Number of essentially different ways of colouring the edges of a regular tetrahedron with n colours ensuring there are monochromatic triangles? I've shown that the number of colourings of the edges of a regular tetrahedron with n different colours when we want to ensure that there is at least one monochromatic triangle is $4n^4 - 6n^2 + 3n$. Now I'm wondering how many of them are essentially different (that is up to rotational symmetry), I can't find out anything... Do you have an idea to give me? Thanks a lot!
 A: This answer follows @Nishant's suggestion to use Burnside's Lemma.
The rotation group of the tetrahedron contains eight rotations of $\pm120^\circ.$  For a coloring to be fixed under one of these, the face through which the rotation axis passes must be monochromatic, and the three edges not on that face must also be of a single color.  So these rotations fix $n^2$ colorings.
The group contains three rotations of $180^\circ$ about an axis joining the centers of a pair of opposite edges.  Call these edges $e$ and $e'.$  One of these edges, say $e,$ must be incident on a monochromatic triangle.  For the coloring to be fixed under rotation about the axis joining $e$ and $e',$ both triangles incident on $e$ must be monochromatic, so the five edges other than $e'$ must have the same color.  This gives $n^2$ colorings fixed by the rotation.  Multiplying by $2$ (because of the choice of $e$ or $e'$) and subtracting $n$ (so as not to double count the colorings where all six edges have the same color) gives $2n^2-n$ colorings fixed by $180^\circ$ rotation.
All colorings are fixed by the identity (and you know how many of these there are).  Now apply Burnside's Lemma.
A: I start by solving for both rotational and reflective symmetry, and then expand to the case with just rotational symmetry.
For rotational and reflective symmetry:
Divide into four cases - 
Case 1: There is only one monochromatic triangle and no other edges have the same color as that triangle.
Case 2: There is only one monochromatic triangle and one other edge has the same color as that triangle.
Case 3: There are exactly two monochromatic triangles.
Case 4: There are exactly four monochromatic triangles (note that just 3 monochromatic triangles is impossible).
Case 1: When there is only one monochromatic triangle and all other edges not of that color, we have three edges which can be colored any of $n-1$ colors. Now it only matters how many of these three are colored with the same colors, because all other cases are rotationally the same. This means that we have 
$$n\bigg[{n-1\choose3}+2{n-1\choose 2}+{n-1\choose 1}\bigg]$$
tetrahedra with one monochromatic triangle. The $n$ on the outside corresponds to choices of color for the monochromatic triangle, the first term inside corresponds to choices of $3$ different colors, the second term to the choices of $2$ different colors (multiplied by $2$  because for each pair of two colors $a$ and $b$, we can have $2$ edges colored by $a$ and one by $b$ or vice versa), and the third term is the number of ways to pick just $1$ other color.
Case 2: One monochromatic triangle, but with an edge that is the same color as that. This leaves $n-1$ choices of color for $2$ edges, giving us a total of 
$$n\bigg[{n-1\choose 2}+{n-1\choose 1}\bigg]$$
colorings for this case where we either have the other two edges different colors or the same color.
Case 3: When there are exactly two monochromatic triangles, we must have $5$ edges of one color and the $6$-th edge different. This gives a total of $n(n-1)$ colorings with exactly two monochromatic triangles.
Case 4: If every triangle is monochromatic then we only have a choice of $1$ color, so we get $n$ colorings with exactly $4$ monochromatic triangles.
All of this results in a total number of colorings of 
$$\frac{1}{6}n^2(n+1)(n+2)$$
under rotational and reflective symmetry.
Under just rotational symmetry, we can just modify the cases above.
Case 1: The only part where things are different from before is when we have three of different color. We now have twice the number of those than before because we can't reflect, giving 
$$n\bigg[2{n-1\choose3}+2{n-1\choose 2}+{n-1\choose 1}\bigg]$$
colorings.
Case 2: Again because we cannot reflect, we get twice the number of colorings here when two edges are the same. This gives
$$n\bigg[2{n-1\choose 2}+{n-1\choose 1}\bigg]$$
colorings.
Case 3: It's the same, hence $n(n-1)$ colorings.
Case 4: Again the same, hence $n$ colorings.
This gives a total of 
$$\frac{1}{3}n^2(n^2+2)$$
colorings under just rotational symmetry.
Thanks to Will Orrick for pointing out that I hadn't accounted for just rotational symmetry.
