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A235919 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, prime(p) - p + 1 and (p^2 - 1)/4 - prime(p) are all prime}|, where phi(.) is Euler's totient function.

%I A235919 #5 Jan 17 2014 03:14:37
%S A235919 0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,3,0,1,2,3,1,0,3,3,2,1,2,1,3,2,4,2,
%T A235919 1,6,2,6,2,3,2,3,6,2,1,7,2,5,4,3,4,3,6,4,5,4,2,1,2,8,2,4,5,5,6,4,5,4,
%U A235919 6,3,3,5,6,5,3,4,8,2,3,7,7,8,5,5,3,3,7,9,3,8,2,4,4,4,9,2,5,8,5,5
%N A235919 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, prime(p) - p + 1 and (p^2 - 1)/4 - prime(p) are all prime}|, where phi(.) is Euler's totient function.
%C A235919 Conjecture: a(n) > 0 for all n > 24.
%C A235919 This implies that there are infinitely many primes p with prime(p) - p + 1 and (p^2 - 1)/4 - prime(p) both prime.
%H A235919 Zhi-Wei Sun, <a href="/A235919/b235919.txt">Table of n, a(n) for n = 1..10000</a>
%e A235919  a(30) = 1 since prime(6) + phi(24)/2 = 13 + 4 = 17, prime(17) - 16 = 59 - 16 = 43 and (17^2 - 1)/4 - prime(17) = 72 - 59 = 13 are all prime.
%e A235919   a(35) = 1 since prime(19) + phi(16)/2 = 67 + 4 = 71, prime(71) - 70 = 353 - 70 = 283 and (71^2 - 1)/4 - prime(71) = 1260 - 353 = 907 are all prime.
%t A235919 PQ[n_]:=PQ[n]=n>0&&PrimeQ[n]
%t A235919 p[n_]:=p[n]=PrimeQ[n]&&PrimeQ[Prime[n]-n+1]&&PQ[(n^2-1)/4-Prime[n]]
%t A235919 f[n_,k_]:=f[n,k]=Prime[k]+EulerPhi[n-k]/2
%t A235919 a[n_]:=a[n]=Sum[If[p[f[n,k]],1,0],{k,1,n-3}]
%t A235919 Table[a[n],{n,1,100}]
%Y A235919 Cf. A000010, A000040, A234695, A235917.
%K A235919 nonn
%O A235919 1,18
%A A235919 _Zhi-Wei Sun_, Jan 17 2014