A179058 -id:A179058 - cp's OEIS frontend

A047659 Number of ways to place 3 nonattacking queens on an n X n board.

0, 0, 0, 0, 24, 204, 1024, 3628, 10320, 25096, 54400, 107880, 199400, 348020, 579264, 926324, 1431584, 2148048, 3141120, 4490256, 6291000, 8656860, 11721600, 15641340, 20597104, 26797144, 34479744, 43915768, 55411720, 69312516, 86004800, 105919940

Offset: 0

Paul Zimmermann

Lucas mentions that the number of ways of placing p <= n non-attacking queens on an n X n chessboard is given by a polynomial in n of degree 2p and attribute the result to Mantel, professor in Delft. Cf. Stanley, exercise 15.

E. Landau, Naturwissenschaftliche Wochenschrift (Aug. 2 1896).
R. P. Stanley, Enumerative Combinatorics, vol. I, exercise 15 in chapter 4 (and its solution) asks one to show the existence of a rational generating function for the number of ways of placing k non-attacking queens on an n X n chessboard.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-queens problem I. General theory, Jan 26 2013. - _N. J. A. Sloane_, Feb 16 2013
S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv:1609.00853 [math.CO], Sep 03 2016.
Christopher R. H. Hanusa, T Zaslavsky, S Chaiken, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv preprint arXiv:1609.00853 [math.CO], 2016-2020.
V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 11.
Edmund Landau, Ueber das Achtdamenproblem und seine Verallgemeinerung, Naturwissenschaftliche Wochenschrift, Aug 02 1896.
Edouard Lucas, Récréations mathématiques, Gauthier-Villars, Paris, 1882-1894, Vol. I, p. 228.
Antal Pinter, Numerical solution of the k=3 Queens problem, 2011, P(3) at p.8-9.
I. Rivin, I. Vardi and P. Zimmermann, The n-queens problem, Amer. Math. Monthly, 101 (1994), 629-639.
Wenxi Wang, Muhammad Usman, Alyas Almaawi, Kaiyuan Wang, Kuldeep S. Meel, Sarfraz Khurshid, A Study of Symmetry Breaking Predicates and Model Counting, National University of Singapore (2020).
Index entries for linear recurrences with constant coefficients, signature (5,-8,0,14,-14,0,8,-5,1).

Cf. A036464, A061994, A108792, A176186, A178721.

Column k=3 of A348129.

[(3*(2*n-1)*(-1)^n +4*n^6 -40*n^5 +158*n^4 -300*n^3 +264*n^2 -86*n +3)/24: n in [0..35]]; // Vincenzo Librandi, Sep 21 2015

f:=n-> n^6/6 - 5*n^5/3 + 79*n^4/12 - 25*n^3/2 + 11*n^2 - 43*n/12 + 1/8 + (-1)^n*(n/4 - 1/8); [seq(f(n),n=1..40)]; # N. J. A. Sloane, Feb 16 2013

Table[If[EvenQ[n],n (n-2)^2 (2n^3-12n^2+23n-10)/12,(n-1)(n-3) (2n^4- 12n^3+25n^2-14n+1)/12],{n,0,30}] (* or *) LinearRecurrence[ {5,-8,0,14,-14,0,8,-5,1},{0,0,0,0,24,204,1024,3628,10320},30] (* Harvey P. Dale, Nov 06 2011 *)

a(n)=if(n%2, (n - 1)*(n - 3)*(2*n^4 - 12*n^3 + 25*n^2 - 14*n + 1), n*(n - 2)^2*(2*n^3 - 12*n^2 + 23*n - 10))/12 \\ Charles R Greathouse IV, Feb 09 2017

