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%I #10 Dec 24 2013 12:42:12
%S 0,0,0,0,0,1,2,1,2,4,2,4,4,3,4,3,6,5,4,6,7,8,6,7,11,7,10,9,9,7,10,11,
%T 8,7,11,10,9,6,11,15,4,14,5,14,11,13,9,13,6,12,10,12,11,10,10,13,9,7,
%U 11,7,11,4,11,9,10,6,11,8,4,10,12,13,9,7,9,6,12,10
%N a(n) = |{2 < k < n-2: 5^{phi(k)} + 5^{phi(n-k)/2} - 1 is prime}|, where phi(.) is Euler's totient function.
%C Conjecture: For any integer a > 1, there is a positive integer N(a) such that if n > N(a) then a^{phi(k)} + a^{phi(n-k)/2} - 1 is prime for some 2 < k < n-2. Moreover, we may take N(2) = N(3) = ... = N(6) = N(8) = 5 and N(7) = 17.
%C Clearly, this conjecture implies that for each a = 2, 3, ... there are infinitely many primes of the form a^{2*k} + a^m - 1, where k and m are positive integers.
%H Zhi-Wei Sun, <a href="/A234359/b234359.txt">Table of n, a(n) for n = 1..2500</a>
%e a(6) = 1 since 5^{phi(3)} + 5^{phi(3)/2} - 1 = 29 is prime.
%e a(11) = 2 since 5^{phi(4)} + 5^{phi(7)/2} - 1 = 149 and 5^{phi(7)} + 5^{phi(4)/2} - 1 = 15629 are both prime.
%t f[n_,k_]:=5^(EulerPhi[k])+5^(EulerPhi[n-k]/2)-1
%t a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,3,n-3}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000010, A000040, A000351, A234309, A234310, A234337, A234344, A234346, A234347
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, Dec 24 2013
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