|
%I #27 Feb 11 2016 09:25:11
%S 0,8,768,30720,917504,23592960,553648128,12213813248,257698037760,
%T 5257039970304,104453604638720,2031897488130048,38843546786070528,
%U 731834939447705600,13618885273168379904,250760427251989217280,4574792530279968800768,82788987402808467652608
%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="/A268148/b268148.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_03">Index entries for linear recurrences with constant coefficients</a>, signature (48,-768,4096).
%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)^2).
%F From _Colin Barker_, Feb 11 2016: (Start)
%F a(n) = 2^(4*n-1)*n*(2*n-1).
%F a(n) = 48*a(n-1)-768*a(n-2)+4096*a(n-3) for n>2.
%F G.f.: 8*x*(1+48*x) / (1-16*x)^3.
%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)^2));
%o (PARI) concat(0, Vec(8*x*(1+48*x)/(1-16*x)^3 + O(x^20))) \\ _Colin Barker_, Feb 11 2016
%Y Cf. A000984, A002894, A166337, A254408.
%K easy,nonn
%O 0,2
%A _Richard P. Brent_, Jan 27 2016
|
