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A268151 A double binomial sum involving absolute values.

%I A268151 #21 Apr 28 2022 15:48:18
%S A268151 0,40,2816,104448,3014656,76021760,1761607680,38520487936,
%T A268151 807453851648,16389595201536,324355930193920,6289206510878720,
%U A268151 119908340078739456,2254051613498933248,41865462136036130816,769575104325070356480,14019525496019259228160,253384476596474400997376
%N A268151 A double binomial sum involving absolute values.
%C A268151 A fast algorithm follows from Theorem 1 of Brent et al. article.
%H A268151 Colin Barker, <a href="/A268151/b268151.txt">Table of n, a(n) for n = 0..800</a>
%H A268151 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 A268151 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (48,-768,4096).
%F A268151 a(n) = Sum_{k=-n..n} (Sum_{l=-n..n} binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k-l)^4).
%F A268151 From _Colin Barker_, Feb 11 2016: (Start)
%F A268151 a(n) = 2^(4*n-1)*n*(6*n-1).
%F A268151 a(n) = 48*a(n-1)-768*a(n-2)+4096*a(n-3) for n>2.
%F A268151 G.f.: 8*x*(5+112*x) / (1-16*x)^3.
%F A268151 (End)
%t A268151 LinearRecurrence[{48,-768,4096},{0,40,2816},20] (* _Harvey P. Dale_, Apr 28 2022 *)
%o A268151 (PARI) a(n) = sum(k=-n,n, sum(l=-n,n, binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k-l)^4));
%o A268151 (PARI) concat(0, Vec(8*x*(5+112*x)/(1-16*x)^3 + O(x^20))) \\ _Colin Barker_, Feb 11 2016
%Y A268151 Cf. A000984, A002894, A166337, A268147, A268149.
%K A268151 easy,nonn
%O A268151 0,2
%A A268151 _Richard P. Brent_, Jan 27 2016