A000539 -id:A000539 - cp's OEIS frontend

A000584 Fifth powers: a(n) = n^5.

0, 1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049, 100000, 161051, 248832, 371293, 537824, 759375, 1048576, 1419857, 1889568, 2476099, 3200000, 4084101, 5153632, 6436343, 7962624, 9765625, 11881376, 14348907, 17210368, 20511149

Offset: 0

N. J. A. Sloane

Totally multiplicative sequence with a(p) = p^5 for prime p. - Jaroslav Krizek, Nov 01 2009
The binomial transform yields A059338. The inverse binomial transform yields the (finite) 0, 1, 30, 150, 240, 120, the 5th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
Equals sum of odd numbers from n^2*(n-1)+1 (A100104) to n^2*(n+1)-1 (A003777). - Bruno Berselli, Mar 14 2014
a(n) mod 10 = n mod 10. - Reinhard Zumkeller, May 10 2014
Numbers of the form a(n) + a(n+1) + ... + a(n+k) are nonprime for all n, k>=0; this can be proved by the method indicated in the comment in A256581. - Vladimir Shevelev and Peter J. C. Moses, Apr 04 2015

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
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).

Franklin T. Adams-Watters, Table of n, a(n) for n = 0..500
Brady Haran and Simon Pampena, Fifth Root Trick, Numberphile video (2014)
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
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Partial sums give A000539.
Cf. A000012, A001477, A000290, A000578, A000583, A062392, A022521 (first differences), A008292, A173018, A123125.
Cf. A013663, A162624, A267316.

a000584 = (^ 5)  -- Reinhard Zumkeller, Nov 11 2012

Range[0,50]^5 (* Vladimir Joseph Stephan Orlovsky, Feb 20 2011 *)

makelist(n^5, n, 0, 30); /* Martin Ettl, Nov 12 2012 */

a(n)=n^5 \\ Charles R Greathouse IV, Jul 28 2015

[n**5 for n in range(30)]  # Zerinvary Lajos, Jun 03 2009

G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (x-1)^6. [Simon Plouffe in his 1992 dissertation]
Multiplicative with a(p^e) = p^(5e). - David W. Wilson, Aug 01 2001
E.g.f.: exp(x)*(x+15*x^2+25*x^3+10*x^4+x^5). - Geoffrey Critzer, Jun 12 2013
a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 120. - Ant King, Sep 23 2013
a(n) = n + Sum_{j=0..n-1}{k=1..4}binomial(5,k)*j^(5-k). - Patrick J. McNab, Mar 28 2016
From Kolosov Petro, Oct 22 2018: (Start)
a(n) = Sum_{k=1..n} A300656(n,k).
a(n) = Sum_{k=0..n-1} A300656(n,k). (End)
a(n) = Sum_{k=1..5} Eulerian(5, k)*binomial(n+5-k, 5), with Eulerian(5, k) = A008292(5, k), the numbers 1, 26, 66, 26, 1, for n >= 0. Worpitzki's identity for powers of 5. See. e.g., Graham et al., eq. (6, 37) (using A173018, the row reversed version of A123125). - Wolfdieter Lang, Jul 17 2019
From Amiram Eldar, Oct 08 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(5) (A013663).
Sum_{n>=1} (-1)^(n+1)/a(n) = 15*zeta(5)/16 (A267316). (End)

More terms from Henry Bottomley, Jun 21 2001

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

Original entry on oeis.org

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

A000540 Sum of 6th powers: 0^6 + 1^6 + 2^6 + ... + n^6.

