OFFSET
1,5
COMMENTS
In terms of triangular chess: Number of nonequivalent ways (mod D_3) to arrange 3 nonattacking rooks on an n X n X n board.
Reflections and rotations of placements are not counted. For numbers if they are to be counted see A193981.
LINKS
Heinrich Ludwig, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-7,3,6,0,-6,-3,7,0,-3,1).
FORMULA
a(n) = (n^6 - 9*n^5 + 27*n^4 - 20*n^3 - 24*n^2 + 24*n)/288 + IF(MOD(n, 2) = 1, n^2 - 3*n - 5)/32 + IF(MOD(n, 3) = 1, 2)/9.
G.f.: x^4*(1 + 2*x + 8*x^2 + 20*x^3 + 16*x^4 + 10*x^5 + 3*x^6) / ((1 - x)^7*(1 + x)^3*(1 + x + x^2)). - Colin Barker, Mar 01 2017
a(n) = ( 2*n^6 - 18*n^5 + 54*n^4 - 40*n^3 - 39*n^2 + 21*n - 45 - 9*(n^2 - 3*n - 5)*(-1)^n + 128*((n mod 3) mod 2) )/576. - Bruno Berselli, Mar 01 2017
EXAMPLE
There is a(4) = 1 way to place 3 points on a 4 X 4 X 4 grid, rotations and reflections ignored:
.
X .
. . X
. X . .
MATHEMATICA
Table[(n^6 - 9 n^5 + 27 n^4 - 20 n^3 - 24 n^2 + 24 n)/288 + Boole[OddQ@ n] (n^2 - 3 n - 5)/32 + Boole[Mod[n, 3] == 1] 2/9, {n, 38}] (* or *)
Rest@ CoefficientList[Series[x^4*(1 + 2 x + 8 x^2 + 20 x^3 + 16 x^4 + 10 x^5 + 3 x^6)/((1 - x)^7*(1 + x)^3*(1 + x + x^2)), {x, 0, 38}], x] (* Michael De Vlieger, Mar 01 2017 *)
LinearRecurrence[{3, 0, -7, 3, 6, 0, -6, -3, 7, 0, -3, 1}, {0, 0, 0, 1, 5, 23, 82, 230, 560, 1208, 2392, 4405}, 40] (* Harvey P. Dale, May 07 2022 *)
PROG
(PARI) concat(vector(3), Vec(x^4*(1 + 2*x + 8*x^2 + 20*x^3 + 16*x^4 + 10*x^5 + 3*x^6) / ((1 - x)^7*(1 + x)^3*(1 + x + x^2)) + O(x^30))) \\ Colin Barker, Mar 01 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Heinrich Ludwig, Mar 01 2017
STATUS
approved