

A000537


Sum of first n cubes; or nth triangular number squared.
(Formerly M4619 N1972)


183



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
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OFFSET

0,3


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.
The nth triangular number T(n) = Sum_{r=1..n} r = n(n+1)/2 satisfies the relations: (i) T(n) + T(n1) = n^2 and (ii) T(n)  T(n1) = n by definition, so that n^2*n = n^3 = {T(n)}^2  {T(n1)}^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 4tuples of integers from {0,1,...,n}, without repetition, whose last component is strictly bigger than the others. Number of 4tuples of integers from {1,...,n}, with repetition, whose last component is greater than or equal to the others.
Number of ordered pairs of twoelement subsets of {0,1,...,n} without repetition.
Number of ordered pairs of 2element 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 ndimensional 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^21)/12 parameters, a smaller number (and a 4dimensional 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 n4 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..n1} A000330(i): this is the case d=1 in the identity n*(n*(d*nd+2)/2)  Sum_{i=0..n1} i*(d*id+2)/2 = n*(n+1)*(2*d*n2*d+3)/6.  Bruno Berselli, Apr 26 2010, Mar 01 2012
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(k1,n)  Sum_{l=1..n} S(k1,l), n >= 1, k >= 1.
This was used for k=4 by Ibn alHaytham 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 nth powers).  Bruno Berselli, Jun 24 2013
A formula for the rth successive summation of k^3, for k = 1 to n, is (6*n^2+r*(6*n+r1)*(n+r)!)/((r+3)!*(n1)!), (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 alHaytham 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
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


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

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].
Marcel Berger, Encounter with a Geometer, Part II, Notices of the American Mathematical Society, Vol. 47, No. 3, (March 2000), pp. 326340. [About the work of Mikhael Gromov.]
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review


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)/(1x)^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
This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k1)*k/6)/((k+3)!/6) at k=1.  Alexander R. Povolotsky, May 17 2008
a(n) = Sum_{j=1..3} j*s(n+1,n+1j)*S(n+3j,n), where s(n,k) and S(n,k) are the Stirling numbers of the first kind and the second kind, respectively.  Mircea Merca, Jan 25 2014
a(n)*((s2)*(s3)/2) = P(3, P(s, n+1))  P(s, P(3, n+1)), where P(s, m) = ((s2)*m^2(s4)*m)/2 is the mth sgonal number. For s=7, 10*a(n) = A000217(A000566(n+1))  A000566(A000217(n+1)).  Bruno Berselli, Aug 04 2015
E.g.f.: x*(4 + 14*x + 8*x^2 + x^3)*exp(x)/4.
Dirichlet g.f.: (zeta(s4) + 2*zeta(s3) + zeta(s2))/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 AshGross 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
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)


EXAMPLE

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


MAPLE

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


MATHEMATICA

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 *)
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 *)


PROG

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


CROSSREFS

Cf. A000332, A000566, A035287, A039623, A053382, A053383, A059376, A059827, A059860, A085582, A127777, A176271.


KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS



STATUS

approved



