A000541 -id:A000541 - 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

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

A000542 Sum of 8th powers: 1^8 + 2^8 + ... + n^8.

Original entry on oeis.org

0, 1, 257, 6818, 72354, 462979, 2142595, 7907396, 24684612, 67731333, 167731333, 382090214, 812071910, 1627802631, 3103591687, 5666482312, 9961449608, 16937207049, 27957167625, 44940730666, 70540730666, 108363590027, 163239463563, 241550448844

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.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
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].
Bruno Berselli, A description of the recursive method in Comments lines: 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.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Row 8 of array A103438.

Cf. A069093.

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^8 od: seq(a[n], n=0..23); # Zerinvary Lajos, Feb 22 2008

lst={};s=0;Do[s=s+n^8;AppendTo[lst, s], {n, 10^2}];lst..or..Table[Sum[k^8, {k, 1, n}], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *)
s = 0; lst = {s}; Do[s += n^8; AppendTo[lst, s], {n, 1, 30, 1}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
Accumulate[Range[0,30]^8] (* Harvey P. Dale, Jun 17 2015 *)

a(n)=n*(n+1)*(2*n+1)*(5*n^6+15*n^5+5*n^4-15*n^3-n^2+9*n-3)/90 \\ Charles R Greathouse IV, Sep 28 2015

A000542_list, m = [0], [40320, -141120, 191520, -126000, 40824, -5796, 254, -1, 0, 0]
for _ in range(24):
    for i in range(9):
        m[i+1] += m[i]
    A000542_list.append(m[-1])
print(A000542_list) # Chai Wah Wu, Nov 05 2014

[bernoulli_polynomial(n,9)/9 for n in range(1, 25)] # Zerinvary Lajos, May 17 2009

a(n) = n*(n+1)*(2*n+1)*(5*n^6 + 15*n^5 + 5*n^4 - 15*n^3 - n^2 + 9*n - 3)/90.
a(n) = n*A000541(n) - Sum_{i=0..n-1} A000541(i). - Bruno Berselli, Apr 26 2010
G.f.: x*(x+1)*(x^6 + 246*x^5 + 4047*x^4 + 11572*x^3 + 4047*x^2 + 246*x + 1)/(x-1)^10. - Colin Barker, May 27 2012
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) + 40320. - Ant King, Sep 24 2013
a(n) = -Sum_{j=1..8} j*Stirling1(n+1,n+1-j)*Stirling2(n+8-j,n). - Mircea Merca, Jan 25 2014
a(n) = Sum_{i = 1..n} J_8(i)*floor(n/i), where J_8 is A069093. - Ridouane Oudra, Jul 17 2025

A323545 a(n) = Product_{k=0..n} (k^7 + (n-k)^7).

Original entry on oeis.org

0, 1, 32768, 79593387129, 328983774635229184, 8781626117710113525390625, 570409595340477623191338982834176, 112244673425189306235795780017831813874289, 49449149324106963036650868175987491957290049732608, 48527312221741371319651099141827554314119977393170380398241

Offset: 0

Vaclav Kotesovec, Jan 17 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..79

Cf. A323535, A323540, A323541, A323542, A323543, A323544, A323546, A320345.

Cf. 2*A000541 (with sum instead of product).

m:=7; [(&*[k^m + (n-k)^m: k in [0..n]]): n in [0..10]]; // G. C. Greubel, Jan 18 2019

Table[Product[k^7+(n-k)^7, {k, 0, n}], {n, 0, 10}]

m=7; vector(10, n, n--; prod(k=0,n, k^m + (n-k)^m)) \\ G. C. Greubel, Jan 18 2019

m=7; [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((8*cos((Pi + arctan(2769*sqrt(3)/239))/6)*Pi/sqrt(21)-6)*n) * n^(7*n+7).

Equivalently, a(n) ~ exp((4*Pi*sqrt(2*(13 + 19*sin(Pi/14) - sin(3*Pi/14))/7)/7 - 6)*n) * n^(7*n+7). - Vaclav Kotesovec, Jan 23 2019

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

Original entry on oeis.org

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

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.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. 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 (1).

