A000542 -id:A000542 - cp's OEIS frontend

A000538 Sum of fourth powers: 0^4 + 1^4 + ... + n^4.

0, 1, 17, 98, 354, 979, 2275, 4676, 8772, 15333, 25333, 39974, 60710, 89271, 127687, 178312, 243848, 327369, 432345, 562666, 722666, 917147, 1151403, 1431244, 1763020, 2153645, 2610621, 3142062, 3756718, 4463999, 5273999, 6197520, 7246096, 8432017, 9768353

Offset: 0

N. J. A. Sloane

This sequence is related to A000537 by the transform a(n) = n*A000537(n) - Sum_{i=0..n-1} A000537(i). - Bruno Berselli, Apr 26 2010
A formula for the r-th successive summation of k^4, for k = 1 to n, is ((12*n^2+(12*n-5)*r+r^2)*(2*n+r)*(n+r)!)/((r+4)!*(n-1)!), (H. W. Gould). - Gary Detlefs, Jan 02 2014
The number of four dimensional hypercubes in a 4D grid with side lengths n. This applies in general to k dimensions. That is, the number of k-dimensional hypercubes in a k-dimensional grid with side lengths n is equal to the sum of 1^k + 2^k + ... + n^k. - Alejandro Rodriguez, Oct 20 2020

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 222.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1991, p. 275.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
J. L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359.
J. L. Bailey, A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359. [Annotated scanned copy]
Bruno Berselli, A description of the transform in Comments lines: website Matem@ticamente (in Italian).
Stefano Capparelli, Notes on Discrete Math, Società Editrice Esculapio SRL (2019) 3-4.
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.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein, MathWorld: Faulhaber's Formula
Wikipedia, Faulhaber's formula
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000217, A000330, A000537, A000539, A000540, A000541, A000542, A007487, A023002, A064538, A101089.
Row 4 of array A103438.
Cf. A000583.
Cf. A254640.

a000538 n = (3 * n * (n + 1) - 1) * (2 * n + 1) * (n + 1) * n `div` 30
-- Reinhard Zumkeller, Nov 11 2012

[n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30: n in [0..35]]; // Vincenzo Librandi, Apr 04 2015

A000538 := n-> n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30;

Accumulate[Range[0,40]^4] (* Harvey P. Dale, Jan 13 2011 *)
CoefficientList[Series[x (1 + 11 x + 11 x^2 + x^3)/(1 - x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 07 2015 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 17, 98, 354, 979}, 35] (* Jean-François Alcover, Feb 09 2016 *)
Table[x^5/5+x^4/2+x^3/3-x/30,{x,40}] (* Harvey P. Dale, Jun 06 2021 *)

A000538(n):=n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30$
makelist(A000538(n),n,0,30); /* Martin Ettl, Nov 12 2012 */

a(n) = n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30 \\ Charles R Greathouse IV, Nov 20 2012

concat(0, Vec(x*(1+11*x+11*x^2+x^3)/(1-x)^6 + O(x^100))) \\ Altug Alkan, Dec 07 2015

A000538_list, m = [0], [24, -36, 14, -1, 0, 0]
for _ in range(10**2):
    for i in range(5):
        m[i+1] += m[i]
    A000538_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014

def A000538(n): return n*(n**2*(n*(6*n+15)+10)-1)//30 # Chai Wah Wu, Oct 03 2024

[bernoulli_polynomial(n,5)/5 for n in range(1, 35)] # Zerinvary Lajos, May 17 2009