Original entry on oeis.org

0, 1, 65, 794, 4890, 20515, 67171, 184820, 446964, 978405, 1978405, 3749966, 6735950, 11562759, 19092295, 30482920, 47260136, 71397705, 105409929, 152455810, 216455810, 302221931, 415601835, 563637724, 754740700, 998881325, 1307797101, 1695217590

Offset: 0

N. J. A. Sloane

This sequence is related to A000539 by a(n) = n*A000539(n)-sum(A000539(i), i=0..n-1). - Bruno Berselli, Apr 26 2010

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.
J. L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, (2008), p. 289.
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, A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359. [Annotated scanned copy]
B. 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.
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
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1)

Cf. A101093, A000539.
Row 6 of array A103438.
Partial sums of A001014.

a000540 n = a000540_list !! n
a000540_list = scanl1 (+) a001014_list -- Reinhard Zumkeller, Dec 04 2011

[n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42: n in [0..30]]; // Vincenzo Librandi, Apr 04 2015

a:=n->sum (j^6,j=0..n): seq(a(n),n=0..27); # Zerinvary Lajos, Jun 27 2007
A000540:=(z+1)*(z**4+56*z**3+246*z**2+56*z+1)/(z-1)**8; # g.f. by Simon Plouffe in his 1992 dissertation, without the leading 0.
A000540 := proc(n) n^7/7+n^6/2+n^5/2-n^3/6+n/42 ; end proc: # R. J. Mathar

Accumulate[Range[0,30]^6] (* Harvey P. Dale, Jul 30 2009 *)
LinearRecurrence[{8, -28, 56, -70, 56, -28, 8, -1}, {0, 1, 65, 794, 4890, 20515, 67171, 184820}, 31] (* Jean-François Alcover, Feb 09 2016 *)

a(n)=n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42 \\ Edward Jiang, Sep 10 2014

a(n)=sum(i=1, n, i^6); \\ Michel Marcus, Sep 11 2014

A000540_list, m = [0], [720, -1800, 1560, -540, 62, -1, 0, 0]
for _ in range(10**2):
    for i in range(7):
        m[i+1] += m[i]
    A000540_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014

[bernoulli_polynomial(n,7)/7 for n in range(1, 29)]# Zerinvary Lajos, May 17 2009

a(n) = n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42.
a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^6). - Alexander Adamchuk, Oct 26 2004
G.f.: A(x) = 3*x/7*G(0); with G(k) = 1 + 2/(k+1+(k+1)/(2*k^2 + 4*k + 1 + 2*(k+1)^2/(3*k + 2 - 9*x*(k+1)*(k+2)^4*(k+3)*(2*k+5)/(3*x*(k+2)^4*(k+3)*(2*k+5)+(k+1)*(2*k+3)/G(k+1))))); (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
G.f.: x*(1+x)*(x^4 + 56*x^3 + 246*x^2 + 56*x + 1) / (x-1)^8 . - R. J. Mathar, Aug 07 2012
a(n) = Sum_{i=1..n} J_6(i)*floor(n/i), where J_6 is A069091. - Enrique Pérez Herrero, Mar 09 2013
a(n) = 7*a(n-1) - 21* a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + 720. - Ant King, Sep 24 2013
a(n) = -Sum_{j=1..6} j*Stirling1(n+1,n+1-j)*Stirling2(n+6-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = 84*Pi*(8*cos(sqrt((sqrt(93) + 9)/6)*Pi) + 15*cos(sqrt((sqrt(93) + 9)/6)*Pi/2) * cosh(sqrt((sqrt(93) - 9)/6)*Pi/2) + 8*cosh(sqrt((sqrt(93) - 9)/6)*Pi) - 7*sqrt(3)*sin(sqrt((sqrt(93) + 9)/6)*Pi/2) * sinh(sqrt((sqrt(93) - 9)/6)*Pi/2)) / (31*(cos(sqrt((sqrt(93) + 9)/6)*Pi) + cosh(sqrt((sqrt(93) - 9)/6)*Pi))) = 0.985708051237101247832970793342271511... . - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)*(n + 1/2)*n*(n + 1/2 + z)*(n + 1/2 - z)*(n + 1/2 + zbar)*(n + 1/2 - zbar)/7, with I^2 = -1 and z = 2^(-3/2)*3^(-1/4)*(sqrt(sqrt(31) + 3*sqrt(3)) + I*sqrt(sqrt(31) - 3*sqrt(3))), and zbar is the complex conjugate of z. See the Graham et al. reference, eq. (6.98), pp. 288-289 (with n -> n+1). (There was a typo in the first edition, which was corrected in the second edition.) - Wolfdieter Lang, Apr 03 2015
a(n+2) = 36*A086020(n+1) + 24*A005585(n+1) + A000330(n+2). - Yasser Arath Chavez Reyes, Apr 16 2024

A000541 Sum of 7th powers: 1^7 + 2^7 + ... + n^7.

