A000537 Sum of first n cubes; or n-th triangular number squared.
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
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
- 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).
Programs
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GAP
List([0..40],n->(n*(n+1)/2)^2); # Muniru A Asiru, Dec 05 2018
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Haskell
a000537 = a000290 . a000217 -- Reinhard Zumkeller, Mar 26 2015
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Magma
[(n*(n+1)/2)^2: n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014
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Maple
a:= n-> (n*(n+1)/2)^2: seq(a(n), n=0..40);
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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 *)
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PARI
a(n)=(n*(n+1)/2)^2
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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) = -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) = 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)
Extensions
Edited by M. F. Hasler, May 02 2015
Comments