For other sequences based upon MagicNKZ(n,k,z):
..... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
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
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

New name from Joerg Arndt, Nov 30 2014

Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

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

A059977 a(n) = binomial(n+2, 2)^4.

Original entry on oeis.org

1, 81, 1296, 10000, 50625, 194481, 614656, 1679616, 4100625, 9150625, 18974736, 37015056, 68574961, 121550625, 207360000, 342102016, 547981281, 855036081, 1303210000, 1944810000, 2847396321, 4097152081, 5802782976, 8100000000, 11156640625, 15178486401

Offset: 0

Robert G. Wilson v, Mar 06 2001

nonn

Number of 4-dimensional cage assemblies.

See Chap. 61, "Hyperspace Prisons", of C. Pickover's book "Wonders of Numbers" for full explanation of "cage numbers."

			1 = (1 + 1)/2, 81 = (33 + 129)/2, 1296 = (276 + 2316)/2, 10000 = (1300 + 18700)/2, ... - _Philippe Deléham_, May 25 2015

Clifford A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Oxford University Press, 2001, p. 325.

Harry J. Smith, Table of n, a(n) for n = 0..1000
Clifford A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000217, A000539, A000541, A059827, A059860.

Cf. A168364 (first differences).

with (combinat):seq(mul(stirling2(n+1,n),k=1..4),n=1..24); # Zerinvary Lajos, Dec 16 2007

m = 4; Table[ ( (n^m)(n + 1)^m )/(2^m), {n, 1, 30} ]

a(n) = { ((n + 1)*(n + 2)/2)^4 } \\ Harry J. Smith, Jun 30 2009

[stirling_number2(n+1,n)^4for n in range(1,25)] # Zerinvary Lajos, Mar 14 2009

L(n) = ((n^m)(n + 1)^m)/(2^m) where m is the dimension, which in this case is 4.
O.g.f.: -(1+72*x+603*x^2+1168*x^3+603*x^4+72*x^5+x^6)/(-1+x)^9. - R. J. Mathar, Mar 31 2008
a(n) = A000217(n+1)^4. - R. J. Mathar, Dec 13 2011
a(n) = (A000539(n+1) + A000541(n+1))/2. - Philippe Deléham, May 25 2015
From Amiram Eldar, May 15 2022: (Start)
Sum_{n>=0} 1/a(n) = 160*Pi^2/3 + 16*Pi^4/45 - 560.
Sum_{n>=0} (-1)^n/a(n) = 560 - 640*log(2) - 96*zeta(3). (End)

Better definition from Zerinvary Lajos, May 23 2006

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

A255177 Second differences of seventh powers (A001015).

Original entry on oeis.org

1, 126, 1932, 12138, 47544, 140070, 341796, 730002, 1412208, 2531214, 4270140, 6857466, 10572072, 15748278, 22780884, 32130210, 44327136, 59978142, 79770348, 104476554, 134960280, 172180806, 217198212, 271178418

Offset: 0

Luciano Ancora, Feb 21 2015

			Second differences:  1, 126, 1932, 12138,  47544, ... (this sequence)
First differences:   1, 127, 2060, 14324,  63801, ... (A152726)
----------------------------------------------------------------------
The seventh powers:  1, 128, 2187, 16384,  78125, ... (A001015)
----------------------------------------------------------------------
First partial sums:  1, 129, 2316, 18700,  96825, ... (A000541)
Second partial sums: 1, 130, 2446, 21146, 117971, ... (A250212)
Third partial sums:  1, 131, 2577, 23723, 141694, ... (A254641)
Fourth partial sums: 1, 132, 2709, 26432, 168126, ... (A254646)
Fifth partial sums:  1, 133, 2842, 29274, 197400, ... (A254684)

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 (6,-15,20,-15,6,-1).

Cf. A000541, A001015, A152726, A250212, A254641, A254646, A254684, A255178, A255179.

