2828924

coloring-cube-sides-and-vertices
A mixed cycle index for faces and vertices combined can  be of use in this problem. We can then apply  Burnside or  Polya as  desired. The group $H$  here are  the  rotations  permuting  six faces  and  eight vertices simultanteously, acting on fourteen slots for the colors. We use $b_q$ for the cycles of the vertices and $a_q$ for the faces.

https://math.stackexchange.com/questions/2828924/coloring-cube-sides-and-vertices/

We proceed to enumerate the permutations of this group. [...]

It follows that the cycle index of $F$ is given by

$$Z(H) = \frac{1}{24} \left(a_1^6 b_1^8 + 6 a_1^2 a_4 b_4^2 + 3 a_1^2 a_2^2 b_2^4 + 8 a_3^2 b_1^2 b_3^2 + 6 a_2^3 b_2^4\right)$$

Using at most $N$ colors for the faces and $M$ for the vertices we get for the number of colorings by Burnside:

$$\bbox[5px,border:2px solid #00A000]H(N, M) = \frac{1}{24}(N^6 M^8 + 6 N^3 M^2 + 3 N^4 M^4 + 8 N^2 M^4 + 6 N^3 M^4)$$

Setting $M=1$ here we should get face colorings. We obtain

$$1, 10, 57, 240, 800, 2226, 5390, 11712, \ldots$$

and we encounter [OEIS A047780](https://oeis.org/A047780) where we see that we have the right values. Setting $N=1$ yields vertex colorings:

$$1, 23, 333, 2916, 16725, 70911, 241913, 701968, \ldots$$

which points  to  [OEIS A000543](https://oeis.org/A000543)  which  is correct as well.

Continuing with the question of colorings that use exactly $N$ colors for the faces and exactly $M$ for the vertices we find using Stirling numbers for set partitions

$$\bbox[5px,border:2px solid #00A000] {\begin{gather} H_X(N, M) = \frac{N! \times M!}{24} \\ \times \left({6\brace N} {8\brace M} + 6 {3\brace N} {2\brace M} + 3 {4\brace N} {4\brace M} + 8 {2\brace N} {4\brace M} + 6 {3\brace N} {4\brace M}\right) \end{gather}}$$

Setting $M=1$ here we get the count of face colorings with exactly $N$ colors:

$$1, 8, 30, 68, 75, 30, 0, \ldots$$

Note that for six colors, which is the maximum,  the orbits have size $24$ because all the  colors are  distinct and  indeed $6!/24  = 30.$ Similarly with $N=1$ we get vertex colorings:

$$1, 21, 267, 1718, 5250, 7980, 5880, 1680, 0, \ldots$$

and once more for eight  colors, the  maximum possible, we  find that $8!/24 = 1680.$

Concluding we get for at most two vertex colors and at most four face colors

$$H(4,2) = 44608$$

and for exactly two vertex colors and four face colors

$$H_X(4,2) = 16552.$$