cp's OEIS Frontend

A236508 a(n) = |{0 < k < n-2: p = 2*phi(k) + phi(n-k)/2 - 1, p + 2, p + 6 and prime(p) + 6 are all prime}|, where phi(.) is Euler's totient function.

%I A236508 #8 Jan 19 2019 04:14:59
%S A236508 0,0,0,0,0,0,0,1,1,0,2,1,1,2,1,3,2,2,0,2,3,1,2,1,3,3,2,2,1,1,1,3,0,2,
%T A236508 3,2,1,3,0,2,0,1,1,1,1,2,0,0,0,0,2,2,0,1,0,0,0,0,0,0,0,0,1,1,0,1,1,2,
%U A236508 1,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1
%N A236508 a(n) = |{0 < k < n-2: p = 2*phi(k) + phi(n-k)/2 - 1, p + 2, p + 6 and prime(p) + 6 are all prime}|, where phi(.) is Euler's totient function.
%C A236508 Conjecture: a(n) > 0 for all n > 146.
%C A236508 We have verified this for n up to 52000.
%C A236508 The conjecture implies that there are infinitely many prime triples {p, p + 2, p + 6} with {prime(p), prime(p) + 6} a sexy prime pair. See A236509 for such primes p.
%H A236508 Zhi-Wei Sun, <a href="/A236508/b236508.txt">Table of n, a(n) for n = 1..10000</a>
%e A236508 a(13) = 1 since 2*phi(3) + phi(10)/2 - 1 = 5, 5 + 2 = 7, 5 + 6 = 11 and prime(5) + 6 = 11 + 6 = 17 are all prime.
%e A236508 a(244) = 1 since 2*phi(153) + phi(244-153)/2 - 1 = 2*96 + 72/2 - 1 = 227, 227 + 2 = 229, 227 + 6 = 233 and prime(227) + 6 = 1433 + 6 = 1439 are all prime.
%t A236508 p[n_]:=PrimeQ[n]&&PrimeQ[n+2]&&PrimeQ[n+6]&&PrimeQ[Prime[n]+6]
%t A236508 f[n_,k_]:=2*EulerPhi[k]+EulerPhi[n-k]/2-1
%t A236508 a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-3}]
%t A236508 Table[a[n],{n,1,100}]
%Y A236508 Cf. A000010, A000040, A022004, A023201, A046117, A236464, A236472, A236509.
%K A236508 nonn
%O A236508 1,11
%A A236508 _Zhi-Wei Sun_, Jan 27 2014