A101097 -id:A101097 - cp's OEIS frontend

A000578 The cubes: a(n) = n^3.

0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507

Offset: 0

N. J. A. Sloane

a(n) is the sum of the next n odd numbers; i.e., group the odd numbers so that the n-th group contains n elements like this: (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), (21, 23, 25, 27, 29), ...; then each group sum = n^3 = a(n). Also the median of each group = n^2 = mean. As the sum of first n odd numbers is n^2 this gives another proof of the fact that the n-th partial sum = (n(n + 1)/2)^2. - Amarnath Murthy, Sep 14 2002
Total number of triangles resulting from criss-crossing cevians within a triangle so that two of its sides are each n-partitioned. - Lekraj Beedassy, Jun 02 2004. See Propp and Propp-Gubin for a proof.
Also structured triakis tetrahedral numbers (vertex structure 7) (cf. A100175 = alternate vertex); structured tetragonal prism numbers (vertex structure 7) (cf. A100177 = structured prisms); structured hexagonal diamond numbers (vertex structure 7) (cf. A100178 = alternate vertex; A000447 = structured diamonds); and structured trigonal anti-diamond numbers (vertex structure 7) (cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {4, 3}.
Least multiple of n such that every partial sum is a square. - Amarnath Murthy, Sep 09 2005
Draw a regular hexagon. Construct points on each side of the hexagon such that these points divide each side into equally sized segments (i.e., a midpoint on each side or two points on each side placed to divide each side into three equally sized segments or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least one side of the polygon. The result is a triangular tiling of the hexagon and the creation of a number of smaller regular hexagons. The equation gives the total number of regular hexagons found where n = the number of points drawn + 1. For example, if 1 point is drawn on each side then n = 1 + 1 = 2 and a(n) = 2^3 = 8 so there are 8 regular hexagons in total. If 2 points are drawn on each side then n = 2 + 1 = 3 and a(n) = 3^3 = 27 so there are 27 regular hexagons in total. - Noah Priluck (npriluck(AT)gmail.com), May 02 2007
The solutions of the Diophantine equation: (X/Y)^2 - X*Y = 0 are of the form: (n^3, n) with n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^2k - XY = 0 are of the form: (m*n^(2k + 1), m*n^(2k - 1)) with m >= 1, k >= 1 and n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^(2k + 1) - XY = 0 are of the form: (m*n^(k + 1), m*n^k) with m >= 1, k >= 1 and n >= 1. - Mohamed Bouhamida, Oct 04 2007
Except for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1). - K.V.Iyer, Mar 16 2009
Totally multiplicative sequence with a(p) = p^3 for prime p. - Jaroslav Krizek, Nov 01 2009
Sums of rows of the triangle in A176271, n > 0. - Reinhard Zumkeller, Apr 13 2010
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Numbers n for which order of torsion subgroup t of the elliptic curve y^2 = x^3 - n is t = 2. - Artur Jasinski, Jun 30 2010
The sequence with the lengths of the Pisano periods mod k is 1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, ... for k >= 1, apparently multiplicative and derived from A000027 by dividing every ninth term through 3. Cubic variant of A186646. - R. J. Mathar, Mar 10 2011
The number of atoms in a bcc (body-centered cubic) rhombic hexahedron with n atoms along one edge is n^3 (T. P. Martin, Shells of atoms, eq. (8)). - Brigitte Stepanov, Jul 02 2011
The inverse binomial transform yields the (finite) 0, 1, 6, 6 (third row in A019538 and A131689). - R. J. Mathar, Jan 16 2013
Twice the area of a triangle with vertices at (0, 0), (t(n - 1), t(n)), and (t(n), t(n - 1)), where t = A000217 are triangular numbers. - J. M. Bergot, Jun 25 2013
If n > 0 is not congruent to 5 (mod 6) then A010888(a(n)) divides a(n). - Ivan N. Ianakiev, Oct 16 2013
For n > 2, a(n) = twice the area of a triangle with vertices at points (binomial(n,3),binomial(n+2,3)), (binomial(n+1,3),binomial(n+1,3)), and (binomial(n+2,3),binomial(n,3)). - J. M. Bergot, Jun 14 2014
Determinants of the spiral knots S(4,k,(1,1,-1)). a(k) = det(S(4,k,(1,1,-1))). - Ryan Stees, Dec 14 2014
One of the oldest-known examples of this sequence is shown in the Senkereh tablet, BM 92698, which displays the first 32 terms in cuneiform. - Charles R Greathouse IV, Jan 21 2015
From Bui Quang Tuan, Mar 31 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ... 2*n-1 as follows. The first column contains all the integers 1, 2, 3, ... 2*n-1. Each succeeding column is the same as the previous column but without the first and last items. The last column contains only n. The sum of all the numbers in the triangle is n^3.
Here is the example for n = 4, where 1 + 2*2 + 3*3 + 4*4 + 3*5 + 2*6 + 7 = 64 = a(4):
  1
  2  2
  3  3  3
  4  4  4  4
  5  5  5
  6  6
  7
(End)
For n > 0, a(n) is the number of compositions of n+11 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
Number of inequivalent face colorings of the cube using at most n colors such that each color appears at least twice. - David Nacin, Feb 22 2017
Consider A = {a,b,c} a set with three distinct members. The number of subsets of A is 8, including {a,b,c} and the empty set. The number of subsets from each of those 8 subsets is 27. If the number of such iterations is n, then the total number of subsets is a(n-1). - Gregory L. Simay, Jul 27 2018
By Fermat's Last Theorem, these are the integers of the form x^k with the least possible value of k such that x^k = y^k + z^k never has a solution in positive integers x, y, z for that k. - Felix Fröhlich, Jul 27 2018

