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A233566
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a(n) = |{0 < p < n: p and p*phi(n-p) - 1 are both prime}|, where phi(.) is Euler's totient function (A000010).
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7
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0, 0, 0, 1, 2, 2, 2, 2, 2, 4, 3, 3, 4, 4, 3, 3, 2, 2, 4, 3, 3, 5, 5, 4, 5, 3, 2, 6, 2, 4, 2, 7, 7, 8, 5, 4, 8, 4, 4, 8, 5, 5, 8, 4, 4, 5, 6, 5, 5, 10, 7, 8, 4, 4, 5, 6, 8, 7, 4, 6, 6, 9, 11, 7, 10, 4, 6, 7, 8, 10, 4, 7, 6, 5, 5, 12, 8, 8, 7, 11, 13, 11, 12, 5, 8, 7, 11, 9, 5, 8, 5, 6, 12, 8, 8, 5, 9, 5, 11, 12
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OFFSET
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1,5
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COMMENTS
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Conjecture: a(n) > 0 for all n > 3. Also, for any n > 2 there is a prime p < n with p^2*phi(n-p) - 1 prime.
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LINKS
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EXAMPLE
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a(4) = 1 since 3 and 3*phi(4-3) - 1 = 2 are both prime.
a(5) = 2 since 2 and 2*phi(5-2) - 1 = 3 are both prime, and also 3 and 3*phi(5-3) - 1 = = 2 are both prime.
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MATHEMATICA
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a[n_]:=Sum[If[PrimeQ[Prime[k]*EulerPhi[n-Prime[k]]-1], 1, 0], {k, 1, PrimePi[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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