PRIME NUMBERS CODE

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2 ABSTRACT In the article "PRIME NUMBERS CODE" I show that the prime numbers are arranged on the number line in order in relation to the products and their occurrence is not random or chaotic, like weeds in the field, and completely subject to strict rules of dependence of prime numbers on the number of integers up to a half of a given quantity ½N. From this dependence and ordering, it is possible to calculate exactly how many primes are in a given interval, knowing the decoding number, half of a given quantity in which they are written from dawn, whether we want it or not, and also predict their number in the next interval of numbers up to half of the given values of ½N, using their constantly increasing by 1.08 ratio of prime numbers to ½N. RATIO OF PRIMES TO PRODUCTS Each natural number is either a prime number or a prime product, i.e. each integer uniquely decomposes into a product of prime numbers. Hence the mutual dependence of prime numbers on the number of integers, i.e. how many primes can be in a given range of numbers.

The ratio of prime numbers to products has already been determined, and it results from the ability to produce identical intermediate sums for a given quantity. Up to ten we have 4 primes (2 + 3 + 5 + 7 = 17), their sum is 17 and they form 4 identical intermediate sums up to 10 [(2 + 8 = 10), (3 + 7 = 10), (5 +5 = 10), (7 + 3 = 10), (8 + 7 + 5 + 3 = 23), and the sum of the complement of 10 is 23. (17 + 23 = 40/4 = 10). According to this scheme, the ratio of prime numbers to their products will be shaped, i.e. 40 odd numbers in a given interval, there can be 17 primes (from 20 - 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 to 100) and 23 of 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), which can be written by the formula π (N) + ∑[p(p')] = ½N, which says that the sum of the number of prime numbers and their products is equal to half of the given amount. From 20 to 100 there are 80 numbers, half of which are 40 = 17 + 23 are prime numbers and their products. To 20/2 = 10 = 8 + 2 we have 8 prime numbers (2, 3, 5, 7, 11, 13, 17, 19) and 2 products (9, 15), and to 100/2 = 50 = [(8 + 17) + (2 + 23)] = 25 + 25 the ratio becomes even.

3 So it is enough to create a table from the numbers that follow regularly every 2 (3, 5, 7, 9, 11, 13, 15, 17, 19), and without Eratosthenes' sieve we have selected all primes and their products. At first glance you can see with what regularity there are products of number 3, every 18 numbers (9-27-45) or products of number 5, every 30 numbers (25-55-85, 35-65-95) or products of number 7, which 42 numbers (49-91-133), or products of 11, every 22 numbers (121-143, 187-209), or products of 13, every 78 numbers (169-247, 403-481), or products of 17 , every 34 numbers (289-323).

4 This table clearly shows that prime numbers together with products greater than 3 form two separate sequences (11-29-47-65-83-101, 13-31-49-67-85-103), next to the series of products of 3 (15-33-51-69) The formula π (N) + ∑p(p')> 3 = 34q + 1 says that the number of primes up to a given quantity N plus the number of products of primes greater than 3 creates a constant sum of 34 + 1, growing exponentially (34q) +1. Also, the products of the number 3 form a constant sum 17 - 1, growing exponentially (17q) -1. These two sums (34 + 1) + (17 - 1) = 51 make up half of the given quantity, which grows exponentially (51q). Hence we write [π(N) + ∑p(p')> 3] + ∑3(p) = ½N, (26 + 9) + 16 = (34 + 1) + (17-1) = 51 = 26 + (9 + 16) = 26 +25 and this is the basic formula for the ratio of prime numbers to products.

Up to 102/2 = 51 only 10 primes (3, 5, 7, 11, 13, 17, 19, 23, 29, 31) produce 25 products (9/3, 15/5, 21/7, 25/5 , 27/9, 33/11, 35/7, 39/13, 45/15, 49/7, 51/17, 55/11, 57/19, 63/7, 65/13, 69/23, 75 / 25, 77/11, 81/27, 85/17, 87/29, 91/13, 93/31, 95/19, 99/33). Each of these numbers knows its place in the series, which means that there must be no more than 25 products in these three sequences, because each product stands in a place which it signifies and is reserved only for him.

