A061994 Number of ways to place 4 nonattacking queens on an n X n board.

0, 0, 0, 0, 2, 82, 982, 7002, 34568, 131248, 412596, 1123832, 2739386, 6106214, 12654614, 24675650, 45704724, 80999104, 138170148, 227938788, 365106738, 569681574, 868289594, 1295775946, 1897176508, 2729909796

Offset: 0

Author

Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), May 31 2001

Keywords

• nonn
• nice
• easy

Comments

An analytical solution for the 4-queens problem permits us to combine six particular cases into a single "unified" expression: a(n) = n(n-1)(45n^6 - 855n^5 + 6945n^4 - 30891n^3 + 78864n^2 - 106226n + 53404)/1080 + (n^3 - 21/2n^2 + 28n - 14)*floor(n/2) + 32/9(n-1)*floor(n/3) + (16/9n-4)*floor((n+1)/3). The method used to derive this formula helps to fine-tune an estimate by E. Lucas for a(n) (see comment to A047659 "3-queens problem"). For any fixed value of k > 1, a(n) = n^(2k)/k! - 5/3n^(2k-1)/(k-2)! + O(n^(2k-2)). - Sergey Perepechko, Sep 16 2005

References

• Vaclav Kotesovec, Between chessboard and computer, 1996, pp. 204-206.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Louis Azemard, Une communication de Vaclav Kotesovec, Echecs et Mathématiques, Rex Multiplex 38/1992.
• Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 12.
• Index entries for linear recurrences with constant coefficients, signature (3,1,-9,0,12,7,-15,-16,16,15,-7,-12,0,9,-1,-3,1).

Crossrefs

Cf. A036464, A047659.

Column k=4 of A348129.

Programs

Mathematica

CoefficientList[Series[x^4*(2 +76*x +734*x^2 +3992*x^3 +13318*x^4 +29356*x^5 +46304*x^6 +53580*x^7 +46890*x^8 +29768*x^9 +13522*x^10 +3804*x^11 +574*x^12)/((1-x)^3*(1-x^2)^4*(1-x^3)^2), {x, 0, 40}], x] (* Vincenzo Librandi, May 12 2013 *)
LinearRecurrence[{3,1,-9,0,12,7,-15,-16,16,15,-7,-12,0,9,-1,-3,1}, {0,0,0,0,2,82, 982,7002,34568,131248,412596,1123832,2739386,6106214,12654614,24675650, 45704724}, 40] (* Harvey P. Dale, Jan 21 2017 *)

SageMath

def p(x): return x^4*(2 +76*x +734*x^2 +3992*x^3 +13318*x^4 +29356*x^5 +46304*x^6 +53580*x^7 +46890*x^8 +29768*x^9 +13522*x^10 +3804*x^11 +574*x^12)/((1-x)^3*(1-x^2)^4*(1-x^3)^2)
def A061994_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( p(x) ).list()
A061994_list(40) # G. C. Greubel, Apr 30 2022

Formula

G.f.: x^4*(2 + 76*x + 734*x^2 + 3992*x^3 + 13318*x^4 + 29356*x^5 + + 46304*x^6 + + 53580*x^7 + 46890*x^8 + 29768*x^9 + 13522*x^10 + 3804*x^11 + 574*x^12)/((1-x)^3*(1-x^2)^4*(1-x^3)^2).

Recurrence: a(n) = 3*a(n-1) + a(n-2) - 9*a(n-3) + 12*a(n-5) + 7*a(n-6) - 15*a(n-7) - 16*a(n-8) + 16*a(n-9) + 15*a(n-10) - 7*a(n-11) - 12*a(n-12) + 9*a(n-14) - a(n-15) - 3*a(n-16) + a(n-17), n >= 17.

Explicit formula (V. Kotesovec, 1992) for n >= 2: a(n) = n^8/24 - 5*n^7/6 + 65*n^6/9 - 1051*n^5/30 + 817*n^4/8 added to one of the following terms:

- 4769*n^3/27 + 1963*n^2/12 - 1769*n/30 if n = 0 (mod 6)

- 9565*n^3/54 + 1013*n^2/6 - 6727*n/90 + 257/27 if n = 1 (mod 6)

- 4769*n^3/27 + 1963*n^2/12 - 5467*n/90 + 28/27 if n = 2 (mod 6)

- 9565*n^3/54 + 1013*n^2/6 - 2189*n/30 + 7 if n = 3 (mod 6)

- 4769*n^3/27 + 1963*n^2/12 - 5467*n/90 + 68/27 if n = 4 (mod 6)

- 9565*n^3/54 + 1013*n^2/6 - 6727*n/90 + 217/27 if n = 5 (mod 6).

a(n) = n^8/24 - 5n^7/6 + 65n^6/9 - 1051n^5/30 + 817n^4/8 - 19103n^3/108 + 3989n^2/24 - 18131n/270 + 253/54 + (n^3/4 - 21n^2/8 + 7n - 7/2)*(-1)^n + 32*(n - 1)/27*cos(2*Pi*n/3) + 40/81*sqrt(3)*sin(2*Pi*n/3). - Vaclav Kotesovec, Feb 11 2010

E.g.f.: (3*(exp(2*x)*(5060 - 4645*x + 1755*x^2 - 590*x^3 + 480*x^4 + 414*x^5 + 870*x^6 + 360*x^7 + 45*x^8) - 135*(28 + 37*x + 15*x^2 + 2*x^3)) - 1920 * exp(x/2) * (2+x) * cos(sqrt(3)*x/2) - 320 * sqrt(3) * exp(x/2) * (6*x-5) * sin(sqrt(3)*x/2)) / (3240 * exp(x)). - Vaclav Kotesovec, Feb 15 2015

Extensions

Minor edits by Vaclav Kotesovec, Feb 15 2015