a(n) = n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30.
The preceding formula is due to al-Kachi (1394-1437). - Juri-Stepan Gerasimov, Jul 12 2009
G.f.: x*(x+1)*(1+10*x+x^2)/(1-x)^6. Simon Plouffe in his 1992 dissertation. More generally, the o.g.f. for Sum_{k=0..n} k^m is x*E(m, x)/(1-x)^(m+2), where E(m, x) is the Eulerian polynomial of degree m (cf. A008292). The e.g.f. for these o.g.f.s is: x/(1-x)^2*(exp(y/(1-x))-exp(x*y/(1-x)))/(exp(x*y/(1-x))-x*exp(y/(1-x))). - Vladeta Jovovic, May 08 2002
a(n) = Sum_{i = 1..n} J_4(i)*floor(n/i), where J_4 is A059377. - Enrique Pérez Herrero, Feb 26 2012
a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 24. - Ant King, Sep 23 2013
a(n) = -Sum_{j=1..4} j*Stirling1(n+1,n+1-j)*Stirling2(n+4-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = -30*(4 + 3/cos(sqrt(7/3)*Pi/2))*Pi/7. - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)*(n + 1/2)*n*(n + 1/2 + sqrt(7/12))*(n + 1/2  - sqrt(7/12))/5, see the Graham et al. reference, p. 275. - Wolfdieter Lang, Apr 02 2015

The general V. Jovovic formula has been slightly changed after his approval by Wolfdieter Lang, Nov 03 2011

A001016 Eighth powers: a(n) = n^8.

Original entry on oeis.org

0, 1, 256, 6561, 65536, 390625, 1679616, 5764801, 16777216, 43046721, 100000000, 214358881, 429981696, 815730721, 1475789056, 2562890625, 4294967296, 6975757441, 11019960576, 16983563041, 25600000000, 37822859361, 54875873536, 78310985281, 110075314176

Offset: 0

N. J. A. Sloane

Besides the first term, this sequence lists the denominators in Pi^8/9450 = 1 + 1/256 + 1/6561 + 1/65536 + 1/390625 + 1/1679616 + ... - Mohammad K. Azarian, Nov 01 2011, edited by M. F. Hasler, Jul 03 2025
For n > 0, a(n) is the largest number k such that k + n^4 divides k^2 + n^4. - Derek Orr, Oct 01 2014
Fourth powers of squares and squares of 4th powers. Squares composed with themselves twice. - Wesley Ivan Hurt, Apr 01 2016

Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), p. 982.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9, -36, 84, -126, 126, -84, 36, -9, 1).

Cf. A000290 (squares), A000583 (fourth powers), A001014 - A001017 (6th - 9th powers), A008454 (10th powers), A010801 (13th powers).
Cf. A000542 (partial sums), A022524 (first differences), A013666 (zeta(8)).
Cf. A003380 - A003390 (sums of 2, ..., 12 eighth powers).

[n^8 : n in [0..50]]; // Wesley Ivan Hurt, Apr 01 2016

A001016:=n->n^8: seq(A001016(n), n=0..50); # Wesley Ivan Hurt, Apr 01 2016

Table[n^8, {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Sep 27 2008 *)

A001016(n):=n^8$
makelist(A001016(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

A001016(n)=n^8 \\ Charles R Greathouse IV, Sep 24 2015

A001016 = lambda n: n**8 # M. F. Hasler, Jul 03 2025

Multiplicative with a(p^e) = p^(8e). - David W. Wilson, Aug 01 2001
Totally multiplicative sequence with a(p) = p^8 for primes p. - Jaroslav Krizek, Nov 01 2009
G.f.: -x*(1+x)*(x^6+246*x^5+4047*x^4+11572*x^3+4047*x^2+246*x+1)/(x-1)^9. - R. J. Mathar, Jan 07 2011
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) + 40320. - Ant King, Sep 24 2013
From Wesley Ivan Hurt, Apr 01 2016: (Start)
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n > 8.
a(n) = A000290(n)^4 = A000290(A000290(A000290(n))).
a(n) = A000583(n)^2. (End)
From Amiram Eldar, Oct 08 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(8) = Pi^8/9450 (A013666).
Sum_{n>=1} (-1)^(n+1)/a(n) = 127*zeta(8)/128 = 127*Pi^8/1209600. (End)
E.g.f.: exp(x)*x*(1 + 127*x + 966*x^2 + 1701*x^3 + 1050*x^4 + 266*x^5 + 28*x^6 + x^7). - Stefano Spezia, Jul 29 2022