Up to 102 can be only 96/6 = 16 products of number 3, to 1.020 and 1.014 / 6 = 169, and to 10.200 and 10.194 / 6 = 1.699 products of number 3. So these are constants that make the sum of the prime numbers and products greater than 3 have the form (34 +1) = 26 + 9, (340 + 1) = 171 + 170, (3400 + 1) = 1252 + 2149. This obviously has a decisive influence on the ratio of prime numbers to as follows: 26/25, (1: 1) ± 0.5, 16 + 9 = 25, 171/339, 169 + 170 = 339, (1: 2) ± 1, 1.252 / 3.848, (1: 3) ± 23 , 1.699 + 2.149 = 3.848.

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As the table above shows, the ratio of prime numbers to products increases steadily by (n: n ') ± n, which results from the equation n + n' = ½N/n, 51/2 = 25.5, 26 - 25.5 = 0.5, 26/25, (1 : 1) ± 0.5, 25.5 + 0.5 = 26, 25.5 - 0.5 = 25, 510/3 = 170, 171 - 170 = 1, 171/339, (1 : 2) ± 1, 170 + 1 = 171, 2(170) - 1 = 339, 5,100/4 = 1275, 1275 - 1252 = 23, 1252/3848, (1 : 3) ± 23, 1275 - 23 = 1252, 3(1275) + 23 = 3848, 51,000/5 = 10,200, 10,200 - 9,766 = 434, 9766/41234, (1: 4) ± 434, 10,200 - 434 = 9,766, 4(10,200) + 434 = 41,234, 510,000/6 = 85,000, 85,000 - 81,749 = 3,251, 81,749/428,251, (1: 5) ± 3,251, 85,000 – 3,251 = 81,749, 5(85,000) + 3,251 = 428,251, 5,100,000/7 = 728,571, 728,571 – 696,575 = 31,996, 696,575/4,403,425, (1 : 6) ± 31,996, 728,571 – 31,996 = 696,575, 6(728.571) + 31,996 = 4,403,425, 51,000,000/8 = 6,375,000, 6,375,000 – 6,068,228 = 306,772, 6,068,228/ 44,931,772, (1 : 7) ± 306,772, 6,375,000 – 306,772 = 6,068,228, 7(6,375,000) + 306,772 = 44,931,772, 510,000,000/9 = 56,666,666, 56,666,666 – 53,757,100 = 2,909,566, 53,757,100/456,242,900, (1:8)±2,909,566, 56,666,666 – 2,909,566 = 53,757,100, 8(56,666,666) + 2,909,566 = 456,242,900, i.e. by how much less ± n there are prime numbers, including 2, 3, 4, 5 , 6, 7, 8, .. times more ± n of their products.

6 And here is the table of prime numbers and their products greater than 3, π(N) + ∑p(p')> 3 expanding exponentially (34q) +1, and the products of the number 3, ∑3(p) in progress (17q) -1.

In such a simple way, using addition and subtraction, we can calculate all the elements of this simple equation based on the constants: [π (N) + ∑p(p')> 3] + ∑3(p) = ½N, (26 + 9 ) + 16 = (34 + 1) + (17-1) = 51 = 26 + (9 + 16), 25 - 9 = 17 - 1 = 16, 26 + 9 = 34 + 1, (34 + 1) - 26 = 9

The ratio of primes to products determined by this method tells rather how many primes there are fewer and more products in the rounded to integer ratio (n : n’) ± n, (1 : 2) ± 1, (1 : 3) ± 23. .

7 PRIMARY NUMBER FUNCTION π(N) Until now, the prime numbers seemed to be arranged quite randomly on the number line. It has been observed that there are fewer primes, the larger the numbers we consider. As far as their distribution is concerned, prime numbers follow one rule that the sum of prime numbers and their products make up half of the given quantity π (N) + ∑p (p ’) = ½N, i.e. they are mutually dependent.

Prime numbers are also subject to the congruence law according to the modulus ≡ 102, hence the number of primes for half of a given quantity decreases asymptotically, while the number of their products increases asymptotically.

(½N)/∑[p(p’)] = (51/25)/25 = (2:1)+1, 2(25) + 1 = 51, (510/339)/170 = (3:2)-1, 2(170) – 1 = 339 (5100/3848)/170 = (30:22)+108, 22(170) + 108 = 3,740 + 108 = 3,848.

