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A234399
a(n) = |{0 < k < n: 2^k*(2^{phi(n-k)} - 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
OFFSET
1,3
COMMENTS
Conjecture: a(n) > 0 for all n > 1.
See also the conjecture in A234388.
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[n-k])-1)+1
a[n_]:=Sum[If[PrimeQ[f[n, k]], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 25 2013
STATUS
approved