

A234359


a(n) = {2 < k < n2: 5^{phi(k)} + 5^{phi(nk)/2}  1 is prime}, where phi(.) is Euler's totient function.


10



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
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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(nk)/2}  1 is prime for some 2 < k < n2. 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

ZhiWei 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[nk]/2)1
a[n_]:=Sum[If[PrimeQ[f[n, k]], 1, 0], {k, 3, n3}]
Table[a[n], {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A000351, A234309, A234310, A234337, A234344, A234346, A234347
Sequence in context: A121339 A307236 A324382 * A099500 A120253 A307018
Adjacent sequences: A234356 A234357 A234358 * A234360 A234361 A234362


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Dec 24 2013


STATUS

approved



