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a(n) = |{0 < k < n: 2^{phi(k)/2} + 3^{phi(n-k)/2} is prime}|, where phi(.) is Euler's totient function.
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%I #10 Dec 24 2013 01:44:37

%S 0,0,0,0,0,1,2,3,4,5,6,7,8,6,8,7,9,12,12,10,10,10,10,16,7,11,9,6,14,

%T 11,17,12,15,15,17,16,15,19,18,12,13,9,20,11,8,17,19,19,12,17,14,16,9,

%U 21,16,13,12,16,19,17,11,21,15,16,15,17,19,16,23,11,20,15

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

%C Conjecture: a(n) > 0 for all n > 5.

%C This implies that there are infinitely many primes of the form 2^k + 3^m, where k and m are positive integers.

%H Zhi-Wei Sun, <a href="/A234344/b234344.txt">Table of n, a(n) for n = 1..7000</a>

%e a(6) = 1 since 2^{phi(3)/2} + 3^{phi(3)/2} = 5 is prime.

%e a(8) = 3 since 2^{phi(3)/2} + 3^{phi(5)/2} = 11, 2^{phi(4)/2} + 3^{phi(4)/2} = 5, and 2^{phi(5)/2} + 3^{phi(3)/2} = 7 are all prime.

%t f[n_,k_]:=2^(EulerPhi[k]/2)+3^(EulerPhi[n-k]/2)

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

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

%Y Cf. A000010, A000040, A000079, A000244, A004051, A234309, A234310, A234337, A234346, A234347

%K nonn

%O 1,7

%A _Zhi-Wei Sun_, Dec 23 2013