Original entry on oeis.org

0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 3297456, 8080425, 18080425, 37567596, 73399404, 136147921, 241561425, 412420800, 680856256, 1091194929, 1703414961, 2597286700, 3877286700, 5678375241, 8172733129, 11577558576, 16164030000, 22267545625

Offset: 0

N. J. A. Sloane

a(n) is divisible by A000537(n) if and only n is congruent to 1 mod 3 (see A016777) - Artur Jasinski, Oct 10 2007

This sequence is related to A000540 by a(n) = n*A000540(n) - Sum_{i=0..n-1} A000540(i). - Bruno Berselli, Apr 26 2010

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 (9,-36,84,-126,126,-84,36,-9,1).

Row 7 of array A103438.

Cf. A000217, A000537, A000539, A069092.

[n^2*(n+1)^2*(3*n^4+6*n^3-n^2-4*n+2)/24: n in [0..30]]; // Vincenzo Librandi, Feb 20 2016

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^7 od: seq(a[n], n=0..25); # Zerinvary Lajos, Feb 22 2008

Table[Sum[k^7, {k, 1, n}], {n, 0, 100}] (* Artur Jasinski, Oct 10 2007 *)
s = 0; lst = {s}; Do[s += n^7; AppendTo[lst, s], {n, 1, 30, 1}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 3297456}, 35] (* Vincenzo Librandi, Feb 20 2016 *)

a(n)=n^2*(n+1)^2*(3*n^4+6*n^3-n^2-4*n+2)/24 \\ Edward Jiang, Sep 10 2014

a(n) = sum(i=1, n, i^7); \\ Michel Marcus, Sep 11 2014

A000541_list, m = [0], [5040, -15120, 16800, -8400, 1806, -126, 1, 0, 0]
for _ in range(10**2):
    for i in range(8):
        m[i+1] += m[i]
    A000541_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014

a(n) = n^2*(n+1)^2*(3*n^4 + 6*n^3 - n^2 - 4*n + 2)/24.
a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^7). - Alexander Adamchuk, Oct 26 2004
Jacobi formula: a(n) = 2(A000217(n))^4 - A000539(n). - Artur Jasinski, Oct 10 2007
G.f.: x*(1 + 120*x + 1191*x^2 + 2416*x^3 + 1191*x^4 + 120*x^5 + x^6)/(1-x)^9. - Colin Barker, May 25 2012
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) + 5040. - Ant King, Sep 24 2013
a(n) = -Sum_{j=1..7} j*Stirling1(n+1,n+1-j)*Stirling2(n+7-j,n). - Mircea Merca, Jan 25 2014
a(n) = 2*A000217(n)^4 - (4/3)*A000217(n)^3 + (1/3)*A000217(n)^2. - Michael Raney, Feb 19 2016
a(n) = 72*A288876(n-2) + 48*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, Apr 27 2024
a(n) = Sum_{i=1..n} J_7(i)*floor(n/i), where J_7 is A069092. - Ridouane Oudra, Jul 17 2025

A323543 a(n) = Product_{k=0..n} (k^5 + (n-k)^5).

Original entry on oeis.org

0, 1, 2048, 64304361, 3995393327104, 775913238525390625, 320224500476333990608896, 273342392644434762426370643281, 429621172463958849019228299940855808, 1175198860360296464427314161342724729270241, 5278148679274118560000000000000000000000000000000

Offset: 0

Vaclav Kotesovec, Jan 17 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..104

Cf. A272248, A323533, A323540, A323541, A323542, A323544, A323545, A323546.

