A089666 a(n) = S2(n,3), where S2(n, t) = Sum_{k=0..n} k^t *(Sum_{j=0..k} binomial(n,j))^2.
0, 4, 137, 2136, 23452, 209840, 1640346, 11648224, 76976048, 481048128, 2874897670, 16564931504, 92584313112, 504313834336, 2687067833492, 14045889333120, 72202366588096, 365713117287680, 1828223537042142, 9032706189007888, 44158716127799240, 213826835772518304
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Jun Wang and Zhizheng Zhang, On extensions of Calkin's binomial identities, Discrete Math., 274 (2004), 331-342.
Crossrefs
Programs
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Maple
S2:= (n, t) -> add(k^t*add(binomial(n, j), j = 0..k)^2, k = 0..n); seq(S2(n, 3), n = 0..40);
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Mathematica
Table[n*(15*n^3+30*n^2+21*n-2)*4^(n-3) -(n-1)^2*n^2*(n+1)*Binomial[2*n,n]/(8*(2*n -1)), {n, 0, 40}] (* G. C. Greubel, May 25 2022 *)
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SageMath
[n*(15*n^3+30*n^2+21*n-2)*4^(n-3) - 9*binomial(n+1, 3)^2 * catalan_number(n-1)/(n+1) for n in (0..40)] # G. C. Greubel, May 25 2022
Formula
a(n) = n*(15*n^3 + 30*n^2 + 21*n - 2)*4^(n-3) - (n-1)^2*n^2*(n+1)*binomial(2*n, n)/(8*(2*n-1)). (See Wang and Zhang, p. 338.)
From G. C. Greubel, May 25 2022: (Start)
a(n) = n*(15*n^3 + 30*n^2 + 21*n - 2)*4^(n-3) - 9*binomial(n+1, 3)^2 * Catalan(n- 1)/(n+1).
G.f.: x*(4*(1 + 15*x + 12*x^2 + 8*x^3) - 3*x*(1 + 6*x - 6*x^2 + 4*x^3)*sqrt(1-4*x))/(1-4*x)^5. (End)
Extensions
Name changed by G. C. Greubel, May 25 2022