A254408 a(n) = 2*n^2*binomial(2*n,n)^2, a closed form for a double binomial sum involving absolute values.
0, 8, 288, 7200, 156800, 3175200, 61471872, 1154305152, 21201523200, 382952512800, 6826955907200, 120427502203008, 2105988385632768, 36562298361680000, 630861905459520000, 10827650254927680000, 184984389244186675200, 3147624998233113895200, 53368036302222346320000
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..500
- Richard P. Brent, Hideyuki Ohtsuka, Judy-anne H. Osborn, and Helmut Prodinger, Some binomial sums involving absolute values, arXiv:1411.1477 [math.CO], 2014-2016.
Programs
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Magma
[(4*Binomial(n+1,2)*Catalan(n))^2/2: n in [0..30]]; // G. C. Greubel, Mar 31 2021
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Maple
A254408:= n-> 2*( n*binomial(2*n, n) )^2; seq(A254408(n), n=0..30); # G. C. Greubel, Mar 31 2021
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Mathematica
a[n_] := 2*n^2*Binomial[2*n, n]^2; Table[a[n], {n, 0, 20}] -
PARI
a(n) = 2*n^2*binomial(2*n,n)^2 \\ Charles R Greathouse IV, May 10 2016
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Sage
[(4*binomial(n+1,2)*catalan_number(n))^2/2 for n in (0..30)] # G. C. Greubel, Mar 31 2021
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)).
From G. C. Greubel, Mar 31 2021: (Start)
a(n) = 8 * binomial(n+1, 2)^2 * C(n)^2, where C(n) = A000108(n) (Catalan numbers).
G.f.: 8*x*Hypergeometric2F1([3/2, 3/2], [1], 16*x) = (16/Pi)*(x/(1-16*x)^2)*( 2*E(16*x) - (1-16*x)*K(16*x) ), where E(x) and K(x) are elliptic functions. (End)
D-finite with recurrence (n-1)^2*a(n) + (n^2-52*n+64)*a(n-1) - 68*(2*n -3)^2*a(n-2) = 0. - R. J. Mathar, Feb 27 2023
a(n) ~ 2^(4*n+1) * n / Pi. - Amiram Eldar, Sep 04 2025