Order on 102

Tel. + 48 668 934 960 lubinaj7@gmail.com

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 “Order on 102”, what laws rule and what order prevails 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 by means of numbers, but how the numbers themselves are ordered, I will try to show in this article “Order on 102”. Let’s look briefly at the table below of prime numbers and their products up to 107. At first glance, we notice a repeating distinct diagonal pattern of 6 product numbers / 17–34–51–68–85–102 /. How it affects the distribution of prime numbers and their products we will find out in a moment.

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 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. So how many distributions creating identical intermediate sums we get from 9 digits of natural numbers arranged in increasing and decreasing sequences (1 + 9) = (2 + 8) = (3 + 7) = (4 + 6) = (5 + 5) = (6 + 4) = (7 + 3) = (8 + 2) = (9 + 1) = (8 + 2) = (7 + 3) = (6 + 4) = (5 + 5) =,..

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/, i.e. all digits, 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 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.

DISTRIBUTION OF NUMBERS FOR 102

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. The sum of the numbers in these six columns is also divided by 17. 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.

From this layout, you can create one column of 17 odd numbers from 5 to 101, in which the sum 901 number 17 falls 53 times, and the second column from 1 to 97, then in sum 833 the number 17 falls 49 times. Summing up both sums 901 + 833 = 1734, we see that the number 17 is 53 + 49 = 102 times. So one can say that the distribution of prime numbers and their products greater than 3 is for one hundred two, that is, the most perfect possible, because the ratio of the sum of the numerical values of the odd numbers from 3 + 5 + 7 + 9 +, .. 101 = 2601 to the sum of the numerical value of prime numbers and their products 3 + 5 + 11 + 19 + 25 + 35, .. = 1734, is always constant and is 2601/1734 = 1.5. Also the ratio of the common factor of odd numbers to primes and their products is constant and amounts to 153/102 = 1.5. As the numbers are arranged at 102 = 6(17), it shows the above table, where 0 + 101 numbers are arranged in 6 columns of 17 numbers. Also the sum of the numerical values of prime numbers and their products greater than 3, consists of 102(17) = 1734 = 901/53 + 833/49 = 53 + 49 = 102 seventeen. And the sum of the numerical values of the products of the number 3 from 8 pairs of homogeneous intermediate sums equal to 102 plus 51. 867 = 8(3 + 99) + (9 + 93) + (15 + 87) + (21 + 81) + (27 +75) + (33 + 69),… + 51

All this happens due to the number 17, which in its total numerical value itself has a common factor of 17 nine times / (1 + 16) (2 + 15) (3 + 14) (4 + 13) (5 + 12) (6 + 11) (7 + 10) (8 + 9) +17 = 153/17 = 9 /. In the next column of even numbers divided by 2 and 3 in their sum of numerical values, common factor 17 will be located 48 times (816/17 = 48). Further, the numerical values of even numbers divisible by two 8(2 + 98) = 8(100) + 50 = 850/17 = 50, the common factor 17 is 50 times. In the whole set of natural numbers consisting of six columns of 17 numbers, the common factor 17, generates increasing arithmetic mean S = 1/n(a + a ‘, ..) 884/17 = 52, which is the total of numerical values 52 times.

This leads to the fact that even the sum of primes and the products falling into them in 17-numeric columns is a number divisible by 17. Thus, we can write π (x) + Σ [p(p’)] = n (17) 26p + 25 = 51/17 = 3, to 101 we have 26 prime numbers and 25 their products, that is, a total of 51 odd numbers. In the next seventeen numbers from 103–203, there are 20 prime numbers and 31 their products, hence the total sum doubles (26 + 20 = 46, 25 + 31 = 56, 46 + 56 = 102/17 = 6.

And this is how it goes on until 1020/17 = 60. The sum of prime numbers and their products is always ½ N, the number to which they occur [π (x) + Σ [p (p ‘)] = ½N, 26 + 25 = 51 = 102/2]. It also arrives evenly in three columns of 17 (3) = 51 numbers, the ratio of prime numbers to their products strives to maintain the ideal ratio of 17/34 = 1: 2, which we can see in the table below.

