A288486 -id:A288486 - cp's OEIS frontend

A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.

1, 8, 75, 400, 1445, 4056, 9583, 20000, 38025, 67240, 112211, 178608, 273325, 404600, 582135, 817216, 1122833, 1513800, 2006875, 2620880, 3376821, 4298008, 5410175, 6741600, 8323225, 10188776, 12374883, 14921200, 17870525, 21268920, 25165831, 29614208

Offset: 0

Daniel Poveda Parrilla, Jun 11 2017

If we add a(n) and A288487(n) graphically we obtain a bigger cuboid which is a square of cubes (a cuboid with dimensions n^2 * n^2 * n).

a(10^n) is a palindrome in base 10.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000584, A002061, A002522, A014206, A059100, A082044, A288486.

Table[(1 + n)*(1 + n^2)^2, {n, 0, 28}] (* or *) CoefficientList[Series[(1 + 2 x + 42 x^2 + 50 x^3 + 25 x^4)/(1 - x)^6, {x, 0, 28}], x] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 8, 75, 400, 1445, 4056}, 29]

Vec((1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6 + O(x^28))

G.f.: (1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = (n + 1)*(n^2 + 1)^2 = (n + 1)*(A002522(n))^2 = (n + 1)*A082044(n).
a(n) = n^5 + A002061(A002061(n+1)).
a(n) = A000584(n+1) - A288486(n)
a(n) = (n + 1)*A059100(n-1) + 4*(n^2 -1)*A014206(n-1) for n > 1.

A006325 4-dimensional analog of centered polygonal numbers.

Original entry on oeis.org

0, 0, 1, 7, 26, 70, 155, 301, 532, 876, 1365, 2035, 2926, 4082, 5551, 7385, 9640, 12376, 15657, 19551, 24130, 29470, 35651, 42757, 50876, 60100, 70525, 82251, 95382, 110026, 126295, 144305, 164176, 186032, 210001, 236215, 264810, 295926

Offset: 0

Albert Rich (Albert_Rich(AT)msn.com)

If X is an n-set and Y and Z disjoint 2-subsets of X then a(n-4) is equal to the number of 6-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
For n>0, a(n+1) is the n-th antidiagonal sum of A213751. - Clark Kimberling, Jun 20 2012
This sequence is the case m=n-1, k=n+3 of b(m,k) = m*(m+1)*((k-2)*m-(k-5))/6, which is the m-th k-gonal pyramidal number. - Luciano Ancora, Apr 11 2015
Starting (1, 7, 26, ...), this is the binomial transform of (1, 6, 13, 12, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 31 2015
Also starting (1, 7, 26, ...), this appears to be the number of magic labelings of the cycle-of-loops graph LOOP X C_4 having magic sum n, where LOOP is the 1-vertex, 1-loop-edge graph. - David J. Seal, Sep 13 2017
The conjecture by David J. Seal is true and easily proved using MacMahon's Omega operators via the "Omega" package for Mathematica authored by Axel Riese (obtaining (up to an offset) the generating function listed in the formula section below). See the second Mathematica program in which the edges of LOOP X C_4 are indexed as in the example below. The Omega package can be downloaded from the link provided in the article by G. E. Andrews et al. - L. Edson Jeffery, Oct 15 2017

			A representation of the LOOP X C_4 graph, with edges and loops indexed as shown, as used in the second Mathematica program below:
.             3         1
.              O_______O
.              |   2   |
.              |4     0|
.              |_______|
.              O   6   O
.             5         7

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
G. E. Andrews, P. Paule and A. Riese, MacMahon's partition analysis III. The Omega package.
David Galvin and Courtney Sharpe, Independent set sequence of linear hyperpaths, arXiv:2409.15555 [math.CO], 2024. See p. 7.
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Milan Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
J. K. Merikoski, R. Kumar and R. A. Rajput, Upper bounds for the largest eigenvalue of a bipartite graph, Electronic Journal of Linear Algebra ISSN 1081-3810, A publication of the International Linear Algebra Society, Volume 26, pp. 168-176, April 2013.
Richard P. Stanley, Examples of Magic Labelings, Unpublished Notes, 1973 [Cached copy, with permission]
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A005900, A169938, A213751, A288486, A019298, A061907, A244497, A244879, A244873, A244880.
Cf. A000027, A000217, A019298, A244497, A244879, A244873, A244880, A293310, A293309 (magic labelings of LOOP X C_k, for k = 1..3,5..10).
Cf. A293311, A293312.

[n*(n-1)*(n^2-n+1)/6: n in [0..40]]; // Vincenzo Librandi, May 22 2011

Table[n*(n-1)*(n^2-n+1)/6, {n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Apr 22 2011 *)
<< Omega.m; n = 4; cond = {}; Do[AppendTo[cond, Sum[a[Mod[2*k - j, 2*n]], {j, 0, 2}] == a[2*n]], {k, 0, n - 1}]; f = OEqSum[Product[x[i]^a[i], {i, 0, 2*n}], cond, u][[1]]; Do[f = OEqR[f, Subscript[u, k]], {k, n}];
(* Generating function: *)
f = Factor[f /. {x[2*n] -> x} /. {x[_] -> 1}]
(* This sequence (with initial zeros dropped): *)
CoefficientList[Series[f, {x, 0, 35}], x] (* L. Edson Jeffery, Oct 15 2017 *)

a(n)=n*(n-1)*(n^2-n+1)/6 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = n*(n-1)*(n^2-n+1)/6.
a(n) = ((n^5 - (n-1)^5) - (n^1 - (n-1)^1))/30 = (n^5 - (n-1)^5 - 1)/30. - Xavier Acloque, Jan 25 2003
The partial sums of the octahedral numbers: a(n+1) = Sum_{i=0..n} A005900(i). - Jonathan Vos Post, Mar 14 2006
G.f.: -x^2*(x+1)^2/(x-1)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009
a(n) = Sum_{i=1..n-1} Sum_{j=1..n-1} min(i,j)^2. - Enrique Pérez Herrero, Jan 15 2013 [Which is just rephrasing the partial sum formula with the Murthy formula in A005900. - R. J. Mathar, Jun 14 2014]
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 4. - Yosu Yurramendi, Sep 03 2013
From Daniel Poveda Parrilla, Sep 09 2017: (Start)
a(n) = A169938(n-1)/6.
a(n+1) = A288486(n)/24. (End)
Sum_{n>=2} 1/a(n) = 12 - 2 * sqrt(3) * tanh(sqrt(3)*Pi/2). - Amiram Eldar, Jun 28 2020
E.g.f.: exp(x)*x^2*(3 + 4*x + x^2)/6. - Stefano Spezia, Dec 12 2021

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A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.

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