More terms from James Sellers, Sep 19 2000

A103438 Square array T(m,n) read by antidiagonals: Sum_{k=1..n} k^m.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 5, 6, 4, 0, 1, 9, 14, 10, 5, 0, 1, 17, 36, 30, 15, 6, 0, 1, 33, 98, 100, 55, 21, 7, 0, 1, 65, 276, 354, 225, 91, 28, 8, 0, 1, 129, 794, 1300, 979, 441, 140, 36, 9, 0, 1, 257, 2316, 4890, 4425, 2275, 784, 204, 45, 10

Offset: 0

Ralf Stephan, Feb 11 2005

For the o.g.f.s of the column sequences for this array, see A196837 and the link given there. - Wolfdieter Lang, Oct 15 2011
T(m,n)/n is the m-th moment of the discrete uniform distribution on {1,2,...,n}. - Geoffrey Critzer, Dec 31 2018
T(1,n) divides T(m,n) for odd m. - Franz Vrabec, Dec 23 2020

			Square array begins:
  0, 1,  2,   3,    4,     5,     6,      7,      8,      9, ... A001477;
  0, 1,  3,   6,   10,    15,    21,     28,     36,     45, ... A000217;
  0, 1,  5,  14,   30,    55,    91,    140,    204,    285, ... A000330;
  0, 1,  9,  36,  100,   225,   441,    784,   1296,   2025, ... A000537;
  0, 1, 17,  98,  354,   979,  2275,   4676,   8772,  15333, ... A000538;
  0, 1, 33, 276, 1300,  4425, 12201,  29008,  61776, 120825, ... A000539;
  0, 1, 65, 794, 4890, 20515, 67171, 184820, 446964, 978405, ... A000540;
Antidiagonal triangle begins as:
  0;
  0, 1;
  0, 1,  2;
  0, 1,  3,  3;
  0, 1,  5,  6,  4;
  0, 1,  9, 14, 10,  5;
  0, 1, 17, 36, 30, 15, 6;

J. Faulhaber, Academia Algebrae, Darinnen die miraculosische inventiones zu den höchsten Cossen weiters continuirt und profitirt werden, Augspurg, bey Johann Ulrich Schönigs, 1631.

G. C. Greubel, Antidiagonals n = 0..50, flattened
José L. Cereceda, Sums of powers of integers and hyperharmonic numbers, arXiv:2005.03407 [math.NT], 2020.
T. A. Gulliver, Divisibility of sums of powers of odd integers, Int. Math. For. 5 (2010) 3059-3066.
T. A. Gulliver, Sums of Powers of Integers Divisible by Three, Int. J. Contemp. Math. Sciences, Vol. 7, 2012, no. 38, pp. 1895-1901. - From _N. J. A. Sloane_, Dec 22 2012
V. J. W. Guo and J. Zeng, A q-analogue of Faulhaber's formula for sums of powers, arXiv:math/0501441 [math.CO], 2005.
H. Helfgott and I. M. Gessel, Enumeration of tilings of diamonds and hexagons with defects, arXiv:math/9810143 [math.CO], 1998.
T. Kim, q-analogues of the sums of powers of consecutive integers, arXiv:math/0502113 [math.NT], 2005.
D. E. Knuth, Johann Faulhaber and sums of powers, Math. Comp. 61 (1993), no. 203, 277-294.
Eric Weisstein's World of Mathematics, Discrete Uniform Distribution.
Wikipedia, Faulhaber's formula

Rows include A000027, A000217, A000330, A000537, A000538, A000539, A000540, A000541, A000542, A007487, A023002.
Columns include A000051, A001550, A001551, A001552, A001553, A001554, A001555, A001556, A001557.
Diagonals include A076015 and A031971.
Antidiagonal sums are in A103439.
Antidiagonals are the rows of triangle A192001.

