

A236968


Number of ordered ways to write n = k + m with k > 0 and m > 0 such that 6*k  1, 6*k + 1 and k + phi(m) are all prime, where phi(.) is Euler's totient function.


3



0, 1, 2, 2, 1, 2, 2, 3, 3, 1, 4, 4, 3, 5, 3, 1, 1, 4, 5, 6, 3, 1, 4, 4, 3, 2, 2, 3, 3, 5, 3, 6, 5, 1, 6, 1, 4, 6, 4, 1, 6, 7, 8, 6, 2, 2, 5, 8, 4, 4, 3, 3, 7, 8, 3, 5, 3, 4, 6, 7, 8, 9, 5, 2, 3, 2, 4, 7, 5, 2, 2, 6, 6, 8, 5, 1, 6, 2, 6, 7, 3, 3, 8, 8, 6, 5, 2, 5, 6, 9, 9, 5, 4, 1, 7, 2, 3, 9, 6, 3
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OFFSET

1,3


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 1. Also, any n > 12 can be written as k + m (k > 0 and m > 2) with 6*k  1, 6*k + 1 and k + phi(m)/2 all prime.
(ii) Each integer n > 34 can be written as p + q (q > 0) with p and p + phi(q) both prime. Also, any integer n > 14 can be written as p + q (q > 2) with p, p + 6 and p + phi(q)/2 all prime.
Clearly, part (i) of the conjecture implies that any integer n > 1 can be written as p + m  phi(m), where p is a prime and m is a positive integer.


LINKS

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


EXAMPLE

a(17) = 1 since 17 = 7 + 10 with 6*7  1 = 41, 6*7 + 1 = 43 and 7 + phi(10) = 7 + 4 = 11 all prime.
a(486) = 1 since 486 = 325 + 161 with 6*325  1 = 1949, 6*325 + 1 = 1951 and 325 + phi(161) = 325 + 132 = 457 all prime.


MATHEMATICA

p[n_, k_]:=PrimeQ[6k1]&&PrimeQ[6k+1]&&PrimeQ[k+EulerPhi[nk]]
a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A001359, A002822, A006512, A182662, A199920, A236531, A236831.
Sequence in context: A057536 A245574 A245573 * A265744 A245588 A014420
Adjacent sequences: A236965 A236966 A236967 * A236969 A236970 A236971


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 02 2014


STATUS

approved



