OFFSET
1,7
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.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..2500
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 24 2013
STATUS
approved