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A double binomial sum involving absolute values.
4

%I #32 Jul 19 2018 15:53:21

%S 0,16,512,12288,262144,5242880,100663296,1879048192,34359738368,

%T 618475290624,10995116277760,193514046488576,3377699720527872,

%U 58546795155816448,1008806316530991104,17293822569102704640,295147905179352825856,5017514388048998039552

%N A double binomial sum involving absolute values.

%C A fast algorithm follows from Theorem 1 of Brent et al. article.

%H Colin Barker, <a href="/A268147/b268147.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_02">Index entries for linear recurrences with constant coefficients</a>, signature (32,-256).

%F 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 From _Colin Barker_, Feb 11 2016: (Start)

%F a(n) = n*16^n.

%F a(n) = 32*a(n-1)-256*a(n-2) for n>1.

%F G.f.: 16*x / (1-16*x)^2.

%F (End)

%p a:= proc(n) option remember;

%p 16*`if`(n<2, n, n*a(n-1)/(n-1))

%p end:

%p seq(a(n), n=0..20); # _Alois P. Heinz_, Jan 29 2016

%t Table[n*16^n, {n, 0, 20}] (* _Jean-François Alcover_, Oct 24 2016 *)

%t LinearRecurrence[{32,-256},{0,16},20] (* _Harvey P. Dale_, Jul 19 2018 *)

%o (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 (PARI) concat(0, Vec(16*x/(1-16*x)^2 + O(x^20))) \\ _Colin Barker_, Feb 11 2016

%o (PARI) a(n)=n*16^n \\ _Charles R Greathouse IV_, May 10 2016

%Y Cf. A000984, A002894, A166337.

%K easy,nonn

%O 0,2

%A _Richard P. Brent_, Jan 27 2016