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%I #32 Sep 05 2012 16:12:35
%S 16,6,323,1079,924,3044,6252,254,21084,4217,42514,48955,63168,101333,
%T 90896,87970,164396,100099,85982,221337,464837,90637,214936,735552,
%U 171600,330425,437845,311632,363522,1972311,38777,202213,414082,1471674,860550,346186
%N 2^(2p-2) modulo p^3 for p=odd primes.
%C 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.
%H Vincenzo Librandi, <a href="/A216160/b216160.txt">Table of n, a(n) for n = 1..1000</a>
%H F. Morley, <a href="http://www.jstor.org/stable/1967516">Note on the Congruence 2^4n == (-1)^n*(2n)!/(n!)^2 where 2n+1 is a prime</a>, Annals of Mathematics, Vol. 9 (1894 - 1895), pp. 168-170.
%p a:= proc(n) local p; p:= ithprime(n+1);
%p 2 &^ (2*p-2) mod p^3
%p end:
%p seq (a(n), n=1..50); # _Alois P. Heinz_, Sep 05 2012
%t Table[Mod[2^(2Prime[n] - 2), Prime[n]^3], {n, 2, 30}] (* _Alonso del Arte_, Sep 03 2012 *)
%o (PARI) a(n) = { local(p); p = prime(n+1); return (2^(2*p-2) % (p^3));}
%Y Cf. A065091.
%K nonn
%O 1,1
%A _Michel Marcus_, Sep 03 2012
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