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a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m + 1 prime}|, where phi(.) is Euler's totient function.
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%I #6 Jan 23 2014 09:50:17

%S 0,0,0,0,0,0,0,1,1,1,1,0,0,0,1,2,1,0,2,3,1,1,0,2,1,0,2,3,3,3,2,3,2,1,

%T 4,1,4,1,4,5,3,5,7,7,8,5,5,4,4,7,7,4,7,3,6,4,5,5,6,7,6,4,5,7,6,9,5,8,

%U 7,7,4,6,5,4,6,9,8,3,6,8,9,8,8,7,8,8,9,8,4,7,4,7,7,5,4,8,6,6,7,11

%N a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m + 1 prime}|, where phi(.) is Euler's totient function.

%C Conjecture: (i) a(n) > 0 for all n > 26.

%C (ii) For any integer n > 37, there is a positive integer k < n such that m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m - 1 prime.

%C We have verified both parts for n up to 50000. Clearly, part (i) implies that there are infinitely many positive integers m with 2*3^m + 1 prime, while part (ii) implies that there are infinitely many positive integers m with 2*3^m - 1 prime.

%H Zhi-Wei Sun, <a href="/A236358/b236358.txt">Table of n, a(n) for n = 1..10000</a>

%e a(36) = 1 since phi(15)/2 + phi(21)/12 = 4 + 1 = 5 with 2*3^5 + 1 = 487 prime.

%t p[n_]:=IntegerQ[n]&&PrimeQ[2*3^n+1]

%t f[n_,k_]:=EulerPhi[k]/2+EulerPhi[n-k]/12

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

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

%Y Cf. A000010, A000040, A000244, A079363, A111974, A234451, A234503, A234504.

%K nonn

%O 1,16

%A _Zhi-Wei Sun_, Jan 23 2014