T:= func< n,k | n eq 0 select k else (&+[j^n: j in [0..k]]) >;
[T(n-k,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 22 2021

seq(print(seq(Zeta(0,-k,1)-Zeta(0,-k,n+1),n=0..9)),k=0..6);
# (Produces the square array from the example.) Peter Luschny, Nov 16 2008
# alternative
A103438 := proc(m,n)
    (bernoulli(m+1,n+1)-bernoulli(m+1))/(m+1) ;
    if m = 0 then
        %-1 ;
    else
        % ;
    end if;
end proc: # R. J. Mathar, May 10 2013
# simpler:
A103438 := proc(m,n)
    (bernoulli(m+1,n+1)-bernoulli(m+1,1))/(m+1) ;
end proc: # Peter Luschny, Mar 20 2024

T[m_, n_]:= HarmonicNumber[m, -n]; Flatten[Table[T[m-n, n], {m, 0, 11}, {n, m, 0, -1}]] (* Jean-François Alcover, May 11 2012 *)

PARI
```
T(m,n)=sum(k=0,n,k^m)
```

from itertools import count, islice
from math import comb
from fractions import Fraction
from sympy import bernoulli
def A103438_T(m,n): return sum(k**m for k in range(1,n+1)) if n<=m else int(sum(comb(m+1,i)*(bernoulli(i) if i!=1 else Fraction(1,2))*n**(m-i+1) for i in range(m+1))/(m+1))
def A103438_gen(): # generator of terms
    for m in count(0):
        for n in range(m+1):
            yield A103438_T(m-n,n)
A103438_list = list(islice(A103438_gen(),100)) # Chai Wah Wu, Oct 23 2024

def T(n,k): return (bernoulli_polynomial(k+1, n+1) - bernoulli_polynomial(1, n+1)) /(n+1)
flatten([[T(n-k,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Dec 22 2021

E.g.f.: e^x*(e^(x*y)-1)/(e^x-1).
T(m, n) = Zeta(-n, 1) - Zeta(-n, m + 1), for m >= 0 and n >= 0, where Zeta(z, v) is the Hurwitz zeta function. - Peter Luschny, Nov 16 2008
T(m, n) = HarmonicNumber(m, -n). - Jean-François Alcover, May 11 2012
T(m, n) = (Bernoulli(m + 1, n + 1) - Bernoulli(m + 1, 1)) / (m + 1). - Peter Luschny, Mar 20 2024
T(m, n) = Sum_{k=0...m-n} B(k)*(-1)^k*binomial(m-n,k)*n^(m-n-k+1)/(m-n-k+1), where B(k) = Bernoulli number A027641(k) / A027642(k). - Robert B Fowler, Aug 20 2024
T(m, n) = Sum_{i=1..n} J_m(i)*floor(n/i), where J_m is the m-th Jordan totient function. - Ridouane Oudra, Jul 19 2025

A254647 Fourth partial sums of eighth powers (A001016).

Original entry on oeis.org

1, 260, 7595, 94360, 723534, 4037712, 17944290, 67127880, 219319815, 642251428, 1718012933, 4258676240, 9892043980, 21721707840, 45414150132, 90930820464, 175208925885, 326205634020, 588861675535, 1033717781096, 1769137540730, 2958360418000, 4842936861750, 7774492635000

Offset: 1

Luciano Ancora, Feb 05 2015

			The eighth powers:   1, 256, 6561, 65536, 390625, ... (A001016)
First partial sums:  1, 257, 6818, 72354, 462979, ... (A000542)
Second partial sums: 1, 258, 7076, 79430, 542409, ... (A253636)
Third partial sums:  1, 259, 7335, 86765, 629174, ... (A254642)
Fourth partial sums: 1, 260, 7595, 94360, 723534, ... (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A000542, A001016, A253636, A254642, A254644, A254645, A254646.