Cf. 2*A000539 (with sum instead of product).

[(&*[(k^5 + (n-k)^5): k in [0..n]]): n in [0..12]]; // Vincenzo Librandi, Jan 18 2019

Table[Product[k^5+(n-k)^5, {k, 0, n}], {n, 0, 12}]

m=5; vector(12, n, n--; prod(k=0,n, k^m +(n-k)^m)) \\ G. C. Greubel, Jan 18 2019

m=5; [product(k^m +(n-k)^m for k in (0..n)) for n in (0..12)] # G. C. Greubel, Jan 18 2019

a(n) ~ exp((2*Pi*sqrt(5 - 2/sqrt(5))/5 - 4)*n) * n^(5*n+5).

A006324 a(n) = n(n + 1)(2n^2 + 2n - 1)/6.

Original entry on oeis.org

1, 11, 46, 130, 295, 581, 1036, 1716, 2685, 4015, 5786, 8086, 11011, 14665, 19160, 24616, 31161, 38931, 48070, 58730, 71071, 85261, 101476, 119900, 140725, 164151, 190386, 219646, 252155, 288145, 327856, 371536, 419441, 471835, 528990, 591186

Offset: 1

Albert Rich (Albert_Rich(AT)msn.com), Jun 14 1998

4-dimensional analog of centered polygonal numbers.
Partial sums of A000447. - Zak Seidov, May 19 2006
From Johannes W. Meijer, Jun 27 2009: (Start)
Equals the absolute values of the coefficients that precede the a(n-1) factors of the recurrence relations RR(n) of A162011.
This sequence enabled the analysis of A162012 and A162013. (End)
Equals the number of integer quadruples (x,y,z,w) such that min(x,y) < min(z,w), max(x,y) < max(z,w), and 0 <= x,y,z,w <= n. - Andrew Woods, Apr 21 2014
For n>3 a(n)=twice the area of an irregular quadrilateral with vertices at the points (C(n,4),C(n+1,4)), (C(n+1,4),C(n+2,4)), (C(n+2,4),C(n+3,4)), and (C(n+3,4),C(n+4,4)). - J. M. Bergot, Jun 14 2014

Delbert L. Johnson, Table of n, a(n) for n = 1..20000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000447, A000539, A000217, A002415, A053134.

Cf. A162011, A162012, a(n-2), and A162013, a(n-3). - Johannes W. Meijer, Jun 27 2009

[ n*(n + 1)*(2*n^2 + 2*n - 1)/6 : n in [1..30] ]; // Wesley Ivan Hurt, Jun 14 2014

A006324:=n->n*(n + 1)*(2*n^2 + 2*n - 1)/6; seq(A006324(n), n=1..30); # Wesley Ivan Hurt, Jun 14 2014

Table[Sum[k^5,{k,n}]/Sum[k,{k,n}], {n,40}] (* Alexander Adamchuk, Apr 12 2006 *)

a(n) = 8*C(n + 2, 4) + C(n + 1, 2).
a(n) = (Sum_{k=1..n} k^5) / (Sum_{k=1..n} k) = A000539(n) / A000217(n). - Alexander Adamchuk, Apr 12 2006
From Johannes W. Meijer, Jun 27 2009: (Start)
Recurrence relation 0 = Sum_{k=0..5} (-1)^k*binomial(5,k)*a(n-k).
G.f.: (1+6*z+z^2)/(1-z)^5. (End)
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Wesley Ivan Hurt, May 02 2021
Sum_{n>=1} 1/a(n) = 6 + 2*sqrt(3)*Pi*tan(sqrt(3)*Pi/2). - Amiram Eldar, Aug 23 2022
a(n) = A053134(n-1) - 4*A002415(n). - Yasser Arath Chavez Reyes, Feb 12 2024

Simpler definition from Alexander Adamchuk, Apr 12 2006

More terms from Zak Seidov

A101092 Second partial sums of fifth powers (A000584).