Other proportions of the ratio of primes to their products result from distributions that form identical indirect sums of 51 (0 + 51 = 51)/17, 0:3:3 (1 + 50 = 51) (2 + 49 = 51) (3 + 48 = 51)/3, 1:16:17, (4 + 47 = 51) (5 + 46 = 51) (6 + 45 = 51)/3, 2:15:17, (7 + 44 = 51) ( 8 + 43 = 51) (9 + 42 = 51)/3, …

And so the first five distributions will be for example among 102 numbers above 1⁰¹², 1⁰²⁴, 1⁰³⁶, 1⁰⁴⁸, 1⁰⁵⁷, 1⁰⁶⁰, 1⁰⁷¹, 1⁰⁷². Here, among the 34 odd numbers, we have only 4, 3, 2, 1, and finally no prime number, only their products, hence the ratio 0 + 51 = 51, (0: 3: 3) will apply here. When the last 26 + 25 = 51 occurs one time in the first circuit to 102.

Further, we can follow further distributions from (9 + 42 = 51)/3, 3:14:17 to 20 + 31 = 51. Also entire blocks of primes and their products can create their own relations, for example (187 + 374 = 561)/ 187, 1: 2: 3, or dividing it by 17, (187 + 374 = 561)/17, we get a ratio of 11:22:33, or (867 + 2 499 = 3366)/51, we get a ratio of 17:49:66. The sum of prime numbers and their products is equal to half of the given quantity to which they occur, and is a divisible number by 17, and also by 3, and then the ratio (141 + 267 = 408)/3 = 47:89:136, (156 + 303 = 459)/3 = 52:101:153. It only testifies to the fact that the primes and their products are arranged perfectly in one hundred and two order, and the number 17 plays a decisive role in the creation of mutual relations.

It is worth noting that not only the sum of primes and their products is a number divisible by 17 π(x) + Σ[p(p’)] = n(17), 26 + 25 = 51 = 3(17), 171 + 339 = 510 = 30(17), 1 229 + 3 769 = 4 998 = 294(17), but also the sum of primes and their products greater than 3, π(x) + Σ[p(p’)]> 3 = n(17), 171 + 169 = 340 = 20(17), 1 229 + 2 103 = 3 332 = 196(17), which means that they are themselves in relation (510/340) /170, 3:2 , (4998/3332)/1666, 3:2. Similarly, the constant ratio prevails between the sum of prime numbers and their products greater than 3, and the given size N, to which they occur and is 1:3, 340/1020, 3 332/9 996, 33 320/99 960, .. Also the number of products number 3, it is constant to a given size and is always N/6, 1020/6 = 170, 9 996/6 = 1 666, 99 960/6 = 16 660, 999906/6 = 166651. As we can see the prime numbers and their products, it arrives in a fixed amount of 51 numbers. Up to 102–51, up to 1 020 — ten times more 10(51) = 510, up to 9 996 ninety — eight times more 98(51) = 4 998.

This testifies to a wonderful order for one hundred and two prevailing among prime numbers and their products, and this makes it easier for us to calculate the number of prime numbers to a given size. In 102 numbers, half is 51 prime numbers and their products. Up to 102, we have 16 products of numbers 3 /9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99/, 6 products of 5 /25, 35, 55, 65, 85, 95/ and 3 products of number 7 /49, 77, 91/. Hence, we can write in 51 numbers can be 51 — (16 + 6 + 3) = 51–25 = 26 prime numbers. Similarly to 1 020, we have 510 numbers, in which 339, as we calculated, the products of prime numbers, i.e. 510 numbers can be 510–339 = 171 prime numbers. These products of prime numbers always consist of the known number of products of number 3, which is N/6, and the number of products greater than 3, which is calculated by subtracting from the whole the fixed number of products of number 3 [339 — (1020/6) = 339–170 = 169]. These products are a part of three times smaller than the given size N, associated with prime numbers. So if from this three times smaller size we subtract the number of products greater than 3, we get the number of prime numbers to a given size N/3 — Σ[p(p’)]> 3 = π(x), 1 020/3 = 340–169 = 171.

