 239 281 323

185

49 29

101 143

133 91

217 175

47 103

331 289 247 205 163 121

301

151 259

125

229

193

131

55 97

95 137

179 221

115

263

269 161

311

203

139 181

157

305

199 241

245 223

283

287 329

265

325

307

„Upon looking at these numbers, one has the feeling of being in the presence of the inexplicable secrets of creation.‖ /D. Zagier/ Are the primes distributed chaotically or can we find some means for computing them?

75 Will it ever be possible to predict with arbitrary accuracy where the next one lies? Yes, here you are! p + 6(7) = p‘, or almost prime ―p‖ The prime numbers are distributed not chaotically. All prime and almost prime numbers to be congruent modulo 7. Because the smallest gap between their equal 2 + 4 = 6, and 6(7) = 42 than is possible to predict with arbitrary accuracy that the next one lies what 42 gap. p + 7(6) = p' 5 -42- 47 -42- 89 -42- 131 -42- 173 -84- 257 -126- 383 -84-467 -42- 509 -,,, "p"+ 7(6) = "p' " 35 -42- 77 -42- 119 -42- 161 -42- 203 - 42- 245 -42- 287 -42- 329 -42- 371 -42- 413 -42- 455 42- 497 -42- 539 .. 545 503 505 463 461 523 527 421 419 485 481 379 377 439 443 335 337 401 397 359 355 295 293 313 317 253 251 275 271 211 209 233 229 169 191 187 167 529 521 487 479 127 149 437 145 125 395353 403445 85 361 83 107 103 319 311 269 61 41 43 65 235277 227 193 185 23 151 143 101 59 19 0 23 67 109 541499457415373331 25 425 509 289247205163121 79 3717 257299341 173215 383 467 5 47 89 131 7 35 77 13 11 294991 119 55 31 71 133 161 97 53 113 175 203 139 73 217 245 155 259 95 115 287 181 301 197 329 137 343 223 371 157 239 385 413 265 179 427 281 455 199 469 307 497 221 323 511 539 349 241 365 263 391 407 283 433 305 449 325 475 491 347 517 367 533 389 409 431 451 473 493 515 535

Primes and almost prime can settle according to their quantity. Such in a row creates fourteen the vertical groups and the innumerable amount of horizontal rows / periods / primes and almost prime. Length period 42 = 7(6) it is product of length of period all natural numbers and seven units about what grow primes and almost prime. In third and eighth group excepting prime 7, have only almost prime numbers. I II III IV V VI VII VIII IX X XI XII XIII XIV \5

7 25

11 29

49 67

131 151

155

193 215

197 217

235 257

199

239

241 263

281

283

209

247

211 233

251 271

289

169 191

229

269 287

167

205

245

127 149

187

227

265

125

163

203

85 107

145

185

223

121

161

43 65

103

143

181

221

259 277

157

119

101

139

179

115

137

175

113

133

173

109

253 275

293

295

76 299

301 319

341

305 323

343

307 325

347

311 329

349

313 331

353

317 335

355

337 359

Number 19, is prime in that case the sum 19 + 42 = 61 is prime too. Number 9 091, is prime in that case the sum 9 091 + 42 = 9 133 is prime too. 9 091 – 19 = 9 072 : 42 = 216 Number 909 091, is prime in that case the sum 909 091 + 42 = 909 133 is prime too. Number 909 090 909 090 909 090 909 090 909 091, is prime so I can predict with arbitrary accuracy that the next one lies in gap 42 = 909 090 909 090 909 090 909 090 909 133 – 43 = 909 090 909 090 909 090 909 090 909 090 : 42 = 21 645 021 645 021 645 021 6 45 021 645 909 1, is prime, because as successor have divisible even number by 4, 9092/4 = 2273 909 091, is prime, because as successor have divisible even number by 4, 909 092/4 = 227273 9 090 909 091 = 11 · 23 · 4093 · 8779 909 090 909 091 = 859 · 1 058 313 049 9 090 909 090 909 091 = 103 · 4013 · 21 993 833 369 909090909090909091, is prime, - 9090909090909090909092/4 = 2272727272727272727273 9 090 909 090 909 090 909 091, is prime – 9 090 909 090 909 090 909 092/4 909 090 909 090 909 090 909 091, is prime, - 909 090 909 090 909 090 909 092/4 9 090 909 090 909 090 909 090 909 091 = 59(154 083 204 930 662 557 781 201 849) 909 090 909 090 909 090 909 090 909 091 + 1 = 227272727272727272727272727273(4) is prime too. They are 4,6,18, 22,24, and 30 digits primes. One from 100 and 1000 million digits prime are 9.090909091e100 000 007 and 9.090909133e1 000 000 007. 8 264 462 809 917 355 371 900 826 446 281 They are 32 digits 90 909 090 909 090 909 090 909 090 909 091 is dividable by 11 and 105 831 304 899 989 415 869 510 001 058 313 049 38 digits numbers by 90 909 090 909 090 909 090 909 090 909 090 909 091 : 859 e32 „p― = e22 + e9 + e1 8 264 462 809 917 355 371 900 826 446 281 + 82 644 628 099 173 553 719 008 264 462 81 90 909 090 909 090 909 090 909 090 909 091 e38―p― = e26 + e10 + e2 Prime number records Last in September 2006 discovered, that the number 232582657-1 is prime. That‘s 9 808 358 digits long. Did you hear that computer scientists discovered the largest known prime number the other day? 2 to the power of 43,112,609 – 1! That‘s 12,978,189 digits long! The largest prime all of whose digits are prime numbers is 7532 (10^1104 – 1) (10^4 – 1)+1, discovered by Dubner in 1988. That`s 1,104 digits long. If 7,5327533E+1103 number is prime, because as predecessor have divisible even number by 4 - 7,5327532E+1103/4 = 1,8831883E+1103, that the largest about 90 580(1 104) = 7,5327533 E+100 000 319 digits long, and 905 798(1 104) = 7,5327533E+1 000 000 991 digits long is prime too. The largest prime with all digits equal to 0 or to 1 is 10^641 (10^640 – 1) / 9 + 1 that‘s 1,281 digits long. If 1,11001E+1280 number is prime, because as predecessor have divisible even number by 4 – 1,11000E+1280/4 = 2,77750E+1280, that the largest about 78 065 (1 281) = 1,11001E+100 001 264 digits long, and 780 641 (1 281) = 1,11001 E+1000 001 120 digits long is prime too.

77 Funnel of prime and almost prime numbers p ± 1 = 2n/4 3 + 1 = 4 5 - 1 = 4 7 + 1 = 8/4 11 + 1 = 12/4 13 - 1 = 12/4 17 - 1 = 16/4 19 + 1 = 20/4 23 + 1 = 24/4 159 151

147 123

119 115

111

107 103

95 91

83 79 71 67 59

75 63 51

47 43

31 23 19 11 7 3

27 15

2 1

143

139 131 127

135

±1 2

₌

161

160 156 152 148 144 140 136 132 128 124 120 116 112 108 104 100 96 92 88 84 80 76 72 68 64 60 56 52 48 44 40 36 32 28 24 20 16 12 8 4

155

157 149

145

129

85 77

105 93 81

61 53

57 45

41 37 29

17 13 5

p(p') ± 1

₌

5 19 25 29 35 49 55 59 65 79

77 85

89 95 109 115 119

9 7

2n/3

1 7 2 11 13 3 17 4 23 5 31 6 37 7 41 43 8 47 9 53 10 61 11 67 12 71 73 13 14 83 15 16 97 17 101 103 18 107 19 113 20

117

113 109 101 97 89

2n/4

141

137

133 125 121

162 158 154 150 146 142 138 134 130 126 122 118 114 110 106 102 98 94 90 86 82 78 74 70 66 62 58 54 50 46 42 38 34 30 26 22 18 14 10 6

153

9 15 21 27 33 39 45 51 57 63 69 75 81 87 91 93 99 105 111 117 121 123

78 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

127

125

131

133 137

139 145

151

143

149 157

155 163

167

169 173

181

175 179 185

191

187

193 197

199 205

211

203 209

129 135 141 147 153 159 161 165 171 177 183 189 195 201 207 213

Here the following odd numbers in rows of three (5, 7, 9) and nine columns are arranged according to the 4 characteristic unit numbers (11 -6 - 17 -6 - 23 -6 - 29) of primes and almost primes (511 527 533 539 545) according to the formula p ± 1 = 2n/4, p(p') ± 1 = 2n/3, shares an almost prime with unit number 5 in funnel-shaped two complementary parts. This formula correctly differences primes of other odd divisible numbers and any business selling prime numbers could realistically peddle their wares under the banner ―satisfaction guaranteed or your money back‖, without too much fear of going bust. Finally I found hidden behind primes and almost prime full secrets structure, after looking for the mathematicians for centuries, and her music can write on to infinity. Who knows this basic interval two four, two four, knows also where what note will come with prime or almost prime numbers. They appear not unpredictably on the number line. From now on we cannot longer talk about their apparent randomness of but rather on their timeless and universal character. 2 + 3 + 5 + 7 + 11,..+ 29 + 31 + 37 = 222

35 + 41 + 43 + 47 + 49,..+ 71 + 73 + 79 = 798

77 + 83 + 85,.. + 113 + 115 + 121 = 1386

119 + 125 + 127,.. + 155 + 157 + 163 = 1986

222 – 48(12) – 798 – 49(12) -1386 – 50(12) – 1986 ,..

14n + [14n + n(12)],..

p + 2(7) = p' + 4(7) = p"

3 + 2 = 5 + 2(7) = 19 + 4(7) = 47 + 2(7) = 61 + 4(7) = 89 + 2(7) = 103 + 4(7) = 131 + 2(7) = 145 + 4(7) = 173

100%

109

80%

115 121 127

101

151

137

179 157 163 169

143

185

60%

175

149

133

107

23 40% 7

20%

41 47

83 89

113 119 125

131

103

181

155 161 167

139

173

145

0% 1

They in this way grow with 14 of primes and almost prime built-up terms, arrange in exquisite mosaic illustrating their row in intervals 2 and 4 in arrangement of sevens. There are two facts about the distribution of prime numbers of which I hope to convince you so overwhelmingly that they will be permanently engraved in your hearts. The first is that, despite their simple definition and role as the building blocks of the natural numbers, the prime numbers same for me a balding blocks, that is to say every prime bigger than 3 the sum their predecessor 2, 3, 5, 11, 13, and 29 is, and n-the multiplicity of prime 7. They grow not like weeds among the natural numbers, seeming to obey no other law than that of chance, and nobody can predict where the next one will sprout. The second fact is even more astonishing, for it states just the opposite: that the prime numbers exhibit stunning regularity, that there are laws governing their behavior, congruence laws modulo 7, and that they obey these laws with military precision. To support the first of these claims, let me begin by showing you a list of the prime up to 100. I hope you will agree that there is apparent reason why one number is prime and another not. 2, 3, 2 + 3 = 5 5+2=7 2(2) + 7 = 11 2(3) + 7 = 13 3 + 2(7) = 17 5 + 2(7) = 19 2 + 3(7) = 23 1 + 4(7) = 29 3 + 4(7) = 31 2 + 5(7) = 37 13 + 4(7) = 41 29 + 2(7) = 43

5 + 6(7) = 47 11 + 6(7) = 53 3 + 8(7) = 59 5 + 8(7) = 61 11 + 8(7) = 67 29 + 6(7) = 71 3 + 10(7) = 73 2 + 11(7) = 79 13 + 10(7) = 83 5 + 12(7) = 89 13 + 12(7) = 97

I 2

II 3

III

IV 5

11 17

VI 7

VII

VIII

13 19

25 31

35 41

49 55

53 59

61 67

77 83 89 97 101

107

85 91 95

103 109 113 115 121 119 127 125

X 4 10 16 22 28 34 40 46 52 58 64 70 76 82 88 94 100 106 112 118 124

XI 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102 108 114 120 126

XII XIII 8 9 14 15 20 21 26 27 32 33 38 39 44 45 50 51 56 57 62 63 68 69 74 75 80 81 86 87 92 93 98 99 104 105 110 111 116 117 122 123 128 129

The numbers 2 and 3 are building blocks all natural numbers. Even indivisible by 2 and 3 prime and almost prime numbers can you from n(2) and n(3) to put together e.g. 2 + 3 = 5 2(2) + 3 = 7 4(2) + 3 = 11 5(2) + 3 = 13 7(2) + 3 = 17 8(2) + 3 = 19 10(2) + 3 = 23 5(2) + 5(3) = 25 9(3) = 27 The periodical table of natural numbers distinguishes 13 groups of even and odd numbers. In columns I - VII we have prime numbers appearing what n(7). e.g. 5 + 2(7) = 19 + 4(7) = 47 + 2(7) = 61 + 4(7) = 89 + 2(7) = 103 + 4(7) = 131 + 6(7) = 173 … In VI column except 7 we have free places on stepping out what n(7) almost prime numbers. e.g. 35 + 2(7) = 49 + 4(7) = 77 + 2(7) = 91 + 4(7) = 119 + 2(7) = 133 + 4(7) = 161 which are in VIII and IX column. 25 + 2(5) = 35 + 4(5) = 55 + 2(5) = 65 + 4(5) = 85 + 2(5) = 95 + 4(5) = 115 + 2(5) = 125 … 121 + 2(11) = 143 + 4(11) = 187 + 2(11) = 209 + 4(11) = 253 + 2(11) = 275 + 4(11) = 319 … In tenth and twelfth column we have even numbers, and in XI and XIII column even and odd numbers divisible by 3, following what 2(3). 101, 1 001=11(91), 100 001=11(9091), 10 000 001=11(909 091), 1.000 001E+99 999 999 103, 1 003=17(59), 100 003, 1 000 003, 1.000 003E+12,+18,+19,+99 999 999,+999 999 999 107, 1 007=19(53), 100 007, 1 000 007, 1.000 007E+99 999 999, E+999 999 999 109, 1 009, 10 009, 100 009, 1 000 009, 1.000 009E+99 999 999, E+999 999 999

81 113, 1 013, 10 013, 100 013, 1 000 013, 1.000 013E+99 999 999, E+999 999 999 115, 1 015, 10 015, 100 015, 1 000 015, 1.000 015E+99 999 999, E+999 999 999 119, 1 019, 10 019, 100 019, 1 000 019, 1.000 019E+99 999 999, E+999 999 999 121, 1 021, 10 021, 100 021, 1 000 021, 1.000 021E+99 999 999, E+999 999 999 125, 1 025, 10 025, 100 025, 1 000 025, 1.000 025E+99 999 999, E+999 999 999 127, 1 027=13(79), 100 027, 1 000 027, 1.000 027E+99 999 999, E+999 999 999 131, 1 031, 10 031, 100 031, 1 000 031, 1.000 031E+99 999 999, E+999 999 999 133, 1 033, 10 033, 100 033, 1 000 033, 1.000 033E+99 999 999, E+999 999 999 137, 1 037=17(61), 10 037, 100 037, 1 000 037, 1.000 037E+13, E+100 000 011 139, 1 039, 10 039, 100 039, 1 000 039, 1.000 039E+12, E+100 000 003, E+1000 000 000 2, 3, 5, 11, 13, 29 + n(7) = p 2 + 15(7) = 107