Original entry on oeis.org

1, 34, 310, 1610, 6035, 18236, 47244, 109020, 229845, 450670, 832546, 1463254, 2465255, 4005080, 6304280, 9652056, 14419689, 21076890, 30210190, 42543490, 58960891, 80531924, 108539300, 144509300, 190244925, 247861926, 319827834, 409004110, 518691535

Offset: 1

Cecilia Rossiter, Dec 14 2004

Vincenzo Librandi, 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.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 90--99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014 (But beware of a typo)
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000539, A101099.

Cf. A254640.

[(n*(1+n)*(2+n)*(-1+n*(2+n))*(1+2*n*(2+n)))/84: n in [1..40]]; // Vincenzo Librandi, Mar 24 2014

f:=n->(2*n^7-7*n^5+7*n^3-2*n)/84;
[seq(f(n),n=0..50)];  # N. J. A. Sloane, Mar 23 2014

CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 24 2014 *)
Nest[Accumulate,Range[30]^5,2] (* or *) LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{1,34,310,1610,6035,18236,47244,109020},30] (* Harvey P. Dale, Aug 22 2022 *)

a(n)=n*(n+1)*(n+2)*(n*(n+2)-1)*(2*n*(2 + n)+1)/84 \\ Charles R Greathouse IV, Apr 21 2015

def A101092(n): return n*(n*(n*(n*(n*(n*(n+7<<1)+35)+35)+7)-7)-2)//84 # Chai Wah Wu, Oct 03 2024

[n*(1+n)*(2+n)*(n*(2+n)-1)*(1+2*n*(2+n))/84 for n in range(1,30)] # Danny Rorabaugh, Apr 21 2015

a(n) = (n*(1 + n)*(2 + n)*(-1 + n*(2 + n))*(1 + 2*n*(2 + n)))/84.
G.f.: x*(1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1-x)^8. - Colin Barker, Apr 16 2012
a(n) = Sum_{i=1..n} i*(n+1-i)^5, by the definition. - Bruno Berselli, Jan 31 2014
a(n) = 2*a(n-1) - a(n-2) + n^5. - Luciano Ancora, Jan 08 2015
E.g.f.: exp(x)*x*(84 + 1344*x + 2954*x^2 + 1995*x^3 + 525*x^4 + 56*x^5 + 2*x^6)/84. - Stefano Spezia, May 04 2024

Edited by Ralf Stephan, Dec 16 2004

A031138 Numbers k such that 1^5 + 2^5 + ... + k^5 is a square.

Original entry on oeis.org

1, 13, 133, 1321, 13081, 129493, 1281853, 12689041, 125608561, 1243396573, 12308357173, 121840175161, 1206093394441, 11939093769253, 118184844298093, 1169909349211681, 11580908647818721, 114639177128975533, 1134810862641936613, 11233469449290390601

Offset: 1

Ignacio Larrosa Cañestro, entry revised Feb 27 2000

Partial sums of A004291 or convolution of A040000 with A054320. - R. J. Mathar, Oct 26 2009
This is a 6th-degree Diophantine equation 12*m^2 = n^2*(n+1)^2*(2*n^2 + 2*n - 1) which reduces to the generalized Pell equation 6*q^2 = (2*n + 1)^2 - 3 where q = 3*m/(n*(n+1)), so there is no surprise that the solutions satisfy a linear recurrent equation. - Charles R Greathouse IV, Max Alekseyev, Oct 22 2012
Also k such that k^2 + (k+1)^2 is equal to the sum of three consecutive squares, for example 13^2 + 14^2 = 10^2 + 11^2 + 12^2. - Colin Barker, Sep 06 2015

			a(2) = 13 because 1^5+2^5+...13^5 = 1001^2; a(1) = 1 because 1^5 = 1^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (11, -11, 1).

Cf. A000539, A006061, A054320, A007667.

