How many ways are there to triangulate a regular convex n-gon, if two triangulations are regarded as being the same if they can be made to coincide by a rotation of the polygon?
I was asked to express this in terms of the Catalan numbers and I got $G_n = \frac{C_{n-2}}{2} + \frac{C_{\frac{n}{2}-1}}{2} +\frac{2C_{\frac{n}{3}-1}}{3}$ where Cn is the Catalan number and if the subscript is not an integer than it is omitted.
Now I have to try and prove that my expression $G_n$ is true but this is where I run into trouble. I tried induction and the inductive step was where it all fell apart.
Consulting the paper by Richard Guy Dissecting a polygon into triangles and using a somewhat similar notation we introduce $D_n$ which is $C_{n-2}$ where $C_n$ is the Catalan number $$C_n = \frac{1}{n+1} {2n\choose n}$$ when $n$ is a positive integer at least two and zero otherwise. We then have consult e.g. Wikipedia on Catalan Numbers that the number of triangulations of the convex $n$-gon is given by $D_n.$
Now make the following observation concerning rotational symmetry of the triangulations of the regular $n$-gon. There are two possibilities for the center of the $n$-gon, either it is located on a diagonal which is in fact a diameter or it is inside one of the triangles. In the first case we can have symmetry only if we map the diameter to itself by a $180$ degree rotation. Call the count of these $F_n$. In the second case we have symmetry only if the triangle containing the center is equilateral and we have two rotations by $120$ and $240$ degrees fixing that triangle. Call the count of these $R_n.$ Finally introduce $U_n$, triangulations having no symmetry. The desired quantity is then given by
$$E_n = F_n + R_n + U_n.$$
We have by inspection that
$$F_n = D_{n/2+1} \quad\text{and}\quad R_n = D_{n/3+1}.$$
Furthermore
$$D_n = \frac{n}{2} F_n + \frac{n}{3} R_n + n U_n$$
where we have computed the sizes of the respective orbits. This yields
$$U_n = \frac{1}{n} D_n - \frac{1}{2} F_n - \frac{1}{3} R_n$$
so that
$$E_n = \frac{1}{n} D_n + \frac{1}{2} F_n + \frac{2}{3} R_n$$
or
$$\bbox[5px,border:2px solid #00A000]{ E_n = \frac{1}{n} D_n + \frac{1}{2} D_{n/2+1} + \frac{2}{3} D_{n/3+1}.}$$
This yields the sequence
$$1, 1, 1, 1, 4, 6, 19, 49, 150, 442, \\ 1424, 4522, 14924, 49536, 167367, \ldots$$
which points us to OEIS A001683, where detailed references await. (We have started at index $2$ rather than $3$ to match the convention that the OEIS uses.)