%I
%S 0,0,1,2,2,3,3,1,3,4,5,2,1,1,3,4,4,3,4,6,5,2,2,6,5,1,2,4,2,2,3,6,5,7,
%T 6,2,3,4,4,2,3,5,1,4,7,4,6,3,9,4,2,5,4,3,9,2,4,3,6,3,5,8,8,5,8,6,2,4,
%U 3,4,1,6,4,3,8,8,6,6,9,11,2,4,2,8,3,4,6,10,5,11,7,8,6,10,4,1,3,1,3,3
%N Number of ways to write n = x + y + z with y <= z such that all the five numbers 6*x1, 6*y1, 6*z1, 6*x*y1 and 6*x*z1 are Sophie Germain primes.
%C Conjecture: a(n) > 0 for all n > 2.
%C This implies that 6*n3 with n > 2 can be expressed as a sum of three Sophie Germain primes (i.e., those primes p with 2*p+1 also prime).
%C We have verified the conjecture for n up to 10^8. Note that any Sophie Germain prime p > 3 has the form 6*k1.
%H ZhiWei Sun, <a href="/A230040/b230040.txt">Table of n, a(n) for n = 1..10000</a>
%H ZhiWei Sun, <a href="http://arxiv.org/abs/1211.1588">Conjectures involving primes and quadratic forms</a>, preprint, arXiv:1211.1588.
%e a(4) = 2, since 4 = 1 + 1 + 2 = 2 + 1 + 1, and 6*11=5 and 6*21=11 are Sophie Germain primes.
%e a(26) = 1, since 26 = 15 + 2 + 9, and all the five numbers 6*151=89, 6*21=11, 6*91=53, 6*15*21=179 and 6*15*9=809 are Sophie Germain primes.
%t SQ[n_]:=PrimeQ[n]&&PrimeQ[2n+1]
%t a[n_]:=Sum[If[SQ[6i1]&&SQ[6j1]&&SQ[6(nij)1]&&SQ[6i*j1]&&SQ[6*i(nij)1],1,0],{i,1,n2},{j,1,(ni)/2}]
%t Table[a[n],{n,1,100}]
%Y Cf. A005384, A219842, A219864, A227938, A229969, A229974, A230037.
%K nonn
%O 1,4
%A _ZhiWei Sun_, Oct 06 2013
