A268152 A double binomial sum involving absolute values.

0, 8, 8832, 1228800, 79364096, 3562536960, 129276837888, 4079413624832, 116608362086400, 3096396542509056, 77661255048888320, 1861218099127123968, 42980384518787039232, 962362945373732864000, 20993511648589057622016, 447858123072052742062080, 9371462498278516088373248

Offset: 0

Author

Richard P. Brent, Jan 27 2016

Keywords

• easy
• nonn

Comments

A fast algorithm follows from Theorem 5 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 (80,-2560,40960,-327680,1048576).

Crossrefs

Cf. A000984, A002894, A166337, A254408, A268148, A268150.

Programs

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)^4));

PARI

concat(0, Vec(8*x*(1+1024*x+67840*x^2+417792*x^3)/(1-16*x)^5 + 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^2 - l^2)^4).

From Colin Barker, Feb 11 2016: (Start)

a(n) = 4^(2*n-1)*n*(36*n^3-84*n^2+67*n-17).

a(n) = 80*a(n-1)-2560*a(n-2)+40960*a(n-3)-327680*a(n-4)+1048576*a(n-5) for n>4.

G.f.: 8*x*(1+1024*x+67840*x^2+417792*x^3) / (1-16*x)^5.

(End)