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a(n) = |{0 < k < n: 6*k - 1, 6*k + 1, 6*k + 5 and prime(n-k) + 6 are all prime}|.
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%I #6 Jan 28 2014 15:26:17

%S 0,0,0,1,2,3,3,3,2,3,2,3,3,4,2,3,2,3,3,4,4,5,4,4,3,4,3,5,5,5,2,2,4,3,

%T 4,3,3,4,5,3,5,4,6,6,6,4,4,1,3,6,5,2,4,4,5,6,8,6,5,4,8,4,4,5,7,6,4,4,

%U 4,6,6,5,5,6,6,7,6,4,3,5,7,3,4,4,3,6,5,5,6,6,7,5,4,3,5,5,3,3,2,2

%N a(n) = |{0 < k < n: 6*k - 1, 6*k + 1, 6*k + 5 and prime(n-k) + 6 are all prime}|.

%C Conjecture: (i) a(n) > 0 for all n > 3.

%C (ii) For any integer n > 4, there is a positive integer k < n such that 6*k - 5, 6*k - 1, 6*k + 1 and prime(n-k) + 6 are all prime.

%C (iii) Any integer n > 7 can be written as p + q with q > 0 such that p, p + 6 and prime(q) + 6 are all prime.

%C (iv) Each integer n > 4 can be written as k*(k+1) + m with k > 0 and m > 0 such that prime(m) + 6 is prime.

%H Zhi-Wei Sun, <a href="/A236552/b236552.txt">Table of n, a(n) for n = 1..10000</a>

%e a(4) = 1 since 6*1 - 1, 6*1 + 1, 6*1 + 5 and prime(4-1) + 6 = 11 are all prime.

%e a(48) = 1 since 6*32 - 1 = 191, 6*32 + 1 = 193, 6*32 + 5 = 197 and prime(48-32) + 6 = 53 + 6 = 59 are all prime.

%t p[n_]:=PrimeQ[6n-1]&&PrimeQ[6n+1]&&PrimeQ[6n+5]

%t q[n_]:=PrimeQ[Prime[n]+6]

%t a[n_]:=Sum[If[p[k]&&q[n-k],1,0],{k,1,n-1}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000040, A022004, A022005, A023201, A046117, A236462, A236464.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Jan 28 2014