			For k=3, b(3) = 2 b(2) - b(1) = 4-1 = 3, so det(S(4,3,(1,1,-1))) = 3*3^2 = 27.
For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27. - _Patrick J. McNab_, Mar 28 2016

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 191.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 43, 64, 81.
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).
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 292.
T. Aaron Gulliver, "Sequences from cubes of integers", International Mathematical Journal, 4 (2003), no. 5, 439 - 445. See http://www.m-hikari.com/z2003.html for information about this journal. [I expanded the reference to make this easier to find. - N. J. A. Sloane, Feb 18 2019]
J. Propp and A. Propp-Gubin, "Counting Triangles in Triangles", Pi Mu Epsilon Journal (to appear).
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 6-7.
D. Wells, You Are A Mathematician, pp. 238-241, Penguin Books 1995.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
H. Bottomley, Illustration of initial terms
British National Museum, Tablet 92698
N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010).
M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/-1(mod m), Involve, Vol. 8 (2015), No. 3, 361-384.
Ralph Greenberg, Math For Poets
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets [Cached version at the Wayback Machine]
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75. - fixed by _Felix Fröhlich_, Jun 16 2014
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (8).
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]
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
James Propp and Adam Propp-Gubin, Counting Triangles in Triangles, arXiv:2409.17117 [math.CO], 25 September 2024.
Kenneth A. Ross, First Digits of Squares and Cubes, Math. Mag. 85 (2012) 36-42. doi:10.4169/math.mag.85.1.36.
Eric Weisstein's World of Mathematics, Cubic Number, and Hex Pyramidal Number
Ronald Yannone, Hilbert Matrix Analyses
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for "core" sequences
Index entries for sequences related to Benford's law

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
For sums of cubes, cf. A000537 (partial sums), A003072, A003325, A024166, A024670, A101102 (fifth partial sums).
Cf. A001158 (inverse Möbius transform), A007412 (complement), A030078(n) (cubes of primes), A048766, A058645 (binomial transform), A065876, A101094, A101097.
Subsequence of A145784.
Cf. A260260 (comment). - Bruno Berselli, Jul 22 2015
Cf. A000292 (tetrahedral numbers), A005900 (octahedral numbers), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Cf. A098737 (main diagonal).

a000578 = (^ 3)
a000578_list = 0 : 1 : 8 : zipWith (+)
   (map (+ 6) a000578_list)
   (map (* 3) $ tail $ zipWith (-) (tail a000578_list) a000578_list)
-- Reinhard Zumkeller, Sep 05 2015, May 24 2012, Oct 22 2011

[ n^3 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014

I:=[0,1,8,27]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jul 05 2014

A000578 := n->n^3;
seq(A000578(n), n=0..50);
isA000578 := proc(r)
    local p;
    if r = 0 or r =1 then
        true;
    else
        for p in ifactors(r)[2] do
            if op(2, p) mod 3 <> 0 then
                return false;
            end if;
        end do:
        true ;
    end if;
end proc: # R. J. Mathar, Oct 08 2013

Table[n^3, {n, 0, 30}] (* Stefan Steinerberger, Apr 01 2006 *)
CoefficientList[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *)
Accumulate[Table[3n^2+3n+1,{n,0,20}]] (* or *) LinearRecurrence[{4,-6,4,-1},{1,8,27,64},20](* Harvey P. Dale, Aug 18 2018 *)

A000578(n):=n^3$
makelist(A000578(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

A000578(n)=n^3 \\ M. F. Hasler, Apr 12 2008

is(n)=ispower(n,3) \\ Charles R Greathouse IV, Feb 20 2012

A000578_list, m = [], [6, -6, 1, 0]
for _ in range(10**2):
    A000578_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

