PERFECT ORDER

ABSTRACT There is a widespread belief bordering on certainly that the prime numbers are arranged on the axis of numerical chaotic, does not given them apparently no law that would allow us to clearly describe them all. I will try to show in this article "Perfect order", what laws rule and what order prevails

2 among primes, which allows all of them to be clearly described. Whoever reads it to the end will find out what extraordinary order and harmony prevails in the world of prime numbers instead of chaos. „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.� "Mathematicians have in vain tried to find some order in the sequence of primes, and we have reason to suppose that this is a mystery that the human mind will never explore."

Leonard Euler – 1751 Since the dawn of time, man has been ordering space and time with numbers, but as the numbers themselves are ordered, I will try to show in this article "Primes in apple pie order". Let's look briefly at the table below of prime numbers and their products up to 1000. At first glance we see a repeating distinct diagonal pattern with 10 numbers of the same rectangles /47-67-89-107-127-145125-103-85-65-/. How it affects the distribution of prime numbers and their products, we'll find out in a moment.

3 BASIC ORDER How they all have each other's numbers results from how they happen one by one. Adding one to the other, we get increasingly larger triangular numbers 1 = (1 * 1), 1 + 2 = 3 = (2 * 1.5), 1 + 2 + 3 = 6 = (3 * 2), 1 + 2 + 3 + 4 = 10 = (4 * 2.5), which can be represented as a product of consecutive numbers and the factor constantly by 0.5 larger / 2 * 1.5 = 3, 3 * 2 = 6 /.

As you know, each natural number can be written as the sum of a certain number of ones, but also as the sum of two components. If the even numbers are simply doubling subsequent natural numbers 2 (1, 2,3,4,5,) = 2k, then the odd numbers that are half of the natural numbers are the sum of the extreme pairs of preceding numbers, as components that have the numbers ability to form identical indirect sums. [1 + (2 + 3) + 4] = 5 = (1 + 4) = (2 + 3)

According to the additive theory of numbers, any odd number may be represented as the sum of two different components which preceded it, numbers, so such distribution creating identical indirect

4 sum of intermediates in this case is three (n - 1) / 2 (7 - 1) / 2 = 3, 7 = {6 + [5 + (4 + 3) + 2] + 1} = (6 + 1) + (5 + 2) + (4 + 3) = 21/3 = 7

Such decreasing and growing sequences of natural numbers /9-8-7-6-5-4-3-2-1-2-3-4-5-6-7-8-9 = 89/ form 17 pairs of extreme elements/1-2-3-4-5-6-7-8-9-8-7-6-5-4-3-2-1 = 81/, which used as factors (9 *1) = 9, (8*2) = 16, (7*3) = 21, (6*4) = 24, (5*5) = 25 give products growing by decreasing odd numbers /9 + 7 = 16 + 5 = 21 + 3 = 24 + 1 = 25/, which proves that these factors, i.e. all natural numbers, form identical indirect sums /9 + 1 = 8 + 2 = 7 + 3 = 6 + 4 = 5 + 5 = 4 + 6 = 3 + 7 = 2 + 8 = 1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 5 + 5 = 6 + 4 = 7 + 3 = 8 + 2 = 9 + 1 = 17(10) = 170 /.

This parabola of numbers clearly shows the mutual dependence between five numbers / 1 3 5 7 9 / and the numbers on its arc / 9 + 7 = 16 + 5 = 21 + 3 = 24 + 1 = 25 /, and the Pythagorean equation 3² + 4² = 5², as well as the numbers of equations 25 + 9 = 34 and 25 - 9 = 16 saying that the sum and

5 difference of numbers with the same parity is an even number. These laws operate in the whole set of natural numbers. This testifies to the perfect order prevailing in the whole sequence of natural numbers, consisting of 50% of even and odd numbers, that is, prime numbers and their products.

Such basic numbers are not determined by nature by accidental coin throwing, or the cube "God does not play with the world in a bone", but based on the ability to create identical indirect sums from the extreme pairs of numbers preceding a given magnitude. The case and chaos are simply unacceptable for mathematics.

COLLECTION OF NATURAL NUMBERS Natural numbers written successively from 1 to 102 in six columns and seventeen lines divide this number system exactly into 51 even numbers, i.e. 2 (17) = 34 numbers divisible by 2 and 17 numbers divisible by 2 and 3, (6, 12, 18, 24, 30, 36, 42), and 51 odd numbers, that is, 2(17) = 34 numbers divisible only by 1 and themselves, and 5 and 7, as well as 17 numbers divisible by 3 (3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99,). Interestingly, multiples of 17 (34, 51, 68, 85) form the main diagonal of this rectangular set of 102 = 17(6) numbers. Numbers 6 and 17 are very important for this set. They break this set into 2 groups of numbers arranged in 3 columns of 17 lines 3 (17) = 51, constantly six times larger. This means that in 100 numbers we have (99 - 3)/6 = 16 products of number 3, six (25 + 95)/20 = 6 products of number 5, and three products of number 7 (49, 77, 91).

