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

Author

Paul Zimmermann

Keywords

• nonn
• easy
• nice

Comments

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.

References

• 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.

Links

• 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).

Crossrefs

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

Column k=3 of A348129.

Programs

Magma

[(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

Maple

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

Mathematica

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 *)

PARI

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

Formula

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

Extensions

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