(define (A000578 n) (* n n n)) ;; Antti Karttunen, Oct 06 2017

a(n) = Sum_{i=0..n-1} A003215(i).
Multiplicative with a(p^e) = p^(3e). - David W. Wilson, Aug 01 2001
G.f.: x*(1+4*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
Dirichlet generating function: zeta(s-3). - Franklin T. Adams-Watters, Sep 11 2005, Amarnath Murthy, Sep 09 2005
E.g.f.: (1+3*x+x^2)*x*exp(x). - Franklin T. Adams-Watters, Sep 11 2005 - Amarnath Murthy, Sep 09 2005
a(n) = Sum_{i=1..n} (Sum_{j=i..n+i-1} A002024(j,i)). - Reinhard Zumkeller, Jun 24 2007
a(n) = lcm(n, (n - 1)^2) - (n - 1)^2. E.g.: lcm(1, (1 - 1)^2) - (1 - 1)^2 = 0, lcm(2, (2 - 1)^2) - (2 - 1)^2 = 1, lcm(3, (3 - 1)^2) - (3 - 1)^2 = 8, ... - Mats Granvik, Sep 24 2007
Starting (1, 8, 27, 64, 125, ...), = binomial transform of [1, 7, 12, 6, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
a(n) = A007531(n) + A000567(n). - Reinhard Zumkeller, Sep 18 2009
a(n) = binomial(n+2,3) + 4*binomial(n+1,3) + binomial(n,3). [Worpitzky's identity for cubes. See. e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n + 6*binomial(n+1,3) = binomial(n,1)+6*binomial(n+1,3). - Ron Knott, Jun 10 2019
A010057(a(n)) = 1. - Reinhard Zumkeller, Oct 22 2011
a(n) = A000537(n) - A000537(n-1), difference between 2 squares of consecutive triangular numbers. - Pierre CAMI, Feb 20 2012
a(n) = A048395(n) - 2*A006002(n). - J. M. Bergot, Nov 25 2012
a(n) = 1 + 7*(n-1) + 6*(n-1)*(n-2) + (n-1)*(n-2)*(n-3). - Antonio Alberto Olivares, Apr 03 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6. - Ant King Apr 29 2013
a(n) = A000330(n) + Sum_{i=1..n-1} A014105(i), n >= 1. - Ivan N. Ianakiev, Sep 20 2013
a(k) = det(S(4,k,(1,1,-1))) = k*b(k)^2, where b(1)=1, b(2)=2, b(k) = 2*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014
For n >= 1, a(n) = A152618(n-1) + A033996(n-1). - Bui Quang Tuan, Apr 01 2015
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Jon Tavasanis, Feb 21 2016
a(n) = n + Sum_{j=0..n-1} Sum_{k=1..2} binomial(3,k)*j^(3-k). - Patrick J. McNab, Mar 28 2016
a(n) = A000292(n-1) * 6 + n. - Zhandos Mambetaliyev, Nov 24 2016
a(n) = n*binomial(n+1, 2) + 2*binomial(n+1, 3) + binomial(n,3). - Tony Foster III, Nov 14 2017
From Amiram Eldar, Jul 02 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(3) (A002117).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*zeta(3)/4 (A197070). (End)
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi.
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)/(3*Pi). (End)
a(n) = Sum_{d|n} sigma_3(d)*mu(n/d) = Sum_{d|n} A001158(d)*A008683(n/d). Moebius transform of sigma_3(n). - Ridouane Oudra, Apr 15 2021

A000537 Sum of first n cubes; or n-th triangular number squared.

Original entry on oeis.org

0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, 4356, 6084, 8281, 11025, 14400, 18496, 23409, 29241, 36100, 44100, 53361, 64009, 76176, 90000, 105625, 123201, 142884, 164836, 189225, 216225, 246016, 278784, 314721, 354025, 396900, 443556, 494209, 549081

Offset: 0

N. J. A. Sloane

Number of parallelograms in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Or, number of orthogonal rectangles in an n X n checkerboard, or rectangles in an n X n array of squares. - Jud McCranie, Feb 28 2003. Compare A085582.
Also number of 2-dimensional cage assemblies (cf. A059827, A059860).
The n-th triangular number T(n) = Sum_{r=1..n} r = n(n+1)/2 satisfies the relations: (i) T(n) + T(n-1) = n^2 and (ii) T(n) - T(n-1) = n by definition, so that n^2*n = n^3 = {T(n)}^2 - {T(n-1)}^2 and by summing on n we have Sum_{ r = 1..n } r^3 = {T(n)}^2 = (1+2+3+...+n)^2 = (n*(n+1)/2)^2. - Lekraj Beedassy, May 14 2004
Number of 4-tuples of integers from {0,1,...,n}, without repetition, whose last component is strictly bigger than the others. Number of 4-tuples of integers from {1,...,n}, with repetition, whose last component is greater than or equal to the others.
Number of ordered pairs of two-element subsets of {0,1,...,n} without repetition.
Number of ordered pairs of 2-element multisubsets of {1,...,n} with repetition.
1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
a(n) is the number of parameters needed in general to know the Riemannian metric g of an n-dimensional Riemannian manifold (M,g), by knowing all its second derivatives; even though to know the curvature tensor R requires (due to symmetries) (n^2)*(n^2-1)/12 parameters, a smaller number (and a 4-dimensional pyramidal number). - Jonathan Vos Post, May 05 2006
Also number of hexagons with vertices in an hexagonal grid with n points in each side. - Ignacio Larrosa Cañestro, Oct 15 2006
Number of permutations of n distinct letters (ABCD...) each of which appears twice with 4 and n-4 fixed points. - Zerinvary Lajos, Nov 09 2006
With offset 1 = binomial transform of [1, 8, 19, 18, 6, ...]. - Gary W. Adamson, Dec 03 2008
The sequence is related to A000330 by a(n) = n*A000330(n) - Sum_{i=0..n-1} A000330(i): this is the case d=1 in the identity n*(n*(d*n-d+2)/2) - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)*(2*d*n-2*d+3)/6. - Bruno Berselli, Apr 26 2010, Mar 01 2012
From Wolfdieter Lang, Jan 11 2013: (Start)
For sums of powers of positive integers S(k,n) := Sum_{j=1..n}j^k one has the recurrence S(k,n) = (n+1)*S(k-1,n) - Sum_{l=1..n} S(k-1,l), n >= 1, k >= 1.
This was used for k=4 by Ibn al-Haytham in an attempt to compute the volume of the interior of a paraboloid. See the Strick reference where the trick he used is shown, and the W. Lang link.
This trick generalizes immediately to arbitrary powers k. For k=3: a(n) = (n+1)*A000330(n) - Sum_{l=1..n} A000330(l), which coincides with the formula given in the previous comment by Berselli. (End)
Regarding to the previous contribution, see also Matem@ticamente in Links field and comments on this recurrences in similar sequences (partial sums of n-th powers). - Bruno Berselli, Jun 24 2013
A rectangular prism with sides A000217(n), A000217(n+1), and A000217(n+2) has surface area 6*a(n+1). - J. M. Bergot, Aug 07 2013, edited with corrected indices by Antti Karttunen, Aug 09 2013
A formula for the r-th successive summation of k^3, for k = 1 to n, is (6*n^2+r*(6*n+r-1)*(n+r)!)/((r+3)!*(n-1)!), (H. W. Gould). - Gary Detlefs, Jan 02 2014
Note that this sequence and its formula were known to (and possibly discovered by) Nicomachus, predating Ibn al-Haytham by 800 years. - Charles R Greathouse IV, Apr 23 2014
a(n) is the number of ways to paint the sides of a nonsquare rectangle using at most n colors. Cf. A039623. - Geoffrey Critzer, Jun 18 2014
For n > 0: A256188(a(n)) = A000217(n) and A256188(m) != A000217(n) for m < a(n), i.e., positions of first occurrences of triangular numbers in A256188. - Reinhard Zumkeller, Mar 26 2015
There is no cube in this sequence except 0 and 1. - Altug Alkan, Jul 02 2016
Also the number of chordless cycles in the complete bipartite graph K_{n+1,n+1}. - Eric W. Weisstein, Jan 02 2018
a(n) is the sum of the elements in the multiplication table [0..n] X [0..n]. - Michel Marcus, May 06 2021