Adding 16 + 6 + 3 = 25 we get that the number of products of prime numbers (3, 5, 7) is equal to 25 prime numbers to 100.

In the table composed of numbers in the natural order (1,2,3,4,5,6,7,8,9,10,11, ...) we see that the prime numbers and their products are subject to strict rules here: Sum and difference two numbers with the same parity, is an even number (9 + 25 = 34, 25 - 9 = 16), and a different parity, is an odd number (4 + 1 = 5, 4 - 1 = 3), and that from number 59, which is 17 with the prime number, they will constantly arrive by 17 (34,51,68,85,102,119,136, ..) in the appropriate proportion to the number of their products, which if the same parity as the number of primes, completes the prime numbers in an even range up to half of the given magnitude (34 + 38 = 72), when different parity as the number of primes, than in an odd range up to half of the given magnitude (51 + 66 = 117) according to the equation π(x) + Σp(p') = ½N - half of the given magnitude is the sum of the number of primes and their products that occur to a given magnitude. And just as the hypotenuse in a triangle is dependent on the parabolas, half of a given magnitude depends on an even or odd multiple of 17 /34, 51, 68, 85,

6 102/ completed by an even or odd number of products of prime numbers /38, 66, 103, 136,177/ for even or odd half of the given size /72, 117, 171, 221, 279 /.

From then on, the prime numbers will always come by the next multiple of 17 (17-17-34-17-51-1768-17-85-17-102), when at 85 = 5(17) the number of products of prime numbers reaches 136 = 8 (17), which in total gives 221 = 13(17) prime numbers and their products in 442 numbers. So 85

7 prime numbers + 136 their products = 221, constitute half of the sequence of 442 numbers among which they occur in a strictly defined ratio 5 + 8 = 13, 5: 8: 13.

If the sequence of prime numbers adheres to such strict rules as the exact ratio of prime numbers to their products, which in the decimal system assumes strictly defined values according to the principles of 17, 34, 51, 68 prime numbers are complemented by their products to the full ten eg: 17 + (23, 33, 43, 63, 73, 83, 93), 34 + (86, 96, 106, 116, 126, 136, 146, 156, 166, 176), 51 + (119, 139, 169, 179 , 199, 209, 229), 68 + (202, 222, 232, 272, 312), depending on the space of the sizes.

However, the rule that the sum of prime numbers and their products is equal to half the given magnitude to which they exist is always preserved. π(x) + Σ[p(p ')] = ½N, 68 + 102 = 170, (2: 3: 5). And this is proof that the prime numbers are not arranged chaotically, but according to strict rules.

THE RELATIONSHIP OF THE FIRST NUMBERS TO THEIR PRODUCTS In fact, the distribution of prime numbers depends on the exact relation to their products, and this is due to the ability to create identical indirect sums to a given magnitude. We have 4 prime numbers up to 10 (2 + 3 + 5 + 7) = 17, they form 4 identical indirect sums up to 10 [2 + 8 = 10, 3 + 7 = 10, 5 + 5 = 10, 7 + 3 = 10, (8 + 7 + 5 + 3) = 23, 17 + 23 = 40/4 = 10]. According to this scheme, the ratio of prime numbers to their products will be shaped, i.e. for 40 odd numbers in a given range; there may be 17 prime numbers and 23 their products.

And this is how it looks in the line chart. Here the sum of 4 prime numbers (2 + 3 + 5 + 7 = 17), complemented by the sum of the differences up to 10 (8 + 7 + 5 + 3 = 23), shows what is the ratio of 17 primes/ from 20 - 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 to 100 / to 23 their products /21, 25, 27, 33, 35, 39, 45 , 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99/ in 17 + 23 = 40 numbers, as half of a given magnitude.

π(x) + Σ[p(p’)] = ½N.

8 As can be seen in the following 10 column table, in the first row there are 4 pairs, i.e. 8 prime numbers (2, 3) (5, 7) (11, 13) (17, 19), and only 2 products of the number 3 (9 and 15 ), (8 + 2 = 10). In further rows this ratio is shaped as follows (4 + 6)= (5 + 5)= (5 + 5)= (3 + 7) = (4 + 5 + 5 + 3) = 17 + 8 = 25, prime numbers up to (6 + 5 + 5 + 7) = 23 + 2 = 25 of their products, so in the fifth row the ratio is equal, and in the sixth row there are evenly 5 first numbers and their products equalizing to 30, that is, the prime numbers to their products are in a 1:1 ratio.