It is difficult to imagine a more even distribution of prime numbers and their products than those resulting from the way they follow one another at constant distances, every 6 numbers, complementing each other in a strictly defined ratio (34 + 1) + (17-1) to half of a given quantity ½N =

8 51, as we can see in the line graph above. [π(N) + Σ[p(p’)> 3] + ∑3(p) = ½N. Therefore, although in the Riemann hypothesis, the prime distribution function π (x) is a gradual function of small serious irregularities, in a triple arithmetic sequence of primes and their products, with a constant interval D = 6 to half a given quantity ½N, we see surprising smoothness . The uniformity with which this graph grows is not due to the number of primes up to half a given quantity ½N, which can be located by a logarithmic function, but to their regular distribution, which comes from the constant difference d = 6 between the members of the triple arithmetic sequence of primes and their products.

We notice a similar regularity when we arrange prime numbers according to the doubly characteristic numbers of one’s 7 - 11 - 13 - 17 - 19 - 23 - 29 - 31. Then they form 8 sequences of prime numbers with a spacing n(30) = n(5*6) .

9 It proves the huge regularity with which the prime numbers are arranged, in blocks of 51(q) numbers, that is up to half of the given quantity ½N, as it can be seen in the table below.

From this dependence and ordering, it is possible to calculate exactly what number of prime numbers is in a given interval, and also predict their number in the next interval of numbers up to half of a given quantity ½N.

10 With the same accuracy as one, you can calculate the number of primes using their constant 1.08 ratio of prime numbers to half of a given quantity ½N. This is because both prime numbers and their products form sequences, and the sum of the sequence is divisible by 16 π(N) + ∑p(p') = ½N/16, 9,766 + 41,234 = 51.000/16 = 3,187.5 when 9,766 primes are paired with 41,234 their products, they form a sequence of 16(3,187.5) = 51,000 numbers. If we now divide the quotient of the sum of the prime numbers and their products by the quotient of the prime numbers alone 9,766/16 = 610.375, we get 3,187.5/610.375 = 5.2221994 what is the ratio of prime numbers to half of the given quantity.

Conversely, dividing the quotient of all numbers by half of a given quantity by the ratio of prime numbers 3,187.5/5.2221994 = 610.375, we obtain the quotient of primes in this sequence, which multiplied by 16(610.375) = 9,766 gives the number of primes in the sequence with the accuracy of one. The easiest way is 51,000/5.2221994 = 9,766, so ½N has been decoded.

31,875/6,2386084 = 5,109.3125(16) = 81,749 p 318,750/7.3215375 = 43,535.9375(16) = 696,575 p 3,187,500/8.4044304 = 379,264.25(16) = 6,068,228 p 31,875,000/9.4871189 = 3,359,818.75(16) = 53,757,100 p 318,750,000/10.5693730 = 30,157,891(16) = 482,526,256 p 3,187,500,000/11.6512593 = 273,575,576.5625(16) = 4,377,209,225 p 31,875,000,000/12.7328246 = 2,503,372,250.5625(16) = 40,053,956,009 p 318,750,000,000/13.8141519 = 23,074,163,410.875(16) = 369,186,614,574 p 3,187,500,000,000/14.8952972 = 213,993,715,426.875(16) = 3,423,899,446,830 p 31,875,000,000,000/15.9762998 = 1,995,142,825,708.4375(16) = 31,922,285,211,335 p 318,750,000,000,000/17.0571887 = 18,687,135,657,310.1875(16) = 298,994,170,516,963 p 3,187,500,000,000,000/18.1379858 = 175,736,161,176,694.75(16) = 2,811,778,578,827,116 p 31,875,000,000,000,000/19.2187076 = 1,658,540,238,158,568.5(16) = 26,536,643,810,537,096 p 31,875(10¹⁸)/25.702018 = 1,240,174,992,729,662,527,120.8125(16) = 19842799883674600433933 p Knowing the steadily increasing ratio of prime numbers to ½N, always by 1.08, we can calculate that in the sequence consisting of 51(10²⁴) numbers there will be 318.750(10¹⁵)/26.7820180 = 11,901,642,363,170,691,618,533 (16) = 190,426,277,810,731,065,896,528 prime numbers, and in the sequence consisting of 510(10²⁴) numbers will be 3,187,500(10¹⁸)/27.862018 = 114,403,055,801,629,300,505,081(16) = 1,830,448,892,826,068,808,081,296 primes, and in the sequence consisting of 5,100(10²⁴) numbers, there will be 31,875(10²¹)/28.942018 = 1,101,339,927,298,780,617,163,599,303(16) = 17,621,438,836,780,489,874,617,588,848 prime numbers. For comparison, below is the function of the number of primes π(x) in the decimal system.

