PRIME NUMBER LAW. DEPENDENCE OF PRIME NUMBERS ON THEIR ORDINAL NUMBERS AND GOLDBACH - EULER BINARY PROBLEM USING COMPUTER Текст научной статьи по специальности «Математика»

DOI: 10.24143/2072-9502-2020-4-80-100 UDC 511.1:004.056

PRIME NUMBER LAW. DEPENDENCE OF PRIME NUMBERS ON THEIR ORDINAL NUMBERS AND GOLDBACH - EULER BINARY PROBLEM USING COMPUTER

S. I. Chermidov

Kuban State University, Krasnodar, Russian Federation

Abstract. The article considers the methods of defining and finding the distribution of composite numbers CN, prime numbers PN, twins of prime numbers Tw and twins of composite numbers TwCN that do not have divisors 2 and 3 in the set of natural numbers - N based on a set of numbers like 0 = {6-k ± 1, k £ N}, which is a semigroup in relation to multiplication. There has been proposed a method of obtaining primes p > 5 by using their ordinal numbers in the set of primes p > 5 and vice versa, as well as a new algorithm for searching and distributing primes based on a closedness of the elements of the set 0. It has been shown that a composite number n e © can be presented in the form of products (6x ± 1) (6y ± 1), where x, y e N - are positive integer solutions of one of the 4 Diophantine equations: P (x,y, 1) = 6xy ± x ± y -1 = 0 . It has been proved that if

there is a parameter 1 of prime twins, then none of Diophantine equations P (x, y, 1) = 0 has positive integer solutions. There has been found the new distribution law of prime numbers n(x) in the segment [1 ^ N]. Any even number Z >8 is comparable to one of the numbers m = (0, 2, -2), i.e.

Z = m (mod6). According to the above remainders m, even numbers Z > 8 are divided into 3 types, each type having its own way of representing sums of 2 elements of the set 0. For any even number Z >8 in a segment [1 ^ v], where v = (Z-m) / 6, v = (Z-m)/6, there is a parameter of an even

number; it is proved that there is always a pair of numbers (11; 12) e[1 ^ v] that are elements of the

united sets of parameters of prime twins nTw and parameters of transition numbers nUPC , i.e. numbers of the form 61 ± 1 with the same 1, if the form 61 -1 is a prime number, then the form 61 +1 is a composite number, and vice versa.

Key words: prime and composite numbers, parameters of primes, Diophantine equations, binary (strong) Goldbach - Euler, algorithm for solving the binary Goldbach - Euler problem.

For citation: Chermidov S. I. Prime number law. Dependence of prime numbers on their ordinal numbers and Goldbach - Euler binary problem using computer. Vestnik of Astrakhan State Technical University. Series: Management, Computer Science and Informatics. 2020;4:80-100. (In Russ.) DOI: 10.24143/2072-9502-2020-4-80-100.

Introduction

Gauss was the first who observed the regularity of the arrangement of primes showing that the

x

probability of the appearance of primes in the int (1 ^ x) is equal to -. Then Legendre improved

ln x

x

the type A (x) =- and introduced the function n (x). The first essential contribution to the

V ' ln x -1.08366 V '

study of n (x) was made by Euler with his Z (s) -function, [1] with the real variables s,

Z (s) = X — = IT - """js > 1 By means of it, we showed that the series ^Tp 1 diverge, which

means that primes are infinite. In 1859 Riemann proposed to consider the variation of the variable s of the Euler function in the complex plane and related the distribution of primes to the nontrivial zeros of the Z (s) - function B. Riemann formulated the problem now known as the Riemann hypothesis, name-

ly, that all the nontrivial zeros of the Z (5) - function are in the strip 0 < Re 5 < 1 and are symmetric with respect to the Z(5) critical line 2+ix , where x e R, i. e. all nontrivial zeros of Z(5) having

a real part 1 are complex numbers. In the work of V. Minayev an attempt was made to describe the

laws of natural numbers, on the basis of the elements of the set 0 with the introduction of concepts of fundamental negative n-s 0- = 6k -1, positive n-s 0+ = 6k +1, composite n-s & primes. But nothing is said about the methods and definitions, i.e. at what values of the parameters k, the numbers 0- & 0+ are composite or prime. In 2013, Itan Chang proved that there are infinitely many pairs of successive primes with a difference < 7 -107 and Terens Tao reduced this diffeerence to 246. In recent decades interest in the laws of prime distribution [1] from the theoretical point of view has been increasingly shifting to the practical one. Their use in cryptography is a particularly important example [2], that is why any results that clarify certain features of the laws of prime distribution immediately become the subject of study of the field cryptographers. Special interest in cryptography in a system with public keys (in particular, in the RSA encryption system) raises the question, whether this particular (large) number is prime or not. Interest in Goldbach - Euler binary problem in mathematics, as well as its application in related sciences and technologies are very important in cryptography [1, 2]. In the last decade the significant advances have been made in the field of additive number theory. For example, in 2013 Harald Helfgott proved Goldbach ternary problem. From the properties of even number Z > 8,

we can see that are Z = m (mod 6), where m = (0, 2, -2) are considered and their representations as sums of 2 numbers of the form 01 = 6X1 ± 1 and 02 = 6X2 ± 1. An even number Z > 8 in the segment

Л =

and Л2 =

^ v

where v = Z-rm, is an even number parameter. It has been proved

1 *

that there always exist a pair of numbers e A1 andX2 e A2, where (X1, X2) e (nTw u niJPC). The

aim of the paper is to study the laws of prime distribution and composite numbers CN, to prove the solvability of the Diophantine equation z = x + y in the domain of integers Z+, z > 2 for any even number and the summands x, y e P .

Method for selecting prime numbers

When searching for primes in a natural series, a large loss of time is necessary for numbers that are divisible by 2 & 3. If in the set N, there is a factorization with base 6, we can simplify the search for primes and study their properties. We divide the set N into 2 and disjoint subsets H and 0 i.

H fl 0 = 0, N=H U 0. Let H include 1 (one) and natural numbers, which after division by 6 give the remainders 0, 2, 3, 4. Obviously, the set H includes two primes 2 and 3. The set 0 contains numbers which when divided by 6 give the remainders 1 and 5, i.e. numbers of the form, 0 = {6k ± 1, k e N}, as in the expression 6m + 5 = 6 (m +1) -1. Numbers with reminders 0, 2, 3, 4 are

composite numbers, since they are divisible by 2 or 3. So, we have embedding the sets P (> 5) c 0 c N, where P (> 5) is a set of primes p > 5 . It is not difficult to verify directly that the elements of a set 0 = {6-k ± 1, k e N}, are closed under the operation of multiplication, i.e. the set 0 is a semigroup with operation of multiplicationumbers. We have relations:

1. n = ( 6x -1)( 6y -1) = 6 (6xy - x - y) +1. 3. n = ( 6x +1)(6 y -1) = 6 (6xy - x + y)-1.

2. n = (6x +1)(6 y +1) = 6 (6xy + x + y) +1. 4. n = (6 x -1)(6 y +1) = 6 (6xy + x - y)-1. (1)

Products of numbers of the forms 01 = 6x ± 1 and 02 = 6 y ± 1 are expressed by numbers of the form, 6xy ±x±y and it is obvious that the secret of composite and prime numbers lies here. Algorithms and programs of obtaining and distributing primes and composite numbers, twins of primes and twins of composite numbers were performed using Computer Software and Computing.

Definition. For a number n, the values of the numerical functions X (x, y) = 6xy ± x ± y V x, y £N represented in (1) will be called the parameters of the number n e 0 .We note that the problem of a single - valued correspondence between the numbers n and their parameters X (x, y) remains open, i. e several different parametric functions X (x, y) can correspond to the same number n . However, from the point of view of later studies, this fact is not significant. We note that the form 6X -1 can be transformed into another form 6X +1 by multiplication (1). Since the set 0 is a semigroup with respect to the operation of multiplication, all its constituent elements V Xi £M will be:

0 = (6X1 ± 1) • (6X2 ± 1) • (6X3 ± 1) •... • (6X. ± 1). (2)

In the set 0 there are elements 0 = 6X ± 1, for which the number of factors in (2) is equal to 1, i. e. they do not decompose into products of other numbers in0 , these numbers are primitive elements of the set 0 , i. e., primes p> 5 in the set N . For each of the expressions (1) we introduce a parameter X = 6xy ± x ± y, then from the link (1) we have:

, (n ± 1)

X = ^~L, (3)

6

where the "-" sign corresponds to the first two expressions (1) and the "+" sign - to the last two expressions (1). That is, we obtain equations P(x,y,X) = 0 for finding positive integer Z + connecting the numbers (x, y) and parameter X:

1. 6xy - x - y - X = 0, P (x, y,X) = 0. 3. 6xy - x + y - X = 0, P3 (x, y,X) = 0.

2. 6xy + x + y -X = 0, P2 (x,y,X) = 0. 4. 6xy + x -y -X = 0, P4 (x,y,X) = 0. (4)

If at least one of the equations in (4) has one solution, then the number n = 6X ± 1 from (1) is a composite number. If for some X the Diophantine equations from (4) do not have positive integer solutions, then the number n from (1) for this X is a prime number. It is quite obvious from representations (1) and (2) that any constituent element 0 = (6X ± 1) £ 0 will ultimately disintegrate into a product of primitive elements. Whence the main theorem of arithmetic follows, that any positive integer numbers n > 1 can be represented as a product: n = pf1,• pf2,•...,• pdk", where p1, p2,...,pk are primes and

d1, d 2, . • •, dk

are the numbers of corresponding primes involved in the decompositionumbers.

