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A234361
a(n) = |{0 < k < n: 2^{phi(k)/2}*3^{phi(n-k)/4} + 1 is prime}|, where phi(.) is Euler's totient function.
9
0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 3, 2, 3, 4, 2, 6, 3, 6, 6, 5, 7, 4, 6, 4, 5, 7, 9, 4, 6, 4, 10, 7, 2, 11, 9, 12, 6, 9, 10, 9, 12, 11, 10, 6, 12, 13, 8, 11, 9, 10, 7, 8, 7, 11, 8, 9, 6, 14, 4, 15, 5, 14, 7, 15, 5, 12, 11, 9, 10, 9, 10, 8, 10, 7, 12, 11, 15, 10
OFFSET
1,11
COMMENTS
Conjecture: a(n) > 0 for all n > 7.
This implies that there are infinitely many primes of the form 2^k*3^m + 1, where k and m are positive integers.
EXAMPLE
a(10) = 1 since 2^{phi(5)/2}*3^{phi(5)/4} + 1 = 13 is prime.
a(12) = 1 since 2^{phi(4)/2}*3^{phi(8)/4} + 1 = 13 is prime.
a(35) = 2 since 2^{phi(3)/2}*3^{phi(32)/4} + 1 = 2*3^4 + 1 = 163 and 2^{phi(5)/2}*3^{phi(30)/4} + 1 = 2^2*3^2 + 1 = 37 are both prime.
MATHEMATICA
f[n_, k_]:=f[n, k]=2^(EulerPhi[k]/2)*3^(EulerPhi[n-k]/4)+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 24 2013
STATUS
approved