You can see that these numbers 10/4, 100/25, 1,000 / 168, .. nothing to do with each other, because there is no visible relationship here.

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Contrary to the table below that shows how prime numbers are padded to ½N.

12 Proposal; accordingly, it is evident that the Riemann zeta function is based on a false hypothesis to represent prime numbers. Therefore, it is impossible to refute or substantiate the proof for the Riemann Hypothesis, since there are no grounds on which it is based.

Thus, the puzzle of the distribution of prime numbers has been solved. Henceforth the sequence of primes is not like a random sequence of numbers, but the ordered progressively decreasing ratio of primes to half a given quantity. Finally, the mysterious structure of prime numbers and their products, sought for centuries by mathematicians, has been discovered and its music can be written endlessly.

THE RATIO OF TWIN NUMBERS TO THE PRIMES Twin numbers are two prime numbers that can be represented as (6n Âą 1), 6(1) - 1 = 5, 6(1) + 1 = 7, 6(2) - 1 = 11, 6(2) + 1 = 13. Here's how primes come together in twin pairs.

We can see that the number of the place on which they rank is a number whose 6(n) Âą 1 product makes it 6(10) - 1 = 59, 6(10) + 1 = 61, 6(12) - 1 = 71, 6(12) + 1 = 73. To 102 we have among 35 numbers there are seven pairs of twin numbers plus one 2(7) + 1 = 15. To 1020 we have among 341 prime numbers and their products greater than 3, there are 2(34 ) + 1 = 69.

13 In the line graph above, we can see exactly how the primes and twins are arranged in 16 sequences with constant spacing n(30), 11-13, 41-43, 71-73, 101-103, 29-31, 59-61, 149 -151, ..

This property can be used when calculating the number of tins to a given magnitude. If the number of prime and twin numbers can be written as the quotient of the number 16, e.g. π(1020) = 16(10.6875) = 171, and π₂ (1020) = 16(4.3125) = 69, it is easy to calculate the ratio of prime numbers to the twin numbers 171/69 = 2.4782608 = 10.6875/4.3125.

14 In the interval 10,000≤p≤10,200, there are 23 prime numbers, i.e. (1229 + 23) = 1252 primes, and the twin numbers 10, i.e. a sum of π₂(10,000) + π₂(10,200) = 408 + 10 = 418 numbers 10007, 10009, 10037, 10039, 10061, 10067, 10069, 10079, 10091, 10093, 10099, 10103, 10111, 10133, 10139, 10141, 10151, 10159, 10163, 10169, 10177, 10181, 10193.

We can see how primes and twins keep a constant spacing n(30), 10007-10009, 10037, 10067-10069, 10061, 10091-10093, 10103, 10111-10133, 10163, 10193, ..

There are 174 prime numbers in the range of π (100,000≤p≤102,000), i.e. (9,592 + 174) = 9,766 in the sum of prime numbers and 29 twins π₂(100,000) + π₂(102,000), it is (2,446 + 29) = 2,475.

