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A235917 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) are all prime}|, where phi(.) is Euler's totient function.

%I A235917 #6 Jan 17 2014 03:14:10
%S A235917 0,0,0,0,0,0,0,1,0,1,1,2,2,0,3,1,4,2,4,3,1,2,2,3,3,1,2,2,1,1,2,4,1,5,
%T A235917 2,2,3,2,6,2,1,3,3,2,4,5,4,2,5,3,4,2,3,4,4,3,3,2,1,4,3,2,3,4,5,7,3,5,
%U A235917 1,6,1,7,3,6,5,3,5,2,3,4,5,3,8,6,4,2,6,4,8,3,7,5,6,6,4,3,5,6,4,3
%N A235917 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) are all prime}|, where phi(.) is Euler's totient function.
%C A235917 Conjecture: (i) a(n) > 0 for all n > 14.
%C A235917 (ii) For any integer n > 12, there is a positive integer k < n such that p = prime(k) + phi(n-k), (p^2 - 1)/2 - prime(p) and (p^2 - 1)/4 - prime(p) are all prime.
%C A235917 Clearly, part (i) of the conjecture implies that there are infinitely many primes p with p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) both prime.
%H A235917 Zhi-Wei Sun, <a href="/A235917/b235917.txt">Table of n, a(n) for n = 1..10000</a>
%e A235917 a(10) = 1 since prime(5) + phi(5)/2 = 11 + 2 = 13, 13^2 - 1 - prime(13) = 168 - 41 = 127 and (13^2 - 1)/2 - prime(13) = 84 - 41 = 43 are all prime.
%e A235917 a(71) = 1 since prime(19) + phi(52)/2 = 67 + 12 = 79, 79^2 - 1 - prime(79) = 6240 - 401 = 5839 and (79^2 - 1)/2 - prime(79) = 3120 - 401 = 2719 are all prime.
%t A235917 PQ[n_]:=n>0&&PrimeQ[n]
%t A235917 p[n_]:=PrimeQ[n]&&PQ[n^2-1-Prime[n]]&&PQ[(n^2-1)/2-Prime[n]]
%t A235917 f[n_,k_]:=Prime[k]+EulerPhi[n-k]/2
%t A235917 a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-3}]
%t A235917 Table[a[n],{n,1,100}]
%Y A235917 Cf. A000010, A000040, A234694, A235912.
%K A235917 nonn
%O A235917 1,12
%A A235917 _Zhi-Wei Sun_, Jan 17 2014