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The hypothesis of the Euler-Goldbach and the opposite sieve of eratosthenes
Section 3. Mathematics
D OI: http://dx.doi.org/10.20534/AJT-17-3.4-11-12
Druzhinin Victor Vladimirovich, Strahov Anton Victorovich, National research nuclear University «MEPhI» Sarov Institute of physics and technology, Department of mathematics Sarov, E-mail: vvdr@newmail.ru
The hypothesis of the Euler-Goldbach and the opposite sieve of eratosthenes
Abstract: The existence of the amount of even number two prime numbers allows us to obtain the sieve of Eratosthenes from very large numbers to decrease. There is a new way to generate primes. Keywords: hypothesis of the Euler-Goldbach, the sieve of Eratosthenes.
The hypothesis of the Euler-Goldbach (HEG) was established by Euler in 1742 in response to trinary hypothesis Goldbach. Goldbach suggested that every odd number is the sum of three primes (PN). Euler said that if any even number are the sum of two PN (binary hypothesis), trinary hypothesis is confirmed automatically [1; 2]. In 2014 Druzhinin first proved binary option using probability theory [3]. In subsequent works on this subject Druzhinin and Lazarev [4; 5] improved this evidence, in particular, they found the supermom that without theory of probability confirms that every even integer greater than "2" is the sum of two primes. There is the formula for calculation of pairs of such numbers, called the pairs of the Euler-Goldbach (PEG), and obtained good agreement with experiment. These works were handed over to the various specialists in the number theory who are doctors of science and members of the SAR, but still negative feedback or detected mistakes in reasoning Druzhinin is not received.
This article discusses two topics related to theory: the inverse sieve of Eratosthenes and development of the original theory to the case of special groups of PN. The sieve of Eratosthenes, which is more than 2000 years, looks like. Taken a consistent set ^i7natural odid numbers beginning with PN pi = 3 is considered. All numbers x 1 = 9 + 6k, i. e. all numbers are multiples of "3", except for the number "3" are removed. After that the first not deleted number, in this case, p2 = 5, is declared the prime and all other numbers are multiples of "5", x 2 = 15 + 10k
are removed. The following not delete number p3 = 7 is the prime, and x 3 = 21 + 14k are removed. Thus, the primes are found sequentially. A characteristic feature of the sieve is that it is not only the first not remote, the number p is PN, but the whole segment I p , p2 -1 I con-
x m |_x m' x m _|
tains the new PN that are not uninstalled the previous arithmetic progressions (AP). For example, the interval [7, 47] contains the PN {11; 13; 17; 19; 23; 29; 31; 37; 41; 43; 47}. This is a direct sieve of Eratosthenes. Reverse the sieve of Eratosthenes, which we offer, consists in the following. Take any even number M = 2N, and his left the neighborhood (smaller numbers) is a sieve that selects a large primes.
The logic of action is based on the works of the author [3; 4; 5] and looks like this. 1. From inequality pm < VM - 3 basis set fond of HR A (M,m) = = {pl;p2;...pm} 2. From the equality \_(N- 1)S2] = n the period of B(M,n) = [1;2;3;.. .;n] found. Here [C] the greatest integer not exceeding "C". 3. From a comparison ofM = nk (mod p^ ,where 0 < ¡uk < pk, we find all remainders modulo M in pk e A (M,m). 4. From the comparison tr + (( + 1)(pk _ 1)S2)] = . = /3k (modpk)find the deductions 1 < /3k < pk. {{3k} - is a complete set of minimal positive deductions. 5. Prepare m AP yk = Pk + tpk for all the PN of A (M,m). Here t = {0;1;2;...} . 6. In the interval B(M,n) delete all numbers from the obtained m AP. Remained not deleted numbers z, give PN p = M -(2z, +1). It is a consistent
Section 3. Mathematics
large PN smaller M.
Example. Take M = 50.
1. A (50,2 ) = {3;5}.
2. n = [24/2] = 12. B = [1,12].
3. // = 2, // = 0.
4. [5] = 2 (mod3),[2 ] = 2 (mod5), /3l = ¡32 = 2.