The distribution of composite numbers of the set 0

From (1) one can see that the primes and composite numbers are formed by values of the following functions:

1 X1 = f11 (x y) = 6xy - x - y. 3. X3 = f21 (x, y) = 6xy - x + y. (5)

2. X2 = /2 (x, y) = 6xy + x + y. 4. X4 = f2 2 (x, y) = 6 xy + x - y,

- composite numbers of the set 0 of the form 6X +1 (we denote them CN +) in view of (1) are generated by values of the symmetric and non-one-to-one functions: f11 (x,y) = 6xy - x - y ,

f12 (x,y) = 6xy + x + y , since for unequal values (x1,y1) ^ (x2,y2) the values of the functions can be

equal to f11(x1,y1) = f11(x2,y2) andf12(x1,y1) = f12(x2,y2). With allowance for (1) we notice that the

numbers of the forms 01 = 6X1 +1 and 02 = 6X2 +1 are composite numbers 015 02 e CN +;

- composite numbers of set 0 of the form 6X -1 (we denote them CN-) in view of (1) are generated by the values of the non-symmetric and non-one-to-one functions f21 (x, y) = 6xy - x + y & f22 (x, y) = 6xy + x - y . It also follows from (1) that the numbers of forms

03 = 6X3 -1 and 04 = 6X4 -1 are composite numbers, 03, 04 £ CN-. Since the numbers 01, 02 are

the composite numbers, then the values of the variables (x, y) are solutions of the corresponding equations P (x, y, X) = 0 or P2 (x,y, X) = 0. And in the same way, for composite numbers 03, 64 values of (x, y) are solutions of the equations P3 (x, y, X) = 0 or P4 (x, y, X) = 0 . We also note that for the factorization (taking into account that prime number is the least divisor [3]) of a composite number, the most effective and the best way is to use expression (2) that is the number n e 0 divided by numbers of the form 6X ± 1, where X = 1, 2, 3, ... (see [4]). From (4) it is easy to deduce that the set of parameters of all 4 types of primes and composite numbers of set 0 in the natural series of numbers is infinite. Indeed, for example, if the range of function f11(x, y) value from (5) is defined as the set

®xy = {6xy - x - y}, then for any specific number y =n eN, x the expression {(6n—1) x—n}^<x> Similarly, other functions are also considered as f j (x, y), 1 < i, j < 2 given in (5), we denote the sets as G1 = {f,j(x, y) | x=1, y> 1}, G2 = {f,j(x,y) | x = 2,y>2}, Gv = {f,j(x,y) | x = v,y>v}, veZ+.

Thus, the set of values of all the functions (5) in Table 1: G = G1 u G2 u... u Gv are infinite as unions of infinite sets. We introduce the notation for the set of parameters of composite numbers of the forms:

1. 6X +1: Xe FN+ = f (x, y)u fl2 (x, y).

2. 6X — 1: Xe FN~ = fa (x, y)u f22 (x, y).

Thus, if all composite numbers of the set 0 consist of a union CN = CN— u CN+ or CN = { 6X — 1, where X e FN—} u { 6X +1, where X e FN+ }, then the set of all parameters of the composite numbers 0 will be a union FN = FN— u FN+. It is obvious that the sets FN, FN+, FN— are infinite as unions of infinite sets. To determine and study the parameters of primes and composite numbers of the set0 , it will be necessary to find all the parametric solutions of the equations (4). However, the solution of Diophantine equations is a complicated problem, therefore, to solve equations (4) we can construct a table of values of the function X (x, y) or functions (5). Then to the number X in the table there corresponds a composite number n, otherwise the number n is prime numbers. To study the parameters of primes and composite numbers of the set 0 , we set any values x, y e N for the values of the functions (5) from 1 to 5 e N, where 5 is the specified table size. Let us construct Table 1 in the dimension 5 x 5 . Note that for the same values of the variables x, y e N in each row (x, y) of Table 1 we have an increasing sequence of functions:

f11 (- y) < f22 (- y ) < f21 (- y ) < f12 (- y) .

Forming strings (x, y) in Table 1 and the search for the values of the functions (5) are carried out, according to the principle:

to s to s

j j f11 (x. y ) . f12 (- y ) . f21 (x. y ) . f22 (- y ) .

from x = 1 from y = x

We choose the value s = 10 to demonstrate the algorithms described below, but the constructions described below can be realized for any 5. Let us find the values of the functions:

fn (x y), f12 (x y), f21 (x y), f22 (x y), where 6 • fu (x y) +1; 6 • f12 (x y)+1; 6 • A (x y)+1;

6 • f22 (x, y) +1 is composite number, since the values of the variables (x, y) are known as predetermined solutions of the equations (4). Let xeN, then the values of the functions (5) in the following row (x, y +1) differ from the values of the previous row (x, y) to the following value functions: for function f11 (x, y) by 6x — 1, for function f12 (x, y) by 6x +1 & for function f21 (x, y) by 6x +1,

specified interval the max parameter: Xmax =

+1 = 26, from the Table 1 we write the parameters of

for function f2 (x, y) by6x-1. Let m1 = f11 (x, y +1) -f11 (x, y) = 6x -1; m2 = f22 (x, y +1) --f 22(x, y) = 6x-1 m3 = f21 (x, y +1)-f21 (x, y) = 6x +1; m4 = f^ (x, y +1)-f^ (x, y) = 6x +1. Since V x e Z + the differences (ml, m2, m3, m4) >0, then the functions (5) are increasing and infinite. Example 1. Find the composite numbers of the set © in the int 1 ^ N = 155. We calculate in the

" N

composite numbers < Xmax . As a result, we have: nCN = { 4, 6, +8, +9, "11, 13, +14, +15, 16, +19, + 20

"21, +22, + " 24, 26}. Symbols in numbers are obtained, according to sets FN + and FN - . On the basis of the definition of the parameters of the composite numbers from the set ® we find their values: CN + : 6 ■ 4 + 1 = 25; 6 ■ 8 + 1 = 49; 6 ■ 9 + 1 = 55; 6 ■ 14 + 1 = 85; 6 ■ 15 + 1 = 91; 6 ■ 19 + 1 = 115; 6 ■ 20 + 1 = 121; 6 ■ 22 + 1 = 133; 6 ■ 24 + 1 = 145.

CN - : 6 ■ 6 - 1 = 35; 6 ■ 11 - 1 = 65; 6 ■ 13 - 1 = 77; 6 ■ 16 - 1 = 95; 6 ■ 20 - 1 = 119; 6 ■ 21 - 1 = 125; 6 ■ 24 - 1 = 143; 6 ■ 26 - 1 = 155. Hence, the complete sequence of composite numbers of ®: CN = CN+ u CN- = {25, 35, 49, 55, 65, 77, 85, 91, 95, 115, 119, 121, 125, 133, 143, 145, 155}.

Table 1

Formation of the parameters of composite numbers in the set 0

G x y Mx, y) /12 (x, У) /21 (x, У) /22 (x, У)

Gi 1 1 4 8 6

2 9 15 13 11

3 14 22 20 16

4 19 29 27 21

5 24 36 34 26

6 29 43 41 31

7 34 50 48 36

8 39 57 55 41

9 44 64 62 46

10 49 71 69 51

G 2 2 20 28 24 24

3 31 41 37 35

4 42 54 50 46

5 53 67 63 57

6 64 80 76 68

7 75 93 89 79

8 86 106 102 90

9 97 119 115 101

10 108 132 128 112

G з 3 3 48 60 54 54

4 65 79 73 71

5 82 98 92 88

6 99 117 111 105

7 116 136 130 122

8 133 155 149 139

9 150 174 168 156

10 167 193 187 173

—> —>

G x y Ун Cx y) у12 (x, У) y2i (x> У) y22 (x, У)

G4 4 4 88 104 96 96

5 111 129 121 119

6 134 154 146 142

7 157 179 171 165

8 180 204 196 188

9 203 229 221 211

10 226 254 246 234

G5 5 5 140 160 150 150

6 169 191 181 179

7 198 222 212 208

8 227 253 243 237

9 256 284 274 266

10 285 315 305 295

G6 6 6 204 228 216 216

7 239 265 253 251

8 274 302 290 286

9 309 339 327 321

10 344 376 364 356

G 7 7 7 280 308 294 294

8 321 351 337 335

9 362 394 380 376

10 403 437 423 417

G 8 8 8 368 400 384 384

9 415 449 433 431

10 462 498 482 478

G 9 9 9 468 504 486 486

10 521 559 541 539

Gio 10 10 580 620 600 600

Distribution of the parameters of primes and composite numbers 0 in N

The distribution of the parameters of primes and composite numbers of the set 0 is the analog of the distribution of primes p > 5 and composite numbers that do not have divisors 2 and 3 in N. It will be necessary to find all primes 0 of the form 6X ± 1. We describe the algorithms for constructing these numbers. Let in the int [1 ^n ) the entries in the file are given according to the structure

Rn = (id. [F1 ]. [F2]), where the field id are the serial numbers of the records and the fields F1 and F2 take the values "+" or "—". Before the beginning of algorithms 1 and 2 below in the interval [1^[n / 6]) the sign "+" are entered line by line in the fields F1 and F2 [5].

1. Algorithm distributing primes of a set 0 of the form 6k — 1. Let (x, y) = 1, 2, 3, ... vary according to the principle of Table 1. Then, by values of the parameters X = 6xy — x — y and X = 6xy + x + y of a composite number of the form 6X +1 e 0 from a file Rn by direct access to the records id=X in the field F2 the sign "+" changes into the sign "-" and the remaining records at the end of the algorithm in the adjacent field Fx = "+" indicate the presence of primes of type:

6X — 1 ePN. We introduce the notation e11 = Jf11 (-,y) s12 = Jf12 (x, y), K—1 = SnJ, then

x, y x, y

PN — = {6X — 1, X e K—1}.

2. Algorithm distributing primes of a set 0 of the form 6k +1. Let (x, y) = 1, 2, 3, ... vary according to the principle of Table 1. Then by values of the parameters X = 6xy — x + y and X = 6xy + x — y of composite numbers of the form 6X — 1e 0 from a file Rn by direct access to the records id = X in the field F\ the sign "+" changes to the sign "-" and the remaining records at the end of the algorithm in adjacent field F2 = "+" indicate the presence of primes of 6X +1 e PN +.

We introduce the notation = J/21 (x, y); E22 = Jf22 (x, y); K+1 = Je22, then

x ,y x ,y

PN + = { 6X + 1, X e K+1}.

So, the primes consist of the union of two sets: P = PN— u PN +. Combining algorithms 1 and 2 into one algorithm [5] we obtain the algorithm for distributing parameters of composite numbers and primes 0 in N. Parameters id = X with the assigned fields Fx and F2 with the value "+", according algorithms to 1 and 2, are parameters of twins of primes and with the value "—" are the parameters of twins of composite numbers.

Theorem 1. At least one of the equations (4) has solutions when the form 6k + 1 or 6k — 1 is composite number.

