A234344 a(n) = |{0 < k < n: 2^{phi(k)/2} + 3^{phi(n-k)/2} is prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 6, 8, 7, 9, 12, 12, 10, 10, 10, 10, 16, 7, 11, 9, 6, 14, 11, 17, 12, 15, 15, 17, 16, 15, 19, 18, 12, 13, 9, 20, 11, 8, 17, 19, 19, 12, 17, 14, 16, 9, 21, 16, 13, 12, 16, 19, 17, 11, 21, 15, 16, 15, 17, 19, 16, 23, 11, 20, 15

Offset: 1

Author

Zhi-Wei Sun, Dec 23 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 5.

This implies that there are infinitely many primes of the form 2^k + 3^m, where k and m are positive integers.

Examples

			a(6) = 1 since 2^{phi(3)/2} + 3^{phi(3)/2} = 5 is prime.
a(8) = 3 since 2^{phi(3)/2} + 3^{phi(5)/2} = 11, 2^{phi(4)/2} + 3^{phi(4)/2} = 5, and 2^{phi(5)/2} + 3^{phi(3)/2} = 7 are all prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A000079, A000244, A004051, A234309, A234310, A234337, A234346, A234347

Programs

Mathematica

f[n_,k_]:=2^(EulerPhi[k]/2)+3^(EulerPhi[n-k]/2)
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]