3 + 14(7) = 101

5 + 14(7) = 103

11 + 14(7) = 109

13 + 18(7) = 139

29 + 12(7) = 113

3 + 148(7) = 1 039

29 + 1430(7) = 10 039

5 + 142 862(7) = 1 000 039

5 + 142 857 142 862(7) = 1 000 000 000 039 5 + 142 857 142 862e99 999 999(7) = 1.000 000 039E+100 000 000 5 + 142 857 142 862e999 999 999(7) = 1.000 000 039E+1000 000 000 3 + 1(7) 2 + 14(7) 6 + 142(7) 4 + 1 428(7) 5 + 14 285(7) 1 + 142 857(7) 3 + 1 428 571(7) 2 + 14 285 714(7) 6 + 142 857 142(7) 4 + 1 428 571 428(7) 5 + 14 285 714 285(7) 1 + 142 857 142 857(7) 3 + 1 428 571 428 571(7) 2 + 14 285 714 285 714(7)

= = = = = = = = = = = = = =

10 100 1 000 10 000 100 000 1,00E+06 `1,00E+07 1,00E+08 1,00E+09 1,00E+10 1,00E+11 1,00E+12 1,00E+13 1,00E+14

82 6 + 142 857 142 857 142(7) 4 + 1 428 571 428 571 428(7) 5 + 14 285 714 285 714 285(7) 4 + 1,428 571 428e99(7) 4 + 1,428 571 428e999(7) 4 + 1,428 571 428e99 999 999(7)

= = = = =

1,00E+15 1,00E+16 1,00E+17 1,00E+100 1,00E+1000

1,00E+100 000 000

4 + 1,428 571 428e999 999 999(7) =

1,00E+1000 000 000

Factorization almost primes in prime factors Factorise large numbers on factors prime, it was in last 2000 years difficult problem. Majority mathematicians‘ is opinion, that factorisation numbers is fundamental extraordinary computational problem. One of main reasons, why the factorise numbers is so difficult; she was alleged fortuity of occurrence of primes. We know since, that primes and almost prime are present not accidentally, but according to rules of congruence of modules 7 and 3, we have also the way on factorise their products. It with theorems about congruence odd numbers results brightly, or number is prime or almost prime, and we for help of binomial formula easily will take apart every odd number on factors prime. We know, that difference among two successive square numbers state always odd number, we have such number from here to write down in some way as difference two squares and take out the common factor. Difference of 2 squares: a(a) – b(b) = (a – b)(a + b) p is common to both terms. Put this common factor outside the brackets. ―p‖ = p(p) ―p‖ = p(p + p`) 25 = 5(2 + 3) „p― = p(p´) = [(p + p‗)/2 – (p‗- p)/2][(p + p‗)/2 + (p‗- p)/2] 1067 = 11(97) = [(11+97)/2 – (97 – 11)/2][(11+97)/2 + (97 – 11)/2] = (54 – 43)(54+43) 19043 = 137(139) = [(137+139)/2 – (139 – 137)/2][(137+139)/2 + (139 – 137)/2] = (138 – 1)(138 + 1) 147 573 952 589 676 412 927 = 193 707 721(761 838 257 287) = [(193 707 721 + 761 838 257 287)/2 – (761 838 257 287 – 193 707 721)/2] [(193 707 721 + 761 838 257 287)/2 + (761 838 257 287 – 193 707 721)/2] (381 015 982 505 – 380 822 274 783)( 381 015 982 505 + 380 822 274 783) 35 = 6(6) – 1(1) = (6 – 1)(6 + 1) = 5(7)

55 = 8(8) – 3(3) = (8 – 3)(8 + 3) = 5(11)

143 = 12(12)–1(1) = (12 – 1)(12+ 1) = 11(13)

221 = 15(15)-2(2) =(15-2)(15 + 2) = 13(17)

253 = 17(17)–6(6) = (17 – 6)(17+ 6) = 11(23)

247 = 16(16)-3(3) =(16-3)(16 + 3) = 13(19)

83 If the difference between one number a, and a prime number is divisible by a prime number, then the number is complex. a p 287  7 e. g. 7(41) = (41 – 1)7 + 7  ( p'1)  40 p(p‘) = (p‘- 1)p + p p 7 The straight is dividing by 3 given number and to subtract from her rounded quotient without the rest. Then we level rounded quotient to the closest third multiplicity of prime numbers or almost prime. About the same number we round off received previously the difference to n the multiplicity of the same prime numbers or almost prime. And so we receive prime factors on what factorizing is number almost prime. 319 : 3 = 106 - 19 = 87 : 3 = 29

343 : 3 = 114 + 33 = 147 : 3 = 49

319 - 106 = 213 + 19 = 232 : 8 = 29

343 - 11 4 = 229 - 33 = 196 : 4 = 49

319 = 11(29)

343 = 7(49)

8051 : 3 = 2683 – 2392 = 291 : 3 = 97 8051 – 2683 = 5368 + 2392 = 7760 : 80 = 97 8051 = 83(97) 9 090 909 091 : 3 = 3 030 303 030 – 550 964 187 = 2 479 338 843 : 3 = 826 446 281 9090909091 – 3030303030 =6060606061 + 550964187 = 6611570248 : 8 = 826446281 9 090 909 091 = 11(826 446 281) 909 090 909 091 : 3 = 303 030 303 030 – 299 855 363 883 = 3174939147 : 3 = 1 058 313 049 909090909091 – 303030303030 = 606060606061 + 299855363883 = 905 915 969 944 : 856= 1 058 313 049 909 090 909 091 = 859(1 058 313 049) 9090909090909091 : 3 = 3030303030303030 – 2765519270373639 = 264783759929391 : 3= 88261253309797 9090909090909091 – 3030303030303030 = 6060606060606061 + 2 765 519 270 373 639 = 8 826 125 330 979 700 : 100 9 090 909 090 909 091 = 103(88 261 253 309 797) 9 090 909 090 909 090 909 090 909 091 : 3 = 3 030 303 030 303 030 303 030 303 030 - 2 568 053 415 511 042 629 686 697 483 462 249 614 791 987 673 343 605 547/3 = 154 083 204 930 662 557 781 201 849 9090909090909090909090909091 – 3030303030303030303030303030 = 6 060 606 060 606 060 606 060 606 061 + 2 568 053 415 511 042 629 686 697 483 8 628 659 476 117 103 235 747 303 544 : 56 = 154 083 204 930 662 557 781 201 849 9 090 909 090 909 090 909 090 909 091 = 59(154 083 204 930 662 557 781 201 849) 8 051 = 90(90) – 7(7) = (90 – 7)(90 + 7) = 83(97)

493=23(23)-6(6)=(23-6)(23+6) = 17(29)

341 = 21(21)-10(10) = (21-10)(21+10) = 11(31)

391 = 20(20)-3(3) =(20-3)(20+3)= 17(23)

84 529 = 23(20+3) 497 = 17(68+3) 1105 = 17(62 + 3) 1309 = 17(74 + 3) 1369 = 37(34 + 3)

25271 = 37(680 + 3)

1147 = 31(34 + 3)

734 591 = 11(66778 + 3)

8453= 79(107) 11111 = 41(271) 120481 = 211(571) 526313=281(1873) 322577= 163(1979) 434779=197(2207)

353357=307(1151) 10 000 043=2089(4787) 10 000 127= 167(59881)

370 267 = 479(773)

370 283 = 379(977)

370 289 = 349(1061)

370 297 = 353(1049)

370 303 = 367(1009)

370 319 = 547(677)

370 327 = 107(3461)

370 339 = 199(1861)

370 351 = 179(2069)

370 361 = 383(967)

370 309 = 67(5527)

370 313 = 47(7879)

370 273 = 43(8611)

370 301 = 29(12769)

370 333 = 37(10009)

370 369 = 23(16103)

370 243 = 17(21779) 370 249 = 11(33659)

370 253 = 13(28481)

370 271 = 11(33661)

370 277 = 17(21781)

370 291 = 19(19489) 370 331 = 13(28487)

370 337 = 11(33667)

370 379 = 17(21787)

370 343 = 59(6277)

370 279 = 7(52897)

370 381 = 11(33671)

370 307 = 7(52901)

370 321 = 7(52903)

370 349 = 7(52907)

370 363 = 7(52909)

9 999 913 = 7(1428559)

9 999 917 = 23(434779)

9 999 941 = 7(1428563)

9 999 947 = 19(526313)

9 999 949 = 31(322579)

9 999 971 =13(769229)

10 000 003 = 13(769 231)

10 000 007 = 941(10 627)

10 000 001 = 11(909 091) 10 000 009 = 23(434 783) 10 000 027 = 37(270 271)

10 000 013 = 421(23 753) 10 000 031 = 227(44 053)

10 000 021 = 97(103 093) 10 000 037 = 43(232 559)

10 000 039 = 7(1 428 577)

10 000 033 = 397(25 189)

10 000 043 = 2 089(4 787)

10 000 049 = 47(212 767)

10 000 061 = 19(526 319)

10 000 067 = 7(1 428 581)

10 000 081 = 7(1 428 583) 10 000 097 = 17(588 241)

10 000 091 = 251(39 841) 10 000 099 = 19(526 321)

10 000 093 = 53(188 681) 10 000 111 = 11(909 101)

10 000 123 = 7(1 428 589)

10 000 127 = 167(59 881)

10 000 133 = 11(909 103)

10 000 129=89(112361)

10 000 171 = 271(36901)

10 000 187 = 41(243907)

4 294 967 297=6 700 417(638+3) 1000001=101(9901) 8 547 008 547(13) = 111 111 111 111 7 709 321 041 217 = 25 271(305 065 927) 7 709321041217=(152 545 599-152520 328)(152 545 599+152 520 328)=25271(305065927)

85 All even numbers composed of two products of prime, they divide by the sum of two primes to prime factors. 2[p(p‘)]/(p + p) = p‘ 2(126 619)/(127 + 127) = 997 253 238/254 = 997 2(1 019 923)/(661 + 661) = 1543 2039846/1322 = 1543 We have with same they also fast way on qualification of primes, necessary to construction of code the RSA. No perceptible order and Riemann‘s Hypothesis Mathematicians since centuries listened intently in sound primes, and they heard unsettled tones only. These numbers resemble accidentally spilled notes on mathematical notes paper, without recognizable melody. Riemann sinusoidal waves what created right away zero Zeta they - showed scenery hidden harmony. Mathematicians despite all could with sure probability to estimate, how many prime numbers is in given interval. Only four in first ten are (2, 3, 5 and 7). It in first hundred is them 25, in first thousand 168 their part comes down from 40 by 25 on 16,8 percentage. Among smaller numbers from billion, 5% is the only just. To describe this down come of frequency of an occurrence in approximation the simple formula. From this however satisfied mathematicians are not. They want to know how far real occurrence numbers first deviates from counted frequency. Riemann in one's famous eight page paper ―On the Number of Prime Numbers less than a Given Quantity /"Über die Anzahl der Primzahlen unter einer gegebenen Größe"/ he wrote: "The known approximating expression F( x) = Li(x) is therefore 1 2

valid up to quantities of the order x and gives somewhat too large a value; But also the increase and decrease in the density of the primes from place to place that is dependent on the periodic terms has already excited attention, without however any law governing this behavior having been observed. In any future count it would be interesting to keep track of the influence of the individual periodic terms in the expression for the density of the prime numbers.‖ Real quantity of prime numbers differs from them counted frequency so alone often, as eagle near repeated throw with coin will fall out tails. Differently saying that is Riemann supposed, that occurrence prime numbers be subject to the rights of case. And he made a mistake here, because prime numbers be subject to the rights of congruence of according to module p‘≡ p (mod.7). He has written: ―One now finds indeed approximately this number of real roots within these limits, and it is very probable that all roots are real. Certainly one would wish for a stricter proof here.‖ Riemann Hipothesis

Riemann zeta- function for s = 0,5 + i * t.

86 The Riemann hypothesis (also called the Riemann zeta-hypothesis), along with suitable generalizations, is considered by many mathematicians to be the most important unresolved problem in pure mathematics. First formulated by Bernhard Riemann in 1859, it has withstood concentrated efforts from many outstanding mathematicians for 150 years (as of 2009). The Riemann hypothesis (RH) is a conjecture about the distribution of the zeros of the Riemann zeta-function ζ(s). The Riemann zeta-function is defined for all complex numbers s ≠ 1. It has zeros at the negative even integers (i.e. at s = −2, s = −4, s = −6, ...). These are called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, and states that: The real part of any non-trivial zero of the Riemann zeta function is ½ Thus the non-trivial zeros should lie on the so-called critical line, ½ + it, where t is a real number and i is the imaginary unit. The Riemann zeta-function along the critical line is sometimes studied in terms of the Z-function, whose real zeros correspond to the zeros of the zeta-function on the critical line. The Riemann hypothesis implies a large body of other important results. Most mathematicians believe the Riemann hypothesis to be true, a $1,000,000 prize has been offered by the Clay Mathematics Institute for the first correct proof. Unsolved problems in mathematics: Does every non-trivial zero of the Riemann zeta function have real part ½? Riemann mentioned the conjecture that became known as the Riemann hypothesis in his 1859 paper On the Number of Primes Less Than a Given Magnitude, but as it was not essential to his central purpose in that paper, he did not attempt a proof. Riemann knew that the nontrivial zeros of the zeta-function were symmetrically distributed about the line s = ½ + it, and he knew that all of its non-trivial zeros must lie in the range 0 ≤ Re(s) ≤ 1. In 1896, Hadamard and de la Vallée-Poussin independently proved that no zeros could lie on the line Re(s) = 1. Together with the other properties of non-trivial zeros proved by Riemann, this showed that all non-trivial zeros must lie in the interior of the critical strip 0 < Re(s) < 1. This was a key step in the first proofs of the prime number theorem. In 1900, Hilbert included the Riemann hypothesis in his famous list of 23 unsolved problems — it is part of Problem 8 in Hilbert's list, along with the Goldbach conjecture. When asked what he would do if awakened after having slept for five hundred years, Hilbert said his first question would be whether the Riemann hypothesis had been proven (Derbyshire 2003:197). The Riemann Hypothesis is one of the Clay Mathematics Institute Millennium Prize Problems. In 1914, Hardy proved that an infinite number of zeros lie on the critical line Re(s) = ½. However, it was still possible that an infinite number (and possibly the majority) of nontrivial zeros could lie elsewhere in the critical strip. Later work by Hardy and Littlewood in 1921 and by Selberg in 1942 gave estimates for the average density of zeros on the critical line.

