OFFSET
1,20
COMMENTS
a(n) = 0 for n = 1, ..., 15, 481, 564, 66641, 70965, 72631, .... If a(n) > 0 infinitely often, then there are infinitely many positive integers m with 2^m + prime(m) prime.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(26) = 1 since phi(5)/2 + phi(21)/12 = 2 + 1 = 3 with 2^3 + prime(3) = 8 + 5 = 13 prime.
a(5907) = 1 since phi(3944)/2 + phi(5907-3944)/12 = 896 + 150 = 1046 with 2^(1046) + prime(1046) = 2^(1046) + 8353 prime.
MATHEMATICA
p[n_]:=IntegerQ[n]&&PrimeQ[2^n+Prime[n]]
f[n_, k_]:=EulerPhi[k]/2+EulerPhi[n-k]/12
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 22 2014
STATUS
approved