And so we have reached the number 9 996, which contains 4 998 odd numbers, including 1 229 prime numbers and 3 769 of their products distributed in 98 fifty one numeric blocks. Up to 99,960 these blocks will be 10 times more, i.e. 980 (51) = 49 980, as can be seen in the table below.

Knowing then these basic constant values, such given size N, its three times smaller size N/3, containing primes with their products greater than 3, and a fixed number of products of number 3 equal to N/6, we can easily calculate the number of prime numbers with accuracy for one.

The distribution of prime numbers and their products follows in order on one hundred two. From 102 to 1020, they arrive at 918/102 = 9 times a hundred and two numbers. Between 1 020 and 9 996 there are 8 976/102 = 88 times 102 numbers. Between 9 996 and 99 960, there are 89 964/102 = 882 times 102 numbers. The same applies to half of a given size (½N), to which the primes are complemented by their products. From 51 = 26 + 25 prime numbers and their products in the first half of a given size up to 510 = 171 + 339 numbers up to the tenth half of a given size, increase comes in, 459/51 = 9 times 51, or 51 + 459 = 510 = (1 + 9) = 10(51). Between 510 and 4 998 there are 4488/51 = 88 times 51, or 510 + 4 488 = 4 998 = (10 + 88) = 98(51). In the next range from 4 998/51 = 98 to 49 980/51 = 980 there will be exactly 10 times more, further 499 953/51 = 9 803, 4 999 530/51 = 98 030, 49 959 453/51 = 980 303,..

The same right of deployment for one hundred and two is subject to three times less sum of prime numbers and their products greater than 3 {N/3 = π(x) + Σ[p(p’)]> 3}, and six times smaller number of products of number 3 {Σ[3(p)] = N/6}. 1020/3 = 340, 9 996/3 = 3 332, 3 332–340 = 2 992/34 = 88, 1020/6 = 170, 9996/6 = 1666, 1666–170 = 1496/17 = 88.

Subtracting from the sum of products of prime numbers, a known six times smaller number of products of the number 3, we get the number of products greater than 3 Σ[p(p)] — Σ[3(p)] = Σ[p(p’)]>3 339 — (1020/6) = 169, and this subtracted from the three times smaller sum of prime numbers and their products greater than 3, gives us the exact number of prime numbers to the half of a given size. N/3 — Σ[p(p’)]> 3 = π(x), 1020/3 = 340–169 = 171, 171 + 339 = 1020/2 = 510.

With the same accuracy of one you can calculate the number of prime numbers using their constantly increasing ratio of 0.8 prime numbers to their sum with products greater than 3. This is because both prime numbers and their products greater than 3 form sequences, and the sum of the sequence is divisible by 34 π (x) + ∑ [p (p ‘)> 3] = n / 34, 9,592 + 23,728 = 33,320 / 34 = 980, when 9,592 prime numbers pair with 23.728 products greater than 3 , form a series of 34 (980) = 33320 numbers. If we now compare the quotient of the sum of prime numbers and their products greater than 3 (980) with the quotient of prime numbers 9.592/34 = 282 only, we get 282/980 = 1/3.4 what is the ratio of prime numbers to their products greater than 3 in this sequence . It is worth noting that the quotient of the sequence 98–980–9803–98039215–980392156–9803921568–98039215686–980392156862 -9803921568627–98039215686274–980392156862745–9803921568627450–98029215686274509 the same 980392156862745098.