List([1..30], n-> Binomial(n+4,5)*(n+2)*((n+2)^2-3)*(2*(n+2)^4 -28*(n+2)^2 +101)/198); # G. C. Greubel, Aug 28 2019

[Binomial(n+4,5)*(n+2)*((n+2)^2-3)*(2*(n+2)^4 -28*(n+2)^2 +101)/198: n in [1..30]]; // G. C. Greubel, Aug 28 2019

seq(binomial(n+4,5)*(n+2)*((n+2)^2-3)*(2*(n+2)^4 -28*(n+2)^2 +101)/198, n=1..30); # G. C. Greubel, Aug 28 2019

Table[n(1+n)(2+n)^2(3+n)(4+n)(1+4n+n^2)(21 -48n +20n^2 +16n^3 +2n^4 )/23760, {n,20}] (* or *)
Accumulate[Accumulate[Accumulate[Accumulate[Range[20]^8]]]] (* or *)
CoefficientList[Series[(1 +247x +4293x^2 +15619x^3 +15619x^4 +4293x^5 + 247x^6 +x^7)/(1-x)^13, {x,0,19}], x]

a(n)=n*(1+n)*(2+n)^2*(3+n)*(4+n)*(1+4*n+n^2)*(21-48*n+20*n^2 +16*n^3+2*n^4)/23760 \\ Charles R Greathouse IV, Sep 08 2015

vector(30, n, m=n+2; binomial(m+2,5)*m*(m^2-3)*(2*m^4-28*m^2 +101)/198)

[binomial(n+4,5)*(n+2)*((n+2)^2-3)*(2*(n+2)^4 -28*(n+2)^2 +101)/198 for n in (1..30)] # G. C. Greubel, Aug 28 2019

G.f.: x*(1 +247*x +4293*x^2 +15619*x^3 +15619*x^4 +4293*x^5 +247*x^6 +x^7)/(1-x)^13.
a(n) = n*(1+n)*(2+n)^2*(3+n)*(4+n)*(1 +4*n +n^2)*(21 -48*n +20*n^2 + 16*n^3 +2*n^4)/23760.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + n^8.

A323546 a(n) = Product_{k=0..n} (k^8 + (n-k)^8).

Original entry on oeis.org

0, 1, 131072, 2843192875329, 94689598336441253888, 30456484467910986480712890625, 24450543078081443740165325452474318848, 74458223912158060479139869222818674648092178561, 545831702006800417886454373052629612732034857946832699392

Offset: 0

Vaclav Kotesovec, Jan 17 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..71

Cf. A323538, A323540, A323541, A323542, A323543, A323544, A323545, A320345.

Cf. 2*A000542 (with sum instead of product).

Table[Product[k^8+(n-k)^8, {k, 0, n}], {n, 0, 10}]

m=8; vector(10, n, n--; prod(k=0,n, k^m + (n-k)^m)) \\ G. C. Greubel, Jan 18 2019

m=8; [product(k^m +(n-k)^m for k in (0..n)) for n in (0..10)] # G. C. Greubel, Jan 18 2019

a(n) ~ exp(((2*sqrt(2+sqrt(2)) - 1/2 - sqrt(2))*Pi - 8)*n) * n^(8*n+8).

A007487 Sum of 9th powers.

Original entry on oeis.org

0, 1, 513, 20196, 282340, 2235465, 12313161, 52666768, 186884496, 574304985, 1574304985, 3932252676, 9092033028, 19696532401, 40357579185, 78800938560, 147520415296, 266108291793, 464467582161, 787155279940, 1299155279940, 2093435326521, 3300704544313

Offset: 0

N. J. A. Sloane

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 815.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
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 (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Row 9 of array A103438.

Cf. A000217, A000542, A069094.

