A089661 a(n) = S1(n,4), where S1(n,t) = Sum_{k=0..n} k^t * Sum_{j=0..k} binomial(n,j).

0, 2, 67, 764, 5492, 30304, 140672, 577920, 2167680, 7577088, 25037056, 79016960, 240028672, 705961984, 2019713024, 5641535488, 15431565312, 41438281728, 109462880256, 284942925824, 732004876288, 1858158460928, 4665915736064, 11600782163968, 28582042664960

Offset: 0

Author

N. J. A. Sloane, Jan 04 2004

Keywords

• nonn
• easy

Links

• Vincenzo Librandi, 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.
• Index entries for linear recurrences with constant coefficients, signature (12,-60,160,-240,192,-64).

Crossrefs

Sequences of S1(n,t): A001792 (t=0), A089658 (t=1), A089659 (t=2), A089660 (t=3), this sequence (t=4), A089662 (t=5), A089663 (t=6).

Programs

Magma

[2^(n-5)*n*(93*n^4+225*n^3+185*n^2+15*n-38)/15: n in [0..30]]; // Vincenzo Librandi, Jun 22 2016

Mathematica

LinearRecurrence[{12,-60,160,-240,192,-64}, {0,2,67,764,5492,30304}, 40] (* Vincenzo Librandi, Jun 22 2016 *)

SageMath

[n*(n+1)*(93*n^3 +132*n^2 +53*n -38)*2^(n-5)/15 for n in (0..40)] # G. C. Greubel, May 24 2022

Formula

a(n) = (1/15)*n*(n+1)*(93*n^3 + 132*n^2 + 53*n - 38)*2^(n-5). (See Wang and Zhang, p. 334)

From Chai Wah Wu, Jun 21 2016: (Start)

a(n) = 12*a(n-1) - 60*a(n-2) + 160*a(n-3) - 240*a(n-4) + 192*a(n-5) - 64*a(n-6) for n > 5.

G.f.: x*(2 + 43*x + 80*x^2 + 24*x^3)/(1 - 2*x)^6. (End)

a(n) = 2^(n-5)*n*(93*n^4 + 225*n^3 + 185*n^2 + 15*n - 38)/15. - Ilya Gutkovskiy, Jun 21 2016

E.g.f.: (1/30)*x*(60 + 885*x + 1930*x^2 + 1155*x^3 + 186*x^4)*exp(2*x). - G. C. Greubel, May 24 2022