Hence, if the number of numbers in a given sequence is known and how many prime numbers with their products greater than 3 fit in it, it is written as the quotient of the number 34, e.g. 333 333 333 333 333 333 333 333 333 302/34 = 9 803 921 568 627 450 980 392 156, as well as the number of prime numbers written in a similar way: 16 352 460 426 841 680 446 427 399 / 34 = 480,954,718,436,520,013,130,217, the ratio to the total number of the first sequence is 480,954,718,436,520,013,130,217/9,803,921,568,627,450,980,392,156 = 1/20.3 and the next string comprising 3 333 333 333 333 333 333 333 333 312/34 number = 98.039.2585.274.509.803.921.568, will contain prime numbers in the ratio 20.3 + 0.8 = 21.1, i.e. dividing this quotient 98.039.2525.268.274.509803.921.568 by the ratio /21.1 = 4.646.408.326.3363.720.843.787.752(34) and multiplying by 34, we get the product of the number of prime numbers = 157 977 883 096 366 508 688 783 569. And so, with the accuracy of one, we can calculate any number of prime numbers in this sequence up to infinity.

The order for the one hundred two that we see here simply means the most excellent order that prevails in the whole set of both odd and even integers, that is, among prime numbers and their products. If one hundred is a whole, for example one hundred percent, and more than one hundred percent cannot be, then exceeding this number gives the feeling of something really excellent, at least one hundred two percent. And in this sense literally, for one hundred and two, all prime numbers are arranged, by an perfect number 6 times the prime number 17 = 102. Has anyone seen any more perfect order than shown here?

DISTRIBUTION OF PRIME NUMBERS

The prime numbers seem to be completely randomly distributed among other numbers. At the same time, it was observed that the prime numbers are the less the larger the numbers we consider. Prime numbers are subject to the only right of deployment, an order of one hundred two, which precisely determines how many primes can be half of a given size. This order serializes prime numbers and their products in such a way that we can read from the middle of a given magnitude the ratio of prime numbers to their products. Up to 1 020 we have 510 prime numbers and their products. If we divide the whole into 3 parts, we will see that the prime numbers 170 + 1 = (171) are complemented by their product 2(170) — 1 = (339) to half of the given size 3(170)= (510) in a ratio of 3: 1 ± 1. To 9 996 we have 4 998 primes and their products, if we divide it by 4, we will see that the prime numbers (1 249.5) — 20.5 = (1229), are complemented by their product 3(1 249.5) + 20.5 = (3 769) to half of the given size 4(1 249.5) = (4998) in a ratio of 4: 1 ± 20.5, i.e. the number of primes, is in a ratio decreasing asymptotically to half of a given size.

The number of prime numbers out of 102 is subject to the only law that says that the number of prime numbers is always supplemented by half a given quantity π(x) + ∑[p(p’)] = ½N. Hence, up to 102 we have 26 prime numbers and 25 their products, which is 51 or half the size of 102/2. Up to 1020 there are 171 and 339 products, which gives half of a given quantity, i.e. 510. Also, there are more of them in a specific relationship, i.e. the number of primes arrived is supplemented by their products to half the increasing size. If the increase is by 918 numbers, then out of 145 prime numbers there will be 314 their products, which in total gives 459 or half of 918/2 with all sums of prime numbers and their products being divisible by (51–459–510–4488–4998 ,) / 51.

From this it follows that the number 3(17) = 51 plays a key role in the even distribution of prime numbers and their products in order of 102, consisting of 19 rows of 17 numbers, which gives 323 plus 18 is equal to 341 numbers in which 171 prime numbers and 170 their products greater than 3. When we add to 341 + 169 products of the number 3, we get 341 + 169 = 510/51 all numbers up to half the size of 1020/2. The number 51 is therefore the basic unit of increase in prime numbers supplemented by their products always half the given quantity divisible by 51/51 = 1, 459/51 = 9, 510/51 = 10, … And so the number of prime numbers increases evenly from number 34 to which we

have 11 so that after 4 (51) = 204/6 = (34) numbers there were 40 more, i.e. 51 to the number 238 and after 238/7 = (34) numbers 91 to the number 476 and after 272/8 = ( 34) numbers by 41 more, i.e. 132 to the number 748 and 39 more by 272/8 = (34) numbers to 1020, where there are 171 prime numbers. So over 30(34) = 1020 numbers there are 171–11 = 160/40 = 4 evenly by 4(40) = 160 prime numbers.