87 The Riemann hypothesis and primes The zeta-function has a deep connection to the distribution of prime numbers. Riemann gave an explicit formula for the number of primes less than a given number in terms of a sum over the zeros of the Riemann zeta function. Helge von Koch proved in 1901 that the Riemann hypothesis is equivalent to the following considerable strengthening of the prime number theorem: for every ε > 0, we have

where π(x) is the prime-counting function, ln(x) is the natural logarithm of x, and the Landau notation is used on the right-hand side.[5] A non-asymptotic version, due to Lowell Schoenfeld, says that the Riemann hypothesis is equivalent to

The zeros of the Riemann zeta-function and the prime numbers satisfy a certain duality property, known as the explicit formulae, which shows that in the language of Fourier analysis the zeros of the Riemann zeta-function can be regarded as the harmonic frequencies in the distribution of primes. The Riemann hypothesis can be generalized by replacing the Riemann zeta-function by the formally similar, but much more general, global L-functions. In this broader setting, one expects the non-trivial zeros of the global L-functions to have real part 1/2, and this is called the generalized Riemann hypothesis (GRH). It is this conjecture, rather than the classical Riemann hypothesis only for the single Riemann zeta-function, which accounts for the true importance of the Riemann hypothesis in mathematics. In other words, the importance of 'the Riemann hypothesis' in mathematics today really stems from the importance of the generalized Riemann hypothesis, but it is simpler to refer to the Riemann hypothesis only in its original special case when describing the problem to people outside of mathematics. For many global L-functions of function fields (but not number fields), the Riemann hypothesis has been proven. For instance, the fact that the Gauss sum, of the quadratic character of a finite field of size q (with q odd), has absolute value

is actually an instance of the Riemann hypothesis in the function field setting. Consequences and equivalent formulations of the Riemann hypothesis The practical uses of the Riemann hypothesis include many propositions which are stated to be true under the Riemann hypothesis, and some which can be shown to be equivalent to the Riemann hypothesis. One is the rate of growth in the error term of the prime number theorem given above.

88 Riemann was particularly interested in feeding imaginary numbers into functions. Usually we can draw a graph of a function where the input runs along the horizontal and the output is the height of the graph. But a graph of an imaginary function consists of a landscape where the output is represented by the height above any point in the world of imaginary numbers. An imaginary landscape Riemann had found one very special imaginary landscape, generated by something called the zeta function, which he discovered held the secret to prime numbers. In particular, the points at sea-level in the landscape could be used to produce these special harmonic waves which changed Gauss's graph into the genuine staircase of the primes. Riemann used the coordinates of each point at sea-level to create one of the prime number harmonics. The frequency of each harmonic was determined by how far north the corresponding point at sea-level was, and how loud each harmonic sounded was determined by the east-west frequency. Riemann‘s sinus – waves what created with zero place Zeta – topography, they showed hidden harmony.

A prime number is a positive whole number greater than one which is divisible only by itself and one. The first few are shown above. If the definition doesn‘t mean much to you, think of prime numbers as follows: If you are presented with a pile of 28 stones, you will eventually deduce that the pile can be divided into 2 equal piles of 14, 4 equal piles of 7, 7 equal piles of 4, etc. However, if one more stone is added to the pile, creating a total of 29, you can spend as long as you like, but you will never be able to divide it into equal piles (other than the trivial 29 piles of 1 stone). In this way, we see that 29 is a prime number, whereas 28 is non-prime or composite. All composites break down uniquely into a product of prime factors: i.e. 28 = 2 x 2 x 7. Note that 2 is the only even prime - all other even numbers are divisible by 2. 1 is neither prime nor composite by convention.

Here the sequence of primes is presented graphically in terms of a step function or counting function which is traditionally denoted

. (Note: this has nothing to do with the value

=3.14159...) The height of the graph at horizontal position x indicates the number of primes less than or equal to x. Hence at each prime value of x we see a vertical jump of one unit. Note that the positions of primes constitute just about the most fundamental, inarguable,

89 nontrivial information available to our consciousness. This transcends history, culture, and opinion. It would appear to exist 'outside' space and time and yet to be accessible to any consciousness with some sense of repetition, rhythm, or counting. The explanation in the previous page involving piles of stones can be used to communicate the concept of prime numbers without the use of spoken language, or to a young child

By zooming out to see the distribution of primes within the first 100 natural numbers, we see that the discrete step function is beginning to suggest a curve.

Zooming out by another factor of 10, the suggested curve becomes even more apparent. Zooming much further, we would expect to see the "granular" nature of the actual graph vanish into the pixilation of the screen.

Now zooming out by a factor of 50, we get the above graph. Senior Max Planck Institute mathematician Don Zagier, in his article "The first 50 million primes" [Mathematical Intelligencer, 0 (1977) 1-19] states: "For me, the smoothness with which this curve climbs is one of the most astonishing facts in mathematics." (Note however that you are not looking at a smooth curve. Sufficiently powerful magnification would reveal that it was made of unit steps. The smoothness to which Zagier refers is smoothness in limit.) The juxtaposition of this property with the apparent 'randomness' of the individual positions of the primes creates a sort of tension which can be witnessed in many popular-mathematical accounts of the distribution of prime numbers. Adjectives such as "surprising", "astonishing", "remarkable", "striking", "beautiful", "stunning" and "breathtaking" have been used. Zagier captures this tension perfectly in the same article:

In 1896, de la Valee Poussin and Hadamard simultaneously proved what had been suspected for several decades, and what is now known as the prime number theorem:

In words, the (discontinuous) prime counting function

is asymptotic to the (smooth)

logarithmic function x/log x. This means that the ratio of to x/log x can be made arbitrarily close to 1 by considering large enough x. Hence x/log x provides an approximation of the number of primes less than or equal to x, and if we take sufficiently large x this approximation can be made as accurate as we would like (proportionally speaking - very simply, as close to 100% accuracy as is desired). The original proofs of the prime number theorem suggested other, better approximations. In the above graph we see that x/log x, despite being asymptotic to

, is far from being the

smooth function which suggests when we zoom out - there is plenty of room for improving the approximation. These improvements turn out to be greatly revealing.

The first improvement on x/log x we consider is the logarithmic integral function Li(x). This is defined to be the area under the curve of the function 1/log u between 2 and x, as illustrated in the left-hand figure. Gauss arrived at this from the empirical fact that the probability of finding a prime number at an integer value near a very large number x is almost exactly 1/log x. l'Hopital's rule can be used to show that the ratio of x/log x to

tends to 1

as x approaches infinity. Thus we may use either expression as an approximation to the statement of the prime number theorem. In the right-hand figure we see that this function provides an excellent fit to the function . As Zagier states, "within the accuracy of our picture, the two coincide exactly."

Zagier goes on to state: "There is one more approximation which I would like to mention. Riemann's research on prime numbers suggests that the probability for a large number x to be prime should be even closer to 1/log x if one counted not only the prime numbers but also the powers of primes, counting the square of a prime as half a prime, the cube of a prime as a third, etc. This leads to the approximation

or, equivalently,

The function on the right side . . . is denoted by R(x), in honor of Riemann. It represents an amazingly good approximation to as the above values show." To be clear about this, it should be pointed out that the explicit definition for the the function R(x) is

where are the Möbius numbers. These are defined to be zero when n is divisible by a square, and otherwise to equal (-1)k where k is the number of distinct prime factors in n. As 1 has no prime factors, it follows that (1) = 1.

It seems, then, that the distribution of prime numbers 'points to' or implies Riemann's function R(x). This function can be thought of as a smooth ideal to which the actual, jagged, prime counting function clings. The next layer of information contained in the primes can be seen above, which is the result of subtracting from R(x). This function relates directly to the local fluctuations of the density of primes from their mean density. In their article "Are prime numbers regularly ordered?", three Argentinian chaos theorists considered this function, treated it as a 'signal', and calculated its Liapunov exponents. These are generally computed for signals originating with physical phenomena, and allow one to decide whether or not the underlying mechanism is chaotic. The authors conclude

93 "...a regular pattern describing the prime number distribution cannot be found. Also, from a physical point of view, we can say that any physical system whose dynamics is unknown but isomorphic to the prime number distribution has a chaotic behavior." A physicist shown the above graph might naturally think to attempt a Fourier analysis - i.e. to see if this noisy signal can be decomposed into a number of periodic sine-wave functions. In fact something very much like this is possible. To understand how, we must look at the Riemann zeta function. Some numbers have the special property that they cannot be expressed as the product of two smaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called prime numbers, and they play an important role, both in pure mathematics and its applications. The distribution of such prime numbers among all natural numbers does not follow any regular pattern; however the German mathematician G.F.B. Riemann (1826 - 1866) observed that the frequency of prime numbers is very closely related to the behavior of an elaborate function ζ(s) = 1 + 1/2s + 1/3s + 1/4s +... called the Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation ζ(s) = 0 lie on a certain vertical straight line. This has been checked for the first 1,500,000,000 solutions. A proof that it is true for every interesting solution would shed light on many of the mysteries surrounding the distribution of prime numbers. ζ(ρ) = 0 0< Re (ρ)< 1

0 0 0

0 0

0 0 0

43 37 41

59 61

67 71

0 0 29 31 0 23 0 19 17 13 0 0 7 11 35 20 00 000000000000000000000000000000000000000000000000000000000000000

Primes seem to be, at the same time very irregularly distributed among all numbers, and yet – if squinted at from a sufficiently far distance – they reveal an astoundingly elegant pattern.

ζ(s) = 0

0 < Re(s) < 1

97 89 83 79 73 71 67 61 59 53 47 43 41 37 31 29 23 19 17 13 11 7 5 3 2

0 0

0,2

0,4

0 0,6

0,8

1,2

He also formulated a conjecture about the location of these zeros, which fall into two classes: the ―obvious zeros‖ -2, -4, -6, etc., and those whose real part lies between 0 and 1. Riemann‘s conjecture was that the real part of the non obvious zeros is exactly ½. That is, they all lie on a specific vertical line in the complex plane. Over 2,300 years ago Euclid proved that the number of primes is infinite, so two possible questions come to mind: 1. How many primes are there less than the number x? 2. There are infinitely many primes, but how big of an infinity? This document will focus on the first question. π(x) is the number of primes less than or equal to x Let x be a positive real number. The question "how many primes are there less than x?" has been asked so frequently that its answer has a name: [π (x) using the Greek letter pi] = π(x) = the number of primes less than or equal to x. The primes under 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23 so π(3) = 2, π(10) = 4 and π25) = 9.

π(10) = 4 = 7(0,57143), π(100) = 25 = 7(3,57143), π(1 000) = 168 =7(0,57143)6(7)

199 197 179 193 191 167 139 149 157 163 173 181 137 127 151 131

109 107113 79 89 97 103 5967 101 73 83 19 293743475361 71 0 0 17 410 0 3 07 0 13 2331 0 0 2 5 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 3 0 0 4 5 6 0 00 0 0 0 0 0 7 8 9 0 0 0 10 11 12 0 13 14 15 16 17 18 19 20 21

Values of π(x) x 10 100 1,000 10,000 100,000 1,000,000 10,000,000 100,000,000 1,000,000,000 10,000,000,000 100,000,000,000 1,000,000,000,000 10,000,000,000,000 100,000,000,000,000 1,000,000,000,000,000 10,000,000,000,000,000 100,000,000,000,000,000 1,000,000,000,000,000,000 10,000,000,000,000,000,000 100,000,000,000,000,000,000 1,000,000,000,000,000,000,000 10,000,000,000,000,000,000,000 100,000,000,000,000,000,000,000

π(x) 4 25 168 1,229 9,592 78,498 664,579 5,761,455 50,847,534 455,052,511 4,118,054,813 37,607,912,018 346,065,536,839 3,204,941,750,802 29,844,570,422,669 279,238,341,033,925 2,623,557,157,654,233 24,739,954,287,740,860 234,057,667,276,344,607 2,220,819,602,560,918,840 21,127,269,486,018,731,928 201,467,286,689,315,906,290 1,925,320,391,606,803,968,923

reference

[LMO85] [LMO85] [LMO85] [DR96] [DR96]

96 On the number of primes less than a given magnitude In 1859 the German mathematician Bernard Riemann proposed a way of understanding and refining that pattern. Its main result is a suggestion, not rigorously proved, for a perfectly precise formula giving the number of primes less than a given quantity. His hypothesis has wide – ranging implications, and this day after 150 years of careful research and exhaustive study we know it is correct. There are 4 primes up to 10 (2, 3, 5, 7), because those they cannot be expressed as the product of two smaller numbers (4 = 2(2), 8 = 4(2), 9 = 3(3), 10 = 5(2). Between 1 and 100 there are 25 primes, and 168 primes up to 1 000. Why 168? Is there a rule, a formula, to tell me how many primes there are less than a given number? Can we find a rule, a law, to describe the thinning out? There is a function π(n) that returns the number of all prime numbers from 2 up to a predetermined limit. But as can be π(n) calculated. The formula is simple: The ratio of half a given number by a given number N, is directly proportional to the quotient of quantity prime numbers by its dual quantity. ½N : N = πx : 2(πx) πx ∝ ½N ½N πx) = N(πx) πx/(2 πx) = y = ½ 2πx(½) = πx N πx 1,0 E+3 168 1,0 E+6 78498 1,0 E+9 50847534 1,0 E+12 37607912018 1,0 E+15 29844570422669 1,0 E+18 24739954287740860 1,0 E+21 2112726946018731928 A two – column table like this is an illustration of a function. The main idea of a function is that some number depends on some other number according to some fixed rule or procedure. Another way to say the same thing is: a function is a way to turn (―maps‖) a number in to another number. The function πx ∝ ½N turns, or maps, the number 1000 in to the number 168 – again, by way of some definite procedure. 500(336) = 1000(168)Therefore primes there are less than a given number sure do thin out, but are directly proportional to the half a given number. Ultimately, it is in the Riemann Hypothesis about the multiplicative basic building blocks of natural numbers to understand: the primes. Can their distribution in the sea of natural numbers mean? How long do you calculate until the next prime coming? Why is the next prime number, such as accidental times already after a few steps, sometimes on the other hand, only after great distance? Is there perhaps a hidden pattern? Theorem: If the quotient of half a given magnitude ½N, by a given magnitude N, is equal to the quotient of quantity prime numbers, by its dual quantity, so in this equation is a proportional relation, that is, in each equation the product of the inner members equal to the product of outer members. Proof:

½N : N = πx : 2(πx) 5/10 = 4/8 = 1/2

πx ∝ ½N 8(1/2) = 4

½N(2 πx) = N(πx) πx/(2 πx) = Re(s) = ½ 2πx (½) = πx 

1 / 2 N 2x  N

97 ½N : N = πx : 2πx = (N – 4)/6 : 2(N – 4)/6 = n(pp‘) : 2n(pp‘) πx + (N - 4)/6 + ,½N - [πx + (N - 4)/6+- = ½N

