%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