[&+[n^9: n in [0..m]]: m in [0..22]]; // Bruno Berselli, Aug 23 2011

[seq(add(i^9,i=1..n),n=0..40)];
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^9 od: seq(a[n], n=0..22); # Zerinvary Lajos, Feb 22 2008

lst={};s=0;Do[s=s+n^9;AppendTo[lst, s], {n, 10^2}];lst..or..Table[Sum[k^9, {k, 1, n}], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *)
Accumulate[Range[0,30]^9] (* Harvey P. Dale, Oct 09 2016 *)

a(n)=n^2*(n+1)^2*(n^2+n-1)*(2*n^4+4*n^3-n^2-3*n+3)/20 \\ Charles R Greathouse IV, Oct 07 2015

A007487_list, m = [0], [362880, -1451520, 2328480, -1905120, 834120, -186480, 18150, -510, 1, 0, 0]
for _ in range(10**2):
    for i in range(10):
        m[i+1]+= m[i]
    A007487_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014

a(n) = n^2*(n+1)^2*(n^2+n-1)*(2*n^4+4*n^3-n^2-3*n+3)/20 (see MathWorld, Power Sum, formula 39). a(n) = n*A000542(n) - Sum_{i=0..n-1} A000542(i). - Bruno Berselli, Apr 26 2010
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)^11. a(n) = a(-n-1). - Bruno Berselli, Aug 23 2011
a(n) = -Sum_{j=1..9} j*Stirling1(n+1,n+1-j)*Stirling2(n+9-j,n). - Mircea Merca, Jan 25 2014
a(n) = (16/5)*A000217(n)^5 - 4*A000217(n)^4 + (12/5)*A000217(n)^3 - (3/5)*A000217(n)^2. - Michael Raney, Mar 14 2016
a(n) = 11*a(n-1) - 55*a(n-2) + 165*a(n-3) - 330*a(n-4) + 462*a(n-5) - 462*a(n-6) + 330*a(n-7) - 165*a(n-8) + 55*a(n-9) - 11*a(n-10) + a(n-11) for n > 10. - Wesley Ivan Hurt, Dec 21 2016
a(n) = 288*A005585(n-1)^2 + 1728*A108679(n-3) + A062392(n)^2. - Yasser Arath Chavez Reyes, May 11 2024
a(n) = Sum_{i=1..n} J_9(i)*floor(n/i), where J_9 is A069094. - Ridouane Oudra, Jul 17 2025

A023002 Sum of 10th powers.

Original entry on oeis.org

0, 1, 1025, 60074, 1108650, 10874275, 71340451, 353815700, 1427557524, 4914341925, 14914341925, 40851766526, 102769130750, 240627622599, 529882277575, 1106532668200, 2206044295976, 4222038196425, 7792505423049, 13923571680850

Offset: 0

David W. Wilson

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).

Sequences of the form Sum_{j=0..n} j^m : A000217 (m=1), A000330 (m=2), A000537 (m=3), A000538 (m=4), A000539 (m=5), A000540 (m=6), A000541 (m=7), A000542 (m=8), A007487 (m=9), this sequence (m=10), A123095 (m=11), A123094 (m=12), A181134 (m=13).
Row 10 of array A103438.
Cf. A215083, A069095.

[&+[n^10: n in [0..m]]: m in [0..19]];  // Bruno Berselli, Aug 23 2011

A023002:= n-> bernoulli(11, n+1)/11; seq(A023002(n), n=0..30); # G. C. Greubel, Jul 21 2021

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 *)

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

a(n)=sum(i=0,10,binomial(11,i)*bernfrac(i)*n^(11-i))/11+n^10 \\ Charles R Greathouse IV, Aug 23 2011

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

[bernoulli_polynomial(n,11)/11 for n in range(2, 21)]# Zerinvary Lajos, May 17 2009

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
a(n) = n*A007487(n) - Sum_{i=0..n-1} A007487(i). - Bruno Berselli, Apr 27 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

A101095 Fourth difference of fifth powers (A000584).

Original entry on oeis.org

1, 28, 121, 240, 360, 480, 600, 720, 840, 960, 1080, 1200, 1320, 1440, 1560, 1680, 1800, 1920, 2040, 2160, 2280, 2400, 2520, 2640, 2760, 2880, 3000, 3120, 3240, 3360, 3480, 3600, 3720, 3840, 3960, 4080, 4200, 4320, 4440, 4560, 4680, 4800, 4920, 5040, 5160, 5280

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original Name: Shells (nexus numbers) of shells of shells of shells of the power of 5.

