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
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).
Programs
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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));
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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)
Comments