Nece55ity. Let a number of the form n = 6X +1 is a composite number, since 0 is a semigroup, where exist 2 numbers 01 = 6x0 ± 1 and 02 = 6y0 ± 1, 01, 02 e 0 with (x0, y0) e N, as n = 6X +1 = (6x0 ± 1)(6y0 ± 1) = 6(x0y0 ± x0 ± y0 )± 1. According to (1), there are only 2 cases for numbers, i. e. 01 = 6x0 — 1; 02 = 6y0 — 1, which gives X = 6x0y0 — x0 — y0 and 01 = 6x0 +1; 02 = 6y0 +1 which gives X = 6x0y0 + x0 + y0. In any case, one of the 2 first equations (4) has positive solu-tionumbers. The same is true for numbers of the form n = 6X — 1, in which case one of the two last equations (4) has positive solutionumbers. This completes the proof.

Adequacy. Suppose, one of the equations (4), for example, equation 6xy — x — y — X = 0, has a solutionumbers. This means that there exists a triple of natural numbers (X, x, y) e N, that is fair X = 6xy — x — y . Then from P1 (x,y,X) = 0 we have n = 6(6xy — x — y) +1 = (6x — 1)(6y — 1), i. e. the

number corresponding to parameter X is a composite number, since both of (6x —1)(6y — 1) are natural

numbers different from 1, because (x, y) ^ 0. The same is true for any of the equations (4).

Solutions of Diophantine equations P(x, y, X) = 0

Example 2.

2.1. We know some solution of one of the equations (4).

1. Let

( x = 11

У = 2

are solution of P1 ( x, y,X ) = 0 ^ X = 6 xy - x - y = 119. Taking (3) into account,

N -1

we have

6

composite n.

= 119 or n = 6 • 119 + 1 = 715. From (1) ^ 715 = (6-11 -1)(6• 2-1) ^ n = 715 is a

2. Let

( x = 2

У = 5

n +1

, is a solution of P3 (X, 2, 5) = 0, which X = 6 • 2 • 5 - 2 + 5 = 63. Taking (3) into ac-

count, we have-= 63 from (1) ^ 377 = (6 • 2 +1) ( 6 • 5 -1), i. e n = 6 • 63 -1 is a composite numbers.

6

2.2. We know that for some X one of the corresponding numbers is prime.

1. Let = 16 — n = 6 -16 +1 = 97 e P, hence, P1 (x, y,X) = 0 and P2 (x, y, X) = 0 have no solutions.

2. Let X = 15 ^ n = 6 -15 -1 = 89 e P , hence, P3 (x, y, X) = 0 & P4 (x, y, X) = 0 have no solutions.

2.3. We know that for some parameter X one of the corresponding numbs is composite. Taking X = 75 89, we know that n = 6 • 7589 +1 = 5 • 9107 is a composite number. Hence, one of the Diophant

equations P1 ( x, y,X ) or P2 ( x, y,X ) has solutions.

7589 + x

1. 6xy - x - y = 7589. If we express y in terms of x, than y =-, which must be an inte-

6 x -1

ger. So, we search for which values of x in 1 ^ (X -1) the above value of y is an integer. Easily found

(X, x1, y1 ) = (7589, 1, 1518),(X, x2, y2) = (7589, 6, 217), (X, x3, y3) = (7589, 217, 6), (X, x4, y4) = = (7589, 1518, 1).

7589 - x

2. 6xy + x + y = 7589. If we express y in terms of x , we have y =-, which must be an

6 x +1

integer. So, we search for which values of x in 1^(X - 1) the above value of y is an integer. We found 2 solutions (X, x1, y1 ) = (7589, 1, 1084), (X, x2, y2) = (7589,1084,1) . Let X = 63, we

know that n = 6 • 63 -1 = 377 = 13 • 29 is a composite numbers Hence, one of P3 (x, y, X) = 0 or

P4 (x, y, X) = 0 has solutions.

3. 6xy - x+y = 63. We rewrite last relation as y =

63 + x 6 x +1

, which must be an integer. So, we

search for which values of x in1 ^ (X -1) the above of y is an integer. One solution of the triplet ( 2, 5, 63 ) is found.

4. 6 xy + x - y = 63. We express y in terms of x, and have y =

63 - x 6 x -1

which must be an integer.

So, we search for which values of x in 1 ^ (X — 1) the above of y is an integer. One solution (2, 5, 63)

is found. According to the distribution of the parameters of prime and composite numbers without divisors 2 and 3, we compose a Table 2.

Table 2

Distribution of the parameters of primes and CN of the set 0 in the N *

Id = 1 Fi F2 m <г> ш О о m О о m О о

1 + + 41 81 + 121 + 161 +

2 + + 42 + - 82 + - 122 - + 162 + -

3 + + 43 + - 83 - + 123 - + 163 + -

4 + - 44 + - 84 + - 124 + - 164 + -

5 + + 45 + + 85 + - 125 - + 165 - +

6 - + 46 - + 86 - - 126 - + 166 - +

7 + + 47 + 87 + + 127 + - 167 - -

8 + - 48 - - 88 - - 128 - + 168 - +

End of table 2

Distribution of the parameters of primes and CN of the set 0 in the M*

Id = 1 F1 F2 m О О m О 0 m О О m О О

9 + - 49 + - 89 - - 129 + - 169 + -

10 + + 50 - 90 - + 130 - - 170 + +

11 - + 51 - + 91 - + 131 - + 171 - -

12 + + 52 + + 92 - - 132 - - 172 + +

13 - + 53 + - 93 +100 - 133 + - 173 - +

14 + - 54 - - 94 + - 134 - - 174 - -

15 + - 55 - + + + 135 + + 175 + +

16 - + 56 - + 96 - + - - 176 - -

17 + + 57 - - 97 - - 137 + + 177 + +

18 + + 58 + + 98 + - 138 + + 178 - +

19 + - 59 + - 99 + - 139 - - 179 - -

20 - - 60 + - 100 + + 140 + - 180 - -

21 - + 61 - + 101 - + 141 - - 181 - +

22 + - 62 - + 102 - + 142 - + 182 + +

23 + + 63 - + 103 + + 143 + + 183 + -

24 - - 64 + - 104 - - 144 + - 184 + -

25 + + 65 + - 105 - + 145 - - 185 + -

26 - + 66 - + 106 - - 146 - + 186 - +

27 - + 67 + - 107 + + 147 +150 + 187 - +

28 + - 68 - + 108 + - 148 + - 188 - +

29 + - 69 - - 109 + - 149 - - 189 - -

30 + + 70 + + 110 + + 150 - - 190 - -

31 - - 71 - - 111 - - 151 - + 191 - -

32 + + 72 + + 112 - + 152 + - 192 + +

33 + + 73 - + 113 + - 153 - + 193 - -

34 - - 74 + - 114 + - 154 - - 194 + -

35 - + 75 + - 115 - + 155 + - 195 - +

36 - - 76 - + 116 - - 156 - + 196 - -

37 - + 77 + + 117 + - 157 + - 197 + -

38 + + 78 + - 118 - + 158 + - 198 + -

39 + - 79 - - 119 - - 159 + - 199 + -

40 +50 + 80 + - 120 + - 160 - - 200 - +

*The values of elements of the set ©, respectively, over the fields F1: 6X - 1 and F2: 6X + 1. By analogy to the filds Fi, F2, the "+" sign corresponds to primes and "-" to composite numbers. It is obvious that these subsets of numbers in the natural series are parameters of the corresponding subsets of the set ©.

Ascribing to serial numbers of records signs "+" or "-", in the fields F1 and F2 form a table of signs (Table 2) and using algorithms 1 and 2, the natural numbers are divided into subsets of numbers according to the combinations of the signs "+" & "-". Consider the relationship between the equations (4) and the twins of primes. From the definition of twins of primes it is known that these are the p1,p2 e P with |p2 - p11 = 2. Note that for the same X the difference of 9+ - 9- = 2 . So, if for this value of the X the numbers 9+ e PN+ & 9- e PN~ are primes, then the pair of these numbers is generating a twin of primes. Hence, for this value of X the numbers 6X ± 1 will be twin primes.

Theorem 2. In order that X e N is a parameter of twins of primes, it is necessary and sufficient that for this particular X none of the Diofantine equations (4) has solutions.

Necessity. Suppose for one and the same X e N all equations (4) have no solutions. Then, by Theorem 1 from equations P1 (x,y,X) and P2 (x, y, X) it follows that 9+ = 6X +1 is a prime number

and from equations P3 (x, y,X) and P4 (x, y,X) it follows that 9- = 6X -1 is a prime number. Since

9+ - 9- = 2, then, by definition, 9+, 9- of twins of primes and this X is a parameter of twins of primes.

Adequacy. Let X be a parameter of twins of primes, i. e. p2 = 6X +1, p1 = 6X -1 are primes by virtue of the definition of the twins of primes p2 - p1 = 2, p1,p2 e P . Hence, from Corollary 1, for

a prime number p2 = 6X +1 it follows that the equations P1 (x, y,X) = 0 & P2 (x, y,X) = 0 have no solutions and from Corollary 2 the equations P3 (x, y,X) = 0 & P4 (x, y,X) = 0 it follows that they have no solutions for prime p1 = 6X -1. Thus, none of the Diophantine equations (4) has solutions.

Definitions of subsets of the set 0

By virtue of the distribution of the parameters of primes and composite numbers of the set 0 in N the serial numbers id = X (see Table 2), which are parameters of the following significant subsets (Tw - twins of primes), (TwCN - twins of composite numbers do not have divisors 2 and 3), (UC - unique composite numbers), (PN - unique primes) and (UPC = UCuPN), of union unique primes and unique composite numbers).

Twin composite numbers (2 composites numbers on the forms 6X ± 1 with the same parameter X, having a difference = 2), i. e. TwCN = {6X ± 1, X e nTwCN}, where

nTwCN = FN "n FN +. (7)

The set nTwCN is parameters X of twins of composite numbers, which lie on non - empty

intersections of the values of two functions (5), one of which belongs to FN + and the other to FN- . The values of the fields F1 = "-" and F2 = "-" in Table 2 correspond to the parameters X, the set of twins of composite numbers.

2. Unique composite numbers (the difference between 2 composite numbers without divisors 2 and 3 is > 2

UC = {6X -1: X e FN- \ n^}u{6X +1: X e FN+ \ n^},

where FN- from (6) is the set of parameters X that are representable in the form {6xy - x + y} or { 6 xy + x - y} for some values (x, y )e N and FN +, is the set of parameters X that are representable in the form { 6xy - x - y } or { 6xy + x + y }. The parameters of unique composite numbers nUC = FN- I nTwCN u FN I nTwCN = (FN u FNT) \ (FN+ D FNT) = FN A FN - we have the symmetric difference of sets

Obviously, the parameters of all composite numbers will be CN = UC u TwCN and

^ nCN =nUC U nTwCN .

