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%I #15 Jul 10 2021 07:04:50
%S 1,40,2008,109120,6173656,357903040,21090174400,1257411781120,
%T 75630327895000,4580277582101440,278915640538355008,
%U 17061127317021130240,1047543937631077672384,64523332938885758410240,3985152917145136901283328,246717298245058901071237120
%N a(n) = Sum_{k=0..n} binomial(2k,k)^3 * binomial(2n-2k,n-k) * 2^(4*(n-k)).
%H Seiichi Manyama, <a href="/A300116/b300116.txt">Table of n, a(n) for n = 0..554</a>
%H Wadim Zudilin, <a href="https://arxiv.org/abs/0712.1332v2">Ramanujan-type formulae for 1/π: A second wind?</a>, arXiv:0712.1332v2 [math.NT], 2008.
%F n^3 * a(n) = 8 * (2*n-1) * (8*n^2-8*n+5) * a(n-1) - 4096 * (n-1)^3 * a(n-2) for n > 1.
%F a(n) ~ Gamma(1/4)^4 * 2^(6*n - 2) / (Pi^(7/2) * sqrt(n)). - _Vaclav Kotesovec_, Jul 10 2021
%o (PARI) {a(n) = sum(k=0, n, binomial(2*k, k)^3*binomial(2*n-2*k, n-k)*2^(4*(n-k)))}
%Y Cf. A036917.
%K nonn
%O 0,2
%A _Seiichi Manyama_, Feb 25 2018
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