9 592 + 16 666 + 23 742 = 50 000 500 000 000 500 000 5000000 50 000 000

500

5 000

50 000

0 1

166

2 105

23 742

254 836

2 668 755 27 571 879

16 16 666

166 666

1666666 16 666 666 166 666 666

9 592

78 498

664 579

282 485 800

166 1 666

25 168

1 229

5 761 455 50 847 534 8

If in several of the same ratios product outer members, is equal to the product of the inner members, then they say about so-called continuous proportion. Therefore, more equal ratios with each other can also be written, as a continuous proportion. ↔ ½N : πx : (N – 4)/6 : n(pp‘) = N : 2πx : 2(N – 4)/6 : 2n(pp‘) ½N - [πx + (N - 4)/6] = n(pp') 50 - [25 + 16] = 9 πx = ½N - [(N - 4)/6 + n(pp')] 168 = 500 - [166 + 166]

0 9

166 2 105

23 742

254 836

2 668 755 27 571 879 282 485 800

4 25 168

1 229

166

1 666

9 592

16 666

78 498

664 579

166 666

1666666 16 666 666 166 666 666

5 761 455 50 847 534

98 The above mentioned bar diagram shows, as from asymptote diminishing / the sum / quantity of primes and odd numbers( the blue column) in half of given magnitude, monotonously grows up the difference among half of given magnitude and quantity of primes and odd numbers, which it is the rest of products of primes( the pink column). If the interval first 10 numbers come before 4 primes, then in the interval of 100 numbers 25 prime numbers occur, and each number in the interval, the proportion ½ is to keep. Observe of proportion ½ in every the block N of numbers is, so the guarantee, that in necessity primes never will disappear because asymptote diminishing quantity of primes and odd numbers in half of given magnitude causes monotonous growth of quantity of products primes. In other words, all zeros of ζ(s) in the half-plane Re(s) > 0 have real part ½ . In mathematics, two quantities are said to be proportional if they vary in such a way that one of the quantities is a constant multiple of the other, or equivalently if they have a constant ratio. Proportion also refers to the equality of two ratios. In proportional quantities is the doubling (tripling, halved) one quantity is always a double (triple, halve) connected to the other quantities. The proportion of ½ means, that is involved in the creation of a half-block of numbers twice the amount of prime numbers. 5/10 = 4/8 50/100 = 25/50 500/1000 = 168/336 1=3–2

5=3+2

7=5+2

9=7+2

11 = 9 + 2 13 = 11 + 2 15 = 13 + 2 17 = 15 + 2 19 = 17 + 2 21 = 19 + 2 23 = 21 + 2 25 = 23 + 2 39 = 37 + 2 53 = 51 + 2 67 = 65 + 2 81 = 79 + 2 95 = 93 + 2

27 = 25 + 2 41 = 39 + 2 55 = 53 + 2 69 = 67 + 2 83 = 81 + 2 97 = 95 + 2

29 = 27 + 2 43 = 41 + 2 57 = 55 + 2 71 = 69 + 2 85 = 83 + 2 99 = 97 + 2

31 = 29 + 2 45 = 43 + 2 59 = 57 + 2 73 = 71 + 2 87 = 85 + 2

33 = 31 + 2 47 = 45 + 2 61 = 59 + 2 75 = 73 + 2 89 = 87 + 2

35 = 33 + 2 49 = 47 + 2 63 = 61 + 2 77 = 75 + 2 91 = 89 + 2

37 = 35 + 2 51 = 49 + 2 65 = 63 + 2 79 = 77 + 2 93 = 91 + 2

It is easy to see that 2, 3 and 5 is the next prime number in the top ten numbers. The next consecutive prime numbers is always divided into odd number, which is the arithmetic average and a multiple of three. It is of great importance for the deployment on both sides of primes. These triplets consecutive odd numbers to 100 (N) is 100 - 4 / 6 = 16, including 25 primes and 9 almost prime, giving a total of half a given quantity ½ N. 50 = 16 + 25 + 9 The question arises whether it is possible to allocate such an elementary function, which give the number of primes of a given size N. How distributed are the prime numbers among natural numbers best describes the function:

½N : N = πx : 2(πx) 5

500

5000

πx ∝ ½N

50000

500000

5000000

50000000 500000000

4 25 168 1229 9592

78498

664579

5761455

50847534

All zeros of ζ(s) in the half-plane Re(s) > 0 have real part ½. ½ N/N = πx/2πx [p + (p+8)]/2 = 2n – 1 [5 + (5 + 8)]/2 = 9 [11 + (11 + 8)]/2 = 15

then πx =

then 2(2n – 1) = [p + (p+8)] 2(9) = [5 + (5 + 8)] 2(15) = [11 + (11 + 8)]

p = [2(2n – 1) – 8]/2 5 = [2(9) – 8]/2 11 = [2(15) – 8]/2

The Riemann Hypothesis is about the distribution of primes in the sea of natural numbers. This sea is defined over the sum, because of numbers will always be number 1 add – just the normal process of counting. The primes, however, are about the multiplication defined, they are about the factorization the prime multiplicative components of the natural numbers. The distribution of primes and the Riemann Hypothesis says something about the changing relationship between addition and multiplication of natural numbers. Both are not problems for themselves, but both together are incredibly complex and still not fully penetrated, such as the lack of evidence for the Riemann Hypothesis impressive displays. All these ideas are based on an analogy, which is easier to describe something like this lets: The primes are ―elementary particles‖, which are about the multiplication in interaction occur and so the composite numbers up. At the same time, ―the particles‖ are arranged through the addition. In the zeta functions are now in the form of a sum – relatively product formula both aspects (additive/natural numbers and multiplicative/primes) linked. 2 + 3 = 5 + 2 = 2(2) + 3 = 7 + 3 + 2 = 12 = 6(2) 2 + 3 = 5 + 4(2) = 13 = 5(2) + 3 + 5 = 18 = 9(2) 4(2) + 3 = 11 + 4(2) = 19 = 8(2) + 3 + 11 = 30 = 15(2)

100 7(2) + 3 = 17 + 4(2) = 25 = 5(2 + 3) + 17 = 42 = 21(2) 10(2) + 3 = 23 + 4(2) = 31 = 14(2) + 3 + 23 = 54 = 27(2) 13(2) + 3 = 29 + 4(2) = 37 = 17(2) + 3 + 29 = 66 = 33(2) πx ∝ ½N

½N(2πx) = N(πx)

2πx(½) = πx 8 · 0,5 = 4 50 · 0,5 = 25 97 89 73 41 17 2 3 11 19 43 59 67 83

103 79 71 47 31 23 7 5 13 29 37 53 61

101

Riemann for help of total numbers translated distribution of prime numbers in mathematical scenery on two-dimensional plane (so called zeta-function). It topography of this scenery contains near this general knowledge about prime numbers. It will suffice, so to know on level of sea points (zero places), to can reconstruct whole scenery. Because zero places contain all information about distribution of prime numbers Riemann created concrete formula, to right away zero to regain distribution of prime numbers. Near what every zero place is how source for spreading wave we which can introduce me as acoustic sound. Sounds of all zero places overlap on me in distribution of prime numbers. Near what zero place is about so many louder, if it lies further eastwards (in right from axis y), and her sound is about so many higher, if it lies further north (over axis - x). Assuming proportional functions graphically in a coordinate system, so you can see that proportional functions are monotonically increasing. The properties of the zeroes out in the complex plane determine the properties of the primes! Riemann conjectured that all the relevant zeroes have real part ½ . The statement that the equation πx/2(πx) = y = ½, is valid for every x with real part equal ½, with the quotient on the right hand side converging, is equivalent to the Riemann hypothesis.

101

πx ∝ ½N ½N(2πx) = N(πx) 8 · 0,5 = 4 50 · 0,5 = 25 336 · 0,5 = 168

πx/(2 πx) = y = ½ 2πx(½) = πx

2,5E+18 y = 0,5x R² = 1

2E+18

1,5E+18 1E+18 5E+17 0 -2E+18

0 -5E+17

2E+18

4E+18

6E+18

They fill with the same gap in thousands theorems basing on legitimacy Riemann‘s hypothesis. Because many mathematicians be obliged for its results such presumption to accept. Primes betrayed their secret, and by this was proved Riemann‘s Hypothesis A solution of the Riemann Hypothesis are huge implications for many other mathematical problems. The transformation of hypothesis the Riemann in theorem, suddenly it proves all the not proved results. Riemann Hypothesis admits to receive, so that really every from infinitely of many, of zero places lies on this straight line then it means, that all sounds in music of prime numbers are alike loud. This would mean, it that was can distribution of prime numbers to me really introduce how even throw dice. Hexahedron dice after line of natural numbers rolls, which what second and fourth wall shows next prime number or almost prime. 5_7__11_13__17_19__23_25=5·5__29_31__35=5·7_37__41_43__47_49=7·7__53_55=11·5 __59_61__65 = 13·5_67__71_73__77 = 11·7_79__83_85 = 17·5__89_91 = (13·7)_95 = 19·5

102 100%

90% 80% 70%

19 13

60% 7 50%

40%

21 0

0 0

99 0 105

95 101

10% 2 0%

15 0

30% 20%

97 103 109

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

The above diagram illustrates what makes a ½ real part of primes for a given quantity π 100. Well, two parallel sequence of 25 primes and 9 of their product (25, 35, 49,) with a constant gap 6, which gives 16 odd numbers divisible by 3, as their mean. In every blocks of natural numbers in proportion ½ that πx primes, almost prime p(p) and odd numbers divisible by 3 (N – 4)/6. And so the ½ ratio is maintained in subsequent blocks of numbers. Look again at the table showing the proportion of primes amongst all odd numbers.

1,0 1,0 1,0 1,0 1,0 1,0

N 100 E+03 E+06 E+09 E+12 E+15 E+18

πx 25 168 78 498 50 847 534 37 607 912 018 29844570422669 24739954287740860

(N- 4)/6 16 166 166 666 166 666 666 166 666 666 666 166666666666666 166666666666666666

n(pp’) Σ½N 9 50 166 500 254 836 500 000 282 485 800 500 000 000 295 725 421 316 500 000 000 000 303488762910665 500000000000000 308593379045592474 500000000000000000

If the quotient of half a given magnitude ½N, by a given magnitude N, is equal to the quotient of quantity prime numbers, by its dual quantity, then half a given magnitude is the sum quotients of common divisor. That is to say, the proportion of odd number in given magnitude is the sum of 3 successive proportion. ½N : N= πx : 2πx =(N – 4)/6 : 2(N – 4)/6=½N – [πx+(N – 4)/6] : 2{½N – [πx+(N – 4)/6]}= k then πx + (N – 4)/6 + [½N – (N – 4)/6] = k{2πx + 2(N – 4)/6 + 2{½N – [πx + (N – 4)/6]} ½

½ ½

50/50 = 25/50 + 16/50 + [50 – (25 +16)]/50

The following table provides an overview of this

103 . πx (N - 4)/6 p(p) 1/2 N 25/50 16/50 9/50 50/50 168/500 166/500 166/500 500/500 1229/5000 1666/5000 2105/5000 5 000/5000 9592/50000 16 666/50000 23742/50000 50 000/50000 78498/500000 166 666/500000 254836/500000 500 000/500000 664579/5000000 1 666666/500000 2668755/500000 5 000000/5000000 5761455/50000000 16666666/50000000 27571879/50000000 50000000/50000000 50847534/500000000 166666666/500000000 282485800/500000000 500000000/500000000 πx + (N - 4)/6 + ,½N - [πx + (N - 4)/6+- = ½N 8 7 6 5 4 3 2

166 666 666

50 847 534 50 000 000 27 571 879 16 666 666 5 761 455 5000000 2 668 755 1666666 664 579 500 000 254 836 166 666 78 498 50 000 23 742 16 666 9 592 5 000 2 105 1 666 1 229 500 166 166 168 50 9 16 25

9 592 + 16 666 + 23 742 = 50 000 500 000 000 282 485 800

Half of given magnitude ½N (purple column) it is sum diminishing quantity of primes πx (the blue column) and the quotient of odd numbers (N - 4) : 6( the pink column), as well as the growing products of primes n( pp') ( the green column) resulting with difference among half of given magnitude and sum of quantity of primes and odd numbers {½N - [πx + (N - 4)/ 6]}. The down mentioned bar diagram shows how does constantly diminishing quantity of primes πx (green bar) and the quotient of odd numbers (N - 4)/6 (pink bar), they in total enlarge growing difference of primes products (blue bar). If the half of given magnitude as member of geometrical sequence is geometrical mean two neighboring terms, then every of these terms is the sum of several components. =

5 000 = 500 + (10²)45

104 50 =

½N = (N – 4)/6 + 2(N – 4)/6 + 2

50 = (100 – 4)/6 + 2(16) + 2

5 000 = (10 000 – 4)/6 + 2(10 000 – 4)/6 + 2 = 1 666 + 2(1 666) + 2 = 1 666 + 3 334 n(pp') + (N - 4)/6 + πx = ½N = 3(N - 4)/6 + 2 = (N - 4)/6 + 2(N - 4)/6 + 2 9 + 16 + 25 = 50 = 3(16) + 2 = 16 + 2(16) + 2 1

30 128 391 582 532 16 666 666 666 666 3 204 941 750 802

2 987 267 896 695

295 725 421 316

29 215 278 521

2 878 280 823

1 666 666 666

282 485 800

166 666 666

27 571 879

16 666 666

2 668 755

1666666

254 836

23 742

1 666 666 666 666

166 666 666 666 37 607 912 018 16 666 666 666 4 118 054 813

166 666 16 666

2 105

1 666

166

1 0

(N – 4)/6 + (10 – 4)/6 + 1 + 10(1,5) + 16 + 10²(1,5) + 166 + 10³(1,5) + 1666 + 10´(1,5) + 16 666 + 10µ(1,5) + 166 666 + 10¶(1,5) + 1 666 666 + 10·(1,5) + 16 666666 + 10¸(1,5) + 166 666666 + 10¹(1,5) + 1666666666+

346 065 536 839

455 052 511

5 000 000 000

50 000 000

664 579

5000000

78 498

500 000

9 592

50 000 5 000 500

n(pp’) = 0 = 0 = 9 = 9 = 157 = 166 = 1939 = 2105 = 21637 = 23 742 = 231094 = 254 836 = 2 413 919= 2 668 755 = 24 903 124 = 27 571 879 = 254910921 = 282 485800 = 2595795023= 2878280823=

50 000 000 000

5 761 455

[πx + [4 + [4 + [21 + [25 + [143 + [168 + [1061 + [1229 + [8363 + [9592 + [68906 + [78 498 + [586 081 + [664 579 + [5 096 876 + [5 761 455 + [45074079 + [50 847534 + [404204977+ [455052511+

500 000 000 000

500 000 000

168

5 000 000 000 000

50 847 534

1 229

166

50 000 000 000 000

2(N – 4)/6 +2] 2(10 – 4)/6+2] 2(1) + 2] (4) 3(10)]+ 2(16) +2] (34) 3(10²)++ 2(166)+2] (334) 3(10³)++ 2(1666)+2](3334) 3(10´)++ 33 334] 3(10µ)++ 333 334] 3(10¶)++ 3 333 334] 3(10·)++ 33 333 334] 3(10¸)++ 333 333334] 3(10¹)++ 3 333 333334]

= ½N = = 5 = = 5 = = 10(4,5) + = 50 = 5(10) = 10²(4,5) + = 500= 5(10²) = 10³(4,5) + = 5 000 = = 10´(4,5) + = 50 000 = = 10µ(4,5) + = 500 000 = = 10¶(4,5) + = 5000000 = = 10·(4,5) + = 50000000 = = 10¸(4,5) + =500000000= = 10¹(4,5) + =5000000000=

3(N – 4)/6 + 2 3(10 – 4)/6+ 2 3(1) + 2 3(16) + 2 3(166) + 2 3(1666) + 2 3(16666) + 2 3(166666) + 2 3(1666666)+2 3(16666666)+2 3(166666666)+2 3(1666666666)+2

If half of the given magnitude is growing in geometrical sequence 5(q), then the sum of difference between prime numbers and almost prime in geometrical sequence 3 (q).