a(n) = n(n - 2)^2(2n^3 - 12n^2 + 23n - 10)/12 if n is even and (n - 1)(n - 3)(2n^4 - 12n^3 + 25n^2 - 14n + 1)/12 if n is odd (Landau, 1896).
a(n) = 5a(n - 1) - 8a(n - 2) + 14a(n - 4) - 14a(n - 5) + 8a(n - 7) - 5a(n - 8) + a(n - 9) for n >= 9.
G.f.: 4(9*x^4 + 35*x^3 + 49*x^2 + 21*x + 6)*x^4/((1 - x)^7*(1 + x)^2).
a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=24, a(5)=204, a(6)=1024, a(7)=3628, a(8)=10320, a(n) = 5*a(n-1)-8*a(n-2)+14*a(n-4)-14*a(n-5)+8*a(n-7)- 5*a(n-8)+ a(n-9). - Harvey P. Dale, Nov 06 2011
a(n) = n^6/6 - 5*n^5/3 + 79*n^4/12 - 25*n^3/2 + 11*n^2 - 43*n/12 + 1/8 + (-1)^n*(n/4 - 1/8) [Chaiken et al.]. - N. J. A. Sloane, Feb 16 2013
a(n) = (3*(2*n-1)*(-1)^n +4*n^6 -40*n^5 +158*n^4 -300*n^3 +264*n^2 -86*n +3)/24. - Antal Pinter, Oct 03 2014
E.g.f.: (exp(2*x)*(3 - 6*x^2 + 8*x^3 + 18*x^4 + 20*x^5 + 4*x^6) -3 - 6*x) / (24*exp(x)). - Vaclav Kotesovec, Feb 15 2015
For n>3, a(n) = A179058(n) -4*(n-2)*A000914(n-2) -2*(n-2)*A002415(n-1) + 2*A008911(n-1) +8*(A001752(n-4) +A007009(n-3)). - Antal Pinter, Sep 20 2015
In general, for m <= n, n >= 3, the number of ways to place 3 nonattacking queens on an m X n board is n^3/6*(m^3 - 3*m^2 + 2*m) - n^2/2*(3*m^3 - 9*m^2 + 6*m) + n/6*(2*m^4 + 20*m^3 - 77*m^2 + 58*m) - 1/24*(39*m^4 - 82*m^3 - 36*m^2 + 88*m) + 1/16*(2*m - 4*n + 1)*(1 + (-1)^(m+1)) + 1/2*(1 + abs(n - 2*m + 3) - abs(n - 2*m + 4))*(1/24*((n - 2*m + 11)^4 - 42*(n - 2*m + 11)^3 + 656*(n - 2*m + 11)^2 - 4518*(n - 2*m + 11) + 11583) - 1/16*(4*m - 2*n - 1)*(1 + (-1)^(n+1))) [Panos Louridas, idee & form 93/2007, pp. 2936-2938]. - Vaclav Kotesovec, Feb 20 2016

The formula given in the Rivin et al. paper is wrong.

Entry improved by comments from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), May 30 2001

A144084 T(n,k) is the number of partial bijections of height k (height(alpha) = |Im(alpha)|) of an n-element set.

Original entry on oeis.org

1, 1, 1, 1, 4, 2, 1, 9, 18, 6, 1, 16, 72, 96, 24, 1, 25, 200, 600, 600, 120, 1, 36, 450, 2400, 5400, 4320, 720, 1, 49, 882, 7350, 29400, 52920, 35280, 5040, 1, 64, 1568, 18816, 117600, 376320, 564480, 322560, 40320

Offset: 0

Abdullahi Umar, Sep 10 2008, Sep 30 2008

T(n,k) is also the number of elements in the Green's J equivalence classes in the symmetric inverse monoid, I sub n.
T(n,k) is also the number of ways to place k nonattacking rooks on an n X n chessboard. It can be obtained by performing P(n,k) permutations of n-columns over each C(n,k) combination of n-rows for the given k-rooks. The rule is also applicable for unequal (m X n) sized rectangular boards. - Antal Pinter, Nov 12 2014
Rows also give the coefficients of the matching-generating polynomial of the complete bipartite graph K_{n,n}. - Eric W. Weisstein, Apr 24 2017
Rows also give the coefficients of the independence polynomial of the n X n rook graph and clique polynomial of the n X n rook complement graph. - Eric W. Weisstein, Jun 13 and Sep 14 2017
T(n,k) is the number of increasing subsequences of length n-k over all permutations of [n]. - Geoffrey Critzer, Jan 08 2023

			T(3,1) = 9 because there are exactly 9 partial bijections (on a 3-element set) of height 1, namely: (1)->(1), (1)->(2), (1)->(3), (2)->(1), (2)->(2), (2)->(3), (3)->(1), (3)->(2), (3)->(3).
Triangle T(n,k) begins:
  1;
  1,  1;
  1,  4,   2;
  1,  9,  18,    6;
  1, 16,  72,   96,   24;
  1, 25, 200,  600,  600,  120;
  1, 36, 450, 2400, 5400, 4320, 720;
  ...

