A101089 Second partial sums of fourth powers (A000583).
1, 18, 116, 470, 1449, 3724, 8400, 17172, 32505, 57838, 97812, 158522, 247793, 375480, 553792, 797640, 1125009, 1557354, 2120020, 2842686, 3759833, 4911236, 6342480, 8105500, 10259145, 12869766, 16011828, 19768546, 24232545, 29506544
Offset: 1
Examples
a(7) = 8400 = 1*(8-1)^4 + 2*(8-2)^4 + 3*(8-3)^4 + 4*(8-4)^4 + 5*(8-5)^4 + 6*(8-6)^4 + 7*(8-7)^4. - _Bruno Berselli_, Jan 31 2014
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.
- Craig L. Knecht, Walter Trump, Daniel ben-Avraham and Robert M. Ziff, Retention Capacity of Random Surfaces, Phys. Rev. Lett., Vol. 108 (2012), 045703.
- 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 p. 13.
- C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev., Vol. 19, No. 1 (1977), pp. 90-99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014
- Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq., Vol. 16 (2013), Article 13.5.7.
- Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Dead link]
- Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
- K. J. Schrenk, N. A. M. Araújo, R. M. Ziff and H. J. Herrmann Retention Capacity of Correlated Surfaces, arXiv:1403.2082 [cond-mat.stat-mech], 2014.
- Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).
Programs
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GAP
List([1..40], n-> (n+1)^2*(2*(n+1)^4-5*(n+1)^2+3)/60); # G. C. Greubel, Jul 31 2019
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Magma
[(1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1): n in [1..40]]; // Vincenzo Librandi, Mar 24 2014
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Maple
f:=n->(2*n^6-5*n^4+3*n^2)/60; [seq(f(n),n=0..50)]; # N. J. A. Sloane, Mar 23 2014
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Mathematica
a[n_] := n(n+1)^2(n+2)(2n(n+2) -1)/60; Table[a[n], {n, 40}] CoefficientList[Series[(1+x)*(1+10*x+x^2)/(1-x)^7, {x,0,40}], x] (* Vincenzo Librandi, Mar 24 2014 *) Nest[Accumulate[#]&,Range[30]^4,2] (* Harvey P. Dale, Aug 13 2024 *)
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PARI
a(n)=n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 \\ Charles R Greathouse IV, Mar 18 2014
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Sage
[n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 for n in range(1,40)] # Danny Rorabaugh, Apr 20 2015
Formula
a(n) = (1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1).
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^7. - Colin Barker, Apr 04 2012
a(n) = Sum_{i=1..n} i*(n+1-i)^4, by the definition. - Bruno Berselli, Jan 31 2014
a(n) = 2*a(n-1) - a(n-2) + n^4. - Luciano Ancora, Jan 08 2015
Sum_{n>=1} 1/a(n) = 85/3 + 10*Pi^2/3 - 20*sqrt(2/3)*Pi*cot(sqrt(3/2)*Pi). - Amiram Eldar, Jan 26 2022
a(n) = (1/2)*Sum_{1 <= i, j <= n+1} (i - j)^4 - Peter Bala, Jun 11 2024
Extensions
Edited by Ralf Stephan, Dec 16 2004
Comments