3. Unique primes

PN = (2 & 3) u {6X +1, X e FN- \ HTwCN u 6X -1, X e FN + \ UTwCN}.

Parameters nPN = FN-1 UTwCN u FN I UTwCN = (FN u FN-) \ (FN u FN-) = FN A FN- i. e. the symmetric difference of sets. Since the set of parameters FN- \ nTwCN are not solutions of equations P1 (x, y,X) = 0, P2 (x, y,X) = 0, then by Corollary 1 number of the form 6X +1 is a prime number and set of parameters FN + \ nTwCN are not solutions of equations P3(x,y,X) = 0, P4(x,y,X) = 0, then by Corollary 2 number of the form 6X -1 is prime.

4. Twins of primes (part 2): Tw = {6X± 1, X enTw}, where

6X -1 e PN-; 6X + 1e PN+.

Parameters of twins of primes are at the intersections of the complements of the sets PN- & PN + i. e. nTw c N \FN. In Table 2 the parameters of twins of primes X correspond to the values of the fields (F = "+"; F2 = "+"). Then, the parameters of all primes nP = nPN u nTw.

5. Transition numbers (Unique Prime Composite)

UPC = {6X ± 1, X e FN + A FN-}

That is, numbers go from composite numbers to primes or vice versa by changing the forms 6X +1 & 6X -1 with the same parameter X.The parameters n№C = nuc unp# corresponded in Table 2 to the values of the fields (F = "-",F2 = "+") or (F = "+", F2 = "-",).

Lemma 1. The set of natural series of numbers M and the set 0 are equivalent sets.

Proof. For two sets M~0 to be equivalent, it is necessary and sufficient that there is a one-to-one correspondence between their elements. Note that the distribution of the parameters of the elements of the set 0 by the design of Table 2 sets to each ordinal number id = X an element from the subsets TwCN, Tw, and UPC . This means that the elements of set natural series can be fully described by the function (5), i. e. set 0 covers set M. Hence, we have the right to assert that Vne N is a parameter of one of the listed subsets of the set 0. So, it holds: N = nTw u nTwCN u nupc.

Lemma 2. ParametersX^ X2 and X3, X4, respectively, of the unique composite numbers of the types 6X + 1 and 6X - 1, then the parameters of unique primes will be numbers of the form 6X -1 and 6X + 1, respectively.

Proof. Excluding the values of parameters nTwCN from Table 1, we have the remaining values of parameters X1, X2 of the composite numbers of the form 6X +1 and values of parameters X3,X4 for composite numbers of the form 6X -1. Since the values of parameters X3, X4 are in most cases different from the parameters X1, X2, (because functions (5) are all different from each other), then, by Corollaries 1-2 they are the parameters of unique primes for the form 6X +1 and likewise the values of the parameters X1, X2 are not representable with the parameterss X3, X4 i. e. are the parameters of unique primes for the form 6X -1. So, the laws of the distribution of unique primes by 6X +1, 6X -1 are the same as for composite numbers, respectively by type6X -1 and 6X +1. And, since the X1, X2 and X3, X4 are the infinity, then the infinity of unique primes follows from this.

Calculation of primes p > 5 by their ordinal numbers & vice versa in the set primes P

From Table 2 of distribution of parameters of primes & composite numbers of 0 in M it is not difficult to see that between the ordinal numbers of primes in the set of primes P (> 5) and parameters X of primes p> 5 shown in Table 2, there are dependencies. Let n be the ordinal number of the prime number p> 5. in the set of primes P (> 5), then the corresponding prime number is found from the

formula: P = 6 • id + (- 1 )¥(n), where id is the line number, y (n) is the index of field , on

which the n-th symbol "+" in Table 2. Counting of the signs "+" is carried out according to the following principle of viewing the lines. At the beginning of the algorithm, the values of the variables S1 = S2 = y(n) = 0.

a. Algorithm for obtaining a prime by its ordinal number in the prime number P.

End of fail Г 0 F = "_" '

n ^ P. > Si = Si + ^ ' 1 ^ S + S2 = n Yes ... л. A=, no ^ id=1 [1, Fi = + J id

1 S 2 Yes, .(n )=1i

I 0, F2 = "-" I ?

^ S2 = S2 + |1, F2 = " +" I d S1 + S2 = n Yes, ¥(n )= 2,

p. Algorithm for obtaining serial numbers by their primes.

' p± 1

P ^ n, m =

6

m I 0, F1 = "-" 1 ?

+ 1, > S, = S, + { J 6i - 1 = ? p Yes: S ,S no ^

£ 1 1 [1, F ="+" J, p =

Yes: n = Sj + S2 ^

Г о, F2 = "-" 1 ?

^ S 2 = S 2 +[1 F = " + " J 6 • i + 1 = PYes, n = S1 + S2 , .

2

Example 3. Let the ordinal number n = 10 of a prime number be p > 5, then from Table 2 in the fields F1 and F2, summing up the number of characters "+" starting from the first line id = 1 and on,

each time we chek the sum S1 + S2 for equality with n. If equal, the algorithm ends, otherwise continues. If equality occurs when calculating Si, then y(n) = 1, otherwise for S2 then y(n) = 2.

Algorithmic way to get a prime number by its ordinal number in the P (> 5)

Table 3

id St = S1 + (0: F2 = "-"), (1: F2 = "+") S1 + S2 = n ? V(n) S2 = S2 + (0: F2 = "-"), (1: F2 = "+") S1 + S2 = n ? V(n)

1 0 + 1=1 1 + 0 =1 0 0 + 1 = 1 1 + 1 = 2 0

2 1 + 1=2 2 + 1 = 3 0 1 + 1 = 2 2 + 2 = 4 0

3 2 + 1=3 3 + 2 = 5 0 2 + 1 = 3 3 + 3 = 6 0

4 3 + 1=4 4 + 3 = 7 0 3 + 0 = 3 4 + 3 = 7 0

5 4 + 1=5 5 + 3 = 8 0 3 + 1 = 4 5 + 4 = 9 0

6 5 + 0=5 5 + 4 = 9 0 4 + 1 = 5 5 + 5 = 10 2

Hence, this prime number with the ordinal number n = 10 in the table of primes P > 5 will be the number P10 = 6 • id + + (-1) = 37. Now, consider the opposite case. Let given a prime

number P = 37, find its serial number n.

Table 4

Algorithmic way of getting a serial number by the value of its prime in the P(>5 )

i S1 = S1 + (0: F1 = "-") (1: F1 = "+") n = S1 + S2 6 i-1 = p ? S2 = S2+(0: F2 = "-") (1:F2 = "+") n =S2 + S1 6i + 1 = p ?

1 0 + 1 = 1 1 + 0 = 1 5 ф p 0 + 1 = 1 1 + 1 = 2 7 ф p

2 1 + 1 = 2 2 + 1 = 3 11 ф p 1 + 1 = 2 2 + 2 = 4 13 ф p

3 2 + 1 = 3 3 + 2 = 5 17 ф p 2 + 1 = 3 3 + 3 = 6 19 ф p

4 3 + 1 = 4 4 + 3 = 7 23 ф p 3 + 0 = 3 3 + 4 = 7 25 ф p

5 4 + 1 = 5 5 + 3 = 8 29 ф p 3 + 1 = 4 4 + 5 = 9 31 ф p

6 5 + 0 = 5 5 + 4 = 9 35 ф p 4 + 1 = 5 5 + 5 = 10 P = 37

Primes distribution and formula for finding n(x)

For the calculation of primes in (1 x) there are functions of 2 kinds: arithmetic formulas

x

n (x)«- and analytical formulas derived from the works of Riemann and Mangoldt. For example,

ln x

dt

the known formula Li(*) *P - ln2 + J

t (t2 - 1 ) ln t

where * > 1, p - non - trivial zeros Z (5)

functions. Since Riemann was a great specialist in the development of complex numbers and was oriented in an imaginary plane better than in real one, the learned people were at an impasse. However, the most accurate and simple formulas are rear. Since we have (Table 2) the distribution of the parameters primes & composite numbers ® in N, then the law of distribution of primes n(x) in interval

(1 - *) will look like for m =

n ( * ) = 2 + £( Si + S2).

(8)

Example 4. Determine the number of primes, n(x) in the numerical interval, (1 ^ N) .

" N'

1. 1 - N = 100 ^ 1 - m =

2. 1 - N = 558 ^ 1 - m =

6

N 6

= 16, ^S1 = 12, £S2 = 11, then n ( * ) = 2 + 12 + 11 = 25.

X=1 <m

X =1 < m

= 93, ^ = 51, £S2 = 47, then n ( * ) = 2 + 51 + 47 = 100, it is

known that in 1 ^ N = 558 the n (x ) = 100. However in the same segment, according to the known

p

*

m

X =1

X=1

X=1

function of the law of distribution of primes n (x) = •

558

90 it is easy to see that the

ln x 6.32435896

result of formula (8) obtained i s closer to the truth. And no matter how large the interval of the value x remains unchanged with deviation of ±1.

An algorithm for finding primes p > 5 in the interval [1 ^ N)

Since the set of primes P(> 5) c 0 c M, it is obvious that the search for primes will go faster in the set 0 than in the set N. The most natural way to remove composite numbers of set 0 in interval (1 ^ N) is by using the properties of closed elements with respect to the multiplication operationumbers Multiplying numbers of the forms 6i -1 & 6i +1 by each element of the set 0,

'n_' _ 6 _

by line in the file 0 = PrmNub1(id. [N]), but in the place the numbers divisible by 2 or 3 are filled with an empty symbol "-". Then, based on the RasPrm algorithm (Fig.) the elements of the file 0' are deleted, those that are elementwise products of numbers 0. = 6i -1 are multiplied by the next (at the begining j = i) number 0 j=j+1 e 0 and deleting for numbers with record numbers id = 0i • 0j, where i = 1, 2, 3, ... (see Example 5 below).

where

i e

1 -

is easy and simple to achieve the goal. First, the natural numbers are entered line

The window of the program that implements the RasPrm algorithm

Each successive new element 0;. erects squared to avoid repetitions of multiplication operation, and then is multiplied by the 0j=j+1. The removal process continues until 0f < n . If the product of the numbers 0i • 0j is greater than n, then performs the next element 0;.=+1 and the above procedure is

repeated. Similarly, it is true for numbers of the form 0;. = 6i +1. The method described for the elimination of composite numbers from the set 0 for numbers of the form 6k ± 1 is easy to use and works more efficiently and faster than such well-known algorithms as the Eratosthenes, the Sundaram and the Atkin sieves. Since in all these algorithms the domain of functioning is the set N, the method of deleting composite numbers from the set 0 by numbers of the form 6k ± 1 allows to obtain the same results as the above mentioned algorithms, but with a much smaller number of multiplication operations.

