A002593 a(n) = n^2*(2*n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3.

0, 1, 28, 153, 496, 1225, 2556, 4753, 8128, 13041, 19900, 29161, 41328, 56953, 76636, 101025, 130816, 166753, 209628, 260281, 319600, 388521, 468028, 559153, 662976, 780625, 913276, 1062153, 1228528, 1413721, 1619100, 1846081

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• nice
• easy

Comments

The m-th term, for m = A065549(n), is perfect (A000396). - Lekraj Beedassy, Jun 04 2002

Partial sums of A016755. - Lekraj Beedassy, Jan 06 2004

Also, the k-th triangular number, where k = 2n^2 - 1 = A056220(n), i.e., a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004

Also, the j-th hexagonal number, where j = n^2 = A000290(n), i.e., a(n) = A000384(A000290(n)) and a(n) = A056220(n) * A000290(n) or j * k. This sequence is a subsequence of the hexagonal number sequence and retains the aspect intrinsic to the hexagonal number sequence that each number in this sequence can be found by multiplying its triangular number by its hexagonal number. - Bruce J. Nicholson, Aug 22 2017

Odd numbers and their squares both having the form 2x-+1, we may write (2r+1)^3 = (2r+1)*(2s-1), where s = centered squares = (r+1)^2 + r^2. Since 2r+1 = (r+1)^2 - r^2, it follows immediately from summing telescopingly over n-1, the product 2*{(r+1)^4 - r^4} - {(r+1)^2 - r^2}, that Sum_{r=0..n-1} (2r+1)^3 = 2*n^4 - n^2 = n^2*(2n^2 - 1). - Lekraj Beedassy, Jun 16 2004

a(n) is also the starting term in the sum of a number M(n) of consecutive cubed integers equaling a squared integer (A253724) for M(n) equal to twice a squared integer (A001105). Numbers a(n) such that a^3 + (a+1)^3 + ... + (a+M-1)^3 = c^2 has nontrivial solutions over the integers for M equal to twice a squared integer (A001105). If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting from a^3 and equaling a squared integer c^2. For n >= 1, M(n) = 2n^2 (A001105), a(n) = M(M-1)/2 = n^2(2n^2 - 1), and c(n) = sqrt(M/2) (M(M^2-1)/2) = n^3(4n^4 - 1). The trivial solutions with M < 1 and a < 2 are not considered. - Vladimir Pletser, Jan 10 2015

Binomial transform of the sequence with offset 1 is (1, 27, 98, 120, 48, 0, 0, 0, ...). - Gary W. Adamson, Jul 23 2015

References

• Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 169, #31.
• F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
• L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 7.
• Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 47.
• 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).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• F. E. Croxton and D. J. Cowden, Applied General Statistics, 2nd Ed., Prentice-Hall, Englewood Cliffs, NJ, 1955. [Annotated scans of just pages 742-743]
• Neslihan Kilar, Abdelmejid Bayad, and Yilmaz Simsek, Finite sums involving trigonometric functions and special polynomials: analysis of generating functions and p-adic integrals, Appl. Anal. Disc. Math., hal-04535748, 2024. See p. 22.
• Vladimir Pletser, File Triplets (M,a,c) for M=2n^2
• Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
• 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.
• R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
• G. Xiao, Sigma Server, Operate on "(2*n-1)^3".
• M. J. Zerger, Proof without words: The sum of consecutive odd cubes is a triangular number, Math. Mag., 68 (1995), 371.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000290, A000384, A000447, A000583, A002309, A253724, A253725, A260810.

Programs

Magma

[n^2*(2*n^2 - 1): n in [0..40]]; // Vincenzo Librandi, Sep 07 2011

Maple

A002593:=-z*(z+1)*(z**2+22*z+1)/(z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation
a:= n-> n^2*(2*n^2-1): seq(a(n), n=0..50);  # Vladimir Pletser, Jan 10 2015

Mathematica

CoefficientList[Series[(-x^4-23x^3-23x^2-x)/(x-1)^5,{x,0, 80}],x]  (* or *)
Table[ n^2 (2n^2-1),{n,0,80}]  (* Harvey P. Dale, Mar 28 2011 *)
Join[{0},Accumulate[Range[1,91,2]^3]] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,28,153,496},40] (* Harvey P. Dale, Mar 22 2017 *)

PARI

a(n) = n^2*(2*n^2 - 1) \\ Charles R Greathouse IV, Feb 07 2017

Formula

a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004

G.f.: (-x^4 - 23*x^3 - 23*x^2 - x)/(x - 1)^5. - Harvey P. Dale, Mar 28 2011

a(n) = n^2*(2n^2 - 1). - Vladimir Pletser, Jan 10 2015

E.g.f.: exp(x)*x*(1 + 13*x + 24*x^2/2! + 12*x^3/3!). - Wolfdieter Lang, Mar 11 2017

a(n) = A000384(A000290(n)) = A056220(n) * A000290(n). - Bruce J. Nicholson, Aug 22 2017

From Amiram Eldar, Aug 25 2022: (Start)

Sum_{n>=1} 1/a(n) = 1 - Pi^2/6 - cot(Pi/sqrt(2))*Pi/sqrt(2).

Sum_{n>=1} (-1)^(n+1)/a(n) = cosec(Pi/sqrt(2))*Pi/sqrt(2) - Pi^2/12 - 1. (End)