To N 20 we have 8 prime numbers /π (20) = 8/ and only 2 products of the number 3 /9, 15/, and in the next N 80 there will be exactly 17 prime numbers, so that at N 100 - π (100) it equals 8 + 17 = 25, supplemented to the full ten by 23 products of numbers 3, 5, 7 (21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99) / 17 + 23 = 40 /, which is half of the 80 numbers among which they are and equalize the number of products with the number of prime numbers 2 + 23 = 25, 25 + 25 = 50 to half of a given magnitude.

9 In further rows 6 to 11 this ratio is 17/33, that is, in the range of 50 numbers (50 = 17 + 33), there are 17 prime numbers and 33 their products, i.e. prime numbers are distributed among their products in a precisely defined ratio. However, the rule that the sum of primes and their products is half of a given magnitude is always preserved (π(x) + Σ[p(p')] = ½N, 68 + 102 = 170, 2: 3: 5 /.

In the next rows / 34 - 46 / the ratio of prime numbers to their products doubles from 17/43 to 34/86 because it covers the range 34 + 86 = 120 numbers. We still have a range of 17 + 43 = 60 numbers, 17 + 53 = 70 numbers, 34 + 66 = 100 numbers, 34 + 96 = 130 numbers and 34 + 106 = 140 numbers.

Such and not the other arrangement of prime numbers and their products results from the natural order in which they occur. Adding one number to the other, we always get a triangular number 1 + 2 = 3/1 + 3 = 6 + 4 = 10/2 + 5 = 15 + 6 = 21/3 + 7 = 28, and when the sum of every second triangle number is received / 1 - 3 - 6 - 10 - 15 - 21 / spreads into the prime factors to the number to be added, not to the smaller one, then the sign (1 + 2) = 3/1 = (3 + 3) = ( 6 + 4) = 10 | 2 = (10 + 5) = (15 +

10 6) = 21/3 = (21 + 7) = 28, that is the prime number, or to a smaller number (21 + 7) = 28 + 8 = 36 | 2 = 18 | 2 = 9 | 3 = 3, which means it is complex.

11

The sum of two triangular numbers is always a quadratic number: t + t'= n^ 1 + 3 = 2^ 3 + 6 = 3^, and the sum of two square numbers 4 + 9 = 13, 9 + 16 = 25, 16 + 25 = 41, is by 12 - 16, i.e. steadily by 4, and their sum is 4 + 8 = 12, 8 + 12 = 20, 12 + 16 = 28, constantly by 12 - 20 - 28 eight more. Such squares consist of two equilateral triangular numbers 6 and 10 = 16 = 4 (4) with an even diagonal = 4^ /(1 + 2 + 3) = 6 + 4 = 10/2 = 5, 1 + 4 = 5, 2 + 3 = 5, 4(4) = 16 = 6 + 10/ 1 + 3 = 4 = 2^, 6 + 10 = 16 = 4^, 15 + 21 = 36 = 6^, 28 + 46 = 64 = 8^, 45 + 55 = 100 = 10^, ...

12

It is thanks to such distribution of denominations of consecutive natural numbers per units that result in triangular numbers /t = n + (n + 1) + (n '+ 1), 3 = 1 + (1 + 1), 6 = 1 + (1 + 1) + [(1 + 1) + 1],...

13

1 1 1 1

1 1

1 1

3 1

1

6 1

10

All pairs of primes and their products greater than 3 occur every 6 numbers from the number of the first / 5 - 11 - 17 /, / 7 - 13 - 19 /, just like the product of the number 3/9 - 15 - 21 /. It is thanks to the perfect number 6 that the prime numbers and their products are arranged in perfect order. The sum of the units of the four prime numbers is 17 = 2 + 3 + 5 + 7 and this number will be a measure of the number of prime numbers and their products that fall within a given magnitude. In 6(17) = 102 numbers, there are 2(17) = 34 prime numbers and their products greater than 3, and 16 products of the number three / 34 + 16 = 50 /.

14 All products of number 3 has always in a given magnitude capacity: in 100 - 16, in 1000 - 166, and thus also in prime numbers and their products greater than 3 cannot be more than 34 to 100, 334 to 1000. We have so 17 prime numbers and their products greater than 3 in one column, 17 prime numbers and their products greater than 3 in the second column, and 16 products number 3 to 100 in the third column, together 17 + 17 + 16 = 50 numbers coming in given magnitude/167 + 167 = 334 + 166 = 500/.