105

f(½N) = 5(10ⁿ⁻¹) f(5000) = 5(10³) f,*πx' - πx) + [n'(pp') - n(pp')+- = 3(10ⁿ⁻¹) f{(1229-168) + (2105-166)- = 3(10³) 9

282485800

50 847 534

27571879

5 761 455

2668755

664 579

254836

78 498

23742

9 592

2105

1 229

166

168

1 0

4 +d(2²) 8 +d(2²) 12 +d(2²) 16 +d(2²) 20 +d(2²) 24 +d(2²) 28 +d(2²) 32 50+q(10) 500+q(10) 5000+q(10)50000+q(10)500000+q(10)5000000+q(10)50000000+q(10)500000000

16270321088

161320740 15949896 1569960 153 472 14 748 1344 100 0%

10%

20%

30%

40%

50%

60%

70%

100

1344

14748

153472

1569960

Serie1

Serie2

168

1229

9592

78 498

664 579

Serie3

500

5 000

50 000

500 000

80%

90%

15949896 161320740 28

100%

162703210 88 32

5 761 455 50 847 534

5 000 000 50 000 000 500 000 00

Although the sum of the differences between prime and almost prime growing in geometrical sequence 3 (q) equals half of the given magnitude growing in geometrical sequence 5 (q), as shown in the above diagram, since the prime and almost prime are components of half a given magnitude, and steadily declining prime sequence (green bar), results in steady increase in the almost prime sequence (blue bar). Here we see the interdependence of all numbers in the half of

106 the given magnitude of asymptotically decreasing number of primes, in hidden arithmetical sequence with constant difference d = 2². In 50 odd half figures / 25 / is the prime number, the rest (16 + 9) is product of primes. This is the (N - 4) / 6 = 16 multiples of 3 (9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99) and 9 (pp ') other products of prime numbers (25, 35, 49, 55, 65, 77, 85, 91, 95). Multiples of the number 3 in given magnitude are always quotient of the number 6 in the difference (N - 4) as there are intervals fixed every 6 numbers. 50 - 16 = 34 = 25 + 9 Subtracting from the half of given magnitude multiple receive the remainder consisting of prime numbers and their product. ½ N - [N - 4] / 6 = πx + n (pp ') This equation is very interesting, because we consider their information content, it is on the right side of an expression, which is made up of all prime numbers and their products and on the left side is an expression that is composed of all odd numbers. This means however, that in the infinite sum the information on primes implicitly built is. So the differences and the sum of the numbers making up half of a given magnitude must be equal. The sum of prime numbers and their products is growing steadily by the difference between the half of given magnitude and number of quotients the number 6 in given magnitude. Asymptotically decreasing the number of primes is always half difference between prime numbers and their products, causes steady increase in products of primes, too by half difference between them. Differences and the sum of these two equations are equal to half the sum of primes and their products, as shown in the diagram below. [πx - n(pp')]/2=n (25-9)/2=8 n(pp')±n=[n(pp') + πx]/2=n ± πx 9 + 8=(9 +25)/2 =25 - 8 166 + 1 = 167 = 168 - 1 2105 - 438 = 1667 = 438 + 1229 9

282 485 800

115 819 233

166 666 667

115 819 233

50 847 534

27 571 879

10 905 212

16 666 667

10 905 212

5 761 455

2 668 755

1 002 088

1 666 667

1 002 088

664 579

254 836

92 169

166 667

92 169

78 498

23 742

2 105

166

2 1 0

7 075 438 1

8 2

1 667 167

17 2

7 075

16 667

438

1 229

1 8

9 592

168 25 4

In this way, proportion ½ of primes, almost prime and odd numbers take up 100% of the area of 100 blocks numbers. ½ N = πx + {½ N – [πx + (N-4)/6]}+(N-4)/6 50 = 25 + [50 – (25 +16)] + 16

107 Theorem: The term ― ‖ of an arithmetic sequence with first term a, and constant difference ―d‖ is given by the explicit formula + (n – 1)d = a + (n – 1)d 11 d = 12 = 11 + (9 – 1)12 = 11 + 8(12) = 107 a = 7 d = 12 = 7 + (9 – 1)12 = 7 + 8(12) = 7 + 96 = 103

N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

2 5 17 29 41 53

3 7 19 31 43 67 79

89 101 113

103

137 149

127 139 151 163

173 197

199 211 223

233 257 269 281 293

11 23

37 47 59 71 83 107

61 73 97 109

131 157 167 179 191

227 239 251 263

271 283 307

181 193

229 241

277

311

313

317 331 347 359

353 367 379 389 401

337 349 373

383 397 409

2n - 1 9 21 33 45 57 69 81 93 105 117 129 141 153 165 177 189 201 213 225 237 249 261 273 285 297 309 321 333 345 357 369 381 393 405

= k 15 27 39 51 63 75 87 99 111 123 135 147 159 171 183 195 207 219 231 243 255 267 279 291 303 315 327 339 351 363 375 387 399 411

n(5)

n(7)

n(11) n(13)

n(17)

n(19) n(23)

25 35 49 55 65 85 95 115 125 145 155

77 91 119 133

121 143

161 175 185 205 215

169 187

203 217

209 221

235 245 265 275 295 305 325 335 355 365 385 395

253

247

259 287 301

289 299 319

329 343

323

341 361

371 377 391 407

403

108 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77

439 449 461

421 433 457

463 487 499

509 521

419 431 443 467 479 491 503

523 541 547

557 569

563 571 587 599

593 617 641 653

577

607 619 631 643

677

647 659

601 613

661 673

683 691

701

709 719 727 739 751

733 743 757 769

761 773 787 797 809 821

811 823

827 839

829 853

857

859

863 877

881

883 907 919

887 911

417 429 441 453 465 477 489 501 513 525 537 549 561 573 585 597 609 621 633 645 657 669 681 693 705 717 729 741 753 765 777 789 801 813 825 837 849 861 873 885 897 909 921

423 435 447 459 471 483 495 507 519 531 543 555 567 579 591 603 615 627 639 651 663 675 687 699 711 723 735 747 759 771 783 795 807 819 831 843 855 867 879 891 903 915 927

415 425 445 455

413 427 437 451 469

475 485 505 515

473

481 493

497 511

517 527

535 545 565 575

539

533

553

551 559

581 595 605 625 635

529

583

589 611

623 637

629 649

655 665 685 695 715 725 745 755 775 785 805 815 835 845 865 875

671

667

679 689

697

707 721

703 713 731 737

749 763

767 781

791

779 793

803

799 817

833 847

841 851 869

871

889 895 905 925

901 913 917

923

893

899

109 78 79 80 81 82 83

929 941 953

937 947 967

971 983

977 991

997

a ''= a' + (n - 1)d

933 945 957 969 981 993

939 951 963 975 987 999

935

931 949

955 965 985 995

943

959 973

961 979 989

a' = 13 n = 8 d = 12 97 = 13 + (8 - 1))12 tn = a + (n - 1)d n = 8 a = 5 tn = 5 + 12n - 12 = 12n - 7 t8 = 12(8) - 7 = 89

120 100 80 60 40 20 0

Serie1

Serie2

Serie3

Serie4

47 37

101

103

107

109

Theorem: Given number ―a‖ is prime, if their even part equal is to the number 2, 4, 8, 10, or continue term in their arithmetic sequence with constant difference d = 12 (2, 4, 8, 10) + n(12) a = (2, 4, 8, 10) + n(12) = p = n(2) + 3 173 – 3 = 170 = 2 + 14(12) 2 + n(12) 2 14 26 38 50

4 + n(12) 4 16 28 40 64 76

86 98

100

8 + n(12) 10+ n(12) 8 10 20 34 44 56 58 68 70 80 94 104 106

191 – 3 = 188 = 8 + 15(12) 2 5 17 29 41 53

3 7 19 31 43 67 79

89 101

103

11 23

13 37

47 59 71 83 107

61 73 97 109

110 110

113 124 136 148 160

134 146 170

128 154 164 176 188

178 190

137 149

127 139 151 163

131 157 167 179 191

173

181 193

Perfect sieve 2

37 47

25 35 49

109

105

111

117

123

129

135

141

147

143

153

159

155

165

171

103 127

137

139

149

151 163

173

107 131

157 167

177

183

191

193

189

195

199

201

207

211

213

219

209 221

257 269

271

281

283

115

119

121 133 145 169

175 185

187 203

205

215

217

227

229

225

231

239

241

237

243

251

249

255

263

261

267

273

279

275

285

291

287

289

297

303

295

299

301

309

315

321

327

319

323

325

277

293 307

161

181

233

125

179

223

317

113

197

= k

89 101

2n - 1

311

313

235 245

247

253

259

265

305

111 331 347 353

337

333

339

329

349

345

351

341

357

363

369

375

365

381

387

377

359 367 379

373 383

335 343 355

361 371 385

389

397

393

399

391

395

401

409

405

411

403

407

419

421

417

423

413

415

431

433

429

435

425

427

441

447

437

453

459

467

465

471

479

477

483

473

487

491

489

495

485

493

499

503

501

507

497

505

513

519

525

531

537

543

533

549

555

545

561

567

573

579

585

591

601

597

603

613

609

615

619

621

627

631

633

639

645

651

661

657

663

673

669

675

681

687

693

699

705

711

717

723

713

729

735

725

741

747

737

753

759

749

439

443

449 461

457 463

509 521

523 541 547

557 569

563 571

577 587

593

599 607

617 641

643

653

647 659

677

683 691

701

709 719 727 739 751

733 743 757

445 451

455 469

475

511 535

481

515

517

527

529

539 551

559

553 565

575 581

583

589

595 605

611

629

623

625

635

637 649

655 665

667

671

679 689

685 695

703

697

707

715

721 731 745 755

112 761

769

765

771

763

767

777

783

775

779

781

789

795

791

793

801

807

803

805

813

819

815

817

825

831

837

843

833

835

849

855

845

847

861

867

873

879

885

891

897

903

893

909

915

905

921

927

917

933

939

931

945

951

943

957

963

955

971

969

975

983

981

987

997

993

999

1009

1005

1011 1001 1003 1007

1019

1021

1017

1023

1031

1033

1029

1035 1025 1027

773 787 797 809

811

821

823

827

829

839 853 857

859

863 877

881

883 907

887 911

919 929

937

941

947

953 967 977 991 1013

785 799

841 851 865

869

871

875 889

895

899

901 913

923

925

935 949 959

965

961 973

979 989

985 995

1015

Perfect sieve similarly how sieve Eratosthenes be bases on principle that all natural numbers congruent to me according module 0 mod. 6. Arranging one after another only odd numbers have begun for primes 2, 3 what second and fourth (5 - 7 - 11 - 13), because third and fifth is multiplicity number 3 always, we will receive four sequences of primes about constant difference d = 12 (17 19 - 23 - 25). We choose 25 = 5(5)now, then the number almost prime, as the multiplicity number 5, second will be about 2(5) = 10 larger, that is 25 + 10 = 35 = 7(5), third about 4(5) = 20 larger, that is 35 + 20 = 55 = 11(5) and fourth again about 2(5) = 10 larger, that is 55 + 10 = 65 = 13(5) and further already in constant space, what 5(12) = 60 from every of them 25 + 60 = 85 35 + 60 = 95 " we sow‖ all multiplicities of prime 5. 49 = 7(7), then number almost prime, as multiplicity number 7, second will be about 4(7) = 28 larger, that is 49 + 28 = 77 = 11(7), third about 2(7) = 14 larger, that is 77 + 14 = 91 = 13(7) and fourth again about 4(7) = 28 larger, 91 + 28 = 119, and further in constant space, what 7(12) = 84 we sow all multiplicities of prime 7. (49 + 84 = 133 77 + 84 = 161 91 + 84 = 175 …) 121 = 11(11), then number almost prime, as multiplicity number 11, second will be about 2(11) = 22 larger, 121 + 22 = 143 = 13(11), third about 4(11) = 44 larger, 143 + 44 = 187 = 17(11), and

113 fourth about 2(11) = 22 larger, 187 + 22 = 209 = 19(11), and further in constant space, what 11(12) = 132 we sow all multiplicities of prime 11. (121 + 132 = 253 143 + 132 = 275 …) 169 = 13(13), then the number almost prime, as the multiplicity number 13, second will be about 4(13) = 52 larger, 169 + 52 = 221 = 17(13), third about 2(13) = 26 larger, 221 + 26 = 247 = 19(13), fourth about 52 larger, 247 + 52 = 299 = 23(13), and further in constant space, what 13(12) = 156 sow all multiplicities of prime13. (169 + 156 = 325 = 25(13) 221 + 156 = 377 = 29(13) …) In the same way we sow all remaining multiplicities of next primes. Theorem: If even half of an odd number after deducting the half even part threefold prime factor 3 · p = (3p – 1)/2 = (7, 10, 16, 19, 25, 28, 34, 43, 46, 55, 61, 64, 70, 79), is divisible by (5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53). This certainly is a complex number. Proof: [p(p‘) – 1]/2 – (3p – 1)/2 = n/p [p(p‘) – 1]/2 –(7, 10, 16, 19, 25, 28, 34, 43, 46, 55, 61, 64, 70, 79) = n/(5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53) 2009 - 1 = 2008 / 2 = 1004 - 10 = 994 / 7 = 142 1067 - 1 = 1066 / 2 = 533 - 16 = 517/11 = 47 437 - 1 = 436 / 2 = 218 - 28 = 190/19 = 10 961 - 1 = 960 / 2 = 480 - 46 = 434/31 = 14 n(3) 4 10 16 22 28 34 40 46 52 58 64 70 76 82 88 94 100 106 112 118 124 130 136

2009 = 7 (287) = 7 (284 + 3) 1067 = 11 (97) = 11 (94 + 3) 437 = 19 (23) = 19 (20 + 3) 961 = 31 (31) = 31 (28 +3)

n(3) n(5) n(7) n(11) n(13) n(17) n(19) n(23) n(29) n(31) 7 7 10 16 19 25 28 34 43 46 13 12 19 17 25 24 31 27 37 32 43 42 38 49 47 45 55 61 57 59 60 67 62 66 73 72 71 79 77 85 80 84 91 87 97 92 93 103 102 101 109 107 108 104 115 110 121 117 127 122 126 123 133 132 129 139 137

114 142 148 154 160 166 172 178 184 190 196 202 208 214 220 226 232 238 244 250 256 262 268 274 280 286 292 298 304 310 316 322 328 334 340 346 352 358 364 370 376 382 388 394

145 151 157 163 169 175 181 187 193 199 205 211 217 223 229 235 241 247 253 259 265 271 277 283 289 295 301 307 313 319 325 331 337 343 349 355 361 367 373 379 385 391 397

147 152 162 167 177 182 192 197

143 150

144 149 159

164 171

170 180

185 188 195 203

207 212 222 227

161

201

206 213 218 225 234

237 242 252 257

236

240 245

248 255

258 263

267 272 282 287

269

266

276

275 279

290 297 302 312 317

264

291

294 305

311 318

314 324

327 332 342 347 357 362 372 377 387 392

335

333

339 344

348

353 360

351 356 365 368

374 381

383 390

395

389 396

115 400 406 412 418 424 430 436 442 448 454 460 466 472 478 484 490 496

403 409 415 421 427 433 439 445 451 457 463 469 475 481 487 493 499

402 407 417 422 432 437

401

399 408

416 423

420 425 434

435

444 447 452 462 467

450

446

451 458 465

461 474

477 482 492 497

449

471

479 481

480 489 494

½N = 25 p + 16(2n - 1)/3 [9,15,21,27,33,39,45,51,57,63,69,75,81,87,93,99]+ 9 p(p) [25,35,49,55,65,77,85,91,95]= 50 100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

Serie9

Serie8 9

Serie6

13 5

Serie4

Serie2

Serie1

95 81

99 97

101 79

37 29

61 53

83 47

17 7

Serie3

11 65

Serie7 Serie5

116 97 79 73 67 61

101 89 83 71

43 37 31 19 13 7 2

595347 41 29 231711 5

8591

0 0

4955

3 9 15 21 27 33 39 45 51 57 63 69 75 81 87 93 99

35 65 77 95

And so prime numbers is disposed in three-dimensional space.

