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%I #16 Dec 24 2013 01:41:00
%S 0,0,0,1,2,3,4,3,3,5,3,5,6,7,2,6,7,11,7,3,6,8,7,4,11,8,8,6,6,10,7,6,8,
%T 5,6,4,8,4,6,6,6,11,10,3,9,6,6,4,10,6,7,3,4,9,8,9,7,9,5,9,7,9,8,4,6,9,
%U 10,7,8,9,10,5,6,12,5,6,9,10,8,9,7,8,8,10
%N a(n) = |{0 < k < n: 3^k + 3^{phi(n-k)/2} - 1 is prime}|, where phi(.) is Euler's totient function.
%C Conjecture: a(n) > 0 for all n > 3.
%C See also the conjecture in A234337.
%H Zhi-Wei Sun, <a href="/A234347/b234347.txt">Table of n, a(n) for n = 1..3000</a>
%e a(4) = 1 since 3^1 + 3^{phi(3)/2} - 1 = 5 is prime.
%e a(5) = 2 since 3^1 + 3^{phi(4)/2} - 1 = 5 and 3^2 + 3^{phi(3)/2} - 1 are both prime.
%t f[n_,k_]:=3^k+3^(EulerPhi[n-k]/2)-1
%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, A000244, A234309, A234310, A234337, A234344, A234346.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Dec 24 2013
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