O. Ganyushkin and V. Mazorchuk, Classical Finite Transformation Semigroups, 2009, page 61.
J. M. Howie, Fundamentals of semigroup theory. Oxford: Clarendon Press, (1995).
Vaclav Kotesovec, Non-attacking chess pieces, 6th ed. (2013), p. 216, p. 218.

Alois P. Heinz, Rows n = 0..140, flattened
Wayne A. Johnson, Exponential Hilbert series of equivariant embeddings, arXiv:1804.04943 [math.RT], 2018.
W. D. Munn, The characters of the symmetric inverse semigroup, Proc. Cambridge Philos. Soc. 53 (1957), 13-18.
Eric Weisstein's World of Mathematics, Clique Polynomial
Eric Weisstein's World of Mathematics, Complete Bipartite Graph
Eric Weisstein's World of Mathematics, Independence Polynomial
Eric Weisstein's World of Mathematics, Matching-Generating Polynomial
Eric Weisstein's World of Mathematics, Rook Complement Graph
Eric Weisstein's World of Mathematics, Rook Graph

T(n,k) = |A021010|. Sum of rows of T(n,k) is A002720. T(n,n) is the order of the symmetric group on an n-element set, n!.
Columns k=2..10 are A163102, A179058, A179059, A179060, A179061, A179062, A179063, A179064, A179065.
Cf. A089231, A105219, A295383.

/* As triangle */ [[(Binomial(n,k)^2)*Factorial(k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jun 13 2017

T:= (n, k)-> (binomial(n, k)^2)*k!:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Dec 04 2012

Table[Table[Binomial[n, k]^2 k!,{k, 0, n}], {n, 0, 6}] // Flatten (* Geoffrey Critzer, Dec 04 2012 *)
Table[ CoefficientList[n!*LaguerreL[n, x], x] // Abs // Reverse, {n, 0, 8}] // Flatten (* Jean-François Alcover, Nov 18 2013 *)
CoefficientList[Table[n! x^n LaguerreL[n, -1/x], {n, 0, 8}], x] // Flatten (* Eric W. Weisstein, Apr 24 2017 *)
CoefficientList[Table[(-x)^n HypergeometricU[-n, 1, -(1/x)], {n, 5}],
  x] // Flatten (* Eric W. Weisstein, Jun 13 2017 *)

T(n,k) = k! * binomial(n,k)^2 \\ Andrew Howroyd, Feb 13 2018

T(n,k) = (C(n,k)^2)*k!.
T(n,k) = A007318(n,k) * A008279(n,k). - Antal Pinter, Nov 12 2014
From Peter Bala, Jul 04 2016: (Start)
G.f.: exp(x*t)*I_0(2*sqrt(x)) = 1 + (1 + t)*x/1!^2 + (1 + 4*t + 2*t^2)*x^2/2!^2 + (1 + 9*t + 18*t^2 + 6*t^3)*x^3/3!^2 + ..., where I_0(x) = Sum_{n >= 0} (x/2)^(2*n)/n!^2 is a modified Bessel function of the first kind.
The row polynomials R(n,t) satisfy R(n,t + u) = Sum_{k = 0..n} T(n,k)*t^k*R(n-k,u).
R(n,t) = 1 + Sum_{k = 0..n-1} (-1)^(n-k+1)*n!/k!*binomial(n,k) *t^(n-k)*R(k,t). Cf. A089231. (End)
From Peter Bala, Oct 05 2019: (Start)
E.g.f.: 1/(1 - t*x)*exp(x/(1 - t*x)).
Recurrence for row polynomials: R(n+1,t) = (1 + (2*n+1)*t)R(n,t) - n^2*t^2*R(n-1,t), with R(0,t) = 1 and R(1,t) = 1 + t.
R(n,t) equals the denominator polynomial of the finite continued fraction 1 + n*t/(1 + n*t/(1 + (n-1)*t/(1 + (n-1)*t/(1 + ... + 2*t/(1 + 2*t/(1 + t/(1 + t/(1)))))))). The numerator polynomial is the (n+1)-th row polynomial of A089231. (End)
Sum_{n>=0} Sum_{k=0..n} T(n,k)*y^k*x^n/A001044(n) = exp(y*x)*E(x) where E(x) = Sum_{n>=0} x^n/A001044(n). - Geoffrey Critzer, Jan 08 2023
Sum_{k=0..n} k*T(n,k) = A105219(n). - Alois P. Heinz, Jan 08 2023
T(n,k) = Sum_{d=0..2*k} c(k,d)*n^d, where c(k,d) = Sum_{j=max(d-k,0)..k} binomial(k,j)*A008275(k+j,d)/j!. - Eder G. Santos, Jan 23 2025

A179059 Number of non-attacking placements of 4 rooks on an n X n board.