Counting all prime numbers occurring to π(20) = 8 (2,3,5,7,11,13,17,19) and 2 products of number 3 (9,15), we get 8 + 2 = 10 numbers coming in given magnitude. Among the next 80 numbers we have 17 prime numbers and 23 their products / 8 + 17 = 25 = 2 + 23 /. And here we find the first constant ratio of prime numbers to their products 17/23 = for 40 numbers to a given magnitude 17 can be prime numbers and 23 their products 17 + 23 = 40. We have an even smaller ratio only once in the previous 60 numbers, where it equals 17 + 13 = 30.

From now on, they come in close relation to their products in groups of 17 + (23, 33, 43, 53, 63, 73, 83, 93) numbers, 34 + (66, 86, 96, 106, 116, 126, 136, 146, 156, 166, 176) numbers, 51 + (119, 139, 169, 179, 199, 229), 68 + 202 = 270 numbers, 68 + 222 = 290 numbers, 68 + 232 = 300 numbers, 68 + 272 = 340 numbers. Σ [p (p ')] + π (x) = ½N We cannot say that prime numbers always occur every 20 numbers, but it is certain that they come in close relation to their products in groups of, 17 + (23, 33, 43, 53, 63, 73, 83, 93) numbers after 34 + (66, 86, 96, 106, 116, 126, 136, 146, 156, 166, 176) of the numbers, the 51 + (119, 139, 169, 179, 199, 229) of the numbers, and the 68 + 202 = 270 numbers, the 68 + 222 = 290 numbers, the 68 + 232 = 300 numbers, the 68 + 272 = 340 numbers, as seen in the following table to 10960 numbers. Σ[p(p')]

π(x)

½N

3057

1023

4080

116

34

150

3173

1057

4230

3181

1059

4240

232

68

300

3413

1127

4540

169

51

220

3582

1178

4760

63

17

80

3645

1195

4840

126

34

160

3771

1229

5000

179

51

230

3950

1280

5230

199

51

250

4149

1331

5480

15 It is difficult to imagine a more even distribution of prime numbers and their products than those resulting from this as one after the other at fixed distances, every 20 numbers, / 3 - 23 - 43 - 63 - 83 103, 131 - 151 - 171 - 191 - 211 - 231 - 251 - 271 - 291 /, complementing each other in a strictly defined ratio (17/23, 17/33, 17/43, ..) to half of a given magnitude /½N/. Then they form 16 columns, which is 4 times 4 for each characteristic number of unity /1 - 3 - 7 - 9/ after 16 and 17 numbers and 2 times two or 4 columns for products of 5 after 16 + 17 = 33(2) = 66. How easily one can count a total of 13 columns of 17 numbers is their /13(17) = 221/ plus 1 = 222 of 16 numeric columns of products of the number 5 and 7 columns of 16 numbers /7(16) = 112 + 222 = 334 numbers. In this there are only prime numbers /7(10) + 6(11) + 2(12) + 8 = 168/, and the product of numbers greater than 3, is /4(7) + 9(6) + 66 + 9 + 5 + 4 = 166/.

The same distance /20/ from each other of all numbers arranged zigzag in the following table /9, 29, 49, 69, 89, 109 / makes in the first row on 20 columns there is room for 8 prime numbers and 2 products of the number 3. /8 + 2 = 10/ In the next rows there will be room for 6 prime numbers and their products greater than 3, and 4 products of the number 3/6 + 4 = 10/, further on 7 prime numbers and their products greater than 3, and 3 products of the number 3 /7 + 3 = 10 / and once again on 7 prime numbers and their products greater than 3, and 3 products of the number 3 /7 + 3 = 10/ and further on the 6 prime numbers and their products greater than 3, and 4 products numbers 3 /6 + 4 = 10/. By adding 8 + 6 + 7 + 7 + 6 = 34 we get 34 places that occupy prime numbers and their products greater than 3, and 2 + 4 + 3 + 3 + 4 = 16, the number of places that occupy the product of the number 3. 34 + 16 = 50, i.e. to 100, we have 34 prime numbers and their products greater than 3, and 16 products of number 3.

34/16/50 is the basic ratios of products from number 3 to primes and their products greater than 3, which will repeat every 50 numbers up to infinity. Here, up to 1000, we see them 10 /16 + 16 + 17 +

16 17 + 16 + 17 + 17 + 16 + 17 + 17/ = 166 products on the number 3, and 10 /34 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33/ = 334 prime numbers and their products greater than 3. From this constant ratio of the products of the number 3 to the remaining numbers, it follows that they have a direct impact on the number of prime numbers to a given magnitude /34 = 9 + 25 - 9 = 16/. It is the constant number of products of the number 3 in a given row of 16 = 3 + 3 + 4 + 3 + 3 and 17 = 4 + 3 + 3 + 4 + 3 makes them complemented by a fixed number (34) of prime numbers and their products greater than 3 in five subsequent rows 34 = 7 + 7 + 6 + 7 + 7.

