A236325 a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with m! + prime(m) prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 2, 1, 2, 3, 4, 3, 4, 4, 5, 2, 4, 3, 4, 5, 5, 6, 5, 6, 8, 7, 9, 8, 6, 6, 5, 8, 9, 4, 8, 7, 7, 5, 5, 7, 7, 8, 8, 6, 7, 8, 7, 10, 5, 8, 9, 8, 7, 7, 6, 7, 8, 12, 10, 6, 8, 9, 9, 12, 9, 8, 7, 13

Offset: 1

Author

Zhi-Wei Sun, Jan 22 2014

Keywords

• nonn

Comments

It might seem that a(n) > 0 for all n > 14, but a(7365) = 0. If a(n) > 0 infinitely often, then there are infinitely many positive integers m with m! + prime(m) prime.

Examples

			a(10) = 1 since phi(1)/2 + phi(9)/12 = 1/2 + 6/12 = 1 with 1! + prime(1) = 1 + 2 = 3 prime.
a(23) = 1 since phi(10)/2 + phi(13)/12 = 2 + 1 = 3 with 3! + prime(3) = 6 + 5 = 11 prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..7000

Crossrefs

Cf. A000010, A000040, A000142, A063499, A064278, A064401, A236241, A236256, A236263, A236265.

Programs

Mathematica

p[n_]:=IntegerQ[n]&&PrimeQ[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}]