

A234399


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


3



0, 1, 2, 2, 3, 3, 2, 3, 5, 4, 4, 5, 3, 6, 5, 3, 6, 8, 4, 5, 5, 6, 4, 6, 7, 4, 5, 6, 4, 3, 4, 9, 5, 3, 8, 5, 4, 3, 8, 8, 3, 8, 6, 7, 7, 8, 8, 9, 4, 5, 8, 9, 7, 6, 10, 11, 4, 6, 6, 8, 8, 10, 4, 4, 7, 4, 12, 8, 6, 4, 9, 7, 4, 6, 10, 9, 8, 7, 7, 7, 5, 4, 10, 5, 6, 7, 9, 15, 7, 8, 10, 7, 4, 8, 6, 10, 3, 3, 10, 11
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OFFSET

1,3


COMMENTS

Conjecture: a(n) > 0 for all n > 1.
See also the conjecture in A234388.


LINKS



EXAMPLE

a(7) = 2 since 2^1*(2^{phi(6)1) + 1 = 2*3 + 1 = 7 and 2^2*(2^{phi(5)}1) + 1 = 4*15 + 1 = 61 are both prime.


MATHEMATICA

f[n_, k_]:=f[n, k]=2^k*(2^(EulerPhi[nk])1)+1
a[n_]:=Sum[If[PrimeQ[f[n, k]], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A000079, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361, A234388


KEYWORD

nonn


AUTHOR



STATUS

approved