[1] cat [14*(-1+n)*(9-22*n+23*n^2-12*n^3+3*n^4): n in [2..30]]; // Vincenzo Librandi, Mar 12 2015

Join[{1}, Table[14 n (3 n^4 + 5 n^2 + 1), {n, 1, 30}], {n, 0, 24}] (* or *)
CoefficientList[Series[(1 + 120 x + 1191 x^2 + 2416 x^3 + 1191 x^4 + 120 x^5 + x^6)/(1 - x)^6, {x, 0, 22}], x]

def A255177(n): return 14*n*(n**2*(3*n**2 + 5) + 1) if n else 1 # Chai Wah Wu, Oct 07 2024

G.f.: (1 + 120*x + 1191*x^2 + 2416*x^3 + 1191*x^4 + 120*x^5 + x^6)/(1 - x)^6.
a(n) = 14*n*(3*n^4 + 5*n^2 + 1) for n>0, a(0)=1.
a(n) = A022523(n)-A022523(n-1). - R. J. Mathar, Jul 16 2015

Edited by Bruno Berselli, Mar 19 2015

A254646 Fourth partial sums of seventh powers (A001015).

Original entry on oeis.org

1, 132, 2709, 26432, 168126, 804552, 3136014, 10459968, 30856839, 82407052, 202678203, 465069696, 1005729452, 2066218896, 4058958828, 7664805504, 13974953853, 24692818836, 42415687153, 71020845504, 116186669130, 186085891160, 292296070170, 450981236160, 684408934755

Offset: 1

Luciano Ancora, Feb 05 2015

			First differences:   1, 127, 2059, 14197,  61741, ...  (A022523)
----------------------------------------------------------------------
The seventh powers:  1, 128, 2187, 16384,  78125, ...  (A001015)
----------------------------------------------------------------------
First partial sums:  1, 129, 2316, 18700,  96825, ...  (A000541)
Second partial sums: 1, 130, 2446, 21146, 117971, ...  (A250212)
Third partial sums:  1, 131, 2577, 23723, 141694, ...  (A254641)
Fourth partial sums: 1, 132, 2709, 26432, 168126, ...  (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. A000541, A001015, A022523, A250212, A254641, A254645.

List([1..30], n-> Binomial(n+4,5)*(3*(n+2)^6 -40*(n+2)^4 +151*(n+2)^2 -108)/198); # G. C. Greubel, Aug 28 2019

[Binomial(n+4,5)*(3*(n+2)^6 -40*(n+2)^4 +151*(n+2)^2 -108)/198: n in [1..30]]; // G. C. Greubel, Aug 28 2019

seq(binomial(n+4,5)*(3*(n+2)^6 -40*(n+2)^4 +151*(n+2)^2 -108)/198, n=1..30); # G. C. Greubel, Aug 28 2019

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (48 - 100 n - 89 n^2 + 160 n^3 + 140 n^4 + 36 n^5 + 3 n^6)/23760, {n, 20}] (* or *)
Accumulate[Accumulate[Accumulate[Accumulate[Range[20]^7]]]] (* or *)
CoefficientList[Series[(1 + 120 x + 1191 x^2 + 2416 x^3 + 1191 x^4 + 120 x^5 + x^6)/(- 1 + x)^12, {x, 0, 19}], x]

a(n)=n*(1+n)*(2+n)*(3+n)*(4+n)*(48-100*n-89*n^2+160*n^3+140*n^4 +36*n^5+3*n^6)/23760 \\ Charles R Greathouse IV, Oct 07 2015

[binomial(n+4,5)*(3*(n+2)^6 -40*(n+2)^4 +151*(n+2)^2 -108)/198 for n in (1..30)] # G. C. Greubel, Aug 28 2019

G.f.: x*(1 +120*x +1191*x^2 +2416*x^3 +1191*x^4 +120*x^5 +x^6)/(1-x)^12.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(48 - 100*n - 89*n^2 + 160*n^3 + 140*n^4 + 36*n^5 + 3*n^6)/23760.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + n^7.

cp's OEIS Frontend

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

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

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A000542 Sum of 8th powers: 1^8 + 2^8 + ... + n^8.

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A323545 a(n) = Product_{k=0..n} (k^7 + (n-k)^7).

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A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

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A023002 Sum of 10th powers.

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