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A089661
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a(n) = S1(n,4), where S1(n,t) = Sum_{k=0..n} k^t * Sum_{j=0..k} binomial(n,j).
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6
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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
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OFFSET
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0,2
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LINKS
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FORMULA
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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)
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
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MATHEMATICA
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LinearRecurrence[{12, -60, 160, -240, 192, -64}, {0, 2, 67, 764, 5492, 30304}, 40] (* Vincenzo Librandi, Jun 22 2016 *)
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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