A101093 Second partial sums of sixth powers (A001014).
1, 66, 860, 5750, 26265, 93436, 278256, 725220, 1703625, 3682030, 7431996, 14167946, 25730705, 44823000, 75305920, 122566056, 193963761, 299373690, 451829500, 668285310, 970507241, 1386109076, 1949746800, 2704487500
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Luciano Ancora, Recurrence relation for the second partial sums of m-th powers
- Luciano Ancora, Second partial sums of the m-th powers
- Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See pp. 12-13.
- C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 90--99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014
- C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
- Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).
Crossrefs
Cf. A000540.
Programs
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GAP
List([2..30], n-> n^2*(3*n^6 -14*n^4 +21*n^2 -10)/168); # G. C. Greubel, Aug 28 2019
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Magma
[n*(1+n)^2*(2+n)*(-1+n*(2+n))*(-2+3*n*(2+n))/168: n in [1..40]]; // Vincenzo Librandi, Mar 24 2014
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Maple
f:=n->(3*n^8-14*n^6+21*n^4-10*n^2)/168; [seq(f(n),n=0..50)]; # N. J. A. Sloane, Mar 23 2014
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Mathematica
CoefficientList[Series[(x+1)(x^4+56x^3+246x^2+56x+1)/(1-x)^9, {x,0,40}], x] (* Vincenzo Librandi, Mar 24 2014 *) Nest[Accumulate,Range[30]^6,2] (* or *) LinearRecurrence[{9,-36,84,-126, 126,-84,36,-9,1},{1,66,860,5750,26265,93436,278256,725220,1703625},30] (* Harvey P. Dale, Jun 05 2019 *)
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PARI
vector(30, n, m=n+1; m^2*(3*m^6 -14*m^4 +21*m^2 -10)/168) \\ G. C. Greubel, Aug 28 2019
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Sage
[n^2*(3*n^6 -14*n^4 +21*n^2 -10)/168 for n in (2..30)] # G. C. Greubel, Aug 28 2019
Formula
a(n) = n*(1 + n)^2*(2 + n)*(-1 + n*(2 + n))*(-2 + 3*n*(2 + n))/168.
G.f.: x*(1+x)*(1 + 56*x + 246*x^2 + 56*x^3 + x^4)/(1-x)^9. - Colin Barker, Dec 18 2012
a(n) = Sum_{i=1..n} i*(n+1-i)^6, by the definition. - Bruno Berselli, Jan 31 2014
a(n) = 2*a(n-1) - a(n-2) + n^6. - Luciano Ancora, Jan 08 2015
Extensions
Edited by Ralf Stephan, Dec 16 2004