A236358 a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m + 1 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, 0, 2, 3, 1, 1, 0, 2, 1, 0, 2, 3, 3, 3, 2, 3, 2, 1, 4, 1, 4, 1, 4, 5, 3, 5, 7, 7, 8, 5, 5, 4, 4, 7, 7, 4, 7, 3, 6, 4, 5, 5, 6, 7, 6, 4, 5, 7, 6, 9, 5, 8, 7, 7, 4, 6, 5, 4, 6, 9, 8, 3, 6, 8, 9, 8, 8, 7, 8, 8, 9, 8, 4, 7, 4, 7, 7, 5, 4, 8, 6, 6, 7, 11

Offset: 1

Author

Zhi-Wei Sun, Jan 23 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 26.

(ii) For any integer n > 37, there is a positive integer k < n such that m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m - 1 prime.

We have verified both parts for n up to 50000. Clearly, part (i) implies that there are infinitely many positive integers m with 2*3^m + 1 prime, while part (ii) implies that there are infinitely many positive integers m with 2*3^m - 1 prime.

Examples

			 a(36) = 1 since phi(15)/2 + phi(21)/12 = 4 + 1 = 5 with 2*3^5 + 1 = 487 prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A000244, A079363, A111974, A234451, A234503, A234504.

Programs

Mathematica

p[n_]:=IntegerQ[n]&&PrimeQ[2*3^n+1]
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}]