17 This basic ratio of the products of number 3 (16/34) to prime numbers and their products greater than 3 is reflected in the exact ratio of prime numbers to their products. How there in a row of 6 or 7 prime numbers and their products greater than 3 could not be products of the number 3, more than 4 or 3, /7 + 7 + 7 + 7 + 6 = 34, 3 + 3 + 3 + 3 + 4 = 16/, /5 (10) = 50 = 34 + 16/, as a complement to the half of a given magnitude (20/2 = 10), the same is true for 8 prime numbers 2 products of the number 3, /9, 15/, for 4 prime numbers there are 6 their products, for 5 prime numbers there are 5 their products, for 3 prime numbers there are 7 their products, which in total 8 + (4 + 5 + 5 + 3) = 8 + 17 = gives 25 numbers first completed by 2 + (6 + 5 + 5 + 7) = 2 + 23 = 25 their products up to 25 + 25 = 50 halves of a given magnitude.

And it looks like the odd concrete figures we find up to 100.

18 Such an equation of the number of prime numbers with their product 25/25 = 1 : 1 will not happen anymore. Arranged among their products in a strictly defined ratio 17/23, 17/33, 17/43, 17/53, they adhere to the tendency that their products are more and more π (340) = 68/102/170 = 2 : 3 : 5, also in other relations π(910) = 280/630/910 = 4 : 9 : 13, π(4220) = 1055/3165/4220 = 1 : 3 : 4. Such an equalization of the 34/16 to 25/25 ratio takes place at the expense of subtracting and adding the same number 34 = 9 + 25 - 9 = 16 to the half of the sum of 34 + 16 = 50/2 = 25, or subtracting from 34 what is too much and adding this to 16 /34 - 9 = 25 = 9 + 16 /.

The same is true in the next hundred numbers from 120 to 220, here the prime numbers are in the strict 17/33/50 ratio.

19 PRIME NUMBER FUNCTION π(x) Up to now, it seemed that the prime numbers are completely randomly distributed among other numbers. However, the prime numbers are constantly less and the more areas we consider, and their number is in a decreasing ratio, both to the half of the numbers in a given magnitude, and their products 25 : 25 : 50 1/1/2, 68 : 102 : 170, 2/3/5, 85 : 136 : 221, 5/8/13. The primes as to their distribution are subject to one principle, that the sum of prime numbers and their products form half of a given magnitude /π(x) + Σ[p(p')] = ½N /, i.e. they are mutually dependent. Interestingly, the number of their products /Σ[p(p')]/ is a number always with the same parity as the number of prime numbers π(x). (25 + 25 = 50, 168 + 332 = 500, 1 229 + 3 771 = 5,000, 9 592 + 40 408 = 50,000, 78 498 + 421 502 = 500,000) The sum and the difference of two numbers with the same parity is always an even number, and therefore divisible by two. The rule of half the sum and the difference of primes and their products, allows us to calculate the number of prime numbers to half of a given magnitude, because this half consists of half of the sum and the difference of numbers with the same parity.

THEOREM: If the number of their products /Σ[p(p')]/, is the same parity as the number of prime numbers π(x), then half of their sum and difference added when their products are more than prime numbers, or subtracted when prime numbers are less than their products, they give the exact value of π(x) to half of the given magnitude.

PROOF: Up to 1,000 we have 168 prime numbers and 332 of their products. Half of the sum and difference of numbers with the same parity summed up, when their products are more than prime numbers [332 + 168]/2 + [332 - 168]/2 = 250 + 82 = 332, or subtracted when prime numbers are less than their product [332 + 168]/2 - [332 - 168]/2 = 250 - 82 = 168, gives the exact value of π(x) to half of the given magnitude.

