A234359 a(n) = |{2 < k < n-2: 5^{phi(k)} + 5^{phi(n-k)/2} - 1 is prime}|, where phi(.) is Euler's totient function.

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

Offset: 1

Author

Zhi-Wei Sun, Dec 24 2013

Keywords

• nonn

Comments

Conjecture: For any integer a > 1, there is a positive integer N(a) such that if n > N(a) then a^{phi(k)} + a^{phi(n-k)/2} - 1 is prime for some 2 < k < n-2. Moreover, we may take N(2) = N(3) = ... = N(6) = N(8) = 5 and N(7) = 17.

Clearly, this conjecture implies that for each a = 2, 3, ... there are infinitely many primes of the form a^{2*k} + a^m - 1, where k and m are positive integers.

Examples

			a(6) = 1 since 5^{phi(3)} + 5^{phi(3)/2} - 1 = 29 is prime.
a(11) = 2 since 5^{phi(4)} + 5^{phi(7)/2} - 1 = 149 and 5^{phi(7)} + 5^{phi(4)/2} - 1 = 15629 are both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..2500

Crossrefs

Cf. A000010, A000040, A000351, A234309, A234310, A234337, A234344, A234346, A234347

Programs

Mathematica

f[n_,k_]:=5^(EulerPhi[k])+5^(EulerPhi[n-k]/2)-1
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,3,n-3}]
Table[a[n],{n,1,100}]