113

107

79 43 41

29 37 23 7 2 3 5 17 19 31

61 103

73 101

53 67 109

Prime numbers are so fundamental to the working mathematician that any breakthrough in understanding their nature have a massive impact. We know today, at the centre of mathematics, the pursuit of order can the men hear sound of harmony, of the most beautiful music of primes, and we are able to master its twists and turns.

117 The Riemann Hypothesis had been proved, and we are able, to answering the severity of the problem of Goldbach to go, whether each grade number as the sum of two primes is represent able. If proportionality factor all primes in a given quantity ½ is, but this means that the equation πx/2πx = ½N/N is the answer to the problem of Goldbach. She says that every even number is composed of two primes. Theorem: If the quotient of quantity prime numbers by its dual quantity, is equal to the quotient of quantity even numbers by a given magnitude, so in this equation is a proportional relation, that is, in each equation the product of the inner members equal to the product of outer members. πx/2πx = 2n/N

Proof:

25/50 = 50/100 = ½

πx/2πx = 1/2N/N 2n = p + p 8

5 3

25/50 = 50/100 4 = 2 + 2

31 90

31 96

19 54

7 18

7 5

13 42

17 46

7 16

100

3 3

4 2 2

5 2

The proportion of ½ in the case of even numbers means that all even numbers in a block made up of two primes. 2 + 2 = 4 3 + 3 = 6 3 + 5 = 8 5 + 5 = 10 That is to say 50 even numbers in a block of 100 numbers, is the sum of 4(25) primes, as shown in the diagram below.

118

n/p + p = 1/2 2n = p + p' 2+2=4 3+3=6 3+5=8 5+5=10 5+7=12 7+7=14 5+11=16 7+11=18 7+13=20... 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0

23 19 17 13 11 7 5 3 2 0 0

0,5

1,5

23 19 17 13 11 7 5 3 2 0 2,5

3,5

How we see on above diagram, sum two prime numbers lies always on parallel straight line to axis - y and it is even numbers that is consisting with two prime numbers. 2n-1=p+p´+p" 7= 2+2+3 9 = 3+3+3 11=3+3+5 13=3+5+5 15=5+5+5 17=5+ 5+7 19=5+7+7 19-16=17-14=15-12=13-10=11-8=9-6=7-4=3

27 25 23 21 19 17 15 13 11 9 7 5 3 2

7 5 3 2 0 1

13 11 7 5 3 0

Legitimacy "strong‖ hypothesis Goldbacha proves legitimacy "weak‖ hypothesis Goldbacha because 3 have will sufficed from given odd larger number since 7 to subtract and to introduce received even number with strong hypothesis Goldbacha peaceably.

119 →

(2n - 1) - 3 = 2n = p + p'

2n - 1 = p + p + p

2n + p = 2n - 1 = p + p + p' 2 + 2 + 3 = 7 3 + 3 + 3 = 9 3 + 5 + 3 = 11 5 + 5 + 3 = 13 5 + 7 + 3 = 15 31 29 27 25 23 21 19 17 15 13 11 9 7

28 26 24 22 20 18 16 14 12 10 8 6 4 0 0

0,5

28 26 24 22 20 18 16 14 12 10 8 6 4

0 1,5

0 2,5

3,5

Also, as you can see, every odd integer greater than 5 is the sum of 3 primes, because the difference between odd and even numbers always of prime numbers 3 is. p + (p +2)/2 = 2n/3 5+7/2=6/3 11+13/2=12/3 17+19/2=18/3 29+31/2=30/3 41+43/2=42/3

149

150

151

137

138

139

107 101

108 102

109 103

17 11 5 0

18 12 6

19 13 7 0

In addition to the familiar question of whether there are infinitely many prime pairs with difference 2 there. The six- wide array further helps to demonstrate the otherwise still unproven conjecture that there must be infinitely many twin primes.

120 Here are the reasons for this: if there are infinite primes, then twin pairs, with even number divisible by 3 shares. 2y+3y=5y/2=2+0,5=3-0,5 3y+7y=10y/2=3+2=7-2 5y+13y=18y/2=5+4=13-4 11y+19y=30y/2=11+4=19-4 19

13 11 9 0

7 5

2,5 0

2 0 2

0 3

Looking closer at the above graph, you will see that half of the following sums of two primes on a straight line parallel to the y - axis with real part ½ y lie. This means that the linear Diophantine equation ax + by - c = 0, with given integer pairs not have common divisor Coefficient a, b, c, always in prime x, y is solvable. 1(2) + 1(3) – 5 = 0 1(3) + 1(7) – 10 = 0 1(5) + 1(13) – 18 = 0 1(11) + 1(19) – 30 = 0 Still, then, we go on listing to that mysterious prime numbers beat: 2, 3, 5, 7, 11, 13, 17, 19. The primes stretch out into the far reaches of the universe of numbers, never running dry. Do we really have to accept that, despite our desire for order and explanation, these fundamental numbers might forever remain out of reach? Sale long we reflect upon with perspective Gauß and Riemann‘s and we should earlier already look for different possibilities, to better to get know these full of secrets numbers. The primes betrayed in end their secret, and remain not an unanswered riddle. I‘m who made the primes sing. And the primes are for the mathematician is so important that everyone breakthrough a better understanding of each and its nature is of fundamental importance. Counting no human invention, because in prime number distribution and the plan of nature, is encrypted so the whole universe. The number reveals the divine thoughts and Order. It can also recognize the basic structure of reality. The figure provides some insight into the innermost mystery of God and the mystery of the world. Who knows a certain number, possessing power. The counting man accomplishes something similar to God himself, by ordering power exercises over the things he is different and shares, he delimits and summarizes. The reality of the existence of the transcendent blueprint justifies the consideration of whether is not hidden behind the processes in space and time, so our history, an invisible, transcendent steering. Thus, because the knowledge is converted to the blueprint into reality, and we are committed behind the Book of Wisdom 11:21; "But you have everything sorted by measure, number and weight."

121 and grasp its profound meaning. The apparent randomness is regulated and thank God that it need not take at least one million years, until we understand the primes. There is a widespread belief bordering on certainty that the prime numbers are arranged on the axis of numerical chaotic, does not govern them apparently no law that would allow us to clearly describe them all. It is true, that there are 4 known arithmetic sequences which generate the prime numbers and n-the member is given in clear designs, allowing you recognize all the prime numbers in a pattern, / p = n (2) + 3 /. We also know that prime numbers is infinite amount. If you see this let's have similarities as chaos (primes) can generate the order (the numbers represent the natural order because we can give knowing antecedent consequent). For the mathematics is obvious that every natural number is a "combination" of some primes. Anyone who has read this book to the end, knows why rare beauty and harmony in the world of numbers there instead of chaos. Finally, we can also live to see the person whose name will live forever as the mathematician who made the primes sing for the greater glory of God.

FOR THE GREATER GLORY OF GOD. „AD MAJOREM DEI GLORIAM SOLI DEO HONOR ET GLORIA.

122 2,3,5,11,13,29, + n(7) = p 2 5 17 29 41 53

3 7 19 31 43 67 79

89 101 113

103

137 149

127 139 151 163

173 197

199 211 223

233 257 269 281 293

p(p) + 6(7) = p‗(p―)

11 23

37 47 59 71 83 107

61 73 97 109

131 157 167 179 191

227 239 251 263

271 283 307

181 193

229 241

277

311

313

317 331 347 359

353 367 379

373 383

389 401

439 449 461

337 349

419 431 443

397 409 421 433 457

463 487 499

467 479 491 503

2n - 1 9 21 33 45 57 69 81 93 105 117 129 141 153 165 177 189 201 213 225 237 249 261 273 285 297 309 321 333 345 357 369 381 393 405 417 429 441 453 465 477 489 501

7(7) + 6(7) = 13(7) = k 15 27 39 51 63 75 87 99 111 123 135 147 159 171 183 195 207 219 231 243 255 267 279 291 303 315 327 339 351 363 375 387 399 411 423 435 447 459 471 483 495 507

5(7) + 6(7) = 11(7)

25 35 49 55 65 77

85 91

115

119

125 143 155 161 185

169 175 187 203 215

209 221 245

121 133 145

235 247 259

295

205 217

253 265 275 287 299

289 301

305 319 329 341

343 355

365 377

413 425 437

325

361 371 385

391 403 415 427

395 407

445 451

473 485 497

323 335

475

455 469 481 493 505

123 509 521

523 541 547

557 569

563 571 587 599

593 617 641 653

577

607 619 631 643

677

647 659

601 613

661 673

683 691

701

709 719 727 739 751

733 743 757 769

761 773 787 797 809 821

811 823

827 839

829 853

857

859

863

881

883

887

877

907 919 929 941 953

911 937 947

967 977

971 983

991 1013

1019 1031 1039

997 1009 1021 1033

513 525 537 549 561 573 585 597 609 621 633 645 657 669 681 693 705 717 729 741 753 765 777 789 801 813 825 837 849 861 873 885 897 909 921 933 945 957 969 981 993 1005 1017 1029 1041

519 531 543 555 567 579 591 603 615 627 639 651 663 675 687 699 711 723 735 747 759 771 783 795 807 819 831 843 855 867 879 891 903 915 927 939 951 963 975 987 999 1011 1023 1035 1047

511 533 545

535

515 527 539 551

559

517 529 553 565

575 581

583 595

605

611 623 635

629

665

655 667 679

689 713 725 737 749

703 715

625 637 649

671 695 707

685 697 721

731 745 763 775

785 799

833 845

589

835 847

755 767 779 791 803 815

781 793 805 817 841

851 865

869 893 905 917

871 895

931 943 955

875 899 923 935 959

965 979 989 1001 1025 1037

1003 1015 1027

889 901 913 925 949 961 973 985

995 1007

1043 1045

124 1049 1061

1051 1063 1087

1097 1109

1069 1091 1103

1093 1117 1129

1123 1151 1163

1153

1171 1181 1193 1217 1229

1187 1201 1213 1223 1231

1237 1249 1259

1277 1289 1301

1279 1291 1303

1283 1297 1307 1319

1321

1327

1361 1373

1367 1381 1399

1409 1423 1433

1481 1493

1447 1459 1471 1483

1427 1439 1451

1487 1499 1511 1523

1531 1543 1553 1567 1579

1429 1453

1489

1549 1559 1571 1583

1053 1065 1077 1089 1101 1113 1125 1137 1149 1161 1173 1185 1197 1209 1221 1233 1245 1257 1269 1281 1293 1305 1317 1329 1341 1353 1365 1377 1389 1401 1413 1425 1437 1449 1461 1473 1485 1497 1509 1521 1533 1545 1557 1569 1581

1059 1071 1083 1095 1107 1119 1131 1143 1155 1167 1179 1191 1203 1215 1227 1239 1251 1263 1275 1287 1299 1311 1323 1335 1347 1359 1371 1383 1395 1407 1419 1431 1443 1455 1467 1479 1491 1503 1515 1527 1539 1551 1563 1575 1587

1073 1085

1075 1099 1111

1121 1133 1145 1157 1169

1135 1147 1159 1183

1195 1205

1207 1219

1241 1253 1265

1243 1255 1267

1313 1325 1337 1349

1315

1385 1397

1055 1057 1067 1079 1081 1105 1115 1127 1139 1141 1165 1175 1177 1189 1199 1211 1225 1235 1247 1261 1271 1273 1285 1295 1309

1411

1331 1333 1343 1345 1355 1357 1369 1379 1391 1393 1403 1405 1415 1417

1435

1441

1339 1351 1363 1375 1387

1421 1445 1457 1469

1505 1517 1529 1541

1463 1465 1475 1477 1495 1507 1519

1555 1565 1577

1501 1513 1525 1535 1537 1547 1561 1573 1585

125

1601 1613

1607 1619 1627

1637 1663

1697 1709 1721 1733

1667

1699 1723 1747 1759 1783

1787 1811 1823

1831 1847

1877 1889 1901 1913

1867 1879

1871

1907 1931

1949

1951

1973 1997

1979 1987 1999 2011

2003 2027 2039 2063

2069 2081

2083

2087 2099 2111

1597 1593 1609 1605 1621 1617 1629 1641 1657 1653 1669 1665 1677 1693 1689 1701 1713 1725 1741 1737 1753 1749 1761 1777 1773 1789 1785 1801 1797 1809 1821 1833 1845 1861 1857 1873 1869 1881 1893 1905 1917 1933 1929 1941 1953 1965 1977 1993 1989 2001 2017 2013 2029 2025 2037 2053 2049 2061 2073 2089 2085 2097 2113 2109 2121

1599 1611 1623 1635 1647 1659 1671 1683 1695 1707 1719 1731 1743 1755 1767 1779 1791 1803 1815 1827 1839 1851 1863 1875 1887 1899 1911 1923 1935 1947 1959 1971 1983 1995 2007 2019 2031 2043 2055 2067 2079 2091 2103 2115 2127

