A089668 a(n) = S2(n,5), where S2(n, t) = Sum_{k=0..n} k^t *(Sum_{j=0..k} binomial(n,j))^2.
0, 4, 521, 17136, 320716, 4356560, 48024786, 456843520, 3893995184, 30487086144, 223052123830, 1544098243424, 10208488021176, 64917814932256, 399310478637476, 2386386863086080, 13906802738650816, 79261768839946496, 442921922267640894
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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Mathematica
Table[(1/2)*(n*(21*n^5+61*n^4+55*n^3+15*n^2-28*n+4)*4^(n-3) -(n-1)*(3*n-5)*(n^2 + 4*n-6)*Binomial[n+1, 3]*CatalanNumber[n-2]), {n, 0, 40}] (* G. C. Greubel, May 25 2022 *) -
SageMath
[(1/2)*(n*(21*n^5 + 61*n^4 + 55*n^3 + 15*n^2 - 28*n + 4)*4^(n-3) - (n-1)*(3*n-5)*(n^2 + 4*n - 6)*binomial(n+1, 3)*catalan_number(n-2)) for n in (0..40)] # G. C. Greubel, May 25 2022
Formula
a(n) = (1/128)*n*(21*n^5 + 61*n^4 + 55*n^3 + 15*n^2 - 28*n + 4)*4^n - (1/48)*n^2*(n-1)^2*(3*n-5)*(n^2 + 4*n - 6)*binomial(2*n, n)/((2*n-1)*(2*n-3)). (See Wang and Zhang, p. 338)
From G. C. Greubel, May 25 2022: (Start)
a(n) = (1/2)*(n*(21*n^5 + 61*n^4 + 55*n^3 + 15*n^2 - 28*n + 4)*4^(n-3) - (n-1)*(3*n-5)*(n^2 + 4*n - 6)*binomial(n+1, 3)*Catalan(n-2)).
G.f.: x*( 4*(1 + 103*x + 1012*x^2 + 1688*x^3 + 512*x^4 - 256*x^5) - 3*x*(1 + 54*x + 26*x^2 - 156*x^3 - 104*x^4 + 320*x^5 -240*x^6)*sqrt(1-4*x) )/(1-4*x)^7. (End)