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%I #22 Feb 11 2016 09:24:26
%S 0,8,8832,1228800,79364096,3562536960,129276837888,4079413624832,
%T 116608362086400,3096396542509056,77661255048888320,
%U 1861218099127123968,42980384518787039232,962362945373732864000,20993511648589057622016,447858123072052742062080,9371462498278516088373248
%N A double binomial sum involving absolute values.
%C A fast algorithm follows from Theorem 5 of Brent et al. article.
%H Colin Barker, <a href="/A268152/b268152.txt">Table of n, a(n) for n = 0..800</a>
%H Richard P. Brent, Hideyuki Ohtsuka, Judy-anne H. Osborn, Helmut Prodinger, <a href="http://arxiv.org/abs/1411.1477">Some binomial sums involving absolute values</a>, arXiv:1411.1477v2 [math.CO], 2016.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (80,-2560,40960,-327680,1048576).
%F a(n) = Sum_{k=-n..n} (Sum_{l=-n..n} binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k^2 - l^2)^4).
%F From _Colin Barker_, Feb 11 2016: (Start)
%F a(n) = 4^(2*n-1)*n*(36*n^3-84*n^2+67*n-17).
%F a(n) = 80*a(n-1)-2560*a(n-2)+40960*a(n-3)-327680*a(n-4)+1048576*a(n-5) for n>4.
%F G.f.: 8*x*(1+1024*x+67840*x^2+417792*x^3) / (1-16*x)^5.
%F (End)
%o (PARI) a(n) = sum(k=-n,n,sum(l=-n,n, binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k^2 - l^2)^4));
%o (PARI) concat(0, Vec(8*x*(1+1024*x+67840*x^2+417792*x^3)/(1-16*x)^5 + O(x^20))) \\ _Colin Barker_, Feb 11 2016
%Y Cf. A000984, A002894, A166337, A254408, A268148, A268150.
%K easy,nonn
%O 0,2
%A _Richard P. Brent_, Jan 27 2016
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