168 number long string of prime numbers from 2 - 997, is complemented by 166 products of number 3 (9 - 999), and 166 products of numbers greater than 3 [168 + (166 + 166)] = 168 + 332 = 500, up to half given magnitude. It is clear from this graphic system that the products of the number 3 growing like half of a given magnitude in predictable geometric progression 16, [166 = 16(10) + 6, 1666 = 166(10) + 6], influence the number of prime numbers. and their products greater than 3. That is, according to the basic ratio for the 16 products of the number 3 there are 34 = 25 prime numbers plus 9 their products greater than 3. /16 + 34 = 50 = 25 + (16 + 6 + 3)/ The higher the products of number 3 is more /16, 166, 1666/, the less space there is for prime numbers in this natural sequence of numbers, developing arithmetically. Here the sum of the products of 3 is 864 = 16/2 (9 + 99) = 8 (108), the products of the number 5 is 360 = 6/2 (25 + 95) = 3 (120), / 360: 60 = 6 /. Here also the prime numbers develop in a double arithmetic with the difference R - n (6); 2 - 5 – 11 – 17 – 23 – 29 – 41 – 47 – 53 – 59 – 71 – 83 – 89 – 101 – 107 – 3 - 7 – 13 - 19 – 31 – 37 – 43 – 61 – 67 – 73 – 79 – 97 - 103 - 109 Such an equation of the ratio 332/168 takes place at the expense of subtracting and adding the same number 332 = 82 + 250 - 82 = 168 to half of the sum of the numbers 332 + 168 = 500/2 = 250, that is, subtracting from 332 what is too much to ¼ of a given magnitude and adding this to 168 /332 - 82 = 250 = 82 + 168/.

20 The prime numbers and their products are arranged so evenly that the same number which is the difference between the number of products and ¼ of a given magnitude, determines how many primes there are less, and their products more than ¼ of a given magnitude. π(x) 25 168 1 229 9 592 78 498 664 579 5 761 455 50 847 534 455 052 511 4 118 054 813

∑ p(p’) – N/4 0 82 1 271 15 408 171 502 1 835 421 19 238 545 199 152 466 2 044 947 489 20 881 945 187 -

N/4 25 250 2 500 25 000 250 000 2 500 000 25 000 000 250 000 000 2 500 000 000 25 000 000 000

∑ p(p’) – N/4 + 0 + 82 + 1 271 + 15 408 + 171 502 + 1 835 421 + 19 238 545 + 199 152 466 + 2 044 947 489 + 20 881 945 187

∑ p(p’) 25 332 3 771 40 408 421 502 4 335 421 44 238 545 449 152 466 4 544 947 489 45 881 945 187

On this basis, we can conclude that the prime numbers to a given magnitude is less so, if there are more products than ¼ of a given magnitude, that is all the prime numbers are on the line in the middle ¼ of a given magnitude, and the surplus of their products over ¼ of a given magnitude.

Therefore, knowing the number of products to a given magnitude by subtracting from them ¼ of a given magnitude, we calculate if there are more of them, because the fewer the prime numbers will be. So we can write a pattern: π(x) = N/4 - [Σp(p') - N/4],

168 = 250 - [332 - 250] = 250 – 82

Hence, with the increase of the difference between their products and ¼ of a given magnitude, the number of prime numbers to ¼ of a given magnitude decreases: 250 000 000 – 199 152 466 = 50 847 534 The line graph shows how asymptotically the number of prime numbers decreases as compared to the geometric growth of ¼ of a given magnitude. f (¼N)q, 25 - 250 – 2500, 250 = 25(q), 2500 = 250(q)

21 We see that the mathematical laws resulting from the number parity apply and in this equation (N)/2 - Σ[p(p ')] = π (x), when we subtract from the half of a given magnitude the products of prime number numbers contained in it, to obtain the number of prime numbers to a given magnitude: where the sum and the difference of two numbers with the same parity is an even number / 500 - 332 = 168 /, while the number of different parity is an odd number / 5000 - 3771 = 1229 /.

N/2 10/2 10²/2 10³/2 10⁴/2 10⁵/2 10⁶/2 10⁷/2 10⁸/2 10⁹/2 10¹⁰/2 10¹¹/2 10¹²/2 10¹³/2 10¹⁴/2 10¹⁵/2 10¹⁶/2 10¹⁷/2 10¹⁸/2 10¹⁹/2 10²⁰/2 10²¹/2 10²²/2 10²³/2 10²⁴/2 10²⁵/2 10²⁶/2 10²⁷/2 10²⁸/2

-

∑[p(p‘)] 1 25 332 3 771 40 408 421 502 4 335 421 44 238 545 449 152 466 4 544 947 489 45 881 945 187 462 392 087 982 4 653 934 463 161 46 795 058 249 198 470 155 429 577 331 4 720 761 658 966 075 47 376 442 842 345 767 475 260 045 712 259 140 4 765 942 332 723 655 393 47 779 180 397 439 081 160 478 872 730 513 981 268 072 4 798 532 713 310 684 093 710 48 074 679 608 393 196 031 077 481 564 400 232 650 799 132 134 4 823 153 690 600 856 230 588 320 48 300 753 249 127 562 858 672 397 483 647 539 573 158 319 553 572 601 3 175 744 192 306 678 560 887 745 047

