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

%I A002593 M5199 N2262 #162 May 24 2025 16:20:06
%S A002593 0,1,28,153,496,1225,2556,4753,8128,13041,19900,29161,41328,56953,
%T A002593 76636,101025,130816,166753,209628,260281,319600,388521,468028,559153,
%U A002593 662976,780625,913276,1062153,1228528,1413721,1619100,1846081
%N A002593 a(n) = n^2*(2*n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3.
%C A002593 The m-th term, for m = A065549(n), is perfect (A000396). - _Lekraj Beedassy_, Jun 04 2002
%C A002593 Partial sums of A016755. - _Lekraj Beedassy_, Jan 06 2004
%C A002593 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
%C A002593 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
%C A002593 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
%C A002593 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
%C A002593 Binomial transform of the sequence with offset 1 is (1, 27, 98, 120, 48, 0, 0, 0, ...). - _Gary W. Adamson_, Jul 23 2015
%D A002593 Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 169, #31.
%D A002593 F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
%D A002593 L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 7.
%D A002593 Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 47.
%D A002593 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A002593 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A002593 Vincenzo Librandi, <a href="/A002593/b002593.txt">Table of n, a(n) for n = 0..10000</a>
%H A002593 F. E. Croxton and D. J. Cowden, <a href="/A000447/a000447.pdf">Applied General Statistics</a>, 2nd Ed., Prentice-Hall, Englewood Cliffs, NJ, 1955. [Annotated scans of just pages 742-743]
%H A002593 Neslihan Kilar, Abdelmejid Bayad, and Yilmaz Simsek, <a href="https://hal.science/hal-04535748">Finite sums involving trigonometric functions and special polynomials: analysis of generating functions and p-adic integrals</a>, Appl. Anal. Disc. Math., hal-04535748, 2024. See p. 22.
%H A002593 Vladimir Pletser, <a href="/A253724/a253724.txt">File Triplets (M,a,c) for M=2n^2</a>
%H A002593 Vladimir Pletser, <a href="http://arxiv.org/abs/1501.06098">General solutions of sums of consecutive cubed integers equal to squared integers</a>, arXiv:1501.06098 [math.NT], 2015.
%H A002593 Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H A002593 Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992.
%H A002593 R. J. Stroeker, <a href="http://www.numdam.org/item?id=CM_1995__97_1-2_295_0">On the sum of consecutive cubes being a perfect square</a>, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
%H A002593 G. Xiao, Sigma Server, <a href="http://wims.unice.fr/~wims/en_tool~analysis~sigma.en.html">Operate on "(2*n-1)^3"</a>.
%H A002593 M. J. Zerger, <a href="http://www.jstor.org/stable/2690925">Proof without words: The sum of consecutive odd cubes is a triangular number</a>, Math. Mag., 68 (1995), 371.
%H A002593 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A002593 a(n) = A000217(A056220(n)). - _Lekraj Beedassy_, Jun 11 2004
%F A002593 G.f.: (-x^4 - 23*x^3 - 23*x^2 - x)/(x - 1)^5. - _Harvey P. Dale_, Mar 28 2011
%F A002593 a(n) = n^2*(2n^2 - 1). - _Vladimir Pletser_, Jan 10 2015
%F A002593 E.g.f.: exp(x)*x*(1 + 13*x + 24*x^2/2! + 12*x^3/3!). - _Wolfdieter Lang_, Mar 11 2017
%F A002593 a(n) = A000384(A000290(n)) = A056220(n) * A000290(n). - _Bruce J. Nicholson_, Aug 22 2017
%F A002593 From _Amiram Eldar_, Aug 25 2022: (Start)
%F A002593 Sum_{n>=1} 1/a(n) = 1 - Pi^2/6 - cot(Pi/sqrt(2))*Pi/sqrt(2).
%F A002593 Sum_{n>=1} (-1)^(n+1)/a(n) = cosec(Pi/sqrt(2))*Pi/sqrt(2) - Pi^2/12 - 1. (End)
%p A002593 A002593:=-z*(z+1)*(z**2+22*z+1)/(z-1)**5; # conjectured by _Simon Plouffe_ in his 1992 dissertation
%p A002593 a:= n-> n^2*(2*n^2-1): seq(a(n), n=0..50);  # _Vladimir Pletser_, Jan 10 2015
%t A002593 CoefficientList[Series[(-x^4-23x^3-23x^2-x)/(x-1)^5,{x,0, 80}],x]  (* or *)
%t A002593 Table[ n^2 (2n^2-1),{n,0,80}]  (* _Harvey P. Dale_, Mar 28 2011 *)
%t A002593 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 *)
%o A002593 (Magma) [n^2*(2*n^2 - 1): n in [0..40]]; // _Vincenzo Librandi_, Sep 07 2011
%o A002593 (PARI) a(n) = n^2*(2*n^2 - 1) \\ _Charles R Greathouse IV_, Feb 07 2017
%Y A002593 Cf. A000290, A000384, A000447, A000583, A002309, A253724, A253725, A260810.
%K A002593 nonn,nice,easy
%O A002593 0,3
%A A002593 _N. J. A. Sloane_