100003, 100019, 100043, 100049, 100057, 100069, 100103, 100109, 100129, 100151, 100153, 100169, 100183, 100189, 100193, 100207, 100213, 100237, 100267, 100271, 100279, 100291, 100297, 100313, 100333, 100343, 100357, 100361, 100363, 100379, 100391, 100393, 100403, 100411, 100417, 100447, 100459, 100469, 100483, 100493, 100501, 100511, 100517, 100519, 100523, 100537, 100547, 100549, 100559, 100591, 100609, 100613, 100621, 100649, 100669, 100673, 100693, 100699, 100703, 100733, 100741, 100747, 100769, 100787, 100799, 100801, 100811, 100823, 100829, 100847, 100853, 100907, 100913, 100927, 100931, 100937, 100943, 100957, 100981, 100987, 100999, 101009, 101021, 101027, 101051, 101063, 101081, 101089, 101107, 101111, 101113, 101117, 101119, 101141, 101149, 101159, 101161, 101173, 101183, 101197, 101203, 101207, 101209, 101221, 101267, 101273, 101279, 101281, 101287, 101293, 101323, 101333, 101341, 101347, 101359, 101363, 101377, 101383, 101399, 101411, 101419, 101429, 101449, 101467, 101477, 101483, 101489, 101501, 101503, 101513, 101527, 101531, 101533, 101537, 101561, 101573, 101581, 101599, 101603, 101611, 101627, 101641, 101653, 101663, 101681, 101693, 101701, 101719, 101723, 101737, 101741, 101747, 101749, 101771, 101789, 101797, 101807, 101833, 101837, 101839, 101863, 101869, 101873, 101879, 101891, 101917, 101921, 101929, 101939, 101957, 101963, 101977, 101987, 101999.

15 Knowing the number of prime numbers written in the form of the quotient of the number 16 and the ratio of the twin numbers to them, then by dividing this quotient by the ratio we obtain the quotient of the twin numbers resulting from this ratio. It is worth noting that this ratio, from π(10,200)/π₂ (10,200) = 1252/418 = 2.9952153, is constantly increasing by 0.945. Thus, if there are 418 twins and 1252 primes to π₂(10,200), their ratio is 2.9952153 and the next 0.945 greater is 3.9458585. So dividing this quotient of the number of primes by the ratio 3.9458585 and multiplying by 16, (9766)/ 16 = 610.375/3.9458585 = 154.6875(16) = 2.475, we get the number of twin numbers in the ratio 3.9458585 to the primes in this sequence. In this way, we can calculate any number of twins from the ratio.

81,749/16 = 5,109.3125/4.8910494 = 1044.625(16) = 16,714 p, p+2 696,575/16 = 43,535.9375/5.8300064 = 7467.5625(16) = 119,481 p, p+2 6,068,228/16 = 379,264.25/6.7700144 = 56,021.1875(16) = 896,339 p, p+2 53,757,100/16 = 3,359,818.75/7.7150019 = 435,491.625(16) = 6,967,866 p, p+2 482,526,256/16 = 30,157,891/8.6600008 = 3,482,435.125(16) = 55,718,962 p, p+2 4,377,209,225/16 = 273,575,576.5625/9.6050000 = 28,482,621.125(16) = 455,721,938 p, p+2 40,053,956,009/16 = 2,503,372,250.5625/10.550000 = 237,286,469.0625(16) = 3,796,583,505 p, p+2 369,186,614,574/16 = 23,074,163,410.875/11.495 = 2,007,321,740.75(16) = 32,117,147,852 p, p+2 3,423,899,446,830/16 = 213,993,715,426.875/12.44 = 17,202067156.3125(16) = 275233074501 p, p’

16 As we can see, up to the number 360/2 = 180 we have 20 pairs, i.e. 40 twin numbers by a further 40 to 80 numbers will increase after 800 numbers with 1160/2 = 580, so that after 840 numbers with 2000/2 = 1000 there are 120, and after 980 numbers at 2980/2 = 1490 there are 40 more, or 160 twin numbers. So evenly distributed twins are constantly 40 more (40-80-120-160-200-240-280-320360-400,).

Hence it follows that both primes and twins are hexadecimal coded at half the given magnitude and therefore we can reconstruct them.

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BIBLIOGRAPHY

Sierpiński Wacław Elementary Theory of Numbers, Ziegler Günter The great prime number record races, Pomerance Carl The Search for Prime Numbers, Peter Bundschuh: Einführung in die Zahlentheorie, Paulo Ribenboim: The New Book of Prime Number Records, Narkiewicz Władysław The Development of Prime Number.

17 TABLES OF PRIME AND TWIN NUMBERS UP TO 1 021

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20 TABLES OF PRIME NUMBERS AND THEIR PRODUCTS FROM 2 TO 1 023

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23 TABLES OF PRIMES AND TWIN NUMBERS FROM 2 TO 10 993

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