= 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 18 435 599 767 349 200 867 866 176 846 309 399 143 769 411 680 1 699 246 750 872 437 141 327 603 16 352 460 426 841 680 446 427 399 157 589 141 026 654 772 445 588 287

And so are the prime numbers and their products, as follows one by one in 20 columns every 40 - 80 numbers and in rows with a fixed number of places for prime numbers and their products greater than 3, supplemented to ½N by the product of the number 3 (4 - 3 - 3) (6 - 7 - 7), that is 6 places in this row 4 can be occupied by prime numbers, and 2 of their products greater than 3, similarly in rows for 7 places, 5 for prime numbers and 2 for their products greater than 3. Only the first row is the place for 8 prime numbers, and because every third number in each row is the product of the number 3 there will be 2 places for the products of the number 3. In the next row among the 6 places, there are 4 places for products of the number 3, and among 7 places there are 3 places, for products the number three.

22

This leads to the balanced distribution of prime numbers and their products greater than 3 included in the geometric sequence 3(q) in numbers 4 - 30 - 34 - 300 - 334 - 3000 – 3334,. (34 = 9 + 25, 334 = 166 + 168, 3 334 = 2 105 + 1 229), i.e. Σ(p) + [p(p')]> 3 = Σ[p(p')]> 3 + π(x) supplemented always up to ½ N by the product of the number 3 (16, 166, 1 666), as the sum of two fixed numbers 34 + 16 = 50, 334 + 166 = 500, 3 334 + 1 666 = 5,000, and this sum grows within geometric 5 (q). This makes that half of the sum of prime numbers and their products greater than 3, (25 + 9) / 2 = 34/2 = 17 one number subtracted or added from half the difference between them / 25 - 9 = 16/2 = 8 / , you can equalize both of these numbers 9 + 8 = 17 = 25 - 8, 168 - 166 = 2/2 = 1, 166 + 1 = 167 = 168 - 1, 2105 1229 = 876/2 = 438, 2105 - 438 = 1667 = 1229 + 438 half of the sum of prime numbers and their products greater than 3. That is, if the prime numbers are less to half the sum of prime numbers and their products greater than 3, there are more of their products greater than 3.

Hence, we can write this system further: Σ[p(p')> 3 = [π(x) + Σp(p')> 3]/2 ± [π(x) - Σp(p')> 3]/2 9 = (25 + 9) / 2 ± (25 - 9) / 2, 9 = 17 ± 8 = 25, 166 = 167 ± 1 = 168, 2105 = 1667 ± 438 = 1 229

23 So there is a close relationship between the number of arriving primes π(x) and their products greater than three /Σ[p(p')]> 3 /, which increase in geometric progression 3(q), as shown in the table below. /3(q) = Σ[p(p')]> 3 + π(x), 30 = 9 + 21, 300 = 157 + 143/ That is, if we have 4 prime numbers in the top ten, up to 100 there cannot be more than 3(10) = 30, that is 21 prime numbers plus 9 their products greater than 3 is equal 30. 4 + 30 = 34 + 300 = 334 + 3000 = 3334, ... ∑p + [p(p’)]> 3 4 30

∑[p(p’)]> 3 0 9

π(x) 4 21

N 10

34

25 143 168 1 061 1 229 8 363 9 592 68 906 78498 586 081 664 579 5 096 876 5 761 455 45 086 079 50 847 534 404 204 977 455 052 511 3 663 002 302 4 118 054 813 33 489 857 205 37 607 912 018 308 457 624 821 346 065 536 839 2 858 876 213 963 3 204 941 750 802 26 639 628 671 867 29 844 570 422 669 249,393,770,611,256 279 238 341 033 925 2,344,318,816,620,308 2 623 557 157 654 233 22,116,397,130,086,627 24 739 954 287 740 860 209,317,712,988,603,747 234 057 667 276 344 607 1,986,761,935,284,574,233 2 220 819 602 560 918 840 18,906,449,883,457,813,088

10²

300 000 000 000 000 000 000

9 157 166 1 939 2 105 21 637 23 742 231 094 254 836 2 413 919 2 668 755 24 903 124 27 571 879 254 913 921 282 485 800 2 595 795 023 2 878 280 823 26 336 997 698 29 215 278 521 266 510 142 795 295 725 421 316 2 691 542 375 179 2 987 267 796 495 27 141 123 786 037 30 128 391 582 532 273 360 371 328 133 303 488 762 910 665 2,750,606,229,388,744 3 054 094 992 299 409 27,655,681,183,379,692 30,709,776,175,679,101 277,883,602,869,913,373 308,593,379,045,592,474 2,790,682,287,011,396,253 3,099,275,666,056,988,727 28,013,238,064,715,425,767 31,112,513,730,772,414,494 281,093,550,116,542,186,912