Example 5. Find all prime numbers of the set 0 in the interval 1 ^ N = 133.

With using numbers of the form 0;. = 6i ± 1, where i e 1 ^ [133/6] = 22, we form the elements of the file 0 = { 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133,...}.

1. Deleting composite numbers from the file 0 having the form 0i = 6i -1.

Let i = 1, then 01 = 5. As a new 01, we find its square: 02 = 25 and if id = 25 < 133 then from the 0 by the record number id = 25 by direct access, this entry is raised and the field value is deleted to [N] = 25 . Then 01 is multiply by the next number of the file 0 and also deleting for numbers with record numbers id = 5 • 7 = 35 , id = 5 • 11 = 55, id = 5 • 13 = 65, id = 5 • 17 = 85, id = 5 • 19 = 95, id = 5 • 23 = 115, id = 5 • 25 = 125. Since id = 5• 29 > 133 we take the next step i. e. 02 = 11 new

element 112 < 133 the record is deleted id = 121. When id = 11 -13 > 133, the next step 03 = 17 new

element 172 > 133 and terminates, the process for numbers 6i -1.

2. Deleting composite numbers from the file 0 having the form 0; = 6i +1. Let i = 1, then 01 = 7 as a new element 9b we find its square id = 72 = 49 < 133 and it is removed in the same way as in the previous examples. That is, by directly accessing the record the following record numbers are also selected id = 7 -11 = 77, id = 7 -13 = 91, id = 7 -17 = 119 and the numbers are deleted. Since id = 7 -19 > 133 next step 92 = 13 ^ 92 > 133.

Representation of even numbers Z > 8 in the form of sums of 2 elements from the 0

Pair (X1, X2) for numbers of forms 91 = 6X1 ± 1 & 92 = 6X2 ± 1 are called the corresponding

pair ofparameters of an even number Z, if X, еЛ, =

1; v

_ 2 _

X2 = (v -X,) е Л2 =

+1; v

Z = 01+02-

Lemma 3. The even numbers Z > 8 are comparable with Z = 0, 2, -2 (mod 6).

Evidence. From the type of even numbers, Z = 2n —2n/6 = n/3, i. e. the number n has the following form 3v + a, where v e N and it is obvious that the residues a = ( 0, 1, 2). Suppose:

1. a = 0 — Z = 2n = 2 (3v + 0) = 6 v + 0 — m = 0 that is, is divisible by 6 with the remainder 0.

2. a = 1 — Z = 2n = 2 (3 v +1) = 6 v + 2—m = 2 that is, is divided by 6 with the remainder 2.

3. a = 2 — Z = 2n = 2 (3 v + 2 ) = 6 v + 4 or (m = v +1) — 6m - 2 i. e divisible 6 with remain -2.

Lemma 4. Any even number Z >8 is representable by the sums of 2 elements of set 0.

Proof. Types of decomposition of even number Z > 8 comparable with the Z = (0, 2, -2) (mod 6):

1. Let us have even numbers of the form Z = 6v + 0 and let v = + X2, then, if we add and subtract 1 (one), we have an even number Z = 6X + 6X2 = (6X + 1) + (6X2 — 1) = 0j+ + 02—.

2. Let us have even numbers of the form Z = 6v + 2 and let v = + X2 — therefore, the even number will be Z = 6X1 + 6X2 + 2 = (6X1 +1) + (6X2 +1) = 0+ + 0+.

3. Let even numbers of the form be Z = 6v - 2 and v = + X2 we have Z = 6X1 + 6X2 - 2 = (6X1 -1) + (6X2 -1) = 0- + 0- where the elements 0+/- and 02+/- lie in the set 0.

Since the parameters of twins composite numbers nTwCN (7) lie on the intersections of the function (5), th en the number of their parameters on the segment [1 ^ N ] by the intersection of the sets will be less or equal than to their union the number of images of the functions themselves. Hence, on the segment [1 ^ N] for the parameters of the subsets: Tw, TwCN,UPC the following inequality is satisfied:

nTwCN < nUPC ^ nTw (9)

consider by analogy with the remainders m = (0, 2, -2) decomposition of even numbers Z > 8 in the form of sums 01 + 02. To do this, for each type of even numbers of the form 6v + m we add and subtract the numbers 6X ± 1.

(Z - 0)

a) Veven number of the form Z = 6v + 0, where v = ---, we add and remove the numbers of

6

type(6X-1), have Z = 6v + [(6X- 1)-(6X- 1)] = (6X-1) + [6 (v-X) +1] and in the same way for numbers of form 6X +1, then Z = 6v + [(6X +1) - (6X +1)] = (6X +1) + [6(v - X) -1] . We denote the parameters of the terms = v - Hence,

Z = (6X1 -1) + (6X2 +1) or Z = (6X1 +1) + (6X2 -1) (10)

is also fulfilled the identity

(6X -1) + [6 (X + 2 ) +1] = ( 6X +1) + [6 (X + 2)-1 . (11)

Let the even number Z = 96, then 6v = 96, v = [(96 - 0) \6] = 16, whenX1 = 1&X2 = 16 -1 = 15

from link (10) ^ 0- = 5 and 0+ = 91 or 0+ = 7 and 0- = 89 .

P) V even number of the form Z = 6v + 2, where v = (Z - 2) / 6 have 6v + 2 + [(6X +1) - (6X +1)] = (6X +1) + (6 (v -X) + 1).

We denote the parameters of the terms respectively, X1 = X &X2 = v - X1, which means that even number Z > 8 of the form

Z = 6v + 2 = (6X + 1) + [6 (v - X ) + 1] = (6X1 + 1) + (6X2 + 1) (12)

also fulfills the identity

(6X + 1) + (6 (X + 3) + 1) = (6 (X + 1) + 1) + 6 (X + 2) + 1). (13)

Let Z = 98, then 6v + 2 = 98, v = [(98 - 2) \6] = 16, with X1 = 2 and X2 = 14, we have from link (12) ^ 0+ = 13 and 0+ = 85, Z = 0+ + 0+.

y) Veven number of the form Z = 6v - 2, where the parameter v = (Z + 2)\6, we have 6v - 2 + [(6X -1)-(6X - 1)] = (6X -1) + [6(v - X)-1]. We denote the parameters of the terms: X1 = X & X2 = v - X1 hence Z > 8 species

Z = 6v - 2 = (6X -1) + [6(v - X) -1] = (6X1 -1) + (6X2 -1) (14)

also fulfills the identity

(6X -1) + [6 (X + 3) -1] = [6 (X + 1) -1] + [6 (X + 2) -1]. (15)

Suppose that Z = 100 ^ 6v - 2 = 100, v = [(100 + 2) \6] = 17 for X1 = 3, we have X2 = 14 from

link (14) = 17 and Z = 0- + 0-, 0- = 83.

From the properties a, P, y it follows that V even number Z < 8 can be represented by sums (6X1 ± 1) + (6X2 ± 1), where X1 e A1 , X2 = (v - X1) e A2 and also if m is the remainder of even numbers

divided by 6, then the parameter of the even number Z < 8 is found by the formula

v = (Z - m) \ 6 . (16)

The binary (strong) Goldbach - Euler problem

The decomposition of even numbers Z < 8 into the sum of 2 primes is verified directly, so consider the solution of binary Goldbach problem for Z > 8. We first introduce the definition and prove a number of lemmas.For example to find the corresponding parameters of the even numbers Z = 30 .

Since, 30 = 0 (mod 6), then the parameter of even numbers v = 3°6 0 = 5. Hence,

X1 e A1 =[1, 2], X2 e A2 =[3, 4, 5].

1. Let X1 = 1, X2 = v - X1 = 5 -1 = 4, ^ 01 = 6 -1 -1 = 5, 02 = 6 • 4 + 1 = 25 or 01 = 61 +1 = 7, 02 = 6 • 4 -1 = 23. Since, 5 + 25 = 7 + 23 = 30 corresponding parameters (X1,X2 ) = (1; 4).

2. Let X1 = 2, then X2 = v - X1 = 5 - 2 = 3, 01 = 6 • 2 -1 = 11, 02 = 6 • 3 +1 = 19 or 01 = 62 +1 = 13 and 02 = 6 • 3 -1 = 17. Since 11 +19 = 13 +17 = 30, then the corresponding pair paramets (X1, X2) = (2; 3). Thus, even Z = 30 has 2 corresponding parameters (X^X2) = {(1; 4),(2; 3)}.

Lemma 5. V even numbers Z > 8 in the segments Л19 Л2 there always exists parameters X1 e Л1 , X2 = (v-X1 )e Л2, which belong to the union of the sets nTw u Пирс and are

corresponding pair parameters of even numbers.

Evidence. The parameters of twins of composite numbers are increasing in segment [1 ^ v],

since the intersection of the functions (5) is increasing, then in the segments Л1 and Л2 naturally, they will be have different elements nTwCN . Since, the natural numbers in the segment Л1 begin with the parameters of twins of primes nTw ={1, 2, 3, 5, 7, 10, 12, 17, 18, 23, 25, ...} (see Table 6 below or Table 5 in [6]) then all elements of the segment Л1, obviously, can not be parameters of twins of composite numbers. Suppose that all elements of the segment Л 2 are parameters of twins of composite numbers, then by (9) we have a contradiction, since the total number of elements of nTwCN on the segm. [1 ^ v], can't be greater than the number of elements П№С иПт>. Hence, the assumption is

TwCN \ < П№С U ПТ»

If in

false, that is, in the segments Aj and A2, the number of parameters |n. segment Aj the number of parameters of twins of composite numbers is equal to kj and in segment

Л 2 is k2 <

, then the number of parameters |ПТм, иП№С| in Л1 is 51 =

- k1 I and in

Л2 :S2 =

- k

Let

k1 < k2.

then it is obvious that

51 > 52.