The (Worpitzky/Euler/Pascal Cube) "MagicNKZ" algorithm is: MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n, with k>=0, n>=1, z>=0. MagicNKZ is used to generate the n-th accumulation sequence of the z-th row of the Euler Triangle (A008292). For example, MagicNKZ(3,k,0) is the 3rd row of the Euler Triangle (followed by zeros) and MagicNKZ(10,k,1) is the partial sums of the 10th row of the Euler Triangle. This sequence is MagicNKZ(5,k-1,2).

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Archive Machine link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Fourth differences of A000584, third differences of A022521, second differences of A101098, and first differences of A101096.
For other sequences based upon MagicNKZ(n,k,z):
...... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7  |  n = 8
--------------------------------------------------------------------------------------
z =  0 | A000007 | A019590 | .......  MagicNKZ(n,k,0) = T(n,k+1) from A008292  .......
z =  1 | A000012 | A040000 | A101101 | A101104 | A101100 | ....... | ....... | .......
z =  2 | A000027 | A005408 | A008458 | A101103 | thisSeq | ....... | ....... | .......
z =  3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | ....... | .......
z =  4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | ....... | .......
z =  5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181 | .......
z =  6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177 | A255182
z =  7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523 | A255178
z =  8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015 | A022524
z =  9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541 | A001016
z = 10 | A000582 | A054333 | A254469 | A254681 | A254644 | A254640 | A250212 | A000542
z = 11 | A001287 | A054334 | A254869 | A254470 | A254682 | A254645 | A254641 | A253636
z = 12 | A001288 | A057788 | ....... | A254870 | A254471 | A254683 | A254646 | A254642
z = 13 | A010965 | ....... | ....... | ....... | A254871 | A254472 | A254684 | A254647
z = 14 | A010966 | ....... | ....... | ....... | ....... | A254872 | ....... | .......
--------------------------------------------------------------------------------------
Cf. A047969.

I:=[1,28,121,240,360]; [n le 5 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, May 07 2015

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 2, 2}, {k, 0, 34}]
CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/(1 - x)^2, {x, 0, 50}], x] (* Vincenzo Librandi, May 07 2015 *)
Join[{1,28,121,240},Differences[Range[50]^5,4]] (* or *) LinearRecurrence[{2,-1},{1,28,121,240,360},50] (* Harvey P. Dale, Jun 11 2016 *)

a(n)=if(n>3, 120*n-240, 33*n^2-72*n+40) \\ Charles R Greathouse IV, Oct 11 2015

[1,28,121]+[120*(k-2) for k in range(4,36)] # Danny Rorabaugh, Apr 23 2015

a(k+1) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n; n = 5, z = 2.
For k>3, a(k) = Sum_{j=0..4} (-1)^j*binomial(4, j)*(k - j)^5 = 120*(k - 2).
a(n) = 2*a(n-1) - a(n-2), n>5. G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (1-x)^2. - Colin Barker, Mar 01 2012

MagicNKZ material edited, Crossrefs table added, SeriesAtLevelR material removed by Danny Rorabaugh, Apr 23 2015

Name changed and keyword 'uned' removed by Danny Rorabaugh, May 06 2015

A254642 Third partial sums of eighth powers (A001016).

Original entry on oeis.org

1, 259, 7335, 86765, 629174, 3314178, 13906578, 49183590, 152191935, 422931613, 1075761505, 2540663307, 5633367740, 11829663860, 23692442292, 45516670332, 84278105421, 150996708135, 262656041515, 444856105561, 735419759634, 1189222877270

Offset: 1

Luciano Ancora, Feb 05 2015

			First differences:   1, 255, 6305, 58975, 325089, ...(A022524)
--------------------------------------------------------------------
The eighth powers:   1, 256, 6561, 65536, 390625, ...(A001016)
--------------------------------------------------------------------
First partial sums:  1, 257, 6818, 72354, 462979, ...(A000542)
Second partial sums: 1, 258, 7076, 79430, 542409, ...(A253636)
Third partial sums:  1, 259, 7335, 86765, 629174, ...(this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Cf. A000542, A001016, A022524, A253636.