333 333 333 333 333 333 334

312,206,063,847,314,601,406

21 127 269 486 018 731 928

3 000 000 000 000 000 000 000

2,819,659,982,796,702,825,638

180,340,017,203,297,174,362

300 334 3 000 3 334 30 000 33 334 300 000 333 334 3 000 000 3 333 334 30 000 000 33 333 334 300 000 000 333 333 334 3 000 000 000 3 333 333 334 30 000 000 000 33 333 333 334 300 000 000 000 333 333 333 334 3 000 000 000 000 3 333 333 333 334 30 000 000 000 000 33 333 333 333 334 300 000 000 000 000 333 333 333 333 334 3 000 000 000 000 000 3 333 333 333 333 334 30 000 000 000 000 000 33 333 333 333 333 334 300 000 000 000 000 000 333 333 333 333 333 334 3 000 000 000 000 000 000 3 333 333 333 333 333 334 30 000 000 000 000 000 000 33 333 333 333 333 333 334

10³ 10⁴ 10⁵ 10⁶ 10⁷ 10⁸ 10⁹ 10¹⁰ 10¹¹ 10¹² 10¹³ 10¹⁴ 10¹⁵ 10¹⁶ 10¹⁷ 10¹⁸ 10¹⁹ 10²⁰ 10²¹

24 3 333 333 333 333 333 333 334

3,131,866,046,644,017,427,044

201 467 286 689 315 906 290

30 000 000 000 000 000 000 000

28,276,146,895,082,511,937,367

1,723,853,104,917,488,062,633

33 333 333 333 333 333 333 334

31,408,012,941,726,529,364,411

1 925 320 391 606 803 968 923

300 000 000 000 000 000 000 000

283,489,720,624,257,603,101,057

16,510,279,375,742,396,898,943

333 333 333 333 333 333 333 334

314,897,733,565,984,132,465,468

18 435 599 767 349 200 867 866

3 000 000 000 000 000 000 000 000

2,841,589,290,368,205,431,456,186

158,410,709,631,794,568,543,814

3 333 333 333 333 333 333 333 334

3,156,487,023,934,189,563,921,654

176 846 309 399 143 769 411 680

30 000 000 000 000 000 000 000 000

28,477,599,558,526,706,628,084,077

1,522,400,441,473,293,371,915,923

33 333 333 333 333 333 333 333 334

31,634,086,582,460,896,192,005,731

1 699 246 750 872 437 141 327 603

300 000 000 000 000 000 000 000 000

285,346,786,324,030,756,694,900,204

14,653,213,675,969,243,305,099,796

333 333 333 333 333 333 333 333 334

316,980,872,906,491,652,886,905,935

16 352 460 426 841 680 446 427 399

3 000 000 000 000 000 000 000 000 000

2,858,763,319,400,186,848,000,839,112

141,236,680,599,813,151,999,160,888

3 333 333 333 333 333 333 333 333 334

3 175 744 192 306 678 560 887 745 047

157 589 141 026 654 772 445 588 287

10²² 10²³ 10²⁴ 10²⁵ 10²⁶ 10²⁷ 10²⁸

Thus, the puzzle of the distribution of prime numbers has been solved. Henceforth, the sequence of prime numbers is not similar to the accidental sequence of numbers, but to the ordered in geometric progression 3(q) of the increasing sequence of primes and their products greater than 5 to the half of a given magnitude. So the sum of primes and their products greater than five π(x) + Σ[p(p')]> 5 equals the difference between half of a given magnitude and the products of the number three ½N - i (3), 25 + 9 = 34 = 50 - 16 and grows in geometric progression 3(q), 34 - 300 - 334 - 3000 - 3334. The ratio of prime numbers to their products is thus determined by the complement to half the given magnitude /π(x) + Σ[p(p')] = ½N and is 17 + (23, 33, 43, 53, 63, 73, 83, 93), 34 + (66, 86, 96, 106, 116, 126, 136, 146, 156, 166, 176), by 51 + (119, 139, 169, 179, 199, 229), after 68 + (102, 202, 222, 232, 272)/ numbers, with additional ratios within these ratios, e.g. 68(p) + 102Σ[p(p')] = 170 (½N), 2 : 3 : 5, 180(p) + 360Σ[p(p')] = 540(½N), 1 : 2 : 3. Total radar chart gives an image of a grid of numerals arranged radials at 10 columns of fixed distance n(20) between them, although one can also see left-

25 handed and dextrorotary spiral structures. Finally sought for centuries by Mathematicians, the mysterious structure of prime numbers and their products has been discovered and their music can be written forever.

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