Let the parameters

(Xj , Xj, ...,£ )eAJ e nTwCN and assume that (to strengthen the assertion of Lemma 5) the corresponding pair parameters (X 2 , X2 , ...,£) e A2 are elements of nTw un№C. Similar reasoning on the parameters is found in segment A2. Then, the numbers 5j or 52 for parameters |nTw u n^J in

the segment Л1 or Л 2 will be

k1 I k2 =

- (k1 + k2), where (k1 +k2) shows the number of

corresponding pair parameters, one of which the parameter is of twins of composite numbers on the segment [J ^ v]. Because the in the segments Aj and A2 the number of parameters of twins of

composite numbers of is less than the number of elements |nTw u n№C|, then naturally in AJ and A2 there is always a pair of numbers XJ eAJ, X2 eA2 such (XJ, X2 )enTw un^ and are the corresponding pairs of parameters of an even number Z > 8.

Lemma 6. In segm. [J ^ v] the number of elements |nTw u FN |> |FN+|\nTw u FN+| > |FN |.

Evidence. From the contents of the procedure for constructing the (see Table J) it is obvious that the ele ments of the sets FN+ = fJJ (x,y) u fj2 (x,y) and FN- = f2J (x,y) u f22 (x,y) in each row give

rise to 2 elements, that is, in the segment [J ^ v] the number of elements FN+ ~ FN" . Hence, the

following inequalities are true nTw u FN > FN+

nTw u FN+ > \FN-

Theorem 3. Any even number Z > 8 is decomposable into a sum of two primes.

Evidence. Since, any even numbers of the form Z = 6v + m and Z > 8 are considered and

Z - m

following from (16) that the values of the parameters of the even numbers v =

> 1, for the

remainders m = (0, 2, -2). Then, by analogy with the remainders m, the even numbers Z > 8 are represented by the sums 0j + 02, where 0j = 6 X1 ± 1 and 02 = 6 X2 ± 1, (X2 = v - X^ by the types of expansions of even numbers Z > 8 (Lemma 4) or by properties a, P, y we have the following structures of the corresponding forms of the decomposition of even numbers Z > 8:

a). Z = 6v + 0: 91 = (6X1 ± 1 | X1 еЛ1} and 92 = {6X2 ± 1 | X2 еЛ2)

b).Z = 6v + 2 : 91 = {6X1 +1 | X1 еЛ1} and 92 = {6X2 +1 | X2 еЛ2}

c). Z = 6v-2 : 91 = (6X1 -1 | X1 еЛ1} and 92 = {6X2 -1 | X2 еЛ2}

(17)

UPC

We investigate the elements of the segment [1 ^ v] for belonging to the sets nTw, nTwCN, n and find out for which values of the parameters X1 e A1 and X2 e A2 from the listed sets the numbers 01 = 6X1 ± 1 and are primes or for which composite numbers. For elements of the set nTwCN they are not considered, since 01, 02 is composite numbers. However, by Lemma 5 in segment [1 ^ v] there are always X1 eA1 and X2 e A2 such that X1, X2 e (nTw u nupc), then (X1, X2) e nTw U (FN A FN-). Hence it remains to verify the elements X1 e A1, X2 e A2, on belonging to the sets nTw and FN A FN-with the removed parameters of set nTwCN in the segment [1 ^ v], we have:

A. Let X1 e nTw, X2 e nTw, then the terms 01, 02 of the even numbers Z > 8 are primes, as the

(p1 = 6X1 -1, ^ (p2 = 6X2 -1, ^

and 92 =

Hence,

constituent elements of twins of primes 01 = .. ..

I P1 = 6X1 + 1 ) y P2 = 6X2 + 1 )

according to the corresponding forms in (17) we have Z = P1 + P2 and infinity of twins of primes I will soon publish.

B. Let ( X1 or X2) e nTw & (X2 or X1 e (FN A FN ) so the term 01 is prime numbers as constituent elements of twins of primes 01 = (p1 = 6X1 -1, p1 = 6X1 +1). And the 2nd term 02 = 6X2 ± 1 by Lemma 2 in one of the variants of the forms: 02 = (p2 = 6X2 -1 or p' = 6X2 +1) is a prime numbers. Then adding to the prime numbers 02 = p2 obtained a prime numbers 01 = p1 (by analogy with the corresponding from (17), we get Z = p1 + p2 and infinity of twins of primes will soon be published.

C. Let (Xi, X2 e (FN+ A FN-), then in order that the term 01 of an even numbers Z > 8, depending on the corresponding form defined in (17) and by Lemma 2 be a prime numbers, it is sufficient to establish whether the parameter X1 belongs to one of the sets ( FN~or FN +). For the term

02 the parameter of which X2 = v - X1 on the segment [1 ^ v] can obviously correspond to the parameters nTw or again the elements from the set (FN + A FN-). If X2 e nTw, then there are no problems, otherwise the following parameter X1 is determined by the identities (11), (13), (15), respectively, by type of an even numbers 6v + m, which is obviously an element of either nTw or (FN + A FN-). But since by Lemma 6 the number of parameters in segments [1 ^ v ]|n Tw u FN > |FN + |and |n Tw u FN + | > |FN , then one of the items (A, B) is repeated in

A1 & A2. Thus, the decomposition of an even number Z > 8 by the sum of 2 numbers 01 + 02, where the numbers of form 01 = 6X1 ± 1, 02 = 6X2 ± 1, whose parameters (X1 and X2) e (nTw u nupc) in any of the listed variants (A, B, C) always lead to the forms 6X1 ± 1 and 6X2 ± 1 of primes and since the sum of the numbers 01 + 02 is even numbers, either 6 (X1 + X2) or 6 (X1 + X2 )± 2, then the Goldbach -Euler binary problem Z = p1 + p2 is solved in any case positively.

Example 6. Let an even number Z = 360. Let us calculate the remainder m from dividing an even numbers Z by 6. By Lemma 3 we find the type of an even number 6v + 0, and by (17), respectively, we have the string (17, a). We establish the forms of the terms in the expansion of the even numbers Z > 8,

we have numbers of the form 0j = 6^ ± 1, 02 = 6X2 ± 1, we calculate the even numbers' parameter by

the v =

Z - m 6

where the remainder m = 0 whence v = 60. We write out the elements of the sets

nTwCN, nTw, nUPC from the corresponding tables on the segment [1 ^ 60] nT 41, 48, 50, 54, 57} (see Table 5 below or Table 6 [3]).

= {20, 24, 31, 34, 36,

Пv

= NFN = {1, 2, 3, 5, 7, 10, 12, 17, 18, 23, 25, 30, 32, 33, 38, 40, 45, 47, 52, 58} (see

Table 6 below or Table 5 [3]).

FN- = {6, 11, 13, 16,*20, 21,*24, 26, 27,*31,*34, 35,*36, 37, *41, 46, *48, *50, 51,*54, 55, 56, *57} (Table 1).

FN + ={4, 8, 9, 14, 15, 19, *20, 22, *24, 28, 29, *31, *34, *36, 39, *41, 42, 43, 44, *48, 49, *50, 53, *54, 55, *57, 59 ,60}. Symbols in numbers according to belonging to sets nTwCN , n Tw , FN , FN +.

We decompose the elements of the corresponding pairs of parameters lieing in e

1 ^ v = 30

e

_ 2 _

and

TwCN ,nTw ,Пирс parameter sets, according to

X2 e[30 ^ 60] taking into account their belonging to n, A1 ={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}. A2={*31, 32, 33, *34, -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}.

The numbers of corresponding pairs of parameters (Xi, X2) lie in segment [1 + v] and, naturally, for the parameters of the numbers nTwCN, the corresponding summands are composite numbers, but we are not interested in such. Then, it will be necessary to choose from A1, and A2 those parameters that belong to the union of the sets nTw U (FN+ A FN-). For example, suppose the parameter values are:

x1 = 2 e Пт», 6X1 - 1 = 11 e pn - 6x 1 + 1 = 13 e pn +,

x2 = 58 e Ит» 6x 2 + 1 = 349 e pn+ 6x2 + 1 = 347 e pn -

x1 = 25 e Ит», 6x1 - 1 = 149 e PN - 6x1 + 1 = 151 e pn +,

x2 = 35 e FN 6x 2 + 1 = 211 e pn+ 6x2 - 1 = 209 e CN -

x1 = 5 e Ит», 6 x1 -1 = 29 e PN - 6x1 + 1 = 31 e pn +,

x2 и 5 Ul e FN + 6X2 + 1 = 331 e CN+ 6x2 - 1 = 329 e pn -

x1 = 46 e FN , 6x1 - 1 = 275 e CN - 6x1 + 1 = 277 e pn +,

x2 = 14 e FN + 6X2 + 1 = 85 e CN + 6x2 - 1 = 83 e pn -

x1 = 39 e FN+, 6x1 - 1 = 233 e PN - 6x1 + 1 = 235 e CN +,

x2 = 21 e FN - 6x2 + 1 = 127 e PN+ 6x2 - 1 = 125 e CN -

Since the parameters ={1,2,5,23,25, ...}enTw on [1 ^ 30], then the numbers of the form 01 = 6X1 ± 1 are primes as of twins of primes, i. e. (5-7, 11-13, 29-31, 137-139, 149-151) and the corresponding parameters X2 e [30 ^ 60] in one of the variants of the forms 02 = 6X2 -1 or 02 = 6X2 +1

are prime numbers, by Lemma 2. Thus, even number 360 is decomposable into sums of primes 11 + 349, 13 + 347, 149 + 211, 29 + 331, 277 + 83, 233+ 127.

Example 7. Let an even number Z = 362. We compute the remainder m from dividing an even number Z by 6, and by Lemma 3 we find the type of an even number 6v + 2 ^ m = 2, according to (17), respectively, we have the string (17, b). We fix the forms of the terms in the expansion of an even number Z > 8, we have numbers of the form 01 = 6X1 +1; 02 = 6X2 +1. We calculate the even numbers' parameter by the formula (16), where v = 60. We write out the elements of the subsets nTwCN, nTw, nUPC from the corresponding tables on the segment [1 + 60]. It is obvious that the values of the

parameters of the sets nTw, nTwCN, FN +, FN- and A1, A2 remain unchanged, as in Example 6, but since the numbers here are of the same type, in order that 01, 02 are primes, it is necessary that the parameters X1, X2 were taken from the sets nTw or FN-\ nTwCN :

X1 = 23 e Ит» x 2 = 37 e FN-

*,!= 27 e FN- x 2 = 33 e Ит»:

X1= 6 e FN- X 2 = 14 e FN+

X1= 47 e Ит» x 2 = 13 e FN-

6 Xi + 1= 139 e PN+ 6 X1 +1 = 163 e PN+ 6 X1 + 1= 277 e PN+ 6 X1 + 1= 283 e PN+

6X 2+ 1 = 223 e PN+ 6X 2+ 1= 199 e PN+ 6X 2+ 1 = 85 e CN+ 6X 2+ 1 = 79 e PN+

Thus, the number 362 is decomposable into sums of 2 primes: 139 + 223, 163 + 199, 283 + 79.

