A253637 Second partial sums of ninth powers (A001017).

1, 514, 20710, 303050, 2538515, 14851676, 67518444, 254402940, 828707925, 2403012910, 6335265586, 15427298614, 35123831015, 75481410200, 154282348760, 301802764056, 567911055849, 1032378638010, 1819533917950, 3118689197890, 5212124524411, 8512829068724, 13614686274500, 21358351020500, 32916713032125, 49904578722726

Offset: 1

Author

Luciano Ancora, Jan 07 2015

Keywords

• nonn
• easy

Comments

The formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m) - F(m+1), where F(m) are the m-th Faulhaber's formulas.

Links

• G. C. Greubel, 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 p. 13.
• Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Crossrefs

Cf. A001017.

Programs

GAP

List([1..30], n-> n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4+24*n^3 +5*n^2-38*n+ 25)/660 ); # G. C. Greubel, Aug 28 2019

Magma

[n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4+24*n^3+5*n^2-38*n+ 25)/660: n in [1..30]]; // G. C. Greubel, Aug 28 2019

Maple

seq(n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4+24*n^3+5*n^2-38*n+ 25)/660, n=1..30); # G. C. Greubel, Aug 28 2019

Mathematica

CoefficientList[Series[(1 +502x +14608x^2 +88234x^3 +156190x^4 +88234x^5 +14608x^6 +502x^7 +x^8)/(1-x)^12, {x, 0, 30}], x] (* Vincenzo Librandi, Jan 19 2015 *)
Nest[Accumulate,Range[30]^9,2] (* Harvey P. Dale, Apr 18 2021 *)

PARI

a(n) = (6*n^11 + 66*n^10 + 275*n^9 + 495*n^8 + 198*n^7 - 462*n^6 - 330*n^5 + 330*n^4 + 231*n^3 - 99*n^2 - 50*n)/660; \\ Michel Marcus, Jan 08 2015

Sage

[n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4+24*n^3+5*n^2-38*n+ 25)/660 for n in (1..30)] # G. C. Greubel, Aug 28 2019

Formula

a(n) = n*(n+1)*(n+2)*(n^2+n-1)*(n^2+3*n+1)*(6*n^4 + 24*n^3 + 5*n^2 - 38*n + 25)/660.

a(n) = 2*a(n-1) - a(n-2) + n^9.

G.f.: x*(1 + 502*x + 14608*x^2 + 88234*x^3 + 156190*x^4 + 88234*x^5 + 14608*x^6 + 502*x^7 + x^8)/(1-x)^12. - Vincenzo Librandi, Jan 19 2015