1589

1591 1603 1615

1625 1649 1661 1673 1685

1639 1651 1675 1687 1711 1735

1745 1757 1769 1781 1793 1805 1817 1829 1841 1853 1865

1925 1937 1961

1771 1795 1807 1819 1843 1855

1891 1903 1915 1927 1939 1963 1975

1985 2009 2021 2033 2045 2057

2093 2105 2117

2023 2035 2047 2059 2071 2095 2107 2119

1595

1631 1633 1643 1645 1655 1679 1691 1703 1715 1727 1739 1751 1763 1775

1681 1705 1717 1729

1765

1799 1813 1825 1835 1837 1849 1859 1883 1885 1895 1897 1909 1919 1921 1943 1945 1955 1957 1967 1969 1981 1991 2005 2015 2041 2051 2065 2075 2077 2101 2123 2125

126 2129 2141 2153

2131 2143

2179 2203

2207

2213 2237

2239 2251

2243 2267

2273 2287 2297 2309

2311

2333 2347

2339 2351

2357 2381 2393

2371 2383 2399 2411 2423

2417 2441

2447 2459 2467

2477 2503

2549

2539 2551

2531 2543

2579 2591 2609 2621 2633 2657

2647 2659

2663

2137 2133 2145 2161 2157 2169 2181 2193 2205 2221 2217 2229 2241 2253 2269 2265 2281 2277 2293 2289 2301 2313 2325 2341 2337 2349 2361 2377 2373 2389 2385 2397 2409 2421 2437 2433 2445 2457 2473 2469 2481 2493 2505 2521 2517 2529 2541 2557 2553 2565 2577 2593 2589 2601 2617 2613 2625 2637 2649 2661

2139 2151 2163 2175 2187 2199 2211 2223 2235 2247 2259 2271 2283 2295 2307 2319 2331 2343 2355 2367 2379 2391 2403 2415 2427 2439 2451 2463 2475 2487 2499 2511 2523 2535 2547 2559 2571 2583 2595 2607 2619 2631 2643 2655 2667

2165 2177 2189 2201 2225 2249 2261

2155 2167 2191 2215 2227

2263 2275

2285 2299 2321

2323 2335

2345 2359 2369

2405 2429 2453 2465 2489 2501 2513 2525 2537 2561 2573 2585 2597

2645

2395 2407 2419 2431 2443 2455 2479 2491 2515 2527

2563 2575 2587 2599 2611 2623 2635

2135 2147 2159 2171 2183 2195

2149 2173 2185 2197 2209

2219 2231 2233 2245 2255 2257 2279 2291 2303 2305 2315 2317 2327 2329 2353 2363 2365 2375 2387 2401 2413 2425 2435 2449 2461 2471 2483 2485 2495 2497 2507 2509 2519 2533 2545 2555 2567 2569 2581 2603 2615 2627 2639 2651

2605 2629 2641 2653 2665

127 2671 2683 2693

2729 2741 2753

2707 2719 2731

2687 2699 2711

2767 2777 2789 2801

2791 2803 2819

2837

2843 2851

2861 2879 2887 2897 2909

2903 2927 2939

2957 2969

2963 2971

3019

2999 3011 3023

3041 3067 3079

3083

3089 3119 3137 3163 3187

3167 3191 3203

2677 2673 2689 2685 2697 2713 2709 2721 2733 2749 2745 2757 2769 2781 2797 2793 2805 2817 2833 2829 2841 2857 2853 2865 2877 2889 2901 2917 2913 2925 2937 2953 2949 2961 2973 2985 3001 2997 3009 3021 3037 3033 3049 3045 3061 3057 3069 3081 3093 3109 3105 3121 3117 3129 3141 3153 3169 3165 3181 3177 3189 3201

2679 2691 2703 2715 2727 2739 2751 2763 2775 2787 2799 2811 2823 2835 2847 2859 2871 2883 2895 2907 2919 2931 2943 2955 2967 2979 2991 3003 3015 3027 3039 3051 3063 3075 3087 3099 3111 3123 3135 3147 3159 3171 3183 3195 3207

2669 2681

2675 2695

2705 2717 2743 2755 2765 2779

2813 2825

2815 2827 2839

2849 2873 2885

2921 2933 2945

2981 2993 3005 3017 3029 3053 3065 3077 3101 3113 3125 3149 3161 3173 3185 3197

2863 2875 2899 2911 2923 2935 2947 2959 2983 2995 3007 3031 3043 3055

3091 3103 3115 3127 3139 3151 3175 3199

2701 2723 2735 2747 2759 2771 2783 2795 2807

2725 2737 2761 2773 2785 2809 2821

2831 2845 2855 2867 2869 2881 2891 2893 2905 2915 2929 2941 2951 2965 2975 2977 2987 2989 3013 3025 3035 3047 3059 3071 3073 3085 3095 3097 3107 3131 3133 3143 3145 3155 3157 3179 3193 3205

128 3209 3221 3251 3257

3259 3271 3299

3329

3389

3307 3319 3331 3343

3323 3347 3359 3371

3391 3407

3413

3449 3461

3463

3467 3491

3499 3511 3527 3539

3533 3557 3581 3593

3547 3559 3571 3583 3607

3617

3623 3631 3643 3659 3671

3677 3691 3701 3719 3727 3739

3217 3213 3229 3225 3237 3253 3249 3261 3273 3285 3301 3297 3313 3309 3321 3333 3345 3361 3357 3373 3369 3381 3393 3405 3417 3433 3429 3441 3457 3453 3469 3465 3477 3489 3501 3517 3513 3529 3525 3541 3537 3549 3561 3573 3585 3597 3613 3609 3621 3637 3633 3645 3657 3673 3669 3681 3697 3693 3709 3705 3717 3733 3729 3741

3219 3231 3243 3255 3267 3279 3291 3303 3315 3327 3339 3351 3363 3375 3387 3399 3411 3423 3435 3447 3459 3471 3483 3495 3507 3519 3531 3543 3555 3567 3579 3591 3603 3615 3627 3639 3651 3663 3675 3687 3699 3711 3723 3735 3747

3233 3245 3269 3281 3293 3305 3317 3341 3353 3365 3377 3401 3425 3437

3473 3485 3497 3509 3521

3211 3223 3235 3247

3283 3295

3225 3335 3337 3349 3355 3367 3379 3403 3415 3427 3439 3451 3475 3487

3383 3385 3395 3397 3409 3419 3421 3431 3443 3445 3455 3479 3481 3493 3503 3505 3515

3523 3535

3569 3595 3605 3619

3655 3667 3679

3689 3713 3725 3737

3263 3265 3275 3277 3287 3289 3311

3545

3629 3641 3653 3665

3215 3227 3239 3241

3703 3715

3551 3563 3575 3587 3599 3611

3553 3565 3577 3589 3601 3625

3635 3647 3649 3661 3683 3685 3695 3707 3721 3731 3743 3745

129

3761

3767 3779

3769 3793

3797 3821 3833

3803 3823 3847

3851 3863

3877 3889

3881

3917 3929

3853

3907 3919 3931 3943

3911 3923 3947

3967 3989 4001 4013

4003

4007 4019

4021

4027 4049

4051

4073

4057 4079 4091

4093

4099 4111 4127 4139

4133

4129 4153

4157

4159 4177 4201 4211

4217 4229 4241 4253

4219 4231 4243 4259 4271 4283

4261 4273

3753 3765 3777 3789 3801 3813 3825 3837 3849 3861 3873 3885 3897 3909 3921 3933 3945 3957 3969 3981 3993 4005 4017 4029 4041 4053 4065 4077 4089 4101 4113 4125 4137 4149 4161 4173 4185 4197 4209 4221 4233 4245 4257 4269 4281

3759 3771 3783 3795 3807 3819 3831 3843 3855 3867 3879 3891 3903 3915 3927 3939 3951 3963 3975 3987 3999 4011 4023 4035 4047 4059 4071 4083 4095 4107 4119 4131 4143 4155 4167 4179 4191 4203 4215 4227 4239 4251 4263 4275 4287

3749 3773 3785 3809

3751 3763 3775 3787 3799 3811 3835

3845 3857 3869 3893 3905

3941 3953 3965 3977

3859 3871 3883 3895

3955 3979 3991

3755 3757 3781 3791 3805 3815 3817 3827 3829 3839 3841 3865 3875 3887 3899 3901 3913 3925 3935 3937 3949 3959 3961 3971 3973 3983 3985 3995 3997 4009

4015 4025 4037 4061 4085 4097 4109 4121

4039 4063 4075 4087

4031 4033 4043 4045 4055 4067 4069 4081 4103 4105 4115 4117

4145

4123 4135 4147

4169 4181 4193 4205

4171 4183 4195 4207

4265 4277

4255 4267 4279

4141 4151 4163 4165 4175 4187 4189 4199 4213 4223 4225 4235 4237 4247 4249

4285

130 4289 4319 4337 4349

4327 4339 4363

4373 4391 4397 4409 4421

4423 4447

4457 4481 4493 4517

4451 4463

4483 4507 4519

4523 4547

4567 4583 4591 4603

4637 4649

4639 4651 4663

4673

4721 4733

4643

4679 4691 4703 4723 4751 4759 4783

4793 4817

4787 4799

4297 4293 4305 4317 4329 4341 4357 4353 4365 4377 4389 4401 4413 4425 4441 4437 4449 4461 4473 4485 4497 4513 4509 4521 4533 4549 4545 4561 4557 4569 4581 4597 4593 4605 4621 4617 4629 4641 4657 4653 4665 4677 4689 4701 4713 4729 4725 4737 4749 4761 4773 4789 4785 4801 4797 4813 4809 4821

4299 4311 4323 4335 4347 4359 4371 4383 4395 4407 4419 4431 4443 4455 4467 4479 4491 4503 4515 4527 4539 4551 4563 4575 4587 4599 4611 4623 4635 4647 4659 4671 4683 4695 4707 4719 4731 4743 4755 4767 4779 4791 4803 4815 4827

4301 4313 4325

4291 4303 4315

4351 4361 4385

4433 4445 4469

4375 4387 4399 4411 4435 4459 4471 4495

4505 4529 4541 4553 4565 4577 4589 4601 4613 4625

4531 4543 4555

4745 4757 4769 4781 4805

4309 4321 4333 4345

4369 4381 4393 4403 4405 4415 4417 4427 4429 4439 4453 4465 4475 4477 4487 4489 4499 4501 4511 4525 4535 4537

4771

4559 4571 4573 4585 4595 4607 4609 4619 4631 4633 4645 4655 4667 4669 4681 4693 4705 4715 4717 4727 4739 4741 4753 4763 4765 4775 4777

4795 4807 4819

4811 4823 4825

4579

4615 4627

4661 4685 4697 4709

4295 4307 4319 4331 4343 4355 4367 4379

4675 4687 4699 4711 4735 4747

131 4831

4871 4877 4889 4903 4919 4931 4943

4937 4951

4967 4973

5009 5021 5033

4987 4999 5011 5023 5047 5059

5081

4991 5003

5039 5051

5087 5099 5107 5119 5147

5153 5167 5179

5171

5189

5227

5231

5237 5261 5273

5279

5297 5309

5303 5323

5333 5347

5351

4837 4833 4845 4861 4857 4869 4881 4893 4909 4905 4917 4933 4929 4941 4957 4953 4969 4965 4977 4993 4989 5001 5013 5025 5041 5037 5049 5061 5077 5073 5085 5101 5097 5113 5109 5121 5133 5145 5157 5169 5181 5197 5193 5209 5205 5217 5233 5229 5241 5253 5265 5281 5277 5289 5301 5313 5325 5337 5349 5361

4839 4851 4863 4875 4887 4899 4911 4923 4935 4947 4959 4971 4983 4995 5007 5019 5031 5043 5055 5067 5079 5091 5103 5115 5127 5139 5151 5163 5175 5187 5199 5211 5223 5235 5247 5259 5271 5283 5295 5307 5319 5331 5343 5355 5367

4829 4841 4853 4865

4901 4913 4925 4949 4961

4843 4855 4867 4879 4891 4915 4927 4939

5093 5105 5117 5129 5141

4963 4975

5035 5047 5071 5083 5095

5131 5143 5155

5165 5177 5201 5213 5225 5249

5285

5321 5345 5357

4945 4955

4985 4997

5033 5045 5057 5069

4835 4837 4847 4849 4859 4873 4883 4885 4895 4897 4907 4921

5191 5203 5215

4979 4981 4991 5005 5015 5017 5027 5029 5053 5063 5065 5075 5089 5111 5123 5125 5135 5137 5149 5159 5161 5173 5183 5185 5195 5207 5219 5221

5239 5243 5251 5255 5263 5267 5275 5287 5291 5299 5311 5315 5327 5335 5339 5359

5245 5257 5269

5293 5305 5317 5329 5341 5353 5363 5365

132

5381 5393 5417 5441

5477 5501

5387 5399 5407 5419 5431 5443

5479 5503

5413 5437 5449 5471 5483 5507 5519

5521

5527 5557 5569 5581

5563 5573 5591

5623

5657 5669

5647 5659

5639 5651

5683

5641 5653

5689 5701

5693 5711 5717 5741

5779 5791 5801 5813

5849 5861

5897

5737 5749

5743

5783 5807 5821

5827 5839 5851

5843 5867 5879 5903

5845 5857 5869 5881

5373 5385 5397 5409 5421 5433 5445 5457 5469 5481 5493 5505 5517 5529 5541 5553 5565 5577 5589 5601 5613 5625 5637 5649 5661 5673 5685 5697 5709 5721 5733 5745 5757 5769 5781 5793 5805 5817 5829 5841 5853 5865 5877 5889 5901

5379 5391 5403 5415 5427 5439 5451 5463 5475 5487 5499 5511 5523 5535 5547 5559 5571 5583 5595 5607 5619 5631 5643 5655 5667 5679 5691 5703 5715 5727 5739 5751 5763 5775 5787 5799 5811 5823 5835 5847 5859 5871 5883 5895 5907

5369

5371 5383 5395

5405 5429 5453 5465

5455 5467

5489

5491

5513 5525 5537 5549 5561

5515

5585 5597 5609 5621 5633 5645

5539 5551 5575 5587 5599 5611

5681

5729 5753 5765 5777 5789

5695 5707 5719 5731 5755 5767

5803 5815 5825 5837

5873 5885

5533 5543 5545 5555 5567 5579 5593 5603 5605 5615 5617 5627 5629

5635

5671

5705

5375 5377 5389 5401 5411 5423 5425 5435 5447 5459 5461 5473 5485 5495 5497 5509

5863 5875 5887 5899

5633 5665 5675 5677 5687 5699 5713 5723 5725 5735 5747 5759 5761 5771 5773 5785 5795 5797 5809 5819 5831 5833 5845 5855