			G.f. = x + 9*x^2 + 36*x^3 + 100*x^4 + 225*x^5 + 441*x^6 + ... - _Michael Somos_, Aug 29 2022

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.
Avner Ash and Robert Gross, Summing it up, Princeton University Press, 2016, p. 62, eq. (6.3) for k=3.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 110ff.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, pp. 36, 58.
Clifford Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Oxford University Press, 2001, p. 325.
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).
H. K. Strick, Geschichten aus der Mathematik II, Spektrum Spezial 3/11, p. 13.
D. Wells, You Are A Mathematician, "Counting rectangles in a rectangle", Problem 8H, pp. 240; 254, Penguin Books 1995.

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].
Luciano Ancora, Sum of cubes of the first "n" natural numbers
Luciano Ancora, The Square Pyramidal Number and other figurate numbers
M. Azaola and F. Santos, The number of triangulations of the cyclic polytope C(n,n-4), Discrete Comput. Geom., 27 (2002), 29-48 (see Prop. 4.2(b)).
Marcel Berger, Encounter with a Geometer, Part II, Notices of the American Mathematical Society, Vol. 47, No. 3, (March 2000), pp. 326-340. [About the work of Mikhael Gromov.]
Bruno Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
blackpenredpen, Math for fun, how many rectangles?, Youtube video (2018).
Bikash Chakraborty, Proof Without Words: Sums of Powers of Natural numbers, arXiv:2012.11539 [math.HO], 2020.
Robert Dawson, On Some Sequences Related to Sums of Powers, J. Int. Seq., Vol. 21 (2018), Article 18.7.6.
Shel Kaphan, How to see that the difference between two successive squared triangular numbers is a cube
Sameen Ahmed Khan, Sums of the powers of reciprocals of polygonal numbers, Int'l J. of Appl. Math. (2020) Vol. 33, No. 2, 265-282.
Seon-Hong Kim and Kenneth B. Stolarsky, Translations and Extensions of the Nicomachean Identity, J. Int. Seq. (2024), Vol. 27, Issue 6, Art. No. 24.6.3. See p. 1.
Wolfdieter Lang, Ibn al-Haytham's trick.
S. Legendre, The Number of Crossings in a Regular Drawing of the Complete Bipartite Graph, JIS 12 (2009) 09.5.5.
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.
Henri Picciotto, Sum of Cubes, Proof without words.
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
C. J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Chordless Cycle
Eric Weisstein's World of Mathematics, Complete Bipartite Graph
Eric Weisstein's World of Mathematics, Faulhaber's Formula
Wikipedia, Faulhaber's formula
Wikipedia, Squared triangular number
G. Xiao, Sigma Server, Operate on "n^3"
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Convolution of A000217 and A008458.
Cf. A000330, A000538, A006003.
Row sums of triangles A094414 and A094415.
Second column of triangle A008459.
Row 3 of array A103438.
Cf. A000578, A002415, A024166, A101102, A101094, A101097.
Cf. A236770 (see crossrefs).
Cf. A000290, A253169, A256188.
Cf. A000217, A000292, A000332, A000566, A035287, A039623, A053382, A053383, A059376, A059827, A059860, A085582, A127777, A176271.

