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A047659 Number of ways to place 3 nonattacking queens on an n X n board. 21
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,5

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).

Edouard Lucas, Recreations mathematiques, Gauthier-Villars, Paris, 1882-1894, Vol. I, p. 228.

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, January 26, 2013. - N. J. A. Sloane, Feb 16 2013

V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 11.

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.

Index entries for linear recurrences with constant coefficients, signature (5,-8,0,14,-14,0,8,-5,1).

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

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 *)

CROSSREFS

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

Sequence in context: A055857 A239574 A223748 * A108671 A097321 A105946

Adjacent sequences:  A047656 A047657 A047658 * A047660 A047661 A047662

KEYWORD

nonn,easy,nice

AUTHOR

Paul Zimmermann

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

STATUS

approved

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Last modified July 30 23:04 EDT 2015. Contains 260145 sequences.