A234504 Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k) + phi(m)/4) - 5 is prime, where phi(.) is Euler's totient function.

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

Offset: 1

Author

Zhi-Wei Sun, Dec 26 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 10.

We have verified this for n up to 50000. The conjecture implies that there are infinitely many primes of the form 2^n - 5.

Examples

			a(15) = 2 since 2^(phi(2) + phi(13)/4) - 5 = 2^4 - 5 = 11 and 2^(phi(3) + phi(12)/4) - 5 = 2^3 - 5 = 3 are both prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A000079, A050522, A156560, A234309, A234451, A234470, A234475, A234503, A236358.

Programs

Mathematica

f[n_,k_]:=2^(EulerPhi[k]+EulerPhi[n-k]/4)-5
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]