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A172077
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a(n) = n*(n+1)*(7*n^2 - n - 4)/4.
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1
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0, 1, 33, 168, 520, 1245, 2541, 4648, 7848, 12465, 18865, 27456, 38688, 53053, 71085, 93360, 120496, 153153, 192033, 237880, 291480, 353661, 425293, 507288, 600600, 706225, 825201, 958608, 1107568, 1273245, 1456845, 1659616, 1882848, 2127873
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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=7 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, Apr 21 2010
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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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G.f.: x*(1 + 28*x + 13*x^2)/(1-x)^5. - R. J. Mathar, Nov 17 2011
E.g.f.: x*(4 + 62*x + 48*x^2 + 7*x^3)*exp(x)/4. - G. C. Greubel, Aug 30 2019
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MAPLE
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seq(n*(n+1)*(7*n^2-n-4)/4, n=0..40); # G. C. Greubel, Aug 30 2019
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MATHEMATICA
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CoefficientList[Series[x(1 +28x +13x^2)/(1-x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jan 01 2014 *)
Table[n*(n+1)*(7*n^2-n-4)/4, {n, 0, 40}] (* G. C. Greubel, Aug 30 2019 *)
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PROG
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(PARI) vector(40, n, n*(n-1)*(7*(n-1)^2-(n-1)-4)/4) \\ G. C. Greubel, Aug 30 2019
(Sage) [n*(n+1)*(7*n^2-n-4)/4 for n in (0..40)] # G. C. Greubel, Aug 30 2019
(GAP) List([0..40], n-> n*(n+1)*(7*n^2-n-4)/4); # 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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STATUS
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approved
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