Original entry on oeis.org

0, 0, 0, 24, 600, 5400, 29400, 117600, 381024, 1058400, 2613600, 5880600, 12269400, 24048024, 44717400, 79497600, 135945600, 224726400, 360561024, 563376600, 859685400, 1284221400, 1881864600, 2709885024, 3840540000, 5364060000

Offset: 1

Thomas Zaslavsky, Jun 27 2010

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.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Column k=4 of A144084.

Cf. A179058 (3 rooks), A179060 (5 rooks).

LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{0,0,0,24,600,5400,29400,117600,381024},40] (* Harvey P. Dale, Feb 19 2013 *)
a[n_] := If[n<4, 0, Coefficient[n!*LaguerreL[n, x], x, n-4] // Abs];
Array[a, 30] (* Jean-François Alcover, Jun 14 2018, after A144084 *)

a(n) = 4! * binomial(n, 4)^2; \\ Andrew Howroyd, Feb 13 2018

a(n) = 4! * binomial(n, 4)^2.
From Colin Barker, Jan 08 2013: (Start)
a(n) = (n^2*(-6+11*n-6*n^2+n^3)^2)/24.
G.f.: -24*x^4*(x^4 +16*x^3 +36*x^2 +16*x +1) / (x -1)^9.
(End)
From Amiram Eldar, Nov 02 2021: (Start)
Sum_{n>=4} 1/a(n) = (20*Pi^2 - 197)/9.
Sum_{n>=4} (-1)^n/a(n) = (64*log(2) - 44)/9. (End)

A303212 Number of minimum total dominating sets in the n X n rook complement graph.

Original entry on oeis.org

0, 1, 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, 98901600

Offset: 1

Eric W. Weisstein, Apr 19 2018

For n > 2, the minimum total dominating sets are any three vertices such that no two are in the same row or column. - Andrew Howroyd, Apr 20 2018

Essentially the same as A179058(n), differing only for n=2. - Eric W. Weisstein, Dec 06 2023

Colin Barker, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Rook Complement Graph.
Eric Weisstein's World of Mathematics, Minimum Total Dominating Set.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A179058, A303209, A347922.

Table[If[n == 2, 1, 6 Binomial[n, 3]^2], {n, 20}]
Join[{0, 1}, LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 0, 6, 96, 600, 2400, 7350}, {3, 20}]]
CoefficientList[Series[x (-1 + x - 75 x^2 - 19 x^3 - 41 x^4 + 21 x^5 - 7 x^6 + x^7)/(-1 + x)^7, {x, 0, 20}], x]

a(n) = if(n<3, n==2, 6*binomial(n,3)^2) \\ Andrew Howroyd, Apr 20 2018

concat(0, Vec(x^2*(1 - x + 75*x^2 + 19*x^3 + 41*x^4 - 21*x^5 + 7*x^6 - x^7) / (1 - x)^7 + O(x^60))) \\ Colin Barker, Apr 20 2018

a(n) = A179058(n) for n > 2. - Andrew Howroyd, Apr 20 2018
From Colin Barker, Apr 20 2018: (Start)
G.f.: x^2*(1 - x + 75*x^2 + 19*x^3 + 41*x^4 - 21*x^5 + 7*x^6 - x^7) / (1 - x)^7.
a(n) = n^2*(2 - 3*n + n^2)^2 / 6 for n > 2.
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) for n > 9.
(End)

a(6)-a(30) from Andrew Howroyd, Apr 20 2018

A291910 Number of 4-cycles in the n X n rook complement graph.

