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A237130
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Number of ordered ways to write n = k + m with k > 0 and m > 0 such that both {3*k - 1, 3*k + 1} and {phi(m) - 1, phi(m) + 1} are twin prime pairs, where phi(.) is Euler's totient function.
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7
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0, 0, 0, 0, 0, 0, 1, 0, 2, 1, 3, 2, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 4, 3, 3, 1, 3, 4, 4, 3, 3, 5, 5, 3, 2, 2, 3, 4, 2, 3, 5, 5, 3, 4, 4, 5, 3, 5, 2, 3, 4, 4, 4, 2, 6, 4, 3, 4, 3, 5, 1, 5, 5, 5, 4, 2, 5, 4, 4, 2, 4, 6, 5, 6, 3, 5, 5, 6, 5, 1, 5, 3, 5, 3, 6, 4, 5, 7, 3, 5, 3, 5, 5, 3, 7, 3, 9, 4, 6, 5
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OFFSET
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1,9
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 8.
(ii) Any integer n > 6 can be written as k + m with k > 0 and m > 0 such that both {prime(k), prime(k) + 2} and {phi(m) - 1, phi(m) + 1} are twin prime pairs.
(iii) Each n = 12, 13, ... can be written as p + q (q > 0) with p, p + 6, phi(q) - 1 and phi(q) + 1 all prime.
(iv) If n > 2 is neither 10 nor 430, then n can be written as k + m with k > 0 and m > 0 such that both {3k - 1, 3*k + 1} and {6*m - 1, 6*m + 1} are twin prime pairs.
Note that each part of the above conjecture implies the twin prime conjecture.
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LINKS
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EXAMPLE
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a(7) = 1 since 7 = 2 + 5 with 3*2 - 1 = 5, 3*2 + 1 =7, phi(5) - 1 = 3 and phi(5) + 1 = 5 all prime.
a(140) = 1 since 140 = 104 + 36 with 3*104 - 1 = 311, 3*104 + 1 = 313, phi(36) - 1 = 11 and phi(36) + 1 = 13 all prime.
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MATHEMATICA
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PQ[n_]:=PrimeQ[EulerPhi[n]-1]&&PrimeQ[EulerPhi[n]+1]
a[n_]:=Sum[If[PrimeQ[3k-1]&&PrimeQ[3k+1]&&PQ[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, A001359, A006512, A014574, A072281, A182662, A236531, A236566, A236831, A236968, A237127.
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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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