A292691 - OEIS

%I #71 Jul 25 2023 02:55:44

%S 1,3,101505,259105757127,1356566605613854774200240267,

%T 1851197466245939272480116323530608949000567215

%N a(n) = C(A001359(n)), n >= 1, with C(n) = (4*((n-1)! + 1) + n)/(n*(n+2)) for n >= 2.

%C Clement's criterion for twin primes is, for integers with n >= 2: n and n + 2 are both primes if and only if 4*((n-1)! + 1) + n == 0 (mod n*(n+2)). See the Clement and Ribenboim links. Like the criteron for primality using Theorem 81 of Hardy and Wright, p. 69, it "is of course quite useless as a practical test".

%C a(n) is an integer because of the necessary part of this twin prime criterion.

%C Thanks to _Wolfdieter Lang_ for comments and helpful advice.

%D G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford Science Publications, 1979.

%D P. Ribenboim, The New Book of Prime Number Records, Springer-Verlag NY 1996, pp. 259-260 (a proof of Clement's theorem).

%H Jaime Gómez, <a href="/A292691/b292691.txt">Table of n, a(n) for n = 1..23</a>

%H P. A. Clement, <a href="http://www.jstor.org/stable/2305816">Congruences for sets of primes</a>, American Mathematical Monthly, vol. 56 (1949), pages 23-25.

%H L. Cong and Z. Li, <a href="https://arxiv.org/abs/math/0408018">On Wilson's Theorem and Polignac's Conjecture</a>, arXiv:math/0408018 [math.NT], 2004.

%F a(n) = (4*((p1(n)-1)! + 1) + p1(n))/(p1(n)*(p1(n) + 2)) with p1(n) = A001359(n), for n >= 1. See the name.

%F From Wilson's theorem (see Hardy and Wright, Theorem 80, p. 68), a(n) = (4*kp1(n) + 1)/(p1(n) + 2) with p1(n) = A000359(n) and kp1(n) = A007619(p1(n)).

%F a(n) = delta(A014574(n)) with delta(n) = (4*(n-2)!+ n + 3)/(n^2 - 1).

%F delta(n) ~ ((4*(n-2)^(n - 2)* sqrt(2*Pi*(n - 2))) / (e^(n - 2)*(n^2 - 1)))+((n + 3) / (n^2 - 1)) for large n-values (using Stirling's approximation for n!).

%e a(2) = 3, because A001359(2) = 5 and C(5) = (4*(4! + 1) + 5)/(5*7) = 3.

%e a(2) = 3 because A014574(2) = 6 and delta(6) = (4*4! + 6 + 3)/35 = 3.

%t p1[1] = 3; p1[n_] := p1[n] = (p = NextPrime[p1[n-1]]; While[!PrimeQ[p + 2], p = NextPrime[p]]; p);

%t a[n_] := (4*((p1[n] - 1)! + 1) + p1[n])/(p1[n]*(p1[n] + 2));

%t Array[a, 6] (* _Jean-François Alcover_, Nov 04 2017 *)

%o (Python 2.7)

%o import math

%o from sympy import *

%o list = []

%o n = 3

%o l = 1   # parameter that indicates the desired length of the list

%o x = 1

%o while x <= l:

%o        y = (4*factorial(n-2))+n+3

%o        z = n**2 - 1

%o        if y % z == 0:

%o               print (y/z)

%o               list.append(y/z)

%o        n+=1

%o        x+=1

%o (PARI) c(n) = (4*(n - 2)! + n + 3) / (n^2 - 1);

%o lista(nn) = forprime(p=2, nn, if (isprime(p+2), print1(c(p+1), ", "));); \\ _Michel Marcus_, Sep 21 2017

%Y Cf. A001359, A007619, A014574.

%K nonn

%O 1,2

%A _Jaime Gómez_, Sep 20 2017

%E Edited by _Wolfdieter Lang_, Oct 25 2017