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%I #8 Apr 21 2013 14:56:56
%S 25,6,323,1079,924,3044,6252,254,21084,4217,42514,48955,63168,101333,
%T 90896,87970,164396,100099,85982,221337,464837,90637,214936,735552,
%U 171600,330425,437835,311632,363522
%N (-1)^((p-1)/2)*Binomial(p-1,(p-1)/2) mod p^3 where p is the n-th prime.
%C This sequence is related to Morley's Congruence which states that, for prime p>2, (-1)^((p-1)/2)*binomial(p-1,(p-1)/2) == 4^(p-1) (mod p^3).
%C It of interest to note that this congruence can only be illustrated in Maple by using the right hand side of the identity a== b (mod m) iff m|(a-b). Checking for values of n^3 that divide ((-1)^((n-1)/2)*binomial(n-1,(n-1)/2) - 4^(n-1)) produces the sequence of primes. Encoding the left hand side produces this sequence.
%C a(n) == 1 (mod p)
%p p:= n-> ithprime(n): seq((-1)^((p(n)-1)/2)*binomial(p(n)-1,(p(n)-1)/2) mod p(n)^3, n=2..30)
%K nonn
%O 2,1
%A _Gary Detlefs_, Apr 18 2013
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