Original entry on oeis.org

0, 0, 9, 576, 6900, 44100, 196245, 686784, 2023056, 5232600, 12224025, 26310240, 52936884, 100663836, 182452725, 317318400, 532407360, 865571184, 1368508041, 2110550400, 3183182100, 4705372980, 6829824309, 9750223296, 13709610000, 19009965000, 26023131225

Offset: 1

Eric W. Weisstein, Sep 05 2017

nonn

Eric Weisstein's World of Mathematics, Graph Cycle
Eric Weisstein's World of Mathematics, Rook Complement Graph
Index entries for linear recurrences with constant coefficients, signature (9, -36, 84, -126, 126, -84, 36, -9, 1).

Cf. A179058 (3-cycles), A291911 (5-cycles), A291912 (6-cycles).

Table[(-2 + n) (-1 + n)^2 n^2 (-4 + 5 n - 4 n^2 + n^3)/8, {n, 20}]
LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {0, 0, 9, 576, 6900, 44100, 196245, 686784, 2023056}, 30]
CoefficientList[Series[-((3 x^2 (3 + 165 x + 680 x^2 + 660 x^3 + 165 x^4 + 7 x^5))/(-1 + x)^9), {x, 0, 20}], x]

a(n) = (n-2)*(n-1)^2*n^2*(-4 + 5*n - 4*n^2 + n^3)/8.
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9).
G.f.: -((3 x^3 (3 + 165 x + 680 x^2 + 660 x^3 + 165 x^4 + 7 x^5))/(-1 + x)^9).

A291911 Number of 5-cycles in the n X n rook complement graph.

Original entry on oeis.org

0, 0, 36, 3456, 77040, 800640, 5265540, 25514496, 99320256, 327836160, 951285060, 2488844160, 5980596336, 13384215936, 28197301860, 56398325760, 107825391360, 198142765056, 351580800996, 604675808640, 1011282960240, 1649187872640, 2628701385156

Offset: 1

Eric W. Weisstein, Sep 05 2017

nonn

Eric Weisstein's World of Mathematics, Graph Cycle
Eric Weisstein's World of Mathematics, Rook Complement Graph
Index entries for linear recurrences with constant coefficients, signature (11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1).

Cf. A179058 (3-cycles), A291910 (4-cycles), A291912 (6-cycles).

Table[(n - 2)^2 (n - 1)^2 n^2 (4 + 2 n + 3 n^2 - 4 n^3 + n^4)/10, {n, 20}]
LinearRecurrence[{11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1}, {0, 0, 36, 3456, 77040, 800640, 5265540, 25514496, 99320256, 327836160, 951285060}, 20]
CoefficientList[Series[-((36 x^2 (1 + 85 x + 1139 x^2 + 3815 x^3 + 3815 x^4 + 1139 x^5 + 85 x^6 + x^7))/(-1 + x)^11), {x, 0, 20}], x]

a(n) = (n - 2)^2*(n - 1)^2*n^2*(4 + 2*n + 3*n^2 - 4*n^3 + n^4)/10.
a(n) = 11*a(n-1) - 55*a(n-2) + 165*(n-3) - 330*a(n-4) + 462*a(n-5) - 462*a(n-6) + 330*a(n-7) - 165*a(n-8) + 55*a(n-9) - 11*a(n-10) + a(n-11).
G.f.: -((36 x^3 (1 + 85 x + 1139 x^2 + 3815 x^3 + 3815 x^4 + 1139 x^5 + 85 x^6 + x^7))/(-1 + x)^11).

A291912 Number of 6-cycles in the n X n rook complement graph.

Original entry on oeis.org

0, 0, 60, 18336, 840800, 14629200, 143939460, 971877760, 5018582016, 21193207200, 76518984300, 243664127520, 699965254560, 1844973808496, 4520720267700, 10403885452800, 22674321863680, 47112768624960, 93845538165276, 180039346960800, 333959821087200, 600947653207440

Offset: 1

Eric W. Weisstein, Sep 05 2017

nonn

Eric Weisstein's World of Mathematics, Graph Cycle
Eric Weisstein's World of Mathematics, Rook Complement Graph
Index entries for linear recurrences with constant coefficients, signature (13, -78, 286, -715, 1287, -1716, 1716, -1287, 715, -286, 78, -13, 1).

Cf. A179058 (3-cycles), A291910 (4-cycles), A291911 (5-cycles).

