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a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with 2^m + prime(m) prime}|, where phi(.) is Euler's totient function.
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%I #10 Jan 23 2014 06:14:42

%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,2,2,1,1,1,2,1,2,3,4,2,4,3,5,2,

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

%U 4,5,4,3,4,5,6,3,6,9,6,9,8,13,5,11,5,6,7,11,4,9,9,5,6,6,11,7,8,9,9,4

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

%C a(n) = 0 for n = 1, ..., 15, 481, 564, 66641, 70965, 72631, .... If a(n) > 0 infinitely often, then there are infinitely many positive integers m with 2^m + prime(m) prime.

%H Zhi-Wei Sun, <a href="/A236344/b236344.txt">Table of n, a(n) for n = 1..10000</a>

%e a(26) = 1 since phi(5)/2 + phi(21)/12 = 2 + 1 = 3 with 2^3 + prime(3) = 8 + 5 = 13 prime.

%e a(5907) = 1 since phi(3944)/2 + phi(5907-3944)/12 = 896 + 150 = 1046 with 2^(1046) + prime(1046) = 2^(1046) + 8353 prime.

%t p[n_]:=IntegerQ[n]&&PrimeQ[2^n+Prime[n]]

%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, A000079, A077375, A236241, A236256.

%K nonn

%O 1,20

%A _Zhi-Wei Sun_, Jan 22 2014