A179058 Number of non-attacking placements of 3 rooks on an n X n board.

0, 0, 6, 96, 600, 2400, 7350, 18816, 42336, 86400, 163350, 290400, 490776, 794976, 1242150, 1881600, 2774400, 3995136, 5633766, 7797600, 10613400, 14229600, 18818646, 24579456, 31740000, 40560000, 51333750, 64393056, 80110296

Offset: 1

Author

Thomas Zaslavsky, Jun 27 2010

Keywords

• easy
• nonn

Comments

Also the number of 3-cycles in the n X n rook complement graph. - Eric W. Weisstein, Sep 05 2017

Also the number of 6-cycles in the complete tripartite graph K_{n,n,n}. - Eric W. Weisstein, Dec 07 2023

Essentially the same as A303212. - Eric W. Weisstein, Dec 06 2023

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..200
• 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.
• Eric Weisstein's World of Mathematics, Complete Tripartite Graph.
• Eric Weisstein's World of Mathematics, Graph Cycle.
• Eric Weisstein's World of Mathematics, Rook Complement Graph.
• Eric Weisstein's World of Mathematics, Rook Graph.
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Crossrefs

Column k=3 of A144084.

Cf. A163102 (2 rooks), A179059 (4 rooks).

Cf. A291910 (4-cycles), A291911 (5-cycles), A291912 (6-cycles).

Programs

Mathematica

(* Start from Eric W. Weisstein, Sep 05 2017 *)
Table[3! Binomial[n, 3]^2, {n, 20}]
3! Binomial[Range[20], 3]^2
LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 0, 6, 96, 600, 2400, 7350}, 20]
CoefficientList[Series[-((6 x^2 (1 + 9 x + 9 x^2 + x^3))/(-1 + x)^7), {x, 0, 20}], x]
(* End *)
a[n_] := If[n<3, 0, Coefficient[n!*LaguerreL[n, x], x, n-3] // Abs];
Array[a, 30] (* Jean-François Alcover, Jun 14 2018, after A144084 *)

PARI

a(n) = 3!*binomial(n, 3)^2; \\ Andrew Howroyd, Feb 13 2018

Formula

a(n) = 3!*binomial(n, 3)^2.

a(n) = (n^2*(2-3*n+n^2)^2)/6. - Colin Barker, Jan 08 2013

G.f.: -6*x^3*(x+1)*(x^2+8*x+1) / (x-1)^7. - Colin Barker, Jan 08 2013

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7). - Eric W. Weisstein, Sep 05 2017

From Amiram Eldar, Nov 02 2021: (Start)

Sum_{n>=3} 1/a(n) = 3*Pi^2/2 - 117/8.

Sum_{n>=3} (-1)^(n+1)/a(n) = 21/8 - Pi^2/4. (End)