Table[(-2 + n) (-1 + n)^2 n^2 (-52 + 12 n + 76 n^2 - 63 n^3 - 2 n^4 + 20 n^5 - 8 n^6 + n^7)/12, {n, 20}]
LinearRecurrence[{13, -78, 286, -715, 1287, -1716, 1716, -1287, 715, -286, 78, -13, 1}, {0, 0, 60, 18336, 840800, 14629200, 143939460, 971877760, 5018582016, 21193207200, 76518984300, 243664127520, 699965254560}, 20]
CoefficientList[Series[(4 x^2 (-15 - 4389 x - 151778 x^2 - 1277962 x^3 - 3535266 x^4 - 3576650 x^5 - 1293586 x^6 - 137682 x^7 - 1883 x^8 + 11 x^9))/(-1 + x)^13, {x, 0, 20}], x]

a(n) = (-2 + n)*(-1 + n)^2*n^2*(-52 + 12*n + 76*n^2 - 63*n^3 - 2*n^4 + 20*n^5 - 8*n^6 + n^7)/12.
a(n) = 13*a(n-1) - 78*a(n-2) + 286*a(n-3) - 715*a(n-4) + 1287*a(n-5) - 1716*a(n-6) + 1716*a(n-7) - 1287*a(n-8) + 715*a(n-9) - 286*a(n-10) + 78*a(n-11) - 13*a(n-12) + a(n-13).
G.f.: (4 x^3 (-15 - 4389 x - 151778 x^2 - 1277962 x^3 - 3535266 x^4 - 3576650 x^5 - 1293586 x^6 - 137682 x^7 - 1883 x^8 + 11 x^9))/(-1 + x)^13.

A307304 Number of inequivalent ways of placing 2 nonattacking rooks on n X n board up to rotations and reflections of the board.

Original entry on oeis.org

0, 1, 4, 13, 31, 66, 123, 214, 346, 535, 790, 1131, 1569, 2128, 2821, 3676, 4708, 5949, 7416, 9145, 11155, 13486, 16159, 19218, 22686, 26611, 31018, 35959, 41461, 47580, 54345, 61816, 70024, 79033, 88876, 99621, 111303, 123994, 137731, 152590, 168610, 185871

Offset: 1

Mo Li, Apr 19 2019

			For n = 4 the a(4) = 13 solutions are
{{1,0,0,0}}  {{1,0,0,0}}  {{1,0,0,0}}
{{0,1,0,0}}  {{0,0,1,0}}  {{0,0,0,1}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
—————————————————————————————————————
{{1,0,0,0}}  {{1,0,0,0}}  {{1,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
{{0,0,1,0}}  {{0,0,0,1}}  {{0,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,1}}
—————————————————————————————————————
{{0,1,0,0}}  {{0,1,0,0}}  {{0,1,0,0}}
{{1,0,0,0}}  {{0,0,1,0}}  {{0,0,0,1}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
—————————————————————————————————————
{{0,1,0,0}}  {{0,1,0,0}}  {{0,1,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
{{0,0,1,0}}  {{0,0,0,1}}  {{0,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,1,0}}
—————————————————————————————————————
{{0,0,0,0}}
{{0,1,0,0}}
{{0,0,1,0}}
{{0,0,0,0}}

Leisure Maths Entertainment Forum, 2 nonattacking rooks on n X n board, Chinese blog.

Cf. A000903, A163102, A035287, A179058, A144084.

Table[
Piecewise[{{(n (n^3 - 2 n^2 + 6 n - 4))/16, Mod[n, 2] == 0},
{((n - 1) (n^3 - n^2 + 5 n - 1))/16, Mod[n, 2] == 1}}],{n, 20}]

a(n) = (1/16)*n*(n^3-2n^2+6n-4) if n is even;
a(n) = (1/16)*(n-1)*(n^3-n^2+5n-1) if n is odd.
G.f.: -x^2*(x^2+1)*(x^2+x+1)/((x+1)^2*(x-1)^5). - Alois P. Heinz, Apr 26 2019

cp's OEIS Frontend

A047659 Number of ways to place 3 nonattacking queens on an n X n board.

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A144084 T(n,k) is the number of partial bijections of height k (height(alpha) = |Im(alpha)|) of an n-element set.

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A179059 Number of non-attacking placements of 4 rooks on an n X n board.

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A303212 Number of minimum total dominating sets in the n X n rook complement graph.

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A291910 Number of 4-cycles in the n X n rook complement graph.

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A291911 Number of 5-cycles in the n X n rook complement graph.

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A291912 Number of 6-cycles in the n X n rook complement graph.

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