List([0..40],n->(n*(n+1)/2)^2); # Muniru A Asiru, Dec 05 2018

a000537 = a000290 . a000217  -- Reinhard Zumkeller, Mar 26 2015

[(n*(n+1)/2)^2: n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014

a:= n-> (n*(n+1)/2)^2:
seq(a(n), n=0..40);

Accumulate[Range[0, 50]^3] (* Harvey P. Dale, Mar 01 2011 *)
f[n_] := n^2 (n + 1)^2/4; Array[f, 39, 0] (* Robert G. Wilson v, Nov 16 2012 *)
Table[CycleIndex[{{1, 2, 3, 4}, {3, 2, 1, 4}, {1, 4, 3, 2}, {3, 4, 1, 2}}, s] /. Table[s[i] -> n, {i, 1, 2}], {n, 0, 30}] (* Geoffrey Critzer, Jun 18 2014 *)
Accumulate @ Range[0, 50]^2 (* Waldemar Puszkarz, Jan 24 2015 *)
Binomial[Range[20], 2]^2 (* Eric W. Weisstein, Jan 02 2018 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 9, 36, 100}, 20] (* Eric W. Weisstein, Jan 02 2018 *)
CoefficientList[Series[-((x (1 + 4 x + x^2))/(-1 + x)^5), {x, 0, 20}], x] (* Eric W. Weisstein, Jan 02 2018 *)

PARI
```
a(n)=(n*(n+1)/2)^2
```

def A000537(n): return (n*(n+1)>>1)**2 # Chai Wah Wu, Oct 20 2023

a(n) = (n*(n+1)/2)^2 = A000217(n)^2 = Sum_{k=1..n} A000578(k), that is, 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
G.f.: (x+4*x^2+x^3)/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = Sum ( Sum ( 1 + Sum (6*n) ) ), rephrasing the formula in A000578. - Xavier Acloque, Jan 21 2003
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j, row sums of A127777. - Alexander Adamchuk, Oct 24 2004
a(n) = A035287(n)/4. - Zerinvary Lajos, May 09 2007
This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=1. - Alexander R. Povolotsky, May 17 2008
G.f.: x*F(3,3;1;x). - Paul Barry, Sep 18 2008
Sum_{k > 0} 1/a(k) = (4/3)*(Pi^2-9). - Jaume Oliver Lafont, Sep 20 2009
a(n) = Sum_{1 <= k <= m <= n} A176271(m,k). - Reinhard Zumkeller, Apr 13 2010
a(n) = Sum_{i=1..n} J_3(i)*floor(n/i), where J_ 3 is A059376. - Enrique Pérez Herrero, Feb 26 2012
a(n) = Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} min(i,j,k). - Enrique Pérez Herrero, Feb 26 2013 [corrected by Ridouane Oudra, Mar 05 2025]
a(n) = 6*C(n+2,4) + C(n+1,2) = 6*A000332(n+2) + A000217(n), (Knuth). - Gary Detlefs, Jan 02 2014
a(n) = -Sum_{j=1..3} j*Stirling1(n+1,n+1-j)*Stirling2(n+3-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*(3-4*log(2)). - Vaclav Kotesovec, Feb 13 2015
a(n)*((s-2)*(s-3)/2) = P(3, P(s, n+1)) - P(s, P(3, n+1)), where P(s, m) = ((s-2)*m^2-(s-4)*m)/2 is the m-th s-gonal number. For s=7, 10*a(n) = A000217(A000566(n+1)) - A000566(A000217(n+1)). - Bruno Berselli, Aug 04 2015
From Ilya Gutkovskiy, Jul 03 2016: (Start)
E.g.f.: x*(4 + 14*x + 8*x^2 + x^3)*exp(x)/4.
Dirichlet g.f.: (zeta(s-4) + 2*zeta(s-3) + zeta(s-2))/4. (End)
a(n) = (Bernoulli(4, n+1) - Bernoulli(4, 1))/4, n >= 0, with the Bernoulli polynomial B(4, x) from row n=4 of A053382/A053383. See, e.g., the Ash-Gross reference, p. 62, eq. (6.3) for k=3. - Wolfdieter Lang, Mar 12 2017
a(n) = A000217((n+1)^2) - A000217(n+1)^2. - Bruno Berselli, Aug 31 2017
a(n) = n*binomial(n+2, 3) + binomial(n+2, 4) + binomial(n+1, 4). - Tony Foster III, Nov 14 2017
Another identity: ..., a(3) = (1/2)*(1*(2+4+6)+3*(4+6)+5*6) = 36, a(4) = (1/2)*(1*(2+4+6+8)+3*(4+6+8)+5*(6+8)+7*(8)) = 100, a(5) = (1/2)*(1*(2+4+6+8+10)+3*(4+6+8+10)+5*(6+8+10)+7*(8+10)+9*(10)) = 225, ... - J. M. Bergot, Aug 27 2022
Comment from Michael Somos, Aug 28 2022: (Start)
The previous comment expresses a(n) as the sum of all of the n X n multiplication table array entries.
For example, for n = 4:
     1  2  3  4
     2  4  6  8
     3  6  9 12
     4  8 12 16
This array sum can be split up as follows:
   +---+---------------+
   | 0 | 1   2   3   4 |    (0+1)*(1+2+3+4)
   |   +---+-----------+
   | 0 | 2 | 4   6   8 |    (1+2)*(2+3+4)
   |   |   +---+-------+
   | 0 | 3 | 6 | 9  12 |    (2+3)*(3+4)
   |   |   |   +---+---+
   | 0 | 4 | 8 |12 |16 |    (3+4)*(4)
   +---+---+---+---+---+
This kind of row+column sums was used by Ramanujan and others for summing Lambert series. (End)
a(n) = 6*A000332(n+4) - 12*A000292(n+1) + 7*A000217(n+1) - n - 1. - Adam Mohamed, Sep 05 2024

