cp's OEIS Frontend

A236511 a(n) = |{0 < k < n: p = 3*phi(k) + phi(n-k) - 1, p + 2, p + 6 and p + 8 are all prime}|, where phi(.) is Euler's totient function.

%I A236511 #8 Jan 27 2014 09:47:42
%S A236511 0,0,0,0,0,0,0,0,0,1,1,1,2,0,1,0,1,1,0,1,1,1,0,1,0,0,0,0,0,0,0,0,0,0,
%T A236511 0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,1,1,1,2,0,0,0,0,1,1,2,1,
%U A236511 1,1,1,0,2,2,2,2,2,2,0,2,0,4,4,2,1,3,4,2,2,3,0,1,3,2,3,1,4,4,3,1
%N A236511 a(n) = |{0 < k < n: p = 3*phi(k) + phi(n-k) - 1, p + 2, p + 6 and p + 8 are all prime}|, where phi(.) is Euler's totient function.
%C A236511 Conjecture: a(n) > 0 for all n > 1075.
%C A236511 We have verified this for n up to 50000.
%C A236511 The above conjecture implies the well-known conjecture that there are infinitely many prime quadruplets (p, p + 2, p + 6, p + 8).
%H A236511 Zhi-Wei Sun, <a href="/A236511/b236511.txt">Table of n, a(n) for n = 1..10000</a>
%e A236511 a(10) = 1 since 3*phi(3) + phi(7) - 1 = 6 + 6 - 1 = 11, 11 + 2 = 13, 11 + 6 = 17 and 11 + 8 = 19 are all prime.
%e A236511 a(57) = 1 since 3*phi(31) + phi(26) - 1 = 90 + 12 - 1 = 101, 101 + 2 = 103, 101 + 6 = 107 and 101 + 8 = 109 are all prime.
%t A236511 p[n_]:=PrimeQ[n]&&PrimeQ[n+2]&&PrimeQ[n+6]&&PrimeQ[n+8]
%t A236511 f[n_,k_]:=3*EulerPhi[k]+EulerPhi[n-k]-1
%t A236511 a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-1}]
%t A236511 Table[a[n],{n,1,100}]
%Y A236511 Cf. A000010, A000040, A007530, A236508.
%K A236511 nonn
%O A236511 1,13
%A A236511 _Zhi-Wei Sun_, Jan 27 2014