A253725 -id:A253725 - cp's OEIS frontend

A002593 a(n) = n^2(2n^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

N. J. A. Sloane

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

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).

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).

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

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

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

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 *)

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

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)

A253724 Numbers c(n) whose squares are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

Original entry on oeis.org

504, 8721, 65472, 312375, 1119528, 3293829, 8388096, 19131147, 39999000, 77947353, 143325504, 250991871, 421651272, 683434125, 1073737728, 1641349779, 2448874296, 3575480097, 5119992000, 7204344903, 9977420904, 13619289621, 18345871872, 24414046875

Offset: 2

Vladimir Pletser, Jan 10 2015

Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for M(n) being twice a squared integer (A001105) and b(n)=(A002593).
If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting at b^3 and equaling to a squared integer c^2. For n>=1, M(n)= 2n^2 (A001105), b(n) = M(M-1)/2 = n^2(2n^2 - 1) (A002593), and c(n)= sqrt(M/2) (M(M^2-1)/2)= n^3(4n^4 - 1) (this sequence).
The trivial solutions with M < 1 and b < 2 are not considered here.

			For n=2, M(n)=8, b(n)=28, c(n)=504.
See "File Triplets (M,b,c) for M=2n^2" link.

Vladimir Pletser, Table of n, a(n) for n = 2..50000
Vladimir Pletser, File Triplets (M,b,c) for M=2n^2
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681, A253707, A253709, A002593, A253725.

[n^3*(4*n^4 - 1): n in [2..30]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 2 to 50000 do a:= n^3*(4*n^4 - 1): print (a); end do:

f[n_] := n^3 (4 n^4 - 1); Rest@Array[f, 32] (* Michael De Vlieger, Jan 28 2015 *)

Vec(-3*x^2*(x^7-8*x^6+27*x^5-216*x^4-1521*x^3-3272*x^2-1563*x-168)/(x-1)^8 + O(x^100)) \\ Colin Barker, Jan 14 2015

a(n) = n^3(4n^4 - 1).

G.f.: -3*x^2*(x^7-8*x^6+27*x^5-216*x^4-1521*x^3-3272*x^2-1563*x-168) / (x-1)^8. - Colin Barker, Jan 14 2015

A218979 Numbers n such that some sum of n consecutive positive cubes is a square.

Original entry on oeis.org

1, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 18, 19, 21, 23, 25, 27, 28, 29, 31, 32, 33, 35, 37, 39, 40, 41, 42, 43, 45, 47, 48, 49, 50, 51, 53, 54, 55, 57, 59, 60, 61, 63, 64, 65, 67, 69, 71, 72, 73, 75, 76, 77, 79, 81, 82, 83, 85, 87, 89, 91, 92, 93, 94, 95, 97, 98, 99

Offset: 1

Michel Marcus, Nov 08 2012

nonn

The trivial solutions with x = 0 and x = 1 are not considered here.
Numbers n such that x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 has nontrivial solutions over the integers.
The nontrivial solutions are found by solving Y^2 = X^3 + d(n)*X with d(n) = n^2*(n^2-1)/4 (A006011), Y = n*y and X = n*x + (1/2)*n*(n-1). [Corrected by Derek Orr, Aug 30 2014]
x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 can also be written as y^2 = n(x + (n-1)/2)(n(x + (n-1)/2) + x(x-1)). - Vladimir Pletser, Aug 30 2014
There are 892 triples (n,x,y), with n and x less than 10^5 (1 < n,x < 10^5), which are nontrivial solutions of x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 (note that (n,x,y) corresponds to (M,a,c) in A253679, A253680, A253681, A253707, A253708, A253709, A253724, A253725). - Vladimir Pletser, Jan 10 2015

			See "Examples of triples" link.

Michel Marcus, Examples of triples up to n=50
Vladimir Pletser, Triplets (n, x, y) with n,x less than 10^5
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, Fundamental solutions of the Pell equation X^2-(sigma^4-delta^4)Y^2=delta^4 for the first 45 solutions of the sums of consecutive cubed integers equalling integer squares, Research Gate, 2015. See Reference 19.
V. Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253681, A253707, A253708, A253709.

a(n)=for(x=2,10^7, /* note this limit only generates the terms in the data section */ X = n*x + (1/2)*n*(n-1); d=n^2*(n^2-1)/4;if(issquare(X^3+d*X),return(x)))
n=1;while(n<100,if(a(n),print1(n,", "));n++) \\ Derek Orr, Aug 30 2014

Name changed, a(1) = 1 prepended and a(39)-a(68) from Derek Orr, Aug 30 2014

More terms for 50Vladimir Pletser, Jan 10 2015

cp's OEIS Frontend

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

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A253724 Numbers c(n) whose squares are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

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A218979 Numbers n such that some sum of n consecutive positive cubes is a square.

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cp's OEIS Frontend

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

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Author

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Magma

Maple

Mathematica

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Formula

A253724 Numbers c(n) whose squares are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A218979 Numbers n such that some sum of n consecutive positive cubes is a square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Extensions

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