Edited by M. F. Hasler, May 02 2015

A001715 a(n) = n!/6.

Original entry on oeis.org

1, 4, 20, 120, 840, 6720, 60480, 604800, 6652800, 79833600, 1037836800, 14529715200, 217945728000, 3487131648000, 59281238016000, 1067062284288000, 20274183401472000, 405483668029440000, 8515157028618240000, 187333454629601280000, 4308669456480829440000

Offset: 3

N. J. A. Sloane

The numbers (4, 20, 120, 840, 6720, ...) arise from the divisor values in the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ... *(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) (which covers the following sequences: A000578, A000537, A024166, A101094, A101097, A101102). - Alexander R. Povolotsky, May 17 2008
a(n) is also the number of decreasing 3-cycles in the decomposition of permutations as product of disjoint cycles, a(3)=1, a(4)=4, a(5)=20. - Wenjin Woan, Dec 21 2008
Equals eigensequence of triangle A130128 reflected. - Gary W. Adamson, Dec 23 2008
a(n) is the number of n-permutations having 1, 2, and 3 in three distinct cycles. - Geoffrey Critzer, Apr 26 2009
From Johannes W. Meijer, Oct 20 2009: (Start)
The asymptotic expansion of the higher order exponential integral E(x,m=1,n=4) ~ exp(-x)/x*(1 - 4/x + 20/x^2 - 120/x^3 + 840/x^4 - 6720/x^5 + 60480/x^6 - 604800/x^7 + ...) leads to the sequence given above. See A163931 and A130534 for more information.
(End)

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).

Vincenzo Librandi, Table of n, a(n) for n = 3..200
Somaya Barati, Beáta Bényi, Abbas Jafarzadeh, and Daniel Yaqubi, Mixed restricted Stirling numbers, arXiv:1812.02955 [math.CO], 2018.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 263.
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), Article 00.2.4.
D. S. Mitrinovic and R. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No. 77 1962, 77 pp.
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7.
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
Index entries for sequences related to factorial numbers.
Index to divisibility sequences.

Cf. A049352, A049458, A049460.
Cf. A034472, A130128.
Cf. A245334, A000142, A111530.
Cf. A001710, A001720, A007759, A094638.
Cf. A094587, A138533, A173333, A213936.

a001715 = (flip div 6) . a000142 -- Reinhard Zumkeller, Aug 31 2014

[Factorial(n)/6: n in [3..30]]; // Vincenzo Librandi, Jun 20 2011

f := proc(n) n!/6; end;
BB:= [S, {S = Prod(Z,Z,C), C = Union(B,Z,Z), B = Prod(Z,C)}, labelled]: seq(combstruct[count](BB, size=n)/12, n=3..20); # Zerinvary Lajos, Jun 19 2008
G(x):=1/(1-x)^4: f[0]:=G(x): for n from 1 to 18 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..16); # Zerinvary Lajos, Apr 01 2009

a[n_]:=n!/6; (*Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)
Range[3,30]!/6 (* Harvey P. Dale, Aug 12 2012 *)

a(n)=n!/6 \\ Charles R Greathouse IV, Jan 12 2012

a(n) = A049352(n-2, 1) (first column of triangle).
E.g.f. if offset 0: 1/(1-x)^4.
a(n) = A173333(n,3). - Reinhard Zumkeller, Feb 19 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(k+4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: W(0), where W(k) = 1 - x*(k+4)/( x*(k+4) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) = A245334(n,n-3) / 4. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, May 22 2017: (Start)
The o.g.f. A(x) satisfies the Riccati equation x^2*A'(x) + (4*x - 1)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - 4*x/(1 - x/(1 - 5*x/(1 - 2*x/(1 - 6*x/(1 - 3*x/(1 - ... - (n + 3)*x/(1 - n*x/(1 - ... ))))))))) (apply Stokes, 1982).
A(x) = 1/(1 - 3*x - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - 6*x/(1 - ... - n*x/(1 - (n+3)*x/(1 - ... ))))))))). (End)
H(x) = (1 - (1 + x)^(-3)) / 3 = x - 4 x^2/2! + 20 x^3/3! - ... is an e.g.f. of the signed sequence (n!/4!), which is the compositional inverse of G(x) = (1 - 3*x)^(-1/3) - 1, an e.g.f. for A007559. Cf. A094638, A001710 (for n!/2!), and A001720 (for n!/4!). Cf. columns of A094587, A173333, and A213936 and rows of A138533.- Tom Copeland, Dec 27 2019
E.g.f.: x^3 / (3! * (1 - x)). - Ilya Gutkovskiy, Jul 09 2021
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=3} 1/a(n) = 6*e - 15.
Sum_{n>=3} (-1)^(n+1)/a(n) = 3 - 6/e. (End)

