cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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Author

N. J. A. Sloane

Keywords

Comments

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

Examples

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

References

  • 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.

Links

Crossrefs

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.

Programs

  • GAP
    List([0..40],n->(n*(n+1)/2)^2); # Muniru A Asiru, Dec 05 2018
  • Haskell
    a000537 = a000290 . a000217  -- Reinhard Zumkeller, Mar 26 2015
  • Magma
    [(n*(n+1)/2)^2: n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014
  • Maple
    a:= n-> (n*(n+1)/2)^2:
seq(a(n), n=0..40);
  • Mathematica
    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
  • Python
    def A000537(n): return (n*(n+1)>>1)**2 # Chai Wah Wu, Oct 20 2023

Formula

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

Extensions

Edited by M. F. Hasler, May 02 2015

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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