OFFSET
1,14
COMMENTS
It seems that a(n) > 0 for all n > 17. (We have verified this for n up to 13000.) If a(n) > 0 infinitely often, then there are infinitely many positive integers m with m! + prime(m) prime.
See also A236265 for a similar sequence.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..8100
EXAMPLE
a(18) = 1 since phi(3)/2 + phi(15)/8 = 1 + 1 = 2 with 2! + prime(2) = 2 + 3 = 5 prime.
a(356) = 1 since phi(203)/2 + phi(153)/8 = 84 + 12 = 96 with 96! + prime(96) = 96! + 503 prime.
a(457) = 1 since phi(7)/2 + phi(450)/8 = 3 + 15 = 18 with 18! + prime(18) = 18! + 61 = 6402373705728061 prime.
MATHEMATICA
q[n_]:=IntegerQ[n]&&PrimeQ[n!+Prime[n]]
f[n_, k_]:=EulerPhi[k]/2+EulerPhi[n-k]/8
a[n_]:=Sum[If[q[f[n, k]], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 21 2014
STATUS
approved