More terms from Harvey P. Dale, Aug 12 2012

A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

Original entry on oeis.org

0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088, 70584, 93993, 123234, 159334, 203434, 256795, 320804, 396980, 486980, 592605, 715806, 858690, 1023526, 1212751, 1428976, 1674992, 1953776, 2268497, 2622522, 3019422

Offset: 0

Clark Kimberling

Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006
a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012
In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013
The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022

			4*a(7) = 6384 = (0*1)^2 + (1*2)^2 + (2*3)^2 + (3*4)^2 + (4*5)^2 + (5*6)^2 + (6*7)^2 + (7*8)^2. - _Bruno Berselli_, Feb 05 2014

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

T. D. Noe, Table of n, a(n) for n = 0..1000
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 pp. 13, 15.
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
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7.
Alexander R. Povolotsky, Problem 1147, Pi Mu Epsilon Fall 2006 Problems.
Alexander R. Povolotsky, Problem, Pi Mu Epsilon Spring 2007 Problems.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Cf. A000292, A000332, A000389, A000579, A000580, A024166, A027555, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086025, A086026, A086027, A086028, A086029, A086030, A087127.

Cf. A000330, A000537 (first diffs), A001286, A003215, A100536, A101094 (partial sums), A101097, A101102, A222716.

a024166 n = sum $ zipWith (*) [n+1,n..0] a000578_list
-- Reinhard Zumkeller, Oct 14 2001

[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: n in [0..30]]; // G. C. Greubel, Nov 21 2017

A024166:=n->n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: seq(A024166(n), n=0..50); # Wesley Ivan Hurt, Nov 21 2017

Nest[Accumulate,Range[0,40]^3,2] (* Harvey P. Dale, Jan 10 2016 *)
Table[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, {n,0,30}] (* G. C. Greubel, Nov 21 2017 *)

a(n)=sum(j=1,n, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1)))) \\ Alexander R. Povolotsky, May 17 2008

for(n=0,30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017

From Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999: (Start)
a(n) = Sum_{i=0..n} A000537(i), partial sums of A000537.
a(n) = n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60. (End)
a(A004772(n)) mod 2 = 0; a(A016813(n)) mod 2 = 1. - Reinhard Zumkeller, Oct 14 2001
a(n) = Sum_{k=0..n} k^3*(n+1-k). - Paul Barry, Sep 14 2003; edited by Jon E. Schoenfield, Dec 29 2014
a(n) = 2*n*(n+1)*(n+2)*((n+1)^2 + 2*n*(n+2))/5!. This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=2. - Alexander R. Povolotsky, May 17 2008
O.g.f.: x*(1 + 4*x + x^2)/(-1 + x)^6. - R. J. Mathar, Jun 06 2008
a(n) = (6*n^2 + 12*n + 2)*(n+2)!/(120*(n-1)!), n > 0. - Gary Detlefs, Mar 01 2013
a(n) = A222716(n+1)/10 = A000292(n)*A100536(n+1)/10. - Jonathan Sondow, Mar 04 2013
4*a(n) = Sum_{i=0..n} A000290(i)*A000290(i+1). - Bruno Berselli, Feb 05 2014
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j*(n - max(i, j) + 1). - Melvin Peralta, May 12 2016
a(n) = n*binomial(n+3, 4) + binomial(n+2, 5). - Tony Foster III, Nov 14 2017
a(n) = Sum_{i=1..n} i*A143037(n,n-i+1). - J. M. Bergot, Aug 30 2022

A101094 a(n) = n(n+1)(n+2)(n+3)(1+3*n+n^2)/120.

Original entry on oeis.org

1, 11, 57, 203, 574, 1386, 2982, 5874, 10791, 18733, 31031, 49413, 76076, 113764, 165852, 236436, 330429, 453663, 612997, 816431, 1073226, 1394030, 1791010, 2277990, 2870595, 3586401, 4445091, 5468617, 6681368, 8110344, 9785336

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Partial sums of A024166. Third partial sums of cubes (A000578).

Antidiagonal sums of the array A213564. - Clark Kimberling, Jun 18 2012

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
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. 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]
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000537, A101097, A101102, A101104.

[n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120 : n in [1..35]]; // Vincenzo Librandi, Apr 23 2015

Table[n*(n + 1)*(n + 2)*(n + 3)*(1 + 3*n + n^2)/120, {n, 31}] (* Michael De Vlieger, Apr 20 2015 *)

[n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120 for n in range(1,32)] # Danny Rorabaugh, Apr 20 2015

This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k)+(k-1)*k/6)/((k+3)!/6) at k=3. - Alexander R. Povolotsky, May 17 2008
G.f.: -x*(1+4*x+x^2) / (x-1)^7. - R. J. Mathar, Dec 06 2011
Sum_{n>0} 1/a(n) = (8/3)*(25-9*sqrt(5)*Pi*tan(sqrt(5)*Pi/2)). - Enrique Pérez Herrero, Dec 02 2014
a(k) = MagicNKZ(3,k,7) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n. (Cf. A101104.) - Danny Rorabaugh, Apr 23 2015

Edited by Ralf Stephan, Dec 16 2004

A101102 Fifth partial sums of cubes (A000578).