5. y1 = 2 + 3t removes from B the number of {2;5;8;1l} . y2 = 2 + 5t removes from B the numbers {2;7;12}. Remain z = (1;3;4;6;9;1^ ,which give PN {47;43; 41; 37; 31;29}. Above we got these numbers from the A rect si eve of Eratosthen es.
Communication with the HEG is that obtained great PN partially belong in the PEG, to calcultte whioh use t specified algorithm [3; 4; 5]. The practical benefit 2s to rapidly obtain a large prime. If we plot f3, in ascending order, you can easily find the first not remote z, and thus the prime closed to M. We will show this on the example with M = 10000. Here it is necessary to account for the 24 PN of the smaller "100". In ascending order fy have AP: {1 + 13t ;2 + 5t ;3 + 3t ;5 + 7t ;10 + 17t;...} .TheseAP leave z = 4, i. e. p = 9991, and z = 8, i. e. p = 9983.
The second question of this article there is a special PEG. Can an even number M = 2N + 2 be the sum oftwo PN of the form p = 4s +1 ? In our opinion it can. The sequence is similar to above, but with some differences.
1. We find A (M ,m ) = {p1; p2;... pm) where
p <VM - 5.
x m *
2. Find the interval B (M, n) = [1; 2; 3;...; n], where n = [N/4].
3. Generate a set of AP xk = ak + tpk and
^ =(ak+pk)+tpk. Here a = ((dk-°)/4). dk is the
smallest Prime or a composite number greater than or equal to "5" and a multiple of pk. PN pk consist of basic pk = 4s +1 and the auxiliary pk ^ 4s +1. For example, pk = {3;7;11;19;....} , and pk = |5;13;17;...} .We are looking for the sum of the number of M from two pk, i. e. M = pa + pb.
5. From t
Table
4. Create another group up yk = + tpk without separating the primes for the main and auxiliary. fik is found from a comparison of |((pk +1) / 2 )k +
= Pt (modpk) . t < pt,t = 0;1;2;... he segment delete numbers all three sequences. The remaining numbers z give special pair of Euler-Goldbach {4z +1;M - 4z -1} . Example.
M = 42, N = 21,n = 21/4 = 5,pmax <^37, B(42,5) = [1,5], A (42,2) = {3;5},a1 =(9 ^ 1)/4 = 2,a2 =(5 e 1) / 4 = 1,
pi = n,t2 = 5, = 2 c nt ,t2 = a c 5t^ e ^ = 2, [0+
c(0 c 1)2 J = /31 (modn), p1 h 2, [2 • 3 e(2 e 1)-1]s
S /32 (mod5),P2 = 4, y1 = 2 + 3t,y2 = 4 + 5t. All preparatory work is done. We remove from the interval [1,5] numbers of these four AP. Deleted AP: {2;5},{-},{2;5},{4}. Left numbers z = {1;3}. The number z = 1 gives the pair {5; 37}, number z = 3 gives pair {13; 29}. Illustration this problem is visually. Take all the odd numbers of the form y = 4s +1 from "5" to "47" and place them and B(42,5) according to the matrix rows ' 5 9 13 17 21^ 1 2 3 4 5
v37 33 29 25 21, The sum of the top and bottom numbers in the columns is equal to the given number M = 42 . Central line is numbers of columns. You can consider the top row and delete columns with the upper part number of {9; 21}. After that, the bottom line remains one composite number "25". Remove this column. The first and the third columns giving a PEG are remained. Our method of the work on secondary column with a set AP gives the same result and, moreover, allows to obtain an analytical proof that for M = 34 and all numbers of the form M = 2N + 2 can be expressed as the sum of two PN from the p = 4s +1. Compute calculations support this conclusion.
Table 1. - Number PEG from M.
M 102 202 302 402 502 602 702 802 902 1002
Number PEG 4 2 3 8 6 13 14 8 4 8
References:
1. Graham R. L. Concrene mathematics, - Moscow, MIR, - 1998.
2. Buhstab A. A. Theory of number, - Moscow, Lan, - 2015.
3. Druzhinin V. V. NTVP, - № 3, - 2014. - 14-17.
4. Druzhinin V. V., Lazarev A. A. AJT, - № 9-10, - 2014. - 19-21.
5. Druzhinin V. V., Lazarev A. A. ESR, -№ 7-8, - 2016. - 22-23.
