A279111 Number of non-equivalent ways to place 2 non-attacking kings on an n X n board.

0, 0, 4, 13, 37, 75, 147, 246, 406, 610, 910, 1275, 1779, 2373, 3157, 4060, 5212, 6516, 8136, 9945, 12145, 14575, 17479, 20658, 24402, 28470, 33202, 38311, 44191, 50505, 57705, 65400, 74104, 83368, 93772, 104805, 117117, 130131, 144571, 159790, 176590, 194250, 213654

Offset

1,3

Comments

Rotations and reflections of placements are not counted. If they are to be counted, see A061995.

Links

Heinrich Ludwig, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).

Formula

a(n) = (n^4 - 2*n^2 - 4*n + IF(MOD(n, 2) = 1, 2*n^2 - 4*n + 7))/16.

a(n) = (2*n^4 - 2*n^2 - 12*n + 7 - (2*n^2 - 4*n + 7)*(-1)^n)/32. - Bruno Berselli, Dec 07 2016

a(n) = 2*a(n-1) + 2*a(n-2) - 6*a(n-3) + 6*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8).

From Colin Barker, Dec 07 2016: (Start)

a(n) = n*(n - 2)*(n^2 + 2*n + 2)/16 for n even.

a(n) = (n - 1)*(n^3 + n^2 + n - 7)/16 for n odd.

G.f.: x^3*(4 + 5*x + 3*x^2 - x^3 + x^4) / ((1 - x)^5*(1 + x)^3).

(End)

Example

There are 4 non-equivalent ways to place 2 non-attacking kings on a 3 X 3 board:
   K.K   K..   K..   .K.
   ...   ..K   ...   ...
   ...   ...   ..K   .K.

Mathematica

Table[(n^4 - 2 n^2 - 4 n + Boole[OddQ@ n] (2 n^2 - 4 n + 7))/16, {n, 43}] (* or *)
Rest@ CoefficientList[Series[x^3*(4 + 5 x + 3 x^2 - x^3 + x^4)/((1 - x)^5*(1 + x)^3), {x, 0, 43}], x] (* Michael De Vlieger, Dec 08 2016 *)

Prog

(PARI) concat(vector(2), Vec(x^3*(4 + 5*x + 3*x^2 - x^3 + x^4) / ((1 - x)^5*(1 + x)^3) + O(x^60))) \\ Colin Barker, Dec 07 2016

Crossrefs

Cf. A061995, A279112 (3 kings), A279113 (4 kings), A279114 (5 kings), A279115 (6 kings), A279116 (7 kings), A279117, A236679.

Sequence in context: A222189 A003727 A103082 * A299111 A324250 A226866

Adjacent sequences: A279108 A279109 A279110 * A279112 A279113 A279114

Keyword

nonn,easy

Author

Heinrich Ludwig, Dec 06 2016

Status

approved