Table 5

Parameters of twins of composite pairs (ПГи,сЛ) in set 0 1 ^ 10 000

020, 024, 031, 034, 036 139, 141, 145, 149, 150. 225, 226, 231, 232, 234 307, 309, 314, 316, 320 372, 376, 377, 384, 386 456, 460, 462, 464, 468 525, 526, 529, 533, 539 587, 592, 594, 596, 598 662, 663, 664, 666, 672 719, 720, 722, 724, 728 783, 785, 786, 790, 791 849, 854, 855, 856, 857 915, 916, 919, 923, 924 985, 986, 989, 991, 993 1040, 1042, 1047, 1049 1093, 1098, 1099, 1102 1141, 1142, 1146, 1148 1194, 1195, 1197, 1200 1237, 1238, 1240, 1241 1296, 1297, 1300, 1301 1341, 1344, 1346, 1351 1397, 1399, 1400, 1401 1442, 1443, 1454, 1461 1499, 1503, 1504, 1506 1551, 1555, 1559, 1560 1606, 1609, 1611, 1612 1660, 1663, 1664, 1665 1703, 1705, 1706, 1713

, 041, 048, 050, 054, 057, 069, 071,079, 086, 154, 160, 167, 171, 174, 176, 179, 180, 189, 236, 244, 246, 251, 253, 256, 265, 272, 274, 321, 323, 324, 326, 327, 328, 337, 339, 341 387, 388, 394, 401, 405, 409, 414, 415, 416, 469, 471, 478, 479, 482, 487, 489, 491, 496, 540, 541, 544, 546, 547, 548, 549, 556, 559, 600, 608, 609, 611, 614, 619, 624, 625, 626, 673, 674, 677, 678, 681, 684, 686, 687, 691 730, 731, 734, 736, 738, 739, 745, 746, 748, 794, 795,796, 801, 804, 806, 807, 808, 809, 860, 864, 867, 869, 870, 874, 875, 876, 878 925, 931, 933, 934, 935, 936, 938, 939, 944, 994, 995, 996, 999, 1000, 1003, 1004, 1010, 1051, 1058, 1064, 1067, 1068, 1069, 1072, 1104, 1105, 1107, 1108, 1111, 1114, 1116, 1149, 1154, 1155, 1156, 1157, 1159, 1168, 1204, 1210, 1211, 1212, 1213, 1215, 1217, 1244, 1245, 1249, 1252, 1259, 1261, 1266, 1302, 1306, 1308, 1310, 1315, 1316, 1319, 1355, 1356, 1357, 1359, 1364, 1366, 1367, 1402, 1406, 1409, 1412, 1413, 1414, 1415, 1462, 1465, 1466, 1469, 1471, 1474, 1476, 1509, 1512, 1513, 1514, 1516, 1519, 1520, 1561, 1564, 1568, 1571, 1574, 1575, 1576, 1614, 1617, 1618, 1619, 1621, 1626, 1627, 1666, 1667, 1669, 1670, 1671, 1672, 1674, 1714, 1716, 1718, 1721, 1725, 1727, 1729,

088, 089, 092, 097, 104, 106, 111, 116, 119, 130, 132, 134, 136, 190, 191, 193, 196, 201, 207, 209, 211, 212, 219, 222, 223, 224, , 275, 279, 280, 281, 284, 286, 288, 294, 295, 299, 301, 303, 306, , 343, 345, 349, 351, 353, 354, 358, 361, 362, 364, 365, 366, 371, ., 418, 419, 421, 427, 428, 429, 431, 433, 434, 438, 440, 442, 454, 497, 498, 499, 501, 505, 509, 512, 516, 517, 519, 521, 522, 524, 561, 563, 564, 566, 567, 570, 571, 573, 574, 579, 580, 581, 584, 629, 631, 635, 636, 638, 640, 643, 645, 649, 650, 656, 659, 660, , 694, 695, 697, 698, 699, 701, 704, 706, 708, 711, 713, 717, 718, :, 750, 755, 756, 757, 759, 762, 763, 768, 769, 771, 772, 778, 781, I, 811, 814, 816, 819, 821, 827, 830, 836, 838, 839, 841, 844, 845, 881, 882, 886, 888, 890, 893, 894, 895, 896, 904. 909, 910, 911, 946, 951, 954, 955, 959, 960, 961, 962, 966, 972, 979, 981, 982, 1014, 1016, 1018, 1021, 1023, 1026, 1028,1030, 1031, 1032, 1039, 1073, 1074, 1076, 1077, 1081, 1083, 1084, 1085, 1086, 1089, 1090, 1119, 1121, 1124, 1125, 1126, 1128, 1129, 1131, 1133, 1135, 1136, 1172, 1175, 1177, 1179, 1181, 1182, 1183, 1186, 1189, 1190, 1191, 1219, 1221, 1223, 1224, 1226, 1227, 1229, 1230, 1231, 1233, 1234, 1269, 1271, 1272, 1276, 1277, 5192, 1285, 1289, 1291,1294, 1295, 1324, 1326, 1328, 1329, 1330, 1331, 1333, 1334, 1337, 1338, 1339, 1369, 1371, 1375, 1376, 1380, 1384, 1387, 1389, 1390, 1391, 1393, 1416, 1418, 1422, 1425, 1426, 1428, 1431,1432, 1434, 1436, 1439, 1470,1480, 1483, 1484, 1485, 1486, 1491, 1493, 1496, 1497, 1498, 1524, 1528, 1532, 1536, 1539, 1541, 1542, 1544, 1545, 1548, 1550, 1581, 1584, 1586, 1588, 1593,1594, 1595, 1596, 1597, 1599, 1601, 1629, 1633, 1637, 1641, 1644, 1646, 1649, 1652, 1653, 1656, 1659, 1675, 1676, 1679, 1681, 1686, 1687, 1688, 1691, 1698, 1700, 1701, 1730, 1731, 1734,1735, 1736, 1737,_

Table 6

Parameters of twins prime pairs (nrw) 1 50 000

001, 002, 003, 005, 007, 010, 012, 017, 018, 023 135, 137, 138, 143, 147, 170, 172, 175, 177, 182. 325, 333, 338, 347, 348, 352, 355, 357, 373, 378. 560, 562, 565, 577, 578, 588, 590, 593, 597, 612.

753, 758, 773, 775, 787, 798, 800, 822, 82: 1015, 1022, 1033, 1045, 1050, 1060, 1075, 1092, 1225, 1243, 1248, 1258, 1260, 1265, 1293, 1313, 1500, 1502, 1507, 1540, 1547, 1557, 1570, 1572, 1722, 1738, 1743, 1750, 1755, 1785, 1810, 1815, 2007, 2012, 2018, 2027, 2040, 2042, 2063, 2090, 2317, 2322, 2333, 2335, 2347, 2375, 2387, 2398, 2623, 2648, 2662, 2677, 2678, 2690, 2698, 2705, 2933, 2943, 2947, 2958, 2965, 2973, 2985, 2987, 3190, 3197, 3202, 3230, 3237, 3238, 3245, 325, 3458, 3462, 3468, 3483, 3497, 3502, 3503, 3510, 3693, 3712, 3713, 3728, 3747, 3757, 3762, 3770, 3948, 3957, 3972, 3985, 4018, 4030, 4062, 4070, 4375, 4377, 4447, 4450, 4452, 4455, 4477, 4480, 4653, 4657, 4683, 4685, 4697, 4713, 4718, 4725 5002, 5015, 5023, 5045, 5065,5078, 5082, 5093, 5308, 5338, 5343, 5353, 5357, 5365, 5383, 5387, 5548, 5555, 5558, 5598, 5600, 5603, 5625, 5628, 580, 5808, 5827, 5842, 5847, 5880, 5908, 5918, 6170, 6200, 6218, 6223, 6227, 6258, 6262, 6265, 6507, 6527, 6538, 6540, 6557, 6562, 6585, 6638, 6920, 6935, 6960, 6975, 6993, 6997, 7003, 7012, 7268, 7275, 7297, 7298, 7315, 7327, 7338, 7348, 7637, 7675 ,7682, 7697, 7712, 7718, 7725, 7740, 7968, 8020, 8052, 8068, 8080, 8090, 8108, 8113,

8255

025, 030, 032, 033, 038. 192, 205, 213, 215, 217, 385, 390, 397, 425, 432, 628, 637, 642, 653, 655.

835, 837, 850, 872, 88 1095, 1110, 1115, 1117, 1325, 1335, 1348, 1370, 1573, 1577, 1605, 1613, 1823, 1843, 1845, 1853, 2102, 2137, 2153, 2167, 2408, 2425, 2427, 2432, 2727, 2742, 2772, 2775, 2993, 2998, 3007, 3008, 3283, 3292, 3307, 3315, 3532, 3553, 3563, 3582, 3773, 3783, 3790, 3810, 4153, 4163, 4172, 4195, 4482, 4492, 4510, 4518, , 4735, 4758, 762, 4770, 5140, 5142, 5145, 5180, 5395, 5402, 5407, 5422, 5635, 5638, 5672, 5688, 5922, 5932, 5955, 5967, 6282, 6297, 6302, 6332, 6640, 6673, 6688, 6692, 7030, 7037, 7047, 7068, 7355, 7367, 7378, 7380, 7765, 7780, 7795, 7803, 8122, 8130, 8137, 8143, , 8258, 8278, 8290, 8298

