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A234344
a(n) = |{0 < k < n: 2^{phi(k)/2} + 3^{phi(n-k)/2} is prime}|, where phi(.) is Euler's totient function.
14
0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 6, 8, 7, 9, 12, 12, 10, 10, 10, 10, 16, 7, 11, 9, 6, 14, 11, 17, 12, 15, 15, 17, 16, 15, 19, 18, 12, 13, 9, 20, 11, 8, 17, 19, 19, 12, 17, 14, 16, 9, 21, 16, 13, 12, 16, 19, 17, 11, 21, 15, 16, 15, 17, 19, 16, 23, 11, 20, 15
OFFSET
1,7
COMMENTS
Conjecture: a(n) > 0 for all n > 5.
This implies that there are infinitely many primes of the form 2^k + 3^m, where k and m are positive integers.
EXAMPLE
a(6) = 1 since 2^{phi(3)/2} + 3^{phi(3)/2} = 5 is prime.
a(8) = 3 since 2^{phi(3)/2} + 3^{phi(5)/2} = 11, 2^{phi(4)/2} + 3^{phi(4)/2} = 5, and 2^{phi(5)/2} + 3^{phi(3)/2} = 7 are all prime.
MATHEMATICA
f[n_, k_]:=2^(EulerPhi[k]/2)+3^(EulerPhi[n-k]/2)
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 23 2013
STATUS
approved