[Round(-1/2 + ((3 - Sqrt(6))/4)*(5 + 2*Sqrt(6))^n + ((3 + Sqrt(6) )/4)*(5 - 2*Sqrt(6))^n): n in [0..50]]; // G. C. Greubel, Nov 04 2017

LinearRecurrence[{11,-11,1},{1,13,133},20 ] (* Harvey P. Dale, Oct 23 2012 *)

isok(n) = issquare(sum(i=1, n, i^5)); \\ Michel Marcus, Dec 28 2013

Vec(x*(1+x)^2/((1-x)*(x^2-10*x+1)) + O(x^40)) \\ Colin Barker, Sep 06 2015

a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = -1/2 + ((3 - sqrt(6))/4)*(5 + 2*sqrt(6))^n + ((3 + sqrt(6))/4)*(5 - 2*sqrt(6))^n.
a(n)^2 + (a(n) + 1)^2 = (b(n) - 1)^2 + b(n)^2 + (b(n) + 1)^2 = c(n) = 3*d(n) + 2; where b(n) is A054320, c(n) is A007667 and d(n) is A006061.
a(n) = 10*a(n-1) - a(n-2) + 4; a(0) = a(1) = 1. Also sum of first a(n) fifth powers is a square m^2, where m has factors A000217{a(n)} and A054320(n). - Lekraj Beedassy, Jul 08 2002
contfrac(sqrt(6)/A054320(n))[4]/2 - Thomas Baruchel, Dec 02 2003
G.f.: x*(1+x)^2/((1-x)*(x^2-10*x+1)). - R. J. Mathar, Oct 26 2009

A215083 Triangle T(n,k) = sum of the k first n-th powers.

Original entry on oeis.org

0, 0, 1, 0, 1, 5, 0, 1, 9, 36, 0, 1, 17, 98, 354, 0, 1, 33, 276, 1300, 4425, 0, 1, 65, 794, 4890, 20515, 67171, 0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 0, 1, 257, 6818, 72354, 462979, 2142595, 7907396, 24684612, 0, 1, 513, 20196, 282340, 2235465, 12313161, 52666768, 186884496, 574304985, 0, 1, 1025, 60074, 1108650, 10874275, 71340451, 353815700, 1427557524, 4914341925, 14914341925

Offset: 0

Olivier Gérard, Aug 02 2012

First term T(0,0) = 0 can be computed as 1 if one starts the sum at j=0 and take the convention 0^0 = 1.

			Triangle starts (using the convention 0^0 = 1, see the first comment):
[0] 1
[1] 0, 1
[2] 0, 1,  5
[3] 0, 1,  9,  36
[4] 0, 1, 17,  98,  354
[5] 0, 1, 33, 276, 1300,  4425
[6] 0, 1, 65, 794, 4890, 20515, 67171

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Row sums are A215083.
A215078 is the product of this array with the binomial array.
T(3,k) is the beginning of A000537.
T(4,k) is the beginning of A000538.
T(5,k) is the beginning of A000539.
Cf. A103438.

A215083 := (n, k) -> add(i^n, i=0..k):
for n from 0 to 8 do seq(A215083(n, k), k=0..n) od; # Peter Luschny, Oct 02 2017

Flatten[Table[Table[Sum[j^n, {j, 1, k}], {k, 0, n}], {n, 0, 10}], 1]
Table[ HarmonicNumber[k, -n], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2013 *)

T(n, k) = Sum_{j=1..k} j^n

Sum_{j=0..n}((-1)^(n-j)/(j+1)*binomial(n+1,j+1)*T(n,j)) are the Bernoulli numbers B(n) = B(n, 1) by a formula of L. Kronecker. - Peter Luschny, Oct 02 2017

cp's OEIS Frontend

A000584 Fifth powers: a(n) = n^5.

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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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A000540 Sum of 6th powers: 0^6 + 1^6 + 2^6 + ... + n^6.

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A000541 Sum of 7th powers: 1^7 + 2^7 + ... + n^7.

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A006324 a(n) = n(n + 1)(2n^2 + 2n - 1)/6.