A279112 Number of non-equivalent ways to place 3 non-attacking kings on an n X n board.
0, 0, 2, 20, 138, 505, 1547, 3759, 8313, 16350, 30344, 52470, 87212, 138255, 212953, 317065, 461827, 655724, 915038, 1251720, 1688414, 2241365, 2941047, 3808915, 4884893, 6196650, 7795332, 9715914, 12022688, 14759115, 18004709, 21812685, 26280007, 31471000, 37502458
Offset: 1
Examples
There are 2 non-equivalent ways to place 3 non-attacking kings on a 3 X 3 board: K.K K.K ... ... ..K .K.
Links
- Heinrich Ludwig, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (3,1,-11,6,14,-14,-6,11,-1,-3,1).
Crossrefs
Programs
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Mathematica
Table[Boole[n > 1] (n^6 - 27 n^4 + 44 n^3 + 146 n^2 - 404 n + 240 + Boole[OddQ@ n] (8 n^3 - 21 n^2 + 16 n - 51))/48, {n, 35}] (* or *) Rest@ CoefficientList[Series[x^3*(2 + 14 x + 76 x^2 + 93 x^3 + 102 x^4 - 17 x^5 - 36 x^6 - x^7 + 8 x^8 - x^9)/((1 - x)^7*(1 + x)^4), {x, 0, 35}], x] (* Michael De Vlieger, Dec 08 2016 *)
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PARI
concat(vector(2), Vec(x^3*(2 + 14*x + 76*x^2 + 93*x^3 + 102*x^4 - 17*x^5 - 36*x^6 - x^7 + 8*x^8 - x^9) / ((1 - x)^7*(1 + x)^4) + O(x^60))) \\ Colin Barker, Dec 07 2016
Formula
a(n) = (n^6 - 27*n^4 + 44*n^3 + 146*n^2 - 404*n + 240 + IF(MOD(n, 2) = 1, 8*n^3 - 21*n^2 + 16*n - 51))/48 for n >=2.
a(n) = 3*a(n-1) + a(n-2) - 11*a(n-3) + 6*a(n-4) + 14*a(n-5) - 14*a(n-6) - 6*a(n-7) + 11*a(n-8) - a(n-9) - 3*a(n-10) + a(n-11).
From Colin Barker, Dec 07 2016: (Start)
a(n) = (n^6 - 27*n^4 + 44*n^3 + 146*n^2 - 404*n + 240)/48 for n>1.
a(n) = (n^6 - 27*n^4 + 52*n^3 + 125*n^2 - 388*n + 189)/48 for n>1.
G.f.: x^3*(2 + 14*x + 76*x^2 + 93*x^3 + 102*x^4 - 17*x^5 - 36*x^6 - x^7 + 8*x^8 - x^9) / ((1 - x)^7*(1 + x)^4).
(End)
Comments