SECRETS OF TWIN NUMBERS

Twin numbers are two prime numbers that can be represented in the form (6n ± 1), 6(1) — 1 = 5, 6(1) + 1 = 7, 6(2) — 1 = 11, 6(2) + 1 = 13. And this is how prime numbers form twin pairs.

We see that the number of the place they are in is the number whose six fold product of 6 (n) ± 1 makes 6 (10) — 1 = 59, 6 (10) + 1 = 61. Up to 102 we have seven pairs of numbers among 35 numbers twins plus one 2 (7) + 1 = 15. Up to 1020 among 341 prime numbers and their products greater than 3, there are 2 (34) + 1 = 69 twin numbers. Subtracting from 341–69 = 272/34 = 8 we get the quantity numbers that do not form pairs of twin numbers. Interestingly, from the number π₂ (9996) = / 3332–408 = 2924/34 = 86 = 98–12, all numbers are divisible by 34. This proves that all prime numbers and their products are arranged in order of 3 (34) = 102 and according to this measure are arranged relative to each other. Hence, we can write the sum of the number of prime numbers and their products greater than 3 divisible by 34 equals the sum of the number of twin numbers and the quantity of numbers that do not form pairs of twin numbers d divisible by 34 {p + ∑[p(p’)>3]/34 = π₂(n)/34 + d/34, 3332/34 = 98 = 408/34 = 12 + (2924/34 = 86) = 12 + 86.

This property can be used to calculate the number of twin numbers for a given quantity. If the number of prime and twin numbers can be written as the product of 34, e.g.: π (1020) = 5(34) + 1, and π₂(1020) = 2(34) + 1, it is easy to calculate the ratio of twin numbers to 69/171 = 2/5 + 1. It is worth noting that this ratio from π₂(9996)/π(9996) = 408/1229 = 1/3 + 5, is constantly increasing by 0.9. Knowing the number of prime numbers written in the form of the quotient of the number 34 and the ratio in which the twin numbers are to them, by dividing this ratio by the ratio we get the ratio of the twin numbers resulting from this ratio. If to π₂(9 999 999 999 999 948) there are 20 608 391 394 576 twin numbers and 279 238 341 033 925 prime numbers then their ratio is 1/13.5 and the next by 0.9 bigger is equal to 1/14.4. Thus, dividing the quotient of the number of prime numbers by the ratio of 14.4 and multiplying by 34 π(99 999 999 999 999 990) = 2 623 557 157 654230/34 = 77 163 445 813 359/14.4 = 5 358 572 625 927(34) = 182 191 469 281 542 we get the number of twin numbers in a ratio of 14.4 to prime numbers in this sequence.

As we can see to the number 408/2 = 204 we have 20 pairs, i.e. 40 twin numbers by a further 40 to 80 numbers will increase after 408 numbers at 1224/2 = 612, so that after 408 numbers at 2040/2 = 1020 there are 122, and after 510 numbers at 3060/2 = 1530 there are 40 more, i.e. 162 twin numbers.

SUMMARY

Thus, the puzzle of the distribution of prime numbers has been solved. From now on, the sequence of prime numbers is not similar to a random sequence of numbers, but rather to an ordered sequence of evenly growing prime numbers and their products to half of a given quantity. So the sum of prime numbers and their products greater than 3 {π(x) + ∑[p(p’)]> 3} equals the difference between half the given quantity N and the products of 3 {½N — ∑[p(3)]}, 171 + 170 = 341 = 510–169. The ratio of prime numbers to their products is therefore determined by completing half the given quantity {π(x) + ∑[p(p’)] = ½N} 1229 + 3769 = 4998, and the sum of prime numbers and their products greater than 3, is always three times smaller than the given value and is π(x) + ∑[p(p’)] = N/3, 1229 + 2103 = 9996/3 = 3332.

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.

REFERENCES LIST

Leonard Euler, Vine Guy, Paul Erdös, Simon Singh.

TABLES OF PRIME AND TWIN NUMBERS UP TO 1 021