Table[n (1 + n) (2 + n) (3 + n) (3 + 2 n) (1 + 36 n - 69 n^2 + 45 n^4 + 18 n^5 + 2 n^6)/3960, {n, 22}]
Accumulate[Accumulate[Accumulate[Range[22]^8]]]
CoefficientList[Series[(1 + 247 x + 4293 x^2 + 15619 x^3 + 15619 x^4 + 4293 x^5 + 247 x^6 + x^7)/(- 1 + x)^12, {x, 0, 22}], x]

a(n)=n*(1+n)*(2+n)*(3+n)*(3+2*n)*(1+36*n-69*n^2+45*n^4+18*n^5+2*n^6)/3960 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (x + 247*x^2 + 4293*x^3 + 15619*x^4 + 15619*x^5 + 4293*x^6 + 247*x^7 + x^8)/(- 1 + x)^12.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(3 + 2*n)*(1 + 36*n - 69*n^2 + 45*n^4 + 18*n^5 + 2*n^6)/3960.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + n^8.

A255178 Second differences of eighth powers (A001016).

Original entry on oeis.org

1, 254, 6050, 52670, 266114, 963902, 2796194, 6927230, 15257090, 30683774, 57405602, 101263934, 170126210, 274309310, 427043234, 644975102, 948713474, 1363412990, 1919399330, 2652834494, 3606422402, 4830154814, 6382097570, 8329217150, 10748247554

Offset: 0

Luciano Ancora, Feb 21 2015

			Second differences:  1, 254, 6050, 52670, 266114, ... (this sequence)
First differences:   1, 255, 6305, 58975, 325089, ... (A022524)
----------------------------------------------------------------------
The eighth powers:   1, 256, 6561, 65536, 390625, ... (A001016)
----------------------------------------------------------------------
First partial sums:  1, 257, 6818, 72354, 462979, ... (A000542)
Second partial sums: 1, 258, 7076, 79430, 542409, ... (A253636)
Third partial sums:  1, 259, 7335, 86765, 629174, ... (A254642)
Fourth partial sums: 1, 260, 7595, 94360, 723534, ... (A254647)

Luciano Ancora, Table of n, a(n) for n = 0..1000
Luciano Ancora, Sums of powers of positive integers and their recurrence relations, section 0.5.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000542, A001016, A022524, A253636, A254642, A254647, A255177.

[n eq 0 select 1 else 2*(28*n^6+70*n^4+28*n^2+1): n in [0..30]]; // Vincenzo Librandi, Mar 12 2015

Join[{1}, Table[2 (28 n^6 + 70 n^4 + 28 n^2 + 1), {n, 1, 30}]]
Join[{1},Differences[Range[0,30]^8,2]] (* Harvey P. Dale, Aug 26 2024 *)

G.f.: (1 + x)*(1 + 246*x + 4047*x^2 + 11572*x^3 + 4047*x^4 + 246*x^5 + x^6)/(1 - x)^7.

a(n) = 2*(28*n^6 + 70*n^4 + 28*n^2 + 1) for n>0, a(0)=1.

Edited by Bruno Berselli, Mar 19 2015

cp's OEIS Frontend

A000538 Sum of fourth powers: 0^4 + 1^4 + ... + n^4.

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A001016 Eighth powers: a(n) = n^8.

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A103438 Square array T(m,n) read by antidiagonals: Sum_{k=1..n} k^m.

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A254647 Fourth partial sums of eighth powers (A001016).

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A323546 a(n) = Product_{k=0..n} (k^8 + (n-k)^8).

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