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%I #14 Jan 07 2013 09:25:16
%S 0,1,1,2,1,3,1,2,1,3,2,3,1,4,1,4,1,3,2,5,4,4,2,4,1,4,2,5,1,4,2,4,2,6,
%T 2,5,4,4,2,5,1,7,1,7,5,3,2,4,2,4,3,6,3,6,4,7,4,8,2,9,2,8,3,2,3,7,4,7,
%U 1,7,4,7,1,7,4,9,7,8,2,9,3,6,2,6,3,7,2,8,3,7,4,6,8,9,4,6,3,9,5,8
%N a(n) = |{0<=k<n: 2k+3, n(n-k)-1, n(n+k)-1 are all prime}|
%C Conjecture: a(n)>0 for all n>1. Moreover, if n>5 is different from 9, 191, 329, 641, 711, 979, then 2k-3, 2k+3, n(n-k)-1, n(n+k)-1 are all prime for some 0<k<n.
%C Zhi-Wei Sun also made the following conjectures:
%C (1) For any integer n>101 there is an integer 0<k<n such that kn-1 is a Sophie Germain prime.
%C (2) For each n=128,129,... there is an integer 0<k<n with kn-1 and kn+1 both prime.
%H Zhi-Wei Sun, <a href="/A187808/b187808.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1211.1588">Conjectures involving primes and quadratic forms</a>, arXiv:1211.1588.
%e a(25) = 1 since 2*17+3 = 37, 25(25-17)-1 = 199, and 25(25+17)-1 = 1049 are all prime.
%t a[n_]:=a[n]=Sum[If[PrimeQ[2k+3]==True&&PrimeQ[n(n-k)-1]==True&&PrimeQ[n(n+k)-1]==True,1,0],{k,0,n-1}]
%t Do[Print[n," ",a[n]],{n,1,100}]
%Y Cf. A185636, A086686, A034693.
%K nonn
%O 1,4
%A _Zhi-Wei Sun_, Jan 07 2013
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