A236344 a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with 2^m + prime(m) prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 3, 4, 2, 4, 3, 5, 2, 3, 4, 3, 3, 4, 6, 5, 6, 6, 7, 7, 5, 4, 6, 6, 5, 7, 5, 3, 3, 3, 7, 4, 5, 5, 8, 4, 6, 5, 5, 5, 6, 4, 5, 4, 5, 4, 3, 4, 5, 6, 3, 6, 9, 6, 9, 8, 13, 5, 11, 5, 6, 7, 11, 4, 9, 9, 5, 6, 6, 11, 7, 8, 9, 9, 4

Offset: 1

Author

Zhi-Wei Sun, Jan 22 2014

Keywords

• nonn

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.

Examples

			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.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000010, A000040, A000079, A077375, A236241, A236256.

Programs

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}]