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a(n) = |{0 < k < n - 2: 4^k + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.
14

%I #16 Dec 25 2013 02:16:55

%S 0,0,0,1,2,3,3,3,5,5,5,7,7,8,8,7,6,8,6,10,8,5,6,7,10,7,6,10,9,6,7,8,

%T 12,5,9,4,9,4,6,3,8,8,11,10,9,7,7,13,12,6,7,8,6,6,13,10,8,9,9,12,6,11,

%U 14,9,5,11,7,7,10,11,7,9,10,5,9,8,8,13,7,13

%N a(n) = |{0 < k < n - 2: 4^k + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.

%C Conjecture: Let a be 2 or 3 or 4. If n > 3, then a^k + a^{phi(n-k)/2} - 1 is prime for some 0 < k < n - 2.

%C This conjecture for a = 4 implies that there are infinitely many terms of the sequence A234310. The conjecture for a = 3 implies that there are infinitely many primes of the form 3^k + 3^m - 1 (cf. A234346), where k and m are positive integers.

%H Zhi-Wei Sun, <a href="/A234337/b234337.txt">Table of n, a(n) for n = 1..4000</a>

%e a(4) = 1 since 4^1 + 2^{phi(3)} - 1 = 7 is prime.

%e a(5) = 2 since 4^1 + 2^{phi(4)} - 1 = 7 and 4^2 + 2^{phi(3)} - 1 = 19 are both prime.

%t f[n_,k_]:=4^k+2^(EulerPhi[n-k])-1

%t a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-3}]

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

%Y Cf. A000010, A000040, A000244, A000302, A234309, A234310, A234344, A234346, A234347, A234359, A234360, A234361

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Dec 23 2013