

A236389


a(n) = {0 < k < n: m = phi(k)/2 + phi(nk)/12 is an integer with 2^m*p(m) + 1 prime}, where p(.) is the partition function (A000041).


2



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

1,38


COMMENTS

Conjecture: a(n) > 0 for all n > 56.
We have verified this for n up to 33000.
The conjecture implies that there are infinitely many positive integers m with 2^m*p(m) + 1 prime. See A236390 for a list of such numbers m.


LINKS

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


EXAMPLE

a(10) = 1 since phi(1)/2 + phi(9)/12 = 1/2 + 6/12 = 1 with 2^1*p(1) + 1 = 2 + 1 = 3 prime.
a(30) = 1 since phi(17)/2 + phi(13)/12 = 8 + 1 = 9 with 2^9*p(9) + 1 = 512*30 + 1 = 15361 prime.
a(8261) = 1 since phi(395)/2 + phi(8261395)/12 = 156 + 198 = 354 with 2^(354)*p(354) + 1 = 2^(354)*363117512048110005 + 1 prime.


MATHEMATICA

q[n_]:=IntegerQ[n]&&PrimeQ[2^n*PartitionsP[n]+1]
f[n_, k_]:=EulerPhi[k]/2+EulerPhi[nk]/12
a[n_]:=Sum[If[q[f[n, k]], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A000041, A000079, A236390.
Sequence in context: A159459 A304095 A276675 * A143078 A106405 A228601
Adjacent sequences: A236386 A236387 A236388 * A236390 A236391 A236392


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 24 2014


STATUS

approved



