A234361 a(n) = |{0 < k < n: 2^{phi(k)/2}*3^{phi(n-k)/4} + 1 is prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 3, 2, 3, 4, 2, 6, 3, 6, 6, 5, 7, 4, 6, 4, 5, 7, 9, 4, 6, 4, 10, 7, 2, 11, 9, 12, 6, 9, 10, 9, 12, 11, 10, 6, 12, 13, 8, 11, 9, 10, 7, 8, 7, 11, 8, 9, 6, 14, 4, 15, 5, 14, 7, 15, 5, 12, 11, 9, 10, 9, 10, 8, 10, 7, 12, 11, 15, 10

Offset: 1

Author

Zhi-Wei Sun, Dec 24 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 7.

This implies that there are infinitely many primes of the form 2^k*3^m + 1, where k and m are positive integers.

Examples

			a(10) = 1 since 2^{phi(5)/2}*3^{phi(5)/4} + 1 = 13 is prime.
a(12) = 1 since 2^{phi(4)/2}*3^{phi(8)/4} + 1 = 13 is prime.
a(35) = 2 since 2^{phi(3)/2}*3^{phi(32)/4} + 1 = 2*3^4 + 1 = 163 and 2^{phi(5)/2}*3^{phi(30)/4} + 1 = 2^2*3^2 + 1 = 37 are both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000010, A000040, A000079, A000244, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360.

Programs

Mathematica

f[n_,k_]:=f[n,k]=2^(EulerPhi[k]/2)*3^(EulerPhi[n-k]/4)+1
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]