OFFSET
1,1
COMMENTS
2^(4*n) == (-1)^n*(2n)!/(n!)^2 (modulo p^3) (with n = (p-1)/2) for odd primes. Except for p = 3 (n = 1), where the second expression = 25 instead of 16.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
F. Morley, Note on the Congruence 2^4n == (-1)^n*(2n)!/(n!)^2 where 2n+1 is a prime, Annals of Mathematics, Vol. 9 (1894 - 1895), pp. 168-170.
MAPLE
a:= proc(n) local p; p:= ithprime(n+1);
2 &^ (2*p-2) mod p^3
end:
seq (a(n), n=1..50); # Alois P. Heinz, Sep 05 2012
MATHEMATICA
Table[Mod[2^(2Prime[n] - 2), Prime[n]^3], {n, 2, 30}] (* Alonso del Arte, Sep 03 2012 *)
Table[PowerMod[2, 2p-2, p^3], {p, Prime[Range[2, 40]]}] (* Harvey P. Dale, Jun 09 2024 *)
PROG
(PARI) a(n) = { local(p); p = prime(n+1); return (2^(2*p-2) % (p^3)); }
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Sep 03 2012
STATUS
approved