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Formula for exact number of pairs of twins of Primes on the segment
Section 5. Mathematics
Drushinin Victor Vladimirovich, Lazarev Alexey Alexandrovich, Khromov Nikolai Olegovich, National Research Nuclear University "MEPHI" Sarov Physics and Technology Institute, Sarov, E-mail: [email protected]
Formula for exact number of pairs of twins of Primes on the segment
Abstract: Using Druzhinin's sieve type of sieve of Eratosthenes for the twin pairs, we consider the question of the exact number of pairs of twin primes on an arbitrary segment.
Keywords: primes, a pair of twins, Sieve of Eratosthenes.
The question of the distribution of pairs of twin (PT) of primes (P) for the whole real axis was raised, apparently, more by Eratosthenes. In the article [1] it was proved that the PT are spread over any distance up to infinity by creating a special type of sieve of Eratosthenes. In this paper, we derive a formula for calculating the exact number of PT on any segment of the real axis. The PT means two primes {p1; p2} .The difference between them is p2 - pl = 2. These sets are {5;7},{n;13},{l7;19},...,{3n;313},... Each PT has a center 6m, m e N — the natural numbers. PT is formed like {6m-1;6m +1}.For indicated PT m = {l;2;3; ...;52}.All indexes m, we divided into two classes: the "good" m that give PT, and "bad" m, which do not allow such a pair. For example, m = {4;6;9} give the following pairs of numbers (6m +1) = {23;25},{35;37},{54;55} .To remove the "bad" m was proposed the type of sieve of Eratosthenes [1]. On every "good" m have four arithmetic sequence (AS)
m = m + (6m-l)n;m = (5m-l) + (6m-l)(n-l); . .
m = m + (6m + l)n;m = (5m + l) + (6m + l)(n -1), where n e N. For example, the first row m = 1 delete m = (4;6;9;11;14;16;19;...) , i.e. all numbers ending in «1», «4», «6», «9», other than «1». Thus there is a rule: when you sequential walk through m, the first m not deleted by previous m " is also a "good" m. For example, m = 1 (1) does not delete m = 2, and therefore this number gives a pair of PT. A second example: the AS for set m = {1;2;3;} delete m = (4;6;8;9;11;13) , so the numbers m = {5;7;10;12} give PT. Thus it is possible to formulate a rule: if we take the all
good mk, k = 1;2;...;N in ascending order, delete all m by (1) and find the first not deletedd number mN+j, then there is a buffer zone — a segment LN = [m n+1, Mn] the all remaining numbers. Moreover these numbers are good.
The index MN is given by MN = (((6mN+1 -1)2 ± l) / 6) -1 . The sign «+» or «-» is taken from the multiplicity of numerator to denominator. Example. Suppose we have pro-
cessed m = 1. Then m2 = 2. M1 =(((6 • 2 -1)2 ± l) /6)-1. We take the sign «-» and M1 = 19. The bad deleted m by the first four AS m = 1 + 5k, m = 4 + 5k,m = 1 + 7k,kk = 6 + 7k onthe segment [2, 19] following
(4;6;8;9;11;13;14;15;16;19) (2)
The centers of PT remain following
m = {2;3;5;7;10;12;17;18;}. (3)
Thus, the processing of a few PT by (1) gives not only the first not deleted new center, but we get large number of them. In this example, the first PT gives eight additional PT. On this condition we build the formula for exact number of PT on any segment [aof the real axis. First, we must ensure that we are in the buffer zone, ie b < MN, a > mN+l. Next, it is necessary to consider the deleted m, included in 4N AS (1) for the all m1 = 1 < m < mN, on this segment. Enumerate mt (st), where t numbers m in ascending order, st indicates how many times this is deleted by different AS. If such indexation carried out, then on [a, fc] there is the following number of PT
c(a: b) = (b-a + 1)-Nab. (4)
The number ofbad m is Nab = Nab (l) - Nab (2) + Nab (3) -
-..(-IT Nab (k) on this segment, where Nab (k) is the number of bad indexes m deleted k times.
There is the general rule for calculating the number Nab (s) — the deleted members of the AS dk =a + P(k-\),keN on the segment [a,b]. If b <a, Nab (l) = 0. If b > m and using the theory of comparisons a =a(modfi) and (or) b = a(modp), then
b - a
(S) =
P
+ 1 .
(5)
In all other cases, there is no «+1» in (5). Square brackets denote the greatest integer number not exceeding the number in brackets. In the above example m = 1 + 5k gives a «3», m = 4 + 5k gives «4», m = 1 + 7k and m = 6 + 7k give «2»
