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Numbers ((binomial(4*p-1,2*p-1) mod p^5)-3)/p^3, where p = prime(n).
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%I #16 Feb 16 2019 06:45:49

%S 0,8,14,22,68,24,174,330,151,235,179,79,406,1566,1261,2396,3044,3662,

%T 3189,1976,831,4783,3291,915,2692,9389,6846,1263,10937,6296,14083,

%U 1399,14988,8322,12865,1681,13655,5901

%N Numbers ((binomial(4*p-1,2*p-1) mod p^5)-3)/p^3, where p = prime(n).

%C Denote T(p) = binomial(4p-1, 2p-1) mod p^5, where p is the n-th prime. Theorem 30 in the link below states that T(p) = binomial(4p, p) - 1 for p > 5. This is difficult to empirically demonstrate as T(n) = 3, 219, 1753, 7549... <> binomial(4p, p) - 1 (binomial(4p-1, 2p-1) - binomial(4p, p)+1)/p^5 = 27/32, 44/27, 87533/3125, 19681560/16807...not integer.

%C Thus the identity seems to violate both the left and right hand sides of the identity a == b (mod m) if and only if m|(a-b).

%C It is of interest to note however that T(p) mod p = 3 for p > 3 and that T(p) - 3 is divisible by p^3 (this sequence).

%H R. Mestrovic, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv preprint arXiv:1111.3057 [math.NT], 2011.

%p p:= n-> ithprime(n): T:= n-> binomial(4*p(n)-1,2*p(n)-1) mod p(n)^5: seq((T(n)-3)/p(n)^3), n=1..40)

%t a[n_] := Module[{p = Prime[n]}, (Mod[Binomial[4p-1, 2p-1], p^5]-3)/p^3]; Array[a, 40] (* _Jean-François Alcover_, Feb 16 2019 *)

%K nonn

%O 1,2

%A _Gary Detlefs_, Apr 20 2013