

A000537


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


108



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.
Also number of 2dimensional cage assemblies (cf. A059827, A059860).
The nth triangular number T(n)=sum_r(1, n)=n(n+1)/2 satisfies the relations: (i) T(n) + T(n1)=n^2 and (ii) T(n)  T(n1)=n from definition, so that n^2*n=n^3={T(n)}^2  {T(n1)}^2 and thus summing telescopingly over 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 two element 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
Sum(k>0,1/a(k))=(4/3)*(Pi^29).  Jaume Oliver Lafont, Sep 20 2009
a(n) = SUM(A176271(m,k)): 1<=k<=m<=n).  Reinhard Zumkeller, Apr 13 2010
The sequence is related to A000330 by a(n) = n*A000330(n)sum(A000330(i), i=0..n1): this is the case d=1 in the identity n*(n*(d*nd+2)/2)sum(i*(d*id+2)/2, i=0..n1) = n*(n+1)*(2*d*n2*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^k,j=1..n) one has the recurrence S(k,n) = (n+1)*S(k1,n)  sum(S(k1,l),l=1..n), 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(A000330(l),l=1..n), which coincides with the formula given in the preceding Berselli comment. (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 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 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


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.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 110ff.
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.]
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].
M. Azaola and F. Santos, The number of triangulations of the cyclic polytope C(n,n4), Discrete Comput. Geom., 27 (2002), 2948 (see Prop. 4.2(b)).
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
Wolfdieter Lang, Ibn alHaytham's trick.
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
Simon Plouffe_, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe_, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
Eric Weisstein's World of Mathematics, Faulhaber's Formula
Wikipedia, Faulhaber's formula
G. Xiao, Sigma Server, Operate on "n^3"
Index entries for sequences related to linear recurrences with constant coefficients


FORMULA

a(n) = (n*(n+1)/2)^2 = A000217(n)^2, 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.
a(n) = Sum [ Sum ( 1 + Sum (6*n) ) ].  Xavier Acloque, Jan 21 2003
triangle(n)*Sum(j=1, n, j).  Jon Perry, Jul 28 2003
a(n) = Sum[Sum[(i*j), {i, 1, n}], {j, 1, n}].  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) + (k1)*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
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} max(i,j,k).  Enrique Pérez Herrero, Feb 26 2013
a(n)= 6*C(n+2,4)+C(n+1,2), (Knuth).  Gary Detlefs, Jan 02 2014
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


MAPLE

(n*(n+1)/2)^2;
A000537:=(1+4*z+z**2)/(z1)**5; [Simon Plouffe in his 1992 dissertation for sequence without initial zero.]


MATHEMATICA

Table[ Sum[(i*j), {i, n}, {j, n}], {n, 0, 38}] (* modified by Robert G. Wilson v, Nov 16 2012 *)
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 *)


PROG

(PARI) a(n)=(n*(n+1)/2)^2
(PARI) t(n)=n*(n+1)/2 for(i=1, 30, print1(", "sum(j=1, i, j*t(i))))


CROSSREFS

Convolution of A000217 and A008458. Cf. A000330, A006003, A000538.
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).
Sequence in context: A169835 A231686 A231688 * A114286 A098928 A139469
Adjacent sequences: A000534 A000535 A000536 * A000538 A000539 A000540


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane


STATUS

approved



