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!

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.
