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A364515 a(n) = (1/2) * (6*n)!*(2*n)!^2/((3*n)!*(4*n)!*n!^3) for n >= 1 with a(0) = 1.

%I A364515 #15 Oct 05 2023 15:20:38
%S A364515 1,10,594,44200,3640210,317678760,28782923400,2677247862432,
%T A364515 253909524661650,24443479528347880,2381221777637046344,
%U A364515 234220654553738055840,23223217841360039079304,2318164676869886054980000,232736024465276636260692000,23482287467017571303962819200
%N A364515 a(n) =  (1/2) * (6*n)!*(2*n)!^2/((3*n)!*(4*n)!*n!^3) for n >= 1 with a(0) = 1.
%C A364515 Row 4 of A364513.
%H A364515 Paolo Xausa, <a href="/A364515/b364515.txt">Table of n, a(n) for n = 0..400</a>
%F A364515 a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(4*n-1, (1 + x)/(1 - x)) for n >= 1.
%F A364515 a(n) = Sum_{k = 0..n} binomial(4*n-1, n-k)^2 * binomial(2*n+k-2, k).
%F A364515 a(n) = (1/2) *  binomial(6*n, 2*n)*binomial(2*n, n)^2 / binomial(3*n, n).
%F A364515 a(n) =  (4*n - 1)!*(3*n - 1/2)!*(n - 1/2)!/((2*n - 1)! * (2*n - 1/2)!^2 * n!^2) for n >= 1 (fractional factorials are defined in terms of the gamma function, for example, (3*n - 1/2)! = gamma(3*n + 1/2)).
%F A364515 a(n) ~ 108^n * sqrt(2)/(4*Pi*n).
%F A364515 P-recursive: a(0) = 1; for n >= 1, a(n) = 12*(6*n-1)*(6*n-5)*(2*n-1)^2 / ((4*n-1)*(4*n-3)*n^2) * a(n-1) with a(1) = 10.
%F A364515 Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.
%e A364515 Examples of supercongruences:
%e A364515 a(7) - a(1) = 2677247862432 - 10 = 2*(7^3)*1999*1952323 == 0 (mod 7^3).
%e A364515 a(2*7) - a(2) = 232736024465276636260692000 - 594 = 2*(3^3)*(7^3)*1097* 11454314520615738059 == 0 (mod 7^3).
%p A364515 seq( (1/2) * (6*n)!*(2*n)!^2/((3*n)!*(4*n)!*n!^3), n = 1..15);
%t A364515 A364515[n_]:=If[n==0,1,(1/2)(6n)!(2n)!^2/((3n)!(4n)!n!^3)];Array[A364515,15,0] (* _Paolo Xausa_, Oct 05 2023 *)
%Y A364515 Cf. A364513.
%K A364515 nonn,easy
%O A364515 0,2
%A A364515 _Peter Bala_, Aug 01 2023