5891 5893 5905

133

5923

5927 5939 5953

5981

5987 6007

6011

6029

6037 6043

6047

6053

6089 6101 6113

6067 6079 6091

6073 6097

6131 6143

6121 6133

6151 6163 6173 6197

6199 6211

6203 6217 6229

6221 6247 6257 6269

6263 6271

6277 6287 6299 6311 6323

6317 6329

6301

6337 6343

6353

6359 6367 6379

6389

6361 6373 6385 6397 6421

6427

5913 5925 5937 5949 5961 5973 5985 5997 6009 6021 6033 6045 6057 6069 6081 6093 6105 6117 6129 6141 6153 6165 6177 6189 6201 6213 6225 6237 6249 6261 6273 6285 6297 6309 6321 6333 6345 6357 6369 6381 6393 6405 6417 6429 6441

5919 5931 5943 5955 5967 5979 5991 6003 6015 6027 6039 6051 6063 6075 6087 6099 6111 6123 6135 6147 6159 6171 6183 6195 6207 6219 6231 6243 6255 6267 6279 6291 6303 6315 6327 6339 6351 6363 6375 6387 6399 6411 6423 6435 6447

5909 5921 5933 5945 5957 5969 5993 6005 6017

5911 5935 5947 5959 5971 5983 5995 6019 6031

6041 6055 6065 6077

6125 6137 6149 6161 6185

6103 6115 6127 6139

6175 6187

6209 6233 6245

6223 6235 6259

6281 6293 6305

6283 6295 6307 6319 6331

6341

5915 5917 5929 5941 5951 5963 5965 5975 5977 5989 5999 6001 6013 6023 6025 6035 6049 6059 6061 6071 6083 6085 6095 6097 6107 6109 6119

6155 6167 6179 6191

6145 6157 6169 6181 6193 6205

6215 6227 6239 6241 6251 6253 6265 6275 6289 6313 6325 6335 6347 6349

6355 6365 6377 6401 6413 6425 6437

6391 6403 6415 6439

6371 6383 6395 6407 6419 6431 6443

6385 6409 6433 6445

134 6449

6451 6469 6481

6473 6491

6521

6529 6547

6569 6581

6551 6563

6571

6553 6577

6599 6607 6619 6637 6653

6689 6701

6659 6679 6691 6703

6661 6673

6709 6719 6733

6737 6761

6763

6823

6779 6791 6803

6781 6793

6827

6829 6841

6833 6857 6869

6863 6871 6883 6907

6899 6911

6917

6967 6977

6947 6959 6971 6983

6949 6961

6453 6465 6477 6489 6501 6513 6525 6537 6549 6561 6573 6585 6597 6609 6621 6633 6645 6657 6669 6681 6693 6705 6717 6729 6741 6753 6765 6777 6789 6801 6813 6825 6837 6849 6861 6873 6885 6897 6909 6921 6933 6945 6957 6969 6981

6459 6471 6483 6495 6507 6519 6531 6543 6555 6567 6579 6591 6603 6615 6627 6639 6651 6663 6675 6687 6699 6711 6723 6735 6747 6759 6771 6783 6795 6807 6819 6831 6843 6855 6867 6879 6891 6903 6915 6927 6939 6951 6963 6975 6987

6461 6485 6497 6509 6533 6545 6557

6593 6605 6617 6629 6641 6665 6677

6713 6725 6749 6773 6785 6797 6809 6821

6463 6475 6487 6499 6511 6523 6535

6455 6457 6467 6479 6493 6503 6505 6515 6517 6527 6539 6541

6559

6565

6583 6595

6631 6643 6655 6667

6715 6727 6739 6751 6775 6787 6799 6811

6845

6835 6847 6859

6881 6893 6905

6895

6929 6941 6953 6965

6919 6931 6943 6955 6979

6575 6587 6589 6601 6611 6613 6623 6625 6635 6647 6649 6671 6683 6685 6695 6697 6707 6721 6731 6743 6745 6755 6757 6767 6769

6805 6815 6817 6839 6851 6853 6865 6875 6877 6887 6889 6901 6913 6923 6925 6935 6937

6973 6985

135 6991 7001 7013

6997 7019

7027 7039

7043 7057 7069 7079 7103

7109 7121

7127

7129

7151 7159 7177 7187 7193 7207 7219

7211

7229

7213 7237

7243

7247

7253 7283 7301

7303

7307 7331

7349

7297 7309 7321 7333

7351 7369 7393 7411

7417

7433 7451 7457

7459

7481

7487 7499 7507

7517

7523

7477 7489

6993 7005 7017 7029 7041 7053 7065 7077 7089 7101 7113 7125 7137 7149 7161 7173 7185 7197 7209 7221 7233 7245 7257 7269 7281 7293 7305 7317 7329 7341 7353 7365 7377 7389 7401 7413 7425 7437 7449 7461 7473 7485 7497 7509 7521

6999 7011 7023 7035 7047 7059 7071 7083 7095 7107 7119 7131 7143 7155 7167 7179 7191 7203 7215 7227 7239 7251 7263 7275 7287 7299 7311 7323 7335 7347 7359 7371 7383 7395 7407 7419 7431 7443 7455 7467 7479 7491 7503 7515 7527

6989 7003 7015 7025 7037 7049 7061 7073 7085 7097

7133 7145 7157 7169 7181

7051 7063 7075 7087 7099 7111 7123 7135 7147 7171 7183 7195

7205 7217 7231 7241 7265 7277 7289 7301 7313 7325 7337 7361 7373 7385 7397 7409 7421 7445 7469 7493 7505

7255 7267 7279 7291 7303 7315 7327 7339 7363 7375 7387 7399 7423 7435 7447 7471 7483 7495 7519

6995 7007 7009 7021 7031 7033 7045 7055 7067 7081 7091 7093 7105 7115 7117 7139 7141 7153 7163 7165 7175 7177 7189 7199 7201 7223 7225 7235 7249 7259 7261 7271 7273 7285 7295 7319 7343 7355 7367 7379 7391 7403 7415 7427 7439

7345 7357 7381 7405

7429 7441 7453 7463 7465 7475 7501 7511 7513 7525

136 7529 7541

7577 7589

7547 7559

7537 7549 7561 7573

7583 7591 7603

7607 7621

7639

7643

7649 7669 7681

7673 7687 7699

7691 7703 7717

7723

7727 7741 7753

7757

7759 7789

7793 7817 7829 7841 7853 7877

7823

7867 7879

7901

7873 7883 7907 7919

7927 7937 7949

7933

7951 7963 7993

8009

8011

8017 8039 8053

8059

7533 7545 7557 7569 7581 7593 7605 7617 7629 7641 7653 7665 7677 7689 7701 7713 7725 7737 7749 7761 7773 7785 7797 7809 7821 7833 7845 7857 7869 7881 7893 7905 7917 7929 7941 7953 7965 7977 7989 8001 8013 8025 8037 8049 8061

7539 7551 7563 7575 7587 7599 7611 7623 7635 7647 7659 7671 7683 7695 7707 7719 7731 7743 7755 7767 7779 7791 7803 7815 7827 7839 7851 7863 7875 7887 7899 7911 7923 7935 7947 7959 7971 7983 7995 8007 8019 8031 8043 8055 8067

7553 7465

7531 7543 7555 7567 7579

7601 7613 7615 7625 7627 7637 7651 7661 7663 7675 7685 7697 7709 7711 7721 7733 7735 7745 7747 7769 7781

7771 7783 7795 7805 7807 7819 7831 7843 7855 7865 7889 7891 7903 7913 7915 7925 7939 7961 7973 7975 7985 7987 7997 7999 8021 8023 8033 8035 8045 8047 8057

7535

7571 7585 7595 7597 7609 7619 7631 7633 7645 7655 7657 7667 7679 7705 7693 7705 7715 7729 7739 7751 7763 7765 7775 7777 7787 7799 7801 7811 7813 7825 7835 7837 7847 7849 7859 7861 7871 7885 7895 7897 7909 7921 7931 7943 7945 7955 7957 7967 7969 7979 7981 7991 8003 8005 8015 8027 8029 8041 8051 8063 8065

137 8069 8081 8093

8087

8089 8101

8111 8123

8117

8147 8161 8167 8179 8191

8171

8219 8231 8243

8237 8263

8209 8221 8233

8269

8273 8287

8291

8293

8297 8311

8317 8329 8353 8363

8369 8387

8429

8419 8431 8443

8377 8389

8423 8447 8461

8467

8501 8513 8537

8521 8527 8539

8543

8563 8573 8597

8581 8599

8073 8085 8097 8109 8121 8133 8145 8157 8169 8181 8193 8205 8217 8229 8241 8253 8265 8277 8289 8301 8313 8325 8337 8349 8361 8373 8385 8397 8409 8421 8433 8445 8457 8469 8481 8493 8505 8517 8529 8541 8553 8565 8577 8589 8601

8079 8091 8103 8115 8127 8139 8151 8163 8175 8187 8199 8211 8223 8235 8247 8259 8271 8283 8295 8307 8319 8331 8343 8355 8367 8379 8391 8403 8415 8427 8439 8451 8463 8475 8487 8499 8511 8523 8535 8547 8559 8571 8583 8595 8607

8071 8083 8095 8105 8107 8119 8129 8131 8141 8143 8153 8155 8165 8177 8189 8201 8203 8213 8215 8225 8227 8239 8249 8251 8261 8275 8285 8299 8309 8321 8323 8333 8335 8345 8347 8357 8359 8371 8381 8383 8393 8395 8405 8407 8417 8441 8453 8465 8477 8489

8455 8479 8491 8503 8515

8525 8549 8561 8585

8551 8575 8587

8075 8077 8099 8113 8125 8135 8137 8149 8159 8173 8183 8185 8195 8197 8207

8245 8255 8257 8267 8279 8281 8303 8305 8315 8327 8339 8341 8351 8365 8375 8399 8401 8411 8413 8425 8435 8437 8449 8459 8471 8473 8483 8485 8495 8497 8507 8509 8519 8531 8533 8545 8555 8557 8567 8569 8579 8591 8593 8603 8605

138 8609 8623

8627

8629 8641

8647 8663 8669 8681 8693

8677 8689 8699 8707 8719 8731

8741 8753

8737 8747 8761

8779 8803

8837 8849 8861

8713

8783 8807 8819 8831

8821

8839 8863

8867

8887

8893

8923

8929 8941

8933 8951 8963 8969

8971

9007 9029 9041

8999 9011

9043

9001 9013

9049 9059

9067 9091 9103 9127 9137

9109 9133

8613 8625 8637 8649 8661 8673 8685 8697 8709 8721 8733 8745 8757 8769 8781 8793 8805 8817 8829 8841 8853 8865 8877 8889 8901 8913 8925 8937 8949 8961 8973 8985 8997 9009 9021 9033 9045 9057 9069 9081 9093 9105 9117 9129 9141

8619 8631 8643 8655 8667 8679 8691 8703 8715 8727 8739 8751 8763 8775 8787 8799 8811 8823 8835 8847 8859 8871 8883 8895 8907 8919 8931 8943 8955 8967 8979 8991 9003 9015 9027 9039 9051 9063 9075 9087 9099 9111 9123 9135 9147

8621 8633 8645 8657

8611

8615 8617

8635

8639 8651 8653 8665 8675 8687 8701 8711 8723 8725 8735 8749 8759 8771 8773 8785 8795 8797 8809

8659 8671 8683 8695

8705 8717 8729

8765 8777 8789 8801 8813 8825

8743 8755 8767 8791 8815 8827 8851

8873 8855 8897 8909 8921 8945 8957

8875 8899 8911 8935 8947 8959

8981 8993 9005 9017

8983 8995

9053 9065 9077 9089 9101 9113 9125

9055

9019 9031

9079

9115 9139

8833 8843 8845 8855 8857 8869 8879 8881 8891 8903 8905 8915 8917 8927 8939 8953 8965 8975 8977 8987 8989

9023 9025 9035 9037 9047 9061 9071 9073 9083 9085 9095 9097 9107 9119 9121 9131 9143 9145

139 9151

9157

9161 9173

9181 9187 9199

9209 9221

9203 9227 9239

9241

9257 9277 9281 9293

9283

9319 9341

9311 9323 9337 9349

9343 9371

9377 9391 9403 9413 9437 9461 9473

9419 9431

9397 9409 9421 9433

9439 9463

9467 9479 9491

9497 9511 9521 9533 9547

9539 9551

9587 9601 9613 9629

9677

9619 9631 9643

9679

9623 9649 9661

9153 9165 9177 9189 9201 9213 9225 9237 9249 9261 9273 9285 9297 9309 9321 9333 9345 9357 9369 9381 9393 9405 9417 9429 9441 9453 9465 9477 9489 9501 9513 9525 9537 9549 9561 9573 9585 9597 9609 9621 9633 9645 9657 9669 9681

9159 9171 9183 9195 9207 9219 9231 9243 9255 9267 9279 9291 9303 9315 9327 9339 9351 9363 9375 9387 9399 9411 9423 9435 9447 9459 9471 9483 9495 9507 9519 9531 9543 9555 9567 9579 9591 9603 9615 9627 9639 9651 9663 9675 9687

9149 9163 9175 9185 9197

9233 9245 9269

9305 9317 9329 9353 9365

9211 9223 9235 9247 9259 9271 9295 9307 9331 9355 9367 9379

9389 9401 9425

9451

9485

9475 9487 9499

9509 9523 9535

9641 9653 9665

9251 9263 9275 9287 9299

9253 9265 9289 9301 9313 9325

9335 9347 9359 9361 9373 9383 9385 9395 9407

9415 9427

9449

9545 9557 9569 9581 9593 9605 9617

9155 9167 9169 9179 9191 9193 9205 9215 9217 9229

9559 9571 9583 9595 9607

9655 9667

9443 9445 9455 9457 9469 9481 9493 9503 9505 9515 9517 9527 9529 9541 9553 9563 9565 9575 9577 9589 9599 9611 9625 9635 9637 9647 9659 9671 9673 9683 9685

140 9689

9697

9719 9739

9743

9749 9767 9787

9791 9803

9811 9833 9857

9839 9851 9859 9871 9883

9887

9907 9923 9929 9941

9931

9967

10007

9693 9699 9705 9711 9721 9717 9723 9733 9729 9735 9741 9747 9753 9759 9769 9765 9771 9781 9777 9783 9789 9795 9801 9807 9817 9813 9819 9829 9825 9831 9837 9843 9849 9855 9861 9867 9873 9879 9885 9891 9901 9897 9903 9909 9915 9921 9927 9933 9939 9949 9945 9951 9957 9963 9973 9969 9975 9981 9987 9993 9999 10009 10005 10011

9701 9713 9725 9737 9761 9773 9785 9797 9809 9821 9845 9869 9881 9893 9905 9917

9691 9703 9715 9727 9751 9763 9775

9695 9707 9709 9731 9745 9755 9757 9779 9793 9805

9799 9823 9835 9847

9895 9919 9943 9955

9953 9965 9977 9979 9989 9991 10001 10003

9815 9827 9841 9853 9863 9865 9875 9877 9889 9899 9911 9913 9925 9935 9937 9947 9959 9961 9971 9983 9985 9995 9997