Dtyjlui

I need to find out the amount of ways to colour truncated triangular dihedron into 3 colours. So, the task will be easier if I had simple triangular dihedron.

First of all, do I understand right that truncated dihedron looks like :
enter image description here
The only things I know, that I must use Burnsside's lemma, calculate simmetries ( rotations ).

By the way, if you know some beginners algebra books with decent amount of examples in practice and more-or-less easy to understand for not english-speaker, thank you.

Note first of all that we obtain a cube when placing vertices at the
centers of the hexagonal faces and connecting those where the faces
share an edge. So we could solve this problem by asking about
colorings of the vertices and the faces of the cube. We will work with
the given solid however. (The cube was solved at this MSE
link).

We evidently require the cycle index. There is the identity and three
types of rotations. With $a_q$ for hexagonal faces and $b_p$ for the
pyramids at the corners we get from the identity:

$$a_1^8 b_1^6.$$

The rotations about an axis passing through opposite pyramidal corners
contribute

$$3 times (2 a_4^2 b_1^2 b_4 + a_2^4 b_1^2 b_2^2).$$

The flips about an axis passing through the centers of two opposite edges
($180$ degree rotation) contribute

$$6 times a_2^4 b_2^3.$$

Finally, the rotations by $60$ degrees and $120$ degrees about an axis
passing through the centers of opposite hexagonal faces yields

$$4 times 2 a_1^2 a_3^2 b_3^2.$$

We thus have for the cycle index

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

We then get by Burnside for colorings with at most $N$ colors shared
between hexagonal faces and pyramidal corners the closed form:

$$frac{1}{24}(N^{14} + 6 N^5 + 3 N^8 + 6 N^7 + 8 N^6).$$

This gives the sequence

$$1, 776, 200961, 11198720, 254387500, 3265470936, 28260264606,
\ 183254654976, 953206454115, 4166682025000, ldots$$

in particular for $N=3$ we obtain

$$bbox[5px,border:2px solid #00A000]{
200961.}$$

The formula when working with exactly $N$ colors as opposed to at most
$N$ colors is (colors are shared between pyramids and hexagons):

$$frac{N!}{24}left({14brace N}
+ 6 {5brace N} + 3 {8brace N}
+ 6 {7brace N} + 8 {6brace N}right).$$

This gives the finite sequence

$$1, 774, 198636, 10399528, 200395755, 1903119150, 10359167700,
\ 35132702640, 77643165600, 113816102400, 109880971200,
\ 67199932800, 23610787200, 3632428800, 0, 0, ldots $$

because there are only fourteen slots available. Note that the last
one is $$14!/24 = 3632428800.$$

Remark. This cycle index also solves the problem where colors are
not shared between hexagons and pyramids. We find:

$$frac{1}{24}
(N^8 M^6 + 6 N^2 M^3 + 3 N^4 M^4
+ 6 N^4 M^3 + 8 N^4 M^2).$$