A268151 A double binomial sum involving absolute values.

0, 40, 2816, 104448, 3014656, 76021760, 1761607680, 38520487936, 807453851648, 16389595201536, 324355930193920, 6289206510878720, 119908340078739456, 2254051613498933248, 41865462136036130816, 769575104325070356480, 14019525496019259228160, 253384476596474400997376

Offset: 0

Author

Richard P. Brent, Jan 27 2016

Keywords

• easy
• nonn

Comments

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

Links

• Colin Barker, Table of n, a(n) for n = 0..800
• Richard P. Brent, Hideyuki Ohtsuka, Judy-anne H. Osborn, Helmut Prodinger, Some binomial sums involving absolute values, arXiv:1411.1477v2 [math.CO], 2016.
• Index entries for linear recurrences with constant coefficients, signature (48,-768,4096).

Crossrefs

Cf. A000984, A002894, A166337, A268147, A268149.

Programs

Mathematica

LinearRecurrence[{48,-768,4096},{0,40,2816},20] (* Harvey P. Dale, Apr 28 2022 *)

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));

PARI

concat(0, Vec(8*x*(5+112*x)/(1-16*x)^3 + O(x^20))) \\ Colin Barker, Feb 11 2016

Formula

a(n) = Sum_{k=-n..n} (Sum_{l=-n..n} binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k-l)^4).

From Colin Barker, Feb 11 2016: (Start)

a(n) = 2^(4*n-1)*n*(6*n-1).

a(n) = 48*a(n-1)-768*a(n-2)+4096*a(n-3) for n>2.

G.f.: 8*x*(5+112*x) / (1-16*x)^3.

(End)