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Number of non-equivalent ways to place 2 non-attacking kings on an n X n board.
10

%I #16 Dec 21 2016 12:15:57

%S 0,0,4,13,37,75,147,246,406,610,910,1275,1779,2373,3157,4060,5212,

%T 6516,8136,9945,12145,14575,17479,20658,24402,28470,33202,38311,44191,

%U 50505,57705,65400,74104,83368,93772,104805,117117,130131,144571,159790,176590,194250,213654

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

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

%H Heinrich Ludwig, <a href="/A279111/b279111.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-6,0,6,-2,-2,1).

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

%F 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

%F 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).

%F From _Colin Barker_, Dec 07 2016: (Start)

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

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

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

%F (End)

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

%e K.K K.. K.. .K.

%e ... ..K ... ...

%e ... ... ..K .K.

%t Table[(n^4 - 2 n^2 - 4 n + Boole[OddQ@ n] (2 n^2 - 4 n + 7))/16, {n, 43}] (* or *)

%t 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 *)

%o (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

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

%K nonn,easy

%O 1,3

%A _Heinrich Ludwig_, Dec 06 2016