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 A002593 a(n) = n^2*(2*n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3. (Formerly M5199 N2262) 14
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 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, k-th triangular number, where k = 2n^2 - 1 = A056220(n), i.e., a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004 Also, 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 subset 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 triangle 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 L. 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. 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] Vladimir Pletser, File Triplets (M,a,c) for M=2n^2 V. 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. Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992 G. Xiao, Sigma Server, Operate on "(2*n-1)^3" R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307. 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). FORMULA 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 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 *) PROG (MAGMA) [n^2*(2*n^2 - 1): n in [0..40]]; // Vincenzo Librandi, Sep 07 2011 (PARI) a(n) = n^2*(2*n^2 - 1) \\ Charles R Greathouse IV, Feb 07 2017 CROSSREFS Cf. A000290, A000384, A000447, A000583, A002309, A253724, A253725, A260810. Sequence in context: A219887 A271636 A188778 * A015881 A026910 A172220 Adjacent sequences:  A002590 A002591 A002592 * A002594 A002595 A002596 KEYWORD nonn,nice,easy AUTHOR STATUS approved

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Last modified April 23 05:34 EDT 2021. Contains 343199 sequences. (Running on oeis4.)