A249460 Number of proper colorings of the cube with at most n colors under rotational symmetry.

0, 0, 0, 1, 10, 55, 230, 770, 2156, 5250, 11460, 22935, 42790, 75361, 126490, 203840, 317240, 479060, 704616, 1012605, 1425570, 1970395, 2678830, 3588046, 4741220, 6188150, 7985900, 10199475, 12902526, 16178085, 20119330

Offset: 0

Author

Marko Riedel, Oct 29 2014

Keywords

• nonn
• easy

Comments

All terms 3 mod 10 end in 1, all terms 8 mod 10 end in 6 and vice versa. - Jon Perry, Oct 29 2014

Also the number of inequivalent ways to color the corners of an octahedron using at most n colors so that no two adjacent corners have the same color. - David Nacin, Feb 22 2017

Examples

			For n = 3 we see there is only a(3) = 1 way to color the faces of a cube with three colors so that no two adjacent sides have the same color. - _David Nacin_, Feb 22 2017
		

Links

• Marko R. Riedel, Proper colorings of the cube, Mathematics Stack Exchange, Oct 29 2014
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Programs

Magma

[(n-2)*(n-1)*n*(n^3-9*n^2+32*n-38)/24: n in [0..30]]; // Vincenzo Librandi, Oct 30 2014

Maple

q := N -> 1/24*(N*(N-1)*(N-2)*(N^3-9*N^2+29*N-32) + 3*N*(N-1)*(N-2)^2);

Mathematica

Table[(n - 2) (n - 1) n (n^3 - 9 n^2 + 32 n - 38)/24, {n, 0, 30}] (* Bruno Berselli, Oct 30 2014 *)
CoefficientList[Series[x^3 (1 + 3 x + 6 x^2 + 20 x^3)/(1 - x)^7, {x, 0, 30}], x] (* Vincenzo Librandi, Oct 30 2014 *)

PARI

a(n)=(n-2)*(n-1)*n*(n^3-9*n^2+32*n-38)/24 \\ Charles R Greathouse IV, Feb 23 2017

Formula

a(n) = ( n*(n-1)*(n-2)*(n^3-9*n^2+29*n-32) + 3*n*(n-1)*(n-2)^2 )/24 = (n-2)*(n-1)*n*(n^3-9*n^2+32*n-38)/24.

G.f.: x^3*(1+3*x+6*x^2+20*x^3)/(1-x)^7. - Vincenzo Librandi, Oct 30 2014