OFFSET
1,6
COMMENTS
In terms of triangular chess: Number of nonequivalent ways (mod D_3) to arrange 4 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 A193982.
LINKS
Heinrich Ludwig, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-6,0,6,8,-12,-9,13,6,-6,-13,9,12,-8,-6,0,6,0,-3,1).
FORMULA
a(n) = (5*n^8 - 100*n^7 + 810*n^6 - 3336*n^5 + 6940*n^4 - 5120*n^3 - 4080*n^2 + 6336*n)/11520 + IF(MOD(n, 2) = 1, 4*n^3 - 38*n^2 + 144*n - 207)/768 + IF(MOD(n, 3) = 1, n^2 - 6*n + 8)/18 + IF(MOD(n, 6) = 1, -1)/6.
G.f.: x^6*(3 + 31*x + 122*x^2 + 330*x^3 + 630*x^4 + 920*x^5 + 1128*x^6 + 1224*x^7 + 1124*x^8 + 924*x^9 + 644*x^10 + 336*x^11 + 117*x^12 + 27*x^13) / ((1 - x)^9*(1 + x)^4*(1 - x + x^2)*(1 + x + x^2)^3). - Colin Barker, Mar 01 2017
EXAMPLE
There are a(6) = 3 ways to place 4 points on an 6 X 6 X 6 grid, rotations and reflections ignored:
. X .
. X . . X .
. . . . . . . . X
. . X . . . X . . . . .
X . . . . . X . . . . X . . .
. . . X . . . . . X . . . . . X . .
MATHEMATICA
Table[(5 n^8 - 100 n^7 + 810 n^6 - 3336 n^5 + 6940 n^4 - 5120 n^3 - 4080 n^2 + 6336 n)/11520 + Boole[OddQ@ n] (4 n^3 - 38 n^2 + 144 n - 207)/768 + Boole[Mod[n, 3] == 1] (n^2 - 6 n + 8)/18 - Boole[Mod[n, 6] == 1]/6, {n, 34}] (* or *)
Rest@ CoefficientList[Series[x^6*(3 + 31 x + 122 x^2 + 330 x^3 + 630 x^4 + 920 x^5 + 1128 x^6 + 1224 x^7 + 1124 x^8 + 924 x^9 + 644 x^10 + 336 x^11 + 117 x^12 + 27 x^13)/((1 - x)^9*(1 + x)^4*(1 - x + x^2) (1 + x + x^2)^3), {x, 0, 34}], x] (* Michael De Vlieger, Mar 01 2017 *)
PROG
(PARI) concat(vector(5), Vec(x^6*(3 + 31*x + 122*x^2 + 330*x^3 + 630*x^4 + 920*x^5 + 1128*x^6 + 1224*x^7 + 1124*x^8 + 924*x^9 + 644*x^10 + 336*x^11 + 117*x^12 + 27*x^13) / ((1 - x)^9*(1 + x)^4*(1 - x + x^2)*(1 + x + x^2)^3) + O(x^40))) \\ Colin Barker, Mar 01 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Heinrich Ludwig, Mar 01 2017
STATUS
approved