A163102 a(n) = n^2*(n+1)^2/2.

0, 2, 18, 72, 200, 450, 882, 1568, 2592, 4050, 6050, 8712, 12168, 16562, 22050, 28800, 36992, 46818, 58482, 72200, 88200, 106722, 128018, 152352, 180000, 211250, 246402, 285768, 329672, 378450, 432450, 492032, 557568, 629442, 708050, 793800, 887112, 988418

Offset: 0

Author

Omar E. Pol, Jul 24 2009

Keywords

• nonn
• easy
• changed

Comments

Row sums of triangle A163282.

Also, the number of nonattacking placements of 2 rooks on an (n+1) X (n+1) board. - Thomas Zaslavsky, Jun 26 2010

If P_{k}(n) is the n-th k-gonal number, then a(n) = P_{s}(n+1)*P_{t}(n+1) - P_{s+1}(n+1)*P_{t-1}(n+1) for s=t+1. - Bruno Berselli, Sep 05 2014

Subsequence of A000982, see formula. - David James Sycamore, Jul 31 2018

Number of edges in the (n+1) X (n+1) rook complement graph. - Freddy Barrera, May 02 2019

Number of paths from (0,0) to (n+2,n+2) consisting of exactly three forward horizontal steps and three upward vertical steps. - Greg Dresden and Snezhana Tuneska, Aug 24 2023

References

• Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-queens problem, in preparation. - Thomas Zaslavsky, Jun 26 2010

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv preprint arXiv:1609.00853 [math.CO], 2016-2020.
• A. Umar, Combinatorial Results for Semigroups of Orientation-Preserving Partial Transformations, J. Int. Seq., Vol. 14 (2011), Article 11.7.5.
• Eric Weisstein's World of Mathematics, Rook Complement Graph.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Column k=2 of A144084.

Cf. A000217, A000537, A000982, A035287.

Cf. A006002, A099903, A163274, A163275, A163276, A163277, A163282.

Cf. A060300, A254371.

Programs

GAP

List([0..40],n->(n*(n+1))^2/2); # Muniru A Asiru, Aug 02 2018

Magma

[n^2*(n+1)^2/2: n in [0..40]]; // Vincenzo Librandi, Mar 26 2012

Mathematica

CoefficientList[Series[2*x*(1+4*x+x^2)/(1-x)^5,{x,0,40}],x] (* Vincenzo Librandi, Mar 26 2012 *)

PARI

a(n)=n^2*(n+1)^2/2 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = 2*A000537(n) = A035287(n+1)/2. - Omar E. Pol, Nov 29 2011

G.f.: 2*x*(1+4*x+x^2)/(1-x)^5. - R. J. Mathar, Nov 30 2011

Let t(n) = A000217(n). Then a(n) = (t(n-1)*(t(n)+t(n+1)) + t(n)*(t(n-1)+t(n+1)) + t(n+1)*(t(n-1)+t(n)))/3. - J. M. Bergot, Jun 21 2012

a(n) = A000982(n*(n+1)). - David James Sycamore, Jul 31 2018

From Amiram Eldar, Nov 02 2021: (Start)

Sum_{n>=1} 1/a(n) = 2*Pi^2/3 - 6.

Sum_{n>=1} (-1)^(n+1)/a(n) = 6 - 8*log(2). (End)

Another identity: ..., a(4) = 200 = 1*(2+4+6+8) + 3*(4+6+8) + 5*(6+8) + 7*(8), a(5) = 450 = 1*(2+4+6+8+10) + 3*(4+6+8+10) + 5*(6+8+10) + 7*(8+10) + 9*(10) = 30+84+120+126+90, and so on. - J. M. Bergot, Aug 25 2022

From Elmo R. Oliveira, Aug 14 2025: (Start)

E.g.f.: x*(2 + x)*(2 + 6*x + x^2)*exp(x)/2.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

a(n) = A254371(n)/4 = A060300(n)/8. (End)