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A172085
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a(n) = n*(27*n^3 + 22*n^2 - 21*n - 16)/12.
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1
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0, 1, 41, 212, 660, 1585, 3241, 5936, 10032, 15945, 24145, 35156, 49556, 67977, 91105, 119680, 154496, 196401, 246297, 305140, 373940, 453761, 545721, 650992, 770800, 906425, 1059201, 1230516, 1421812, 1634585, 1870385, 2130816, 2417536
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OFFSET
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0,3
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COMMENTS
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This is the case d=9 in the identity n^2*(n+1)*(2*d*n -2*d +3)/6 - Sum_{k=0..n-1} k*(k+1)*(2*d*k -2*d +3)/6 = n*(n+1)*(3*d*n^2 - d*n + 4*n - 2*d + 2)/12. - Bruno Berselli, May 07 2010, Jan 28 2011
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LINKS
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B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
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FORMULA
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a(n) = n*(n+1)*(27*n^2 -5*n -16)/12.
G.f.: x*(1 +36*x +17*x^2)/(1-x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
E.g.f.: x*(12 + 234*x + 184*x^2 + 27*x^3)*exp(x)/12. - G. C. Greubel, Aug 30 2019
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MAPLE
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seq(n*(n+1)*(27*n^2 -5*n -16)/12, n=0..40); # G. C. Greubel, Aug 30 2019
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MATHEMATICA
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CoefficientList[Series[x(1 +36x +17x^2)/(1-x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jan 02 2014 *)
Table[n*(n+1)*(27*n^2-5*n-16)/12, {n, 0, 40}] (* G. C. Greubel, Aug 30 2019 *)
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PROG
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(Magma) [n*(n+1)*(27*n^2-5*n-16)/12: n in [0..40]]; // Vincenzo Librandi, Jan 02 2014
(PARI) vector(40, n, m=n-1; n*m*(27*m^2 -5*m -16)/12) \\ G. C. Greubel, Aug 30 2019
(Sage) [n*(n+1)*(27*n^2 -5*n -16)/12: n in (0..40)] # G. C. Greubel, Aug 30 2019
(GAP) List([0..40], n-> n*(n+1)*(27*n^2 -5*n -16)/12); # G. C. Greubel, Aug 30 2019
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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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EXTENSIONS
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Librandi's contribution restored and rewritten from Bruno Berselli, Feb 29 2012
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STATUS
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approved
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