A023002 Sum of 10th powers.
0, 1, 1025, 60074, 1108650, 10874275, 71340451, 353815700, 1427557524, 4914341925, 14914341925, 40851766526, 102769130750, 240627622599, 529882277575, 1106532668200, 2206044295976, 4222038196425, 7792505423049, 13923571680850
Offset: 0
Links
- T. D. Noe, Table of n, a(n) for n = 0..1000
- Bruno Berselli, A description of the recursive method in Formula lines (second formula): website Matem@ticamente (in Italian).
- 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.
- Eric Weisstein's World of Mathematics, Power Sum.
- Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).
Crossrefs
Programs
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Magma
[&+[n^10: n in [0..m]]: m in [0..19]]; // Bruno Berselli, Aug 23 2011
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Maple
A023002:= n-> bernoulli(11, n+1)/11; seq(A023002(n), n=0..30); # G. C. Greubel, Jul 21 2021
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Mathematica
Table[Sum[k^10, {k, n}], {n, 0, 30}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *) Accumulate[Range[0,20]^10] (* Harvey P. Dale, Aug 23 2011 *)
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PARI
a(n)=(6*x^11+33*x^10+55*x^9-66*x^7+66*x^5-33*x^3+5*x)/66 \\ Charles R Greathouse IV, Aug 23 2011
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PARI
a(n)=sum(i=0,10,binomial(11,i)*bernfrac(i)*n^(11-i))/11+n^10 \\ Charles R Greathouse IV, Aug 23 2011
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Python
A023002_list, m = [0], [3628800, -16329600, 30240000, -29635200, 16435440, -5103000, 818520, -55980, 1022, -1, 0 , 0] for _ in range(20): for i in range(11): m[i+1]+= m[i] A023002_list.append(m[-1]) print(A023002_list) # Chai Wah Wu, Nov 05 2014
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Sage
[bernoulli_polynomial(n,11)/11 for n in range(2, 21)]# Zerinvary Lajos, May 17 2009
Formula
a(n) = n*(n+1)*(2*n+1)*(n^2+n-1)(3*n^6 +9*n^5 +2*n^4 -11*n^3 +3*n^2 +10*n -5)/66 (see MathWorld, Power Sum, formula 40). - Bruno Berselli, Apr 26 2010
From Bruno Berselli, Aug 23 2011: (Start)
a(n) = -a(-n-1).
G.f.: x*(1+x)*(1 +1012*x +46828*x^2 +408364*x^3 +901990*x^4 +408364*x^5 +46828*x^6 +1012*x^7 +x^8)/(1-x)^12. (End)
a(n) = (-1)*Sum_{j=1..10} j*Stirling1(n+1,n+1-j)*Stirling2(n+10-j,n). - Mircea Merca, Jan 25 2014
a(n) = Sum_{i=1..n} J_10(i)*floor(n/i), where J_10 is A069095. - Ridouane Oudra, Jul 17 2025