A278682 Number of non-equivalent ways to place 3 non-attacking ferses on an n X n board.

0, 0, 7, 45, 225, 709, 1974, 4524, 9614, 18382, 33425, 56895, 93447, 146715, 224280, 331814, 480844, 679724, 945099, 1288737, 1733725, 2296065, 3006762, 3886960, 4977210, 6304794, 7921589, 9862099, 12191459, 14952567, 18225900, 22064010, 26564952, 31792280

Offset: 1

Author

Heinrich Ludwig, Nov 26 2016

Keywords

• nonn
• easy

Comments

A fers is a leaper [1, 1].

Rotations and reflections of placements are not counted. If they are to be counted, see A201244.

Examples

			There are 7 ways to place 3 non-attacking ferses "X" on a 3 X 3 board, rotations and reflections being ignored
   XXX   XX.   X.X   ...   X..   X..   X..
   ...   ...   ...   XXX   X.X   ...   ...
   ...   ..X   .X.   ...   ...   XX.   X.X
		

Links

• Heinrich Ludwig, Table of n, a(n) for n = 1..1000
• Wikipedia, Fairy chess piece
• Index entries for linear recurrences with constant coefficients, signature (3,1,-11,6,14,-14,-6,11,-1,-3,1).

Crossrefs

Cf. A201244, A232567 (2 ferses), A278683 (4 ferses), A278684 (5 ferses), A278685 (6 ferses), A278686 (7 ferses), A278687, A278688.

Programs

Mathematica

Table[Boole[n > 2] ((n^6 - 15 n^4 + 32 n^3 + 14 n^2 - 116 n + 96) + Boole[OddQ@ n] (8 n^3 - 9 n^2 - 20 n + 9))/48, {n, 34}] (* Michael De Vlieger, Nov 30 2016 *)

PARI

concat(vector(2), Vec(x^3*(7 + 24*x + 83*x^2 + 66*x^3 + 75*x^4 - 15*x^6 - 2*x^7 + 2*x^8) / ((1 - x)^7*(1 + x)^4) + O(x^40))) \\ Colin Barker, Dec 07 2016

Formula

a(n) = ((n^6 - 15*n^4 + 32*n^3 + 14*n^2 - 116*n + 96) + IF(MOD(n, 2) = 1, 8*n^3 - 9*n^2 - 20*n + 9))/48.

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 - 15*n^4 + 32*n^3 + 14*n^2 - 116*n + 96)/48 for n even.

a(n) = (n^6 - 15*n^4 + 40*n^3 + 5*n^2 - 136*n + 105)/48 for n odd.

G.f.: x^3*(7 + 24*x + 83*x^2 + 66*x^3 + 75*x^4 - 15*x^6 - 2*x^7 + 2*x^8) / ((1 - x)^7*(1 + x)^4).

(End)