040, 045, 047, 052, 058, 070, 072, 077, 087, 095, 100, 103, 107, 110, 220, 238, 242, 247, 248, 268, 270, 278, 283, 287, 298, 312, 313, 322, 443, 448, 452, 455, 465, 467, 495, 500, 520, 528, 542, 543, 550, 555, 667, 670, 675, 682, 688, 693, 703, 705, 707, 710, 712, 723, 737, 747, », 903, 907, 913, 917, 920, 940, 942, 943, 957, 975, 978, 980, 1127, 1130, 1132, 1138, 1145, 1158, 1160, 1188, 1202, 1218, 1222, 1372, 1382, 1398, 1405, 1423, 1433, 1438, 1470, 1473, 1477, 1495, 1620, 1628, 1643, 1655, 1668, 1673, 1678, 1682, 1690, 1712, 1717, 1860, 1862, 1892, 1915, 1925, 1950, 1953, 1963, 1972, 1990, 1995, 2168, 2203, 2223, 2233, 2280, 2282, 2285, 2287, 2293, 2305, 2313, 2438, 2478, 2523, 2545, 2548, 2555, 2560, 2597, 2607, 2608, 2622, 2782, 2805, 2817, 2830, 2838, 2865, 2882, 2898, 2903, 2915, 2930, 3010, 3020, 3022, 3042, 3048, 3052, 3087, 3090, 3152, 3153, 3180, 3332, 3337, 3358, 3372, 3393, 3407, 3413, 3418, 3425, 3440, 3453, 3587, 3593, 3598, 3600, 3602, 3608, 3623, 3640, 3673, 3682, 3685, 3827, 3838, 3840, 3843, 3867, 3882, 3895, 3923, 3927, 3938, 3945, 4217, 4218, 4235, 4245, 4263, 4267, 4300, 4308, 4322, 4333, 4352, 4540, 4547, 4568, 4580, 4588, 4590, 4597, 4615, 4623, 4625, 4632, 4777, 4792, 4837, 4855, 4868, 4898, 4900, 4928, 4945, 4960, 4980, 5187, 5192, 5197, 5208, 5220, 5232, 5252, 5257, 5287, 5288,5295, 5427, 5435, 5453, 5467, 5472, 5485, 5490, 5495, 5512, 5525, 5530, 5693, 5702, 5710, 5717, 5728, 5745, 5750, 5752, 5765, 5775, 5793, 5973, 5983, 6002, 6018, 6057, 6078, 6088, 6130, 6132, 6150, 6155, 6373, 6388, 6408, 6410, 6428, 6435, 6442, 6445, 6452, 6458, 6487, 6738, 6755, 6773, 6783, 6808, 6857, 6863, 6867, 6872, 6898, 6902, 7077, 7095, 7107, 7117, 7140, 7150, 7175, 7220, 7233, 7257, 7263, 7397, 7422, 7437, 7450, 7462, 7520, 7523, 7530, 7553, 7557, 7598, 7805, 7843, 7858, 7892, 7898, 7903, 7943, 7950, 7952, 7957, 7963, 8145, 8165, 8172, 8187, 8195, 8200, 8213, 8222, 8228, 8232, 8235, 8320, 8323, 8332, ....

Example 8. Let an even number Z = 364. We compute the remainder m from dividing an even number Z by 6 and by Lemma 3, we find the type of even numbers 6v - 2 ^ m = -2 and by (17), respectively, we have the string (17, c). We fix the forms of the terms in the expansion of an even number Z = 8 , we have numbers of the form

0 = 6Xj -1

and 02 = 6X2 -1. We calculate the even

numbers' parameter by the formula (16), whence v = 61. Then the values of the parameters of the sets nTw, nTwCN, FN+,FN- and A1, A2 remain almost unchanged, as in Example 6, only the parameters 61 are added, that is, [1 + 61]. Since here are again the numbers of one kind, then, in order for numbers of the form terms 9i, 02 to be primes obviously by Lemma 3 the parameters X1, X2 must belong to one of the sets: nTw or FN-\ nTwCN:

X2 _ 2 G П7Ж X2 = 19 G Пш X2 = 46 G FN

X2 - 1 = 59 G FN h - 1 = 42 G FN X2 - 1 = 15 G FN

6X2 - 1 = 11 G PN 6X1 - 1 = 113 G PN 6 X1 - 1 = 275 G CN -

6X2 - 1 = 353 G PN 6X1 - 1 = 25 G PN 6X2 - 1 = 89 G PN

Thus, the number 362 is decomposable into sums of 2 primes: 11 + 353, 113 + 251. In Examples 7 and 8 we have the same type of terms 9i, and 92. And since inequality (17) is always feasible on A1 and A2, the elements of the sets nTw, FN+,FN- are always present in X1, X2. Therefore, the validity of the Goldbach - Euler binary problem as a whole remains valid for all even numbers

Z > 8, Л1 =

1 h

and Л2 =

hv

It is proved that any even numbers Z > 8 decomposes into

a sum of 2 primes, & since the expansion of even numbers Z ^ 8 is verified directly, the binary Goldbach - Euler problem for even numbers Z > 2 is proved.

Conclusion

A comprehensive study of the problem of finding and distributing primes and composite numbers, twins of primes, twins of composite numbers, including a theoretical study by means of computer software made it possible to obtain great results, to solve the secrets of the structures of subsets of the set of natural numbers and to give the definitions, to obtain a new algorithm for finding and distributing primes P (>5), to fulfil the calculation of the exact number of primes and to obtain the

method of distribution primes p > 5, pn = 6 • id + (-1)v(n) by their ordinal numbers n and vice versa in

N 6

in interval

the set of primes. The method of their distribution law n(x) = 2 + S1 + S2), m =

x=1

[1 ^ N) has been found.

Any even number Z > 8 has the form Z = 6v + m, where v = (Z - m) / 6 v = (Z - m) /6 is an even parameter.

Any even number Z > 8 can be represented by the sums of 2 elements X1 = 6X ± 1, X2 = 6(v - X) ± 1, respectively, over the remainders m = (0, 2, -2).

X1 e

For any even numbers Z > 8 on the segment [1 + v] there always exists a pair of numbers

v 2

1 h

and h2 e

H v

such that both (X1, X2) G nTw U (FN + AFN-).

REFERENCES

1. Prashar K. Primzahlverteilung. Berlin, Springer, 1957. 527 p.

2. Crandall R., Pomerance C. Primes: A Computational Perspective. N. York, Springer-Verlag, 2001. 545 p.

3. Bukhshtab A. A. Teoriia chisel [Number theory]. Moscow, Prosveshchenie Publ., 1966. 384 p.

4. Chermidov S. I. O faktorizatsii natural'nykh chisel [On factorization of natural numbers]. Dialogi o nauke, 2011, no. 2, pp. 68-69.

5. Chermidov S. I. Raspredelenie prostykh chisel. Algoritm chisel-bliznetsov i ikh beskonechnost' [Distribution of prime numbers. Twin numbers algorithm and their infinity]. Politematicheskii setevoi elektronnyi nauch-nyi zhurnal Kubanckogo gosudarstvennogo agrarnogo universiteta, 2015, no. 06 (110), pp. 414-436. Available at: http://ej.kubagro.ru/a/viewaut.asp?id=4701 (accessed: 12.12.2019).

6. Chermidov S. I. Raspredelenie prostykh i sostavnykh chisel i ikh algoritmicheskie prilozheniia [Distribution of prime and composite numbers and their algorithmic applications]. Vestnik Astrakhanskogo gosudarstvennogo tekhnicheskogo universiteta. Seriia: Upravlenie, vychislitel'naia tekhnika i informatika, 2017, no. 3, pp. 48-64.

The article submitted to the editors 23.01.2020

INFORMA TION ABOUT THE A UTHOR

Chermidov Sergey Ivanovich - Russia, 350040, Krasnodar, Kuban State University, Competitor for a Scientific Degree of the Department of Applied Mathematics; chermidov. sergei@mail.ru.

ЗАКОН РАСПРЕДЕЛЕНИЯ ПРОСТЫХ ЧИСЕЛ. ЗАВИСИМОСТЬ ПРОСТЫХ ЧИСЕЛ ОТ ИХ ПОРЯДКОВЫХ НОМЕРОВ И БИНАРНАЯ ЗАДАЧА ГОЛЬДБАХА - ЭЙЛЕРА С ИСПОЛЬЗОВАНИЕМ

ВЫЧИСЛИТЕЛЬНОЙ МАШИНЫ

С. И. Чермидов

Кубанский государственный университет, Краснодар, Российская Федерация

В статье рассматриваются методы определения и нахождения распределения составных чисел CN, простых чисел PN, двойников простых чисел Tw и двойников составных чисел TwCN, не имеющих делителей 2 и 3 в М, основанные на множестве чисел типа

0 = {6к ± 1, к е N}, являющемся полугруппой по отношению к умножению. Предложен метод получения простых чисел p > 5 по их порядковым номерам в множестве простых чисел p > 5 и наоборот, а также новый алгоритм поиска и распределения простых чисел на основе замкнутости элементов множества 0. В статье показано, что составное число n е 0 пред-ставимо в виде произведений (6х ±1) (6y ±1), где х, y е N - целочисленные положительные

решения одного из 4-х диофантовых уравнений: P(х,y, X) = 6xy ± х ± y - X = 0 . Доказано, что

если существует параметр X двойников простых чисел, то ни одно из диофантовых уравнений P (х, y, X) = 0 не имеет положительных целых решений. Найден новый закон распределения простых чисел п (х) в сегменте [1 ^ N]. Любое четное число Z >8 сравнимо с одним из

чисел m = (0, 2, -2), т. е. Z = m (mod6) . Согласно вышеупомянутым остаткам m, четные числа Z > 8 делятся на 3 типа, и каждый вид имеет свой собственный способ представления сумм из 2-х элементов множества 0. Для любого четного числа Z >8 на сегменте [1 ^ v], где v = (Z—m )/6, есть параметр четного числа; доказано, что всегда существует пара чисел (Xj, X2 )е[1 ^ v], являющихся элементами объединения множеств параметров двойников

простых чисел nTw и параметров переходных чисел П№С , т. е. числа вида 6X ± 1 с одинаковым X, если форма 6X -1 является простым числом, то форма 6X +1 является составным числом, и наоборот.

Ключевые слова: простые и составные числа, параметры простых чисел, диофантовы уравнения, бинарные (сильная) задача Гольдбаха - Эйлера, алгоритм решения бинарной задачи Гольдбаха - Эйлера.

Для цитирования: Чермидов С. И. Закон распределения простых чисел. Зависимость простых чисел от их порядковых номеров и бинарная задача Гольдбаха - Эйлера с использованием вычислительной машины // Вестник Астраханского государственного технического университета. Серия: Управление, вычислительная техника и информатика. 2020. № 4. С. 80-100. DOI: 10.24143/2072-9502-2020-4-80-100.

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Статья поступила в редакцию 23.01.2020

ИНФОРМАЦИЯ ОБ АВТОРЕ

Чермидов Сергей Иванович - Россия, 350040, Краснодар; Кубанский государственный университет; соискатель кафедры прикладной математики; chermidov.sergei@mail.ru.