

A237130


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.


7



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

1,9


COMMENTS

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.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000


EXAMPLE

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.


MATHEMATICA

PQ[n_]:=PrimeQ[EulerPhi[n]1]&&PrimeQ[EulerPhi[n]+1]
a[n_]:=Sum[If[PrimeQ[3k1]&&PrimeQ[3k+1]&&PQ[nk], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A001359, A006512, A014574, A072281, A182662, A236531, A236566, A236831, A236968, A237127.
Sequence in context: A327983 A205784 A066272 * A058773 A122805 A103981
Adjacent sequences: A237127 A237128 A237129 * A237131 A237132 A237133


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 04 2014


STATUS

approved