Original entry on oeis.org

1, 13, 82, 354, 1200, 3432, 8646, 19734, 41613, 82225, 153868, 274924, 472056, 782952, 1259700, 1972884, 3016497, 4513773, 6624046, 9550750, 13550680, 18944640, 26129610, 35592570, 47926125, 63846081, 84211128, 110044792, 142559824, 183185200, 233595912

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Colin Barker, Table of n, a(n) for n = 1..1000
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. 15.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Archive 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 (9,-36,84,-126,126,-84,36,-9,1).

Partial sums of A101097.

Cf. A000537, A024166, A101094.

[Binomial(n+5,6)*(3*n^2+15*n+10)/28: n in [1..30]]; // G. C. Greubel, Dec 01 2018

Table[Binomial[n+5,6]*(3*n^2+15*n+10)/28, {n,1,30}] (* G. C. Greubel, Dec 01 2018 *)
Nest[Accumulate,Range[40]^3,5] (* Harvey P. Dale, Feb 06 2023 *)

a(n)=sum(t=1,n,sum(s=1,t,sum(l=1,s,sum(j=1,l, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2,(2*i-1))))))) \\ Alexander R. Povolotsky, May 17 2008

Vec(-x*(x^2+4*x+1)/(x-1)^9 + O(x^100)) \\ Colin Barker, Apr 23 2015

a(n) = binomial(n+5,6)*(3*n^2+15*n+10)/28 \\ Charles R Greathouse IV, Apr 23 2015

[binomial(n+5,6)*(3*n^2+15*n+10)/28 for n in  (1..30)] # G. C. Greubel, Dec 01 2018

a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(10 + 3*n*(n+5))/20160.
This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=5. - Alexander R. Povolotsky, May 17 2008
G.f.: x*(x^2+4*x+1) / (1-x)^9. - Colin Barker, Apr 23 2015
Sum_{n>=1} 1/a(n) = -162*sqrt(21/5)*Pi*tan(sqrt(35/3)*Pi/2) - 136269/100. - Amiram Eldar, Jan 26 2022

Edited by Ralf Stephan, Dec 16 2004

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

A254469 Sixth partial sums of cubes (A000578).

Original entry on oeis.org

1, 14, 96, 450, 1650, 5082, 13728, 33462, 75075, 157300, 311168, 586092, 1058148, 1841100, 3100800, 5073684, 8090181, 12603954, 19228000, 28778750, 42329430, 61274070, 87403680, 122996250, 170922375, 234768456, 318979584, 429024376, 571584200, 754769400

Offset: 1

Luciano Ancora, Feb 15 2015

			First differences:   1,  7, 19,  37,   61,   91, ... (A003215)
-------------------------------------------------------------------------
The cubes:           1,  8, 27,  64,  125,  216, ... (A000578)
-------------------------------------------------------------------------
First partial sums:  1,  9, 36, 100,  225,  441, ... (A000537)
Second partial sums: 1, 10, 46, 146,  371,  812, ... (A024166)
Third partial sums:  1, 11, 57, 203,  574, 1386, ... (A101094)
Fourth partial sums: 1, 12, 69, 272,  846, 2232, ... (A101097)
Fifth partial sums:  1, 13, 82, 354, 1200, 3432, ... (A101102)
Sixth partial sums:  1, 14, 96, 450, 1650, 5082, ... (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 .
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. 15.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000537, A000578, A003215, A024166, A101094, A101097, A101102, A254470, A254471, A254472.

[n*(1+n)^2*(2+n)*(3+n)*(4+n)*(5+n)^2*(6+n)/60480: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015

Table[n (1 + n)^2 (2 + n) (3 + n) (4 + n) (5 + n)^2 (6 + n)/60480, {n, 27}] (* or *) CoefficientList[Series[(1 + 4 x + x^2)/(- 1 + x)^10, {x, 0, 26}], x]
Nest[Accumulate,Range[30]^3,6] (* or *) LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,14,96,450,1650,5082,13728,33462,75075,157300},30] (* Harvey P. Dale, Sep 03 2016 *)

a(n)=n*(1+n)^2*(2+n)*(3+n)*(4+n)*(5+n)^2*(6+n)/60480 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (x + 4*x^2 + x^3)/(- 1 + x)^10.
a(n) = n*(1 + n)^2*(2 + n)*(3 + n)*(4 + n)*(5 + n)^2*(6 + n)/60480.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) + n^3.
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 217/200.
Sum_{n>=1} (-1)^(n+1)/a(n) = 223769/200 - 8064*log(2)/5. (End)

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

cp's OEIS Frontend

A000578 The cubes: a(n) = n^3.

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A000537 Sum of first n cubes; or n-th triangular number squared.

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A001715 a(n) = n!/6.

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A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

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A101094 a(n) = n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120.

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A101094 a(n) = n(n+1)(n+2)(n+3)(1+3*n+n^2)/120.