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A268147 A double binomial sum involving absolute values.

%I A268147 #32 Jul 19 2018 15:53:21
%S A268147 0,16,512,12288,262144,5242880,100663296,1879048192,34359738368,
%T A268147 618475290624,10995116277760,193514046488576,3377699720527872,
%U A268147 58546795155816448,1008806316530991104,17293822569102704640,295147905179352825856,5017514388048998039552
%N A268147 A double binomial sum involving absolute values.
%C A268147 A fast algorithm follows from Theorem 1 of Brent et al. article.
%H A268147 Colin Barker, <a href="/A268147/b268147.txt">Table of n, a(n) for n = 0..800</a>
%H A268147 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 A268147 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (32,-256).
%F A268147 a(n) = Sum_{k=-n..n} (Sum_{l=-n..n} binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k-l)^2).
%F A268147 From _Colin Barker_, Feb 11 2016: (Start)
%F A268147 a(n) = n*16^n.
%F A268147 a(n) = 32*a(n-1)-256*a(n-2) for n>1.
%F A268147 G.f.: 16*x / (1-16*x)^2.
%F A268147 (End)
%p A268147 a:= proc(n) option remember;
%p A268147       16*`if`(n<2, n, n*a(n-1)/(n-1))
%p A268147     end:
%p A268147 seq(a(n), n=0..20);  # _Alois P. Heinz_, Jan 29 2016
%t A268147 Table[n*16^n, {n, 0, 20}] (* _Jean-François Alcover_, Oct 24 2016 *)
%t A268147 LinearRecurrence[{32,-256},{0,16},20] (* _Harvey P. Dale_, Jul 19 2018 *)
%o A268147 (PARI) a(n) = sum(k=-n,n, sum(l=-n,n,binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k-l)^2));
%o A268147 (PARI) concat(0, Vec(16*x/(1-16*x)^2 + O(x^20))) \\ _Colin Barker_, Feb 11 2016
%o A268147 (PARI) a(n)=n*16^n \\ _Charles R Greathouse IV_, May 10 2016
%Y A268147 Cf. A000984, A002894, A166337.
%K A268147 easy,nonn
%O A268147 0,2
%A A268147 _Richard P. Brent_, Jan 27 2016