

A124924


Primes p such that p^2 divides A124923((3p1)/2) = ((3p1)/2)^(3(p1)/2) + 1.


2




OFFSET

1,1


COMMENTS

p divides A124923((3p1)/2) for primes p in A003628. Hence this sequence is a subsequence of A003628.
Also, primes p such that (2)^((p1)/2) == 13p/2 (mod p^2).
No other terms below 10^11.


LINKS

Table of n, a(n) for n=1..4.


EXAMPLE

5 is in this sequence because A124923((3*51)/2) = A124923(7) = 7^8 + 1 = 117650 is divisible by 5^2 = 25.


MATHEMATICA

Do[ p = Prime[n]; m = (3p1)/2; f = PowerMod[ m, m1, p^2 ] + 1; If[ IntegerQ[ f/p^2 ], Print[p] ], {n, 2, 10000} ]


CROSSREFS

Cf. A124923, A003628.
Sequence in context: A214591 A159261 A117077 * A209271 A124878 A085554
Adjacent sequences: A124921 A124922 A124923 * A124925 A124926 A124927


KEYWORD

hard,more,nonn


AUTHOR

Alexander Adamchuk, Nov 12 2006


EXTENSIONS

Edited by Max Alekseyev, Jan 28 2012


STATUS

approved



