cp's OEIS Frontend

A101089 Second partial sums of fourth powers (A000583).

%I A101089 #107 Feb 03 2025 09:38:59
%S A101089 1,18,116,470,1449,3724,8400,17172,32505,57838,97812,158522,247793,
%T A101089 375480,553792,797640,1125009,1557354,2120020,2842686,3759833,4911236,
%U A101089 6342480,8105500,10259145,12869766,16011828,19768546,24232545,29506544
%N A101089 Second partial sums of fourth powers (A000583).
%C A101089 a(n) is the n-th antidiagonal sum of the convolution array A213553. - _Clark Kimberling_, Jun 17 2012
%C A101089 a(n-1)/n^5 is the "retention" of water on a 3 X 3 random surface of n levels - see Knecht et al., 2012, Schrenk et al., 2014. - _Robert M. Ziff_, Mar 08 2014
%C A101089 The general formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m) - F(m+1), where F(m) is the m-th Faulhaber’s polynomial. - _Luciano Ancora_, Jan 26 2015
%H A101089 Vincenzo Librandi, <a href="/A101089/b101089.txt">Table of n, a(n) for n = 1..1000</a>
%H A101089 Luciano Ancora, <a href="/A253636/a253636_3.pdf">Recurrence relation for the second partial sums of m-th powers</a>.
%H A101089 Luciano Ancora, <a href="/A253636/a253636_4.pdf">Second partial sums of the m-th powers</a>.
%H A101089 Craig L. Knecht, Walter Trump, Daniel ben-Avraham and Robert M. Ziff, <a href="http://dx.doi.org/10.1103/PhysRevLett.108.045703">Retention Capacity of Random Surfaces</a>, Phys. Rev. Lett., Vol. 108 (2012), 045703.
%H A101089 Feihu Liu, Guoce Xin, and Chen Zhang, <a href="https://arxiv.org/abs/2412.18744">Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS</a>, arXiv:2412.18744 [math.CO], 2024. See p. 13.
%H A101089 C. P. Neuman and D. I. Schonbach, <a href="http://dx.doi.org/10.1137/1019006">Evaluation of sums of convolved powers using Bernoulli numbers</a>, SIAM Rev., Vol. 19, No. 1 (1977), pp. 90-99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014
%H A101089 Claudio de J. Pita Ruiz V., <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Pita/pita19.html">Some Number Arrays Related to Pascal and Lucas Triangles</a>, J. Int. Seq., Vol. 16 (2013), Article 13.5.7.
%H A101089 Cecilia Rossiter, <a href="http://noticingnumbers.net/300SeriesCube.htm">Depictions, Explorations and Formulas of the Euler/Pascal Cube</a>. [Dead link]
%H A101089 Cecilia Rossiter, <a href="/A101089/a101089_1.pdf">Depictions, Explorations and Formulas of the Euler/Pascal Cube</a>. [Cached copy, May 15 2013]
%H A101089 K. J. Schrenk, N. A. M. Araújo, R. M. Ziff and H. J. Herrmann <a href="http://arxiv.org/abs/1403.2082">Retention Capacity of Correlated Surfaces</a>, arXiv:1403.2082 [cond-mat.stat-mech], 2014.
%H A101089 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).
%F A101089 a(n) = (1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1).
%F A101089 G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^7. - _Colin Barker_, Apr 04 2012
%F A101089 a(n) = Sum_{i=1..n} i*(n+1-i)^4, by the definition. - _Bruno Berselli_, Jan 31 2014
%F A101089 a(n) = 2*a(n-1) - a(n-2) + n^4. - _Luciano Ancora_, Jan 08 2015
%F A101089 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
%F A101089 a(n) = (1/2)*Sum_{1 <= i, j <= n+1} (i - j)^4 - _Peter Bala_, Jun 11 2024
%e A101089 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
%p A101089 f:=n->(2*n^6-5*n^4+3*n^2)/60;
%p A101089 [seq(f(n),n=0..50)]; # _N. J. A. Sloane_, Mar 23 2014
%t A101089 a[n_] := n(n+1)^2(n+2)(2n(n+2) -1)/60; Table[a[n], {n, 40}]
%t A101089 CoefficientList[Series[(1+x)*(1+10*x+x^2)/(1-x)^7, {x,0,40}], x] (* _Vincenzo Librandi_, Mar 24 2014 *)
%t A101089 Nest[Accumulate[#]&,Range[30]^4,2] (* _Harvey P. Dale_, Aug 13 2024 *)
%o A101089 (PARI) a(n)=n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 \\ _Charles R Greathouse IV_, Mar 18 2014
%o A101089 (Magma) [(1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1): n in [1..40]]; // _Vincenzo Librandi_, Mar 24 2014
%o A101089 (Sage) [n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 for n in range(1,40)] # _Danny Rorabaugh_, Apr 20 2015
%o A101089 (GAP) List([1..40], n-> (n+1)^2*(2*(n+1)^4-5*(n+1)^2+3)/60); # _G. C. Greubel_, Jul 31 2019
%Y A101089 Partial sums of A000538.
%Y A101089 Cf. A000583, A002415, A101090, A201126, A213553.
%K A101089 nonn,easy
%O A101089 1,2
%A A101089 Cecilia Rossiter, Dec 14 2004
%E A101089 Edited by _Ralf Stephan_, Dec 16 2004