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A234504
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Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k) + phi(m)/4) - 5 is prime, where phi(.) is Euler's totient function.
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4
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0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 2, 1, 1, 3, 2, 3, 2, 3, 2, 3, 4, 5, 5, 4, 5, 6, 7, 4, 5, 6, 7, 6, 5, 7, 8, 5, 7, 9, 8, 8, 6, 8, 7, 10, 7, 10, 10, 9, 9, 8, 9, 10, 5, 10, 10, 9, 10, 10, 9, 10, 9, 7, 12, 14, 10, 9, 5, 11, 7, 13, 8, 13, 6, 9, 11, 11, 14, 15, 9, 13
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OFFSET
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1,11
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COMMENTS
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Conjecture: a(n) > 0 for all n > 10.
We have verified this for n up to 50000. The conjecture implies that there are infinitely many primes of the form 2^n - 5.
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LINKS
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EXAMPLE
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a(15) = 2 since 2^(phi(2) + phi(13)/4) - 5 = 2^4 - 5 = 11 and 2^(phi(3) + phi(12)/4) - 5 = 2^3 - 5 = 3 are both prime.
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MATHEMATICA
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f[n_, k_]:=2^(EulerPhi[k]+EulerPhi[n-k]/4)-5
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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Cf. A000010, A000040, A000079, A050522, A156560, A234309, A234451, A234470, A234475, A234503, A236358.
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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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