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A234337
a(n) = |{0 < k < n - 2: 4^k + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.
14
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, 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, 14, 9, 5, 11, 7, 7, 10, 11, 7, 9, 10, 5, 9, 8, 8, 13, 7, 13
OFFSET
1,5
COMMENTS
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.
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.
EXAMPLE
a(4) = 1 since 4^1 + 2^{phi(3)} - 1 = 7 is prime.
a(5) = 2 since 4^1 + 2^{phi(4)} - 1 = 7 and 4^2 + 2^{phi(3)} - 1 = 19 are both prime.
MATHEMATICA
f[n_, k_]:=4^k+2^(EulerPhi[n-k])-1
a[n_]:=Sum[If[PrimeQ[f[n, k]], 1, 0], {k, 1, n-3}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 23 2013
STATUS
approved