OFFSET
1,22
COMMENTS
It seems that a(n) > 0 for all n > 21. If a(n) > 0 infinitely often, then there are infinitely many positive integers m with m! - prime(m) prime.
See also A236263 for a similar sequence.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..7000
EXAMPLE
a(23) = 1 since phi(7)/2 + phi(16)/8 = 3 + 1 = 4 with 4! - prime(4) = 24 - 7 = 17 prime.
a(26) = 1 since phi(9)/2 + phi(17)/8 = 3 + 2 = 5 with 5! - prime(5) = 120 - 11 = 109 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