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A234344
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a(n) = |{0 < k < n: 2^{phi(k)/2} + 3^{phi(n-k)/2} is prime}|, where phi(.) is Euler's totient function.
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14
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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
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OFFSET
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1,7
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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