# Number of isomorphism classes of triangulations of a convex polygon

The number of triangulations of a convex $n$-gon is $C_{n-2}$ the $n-2$nd Catalan number. What I am wondering, is if there is a way to enumerate the isomorphism types of these as graphs? I am currently working an a project where I want to only consider all possible unlabeled triangulations of $n$-gons as a way to classify an $n$-gon. Clearly $C_{n-2}$ is an upper bound, but I haven't been able to make or find any progress on that thus far.

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This may be the same as oeis.org/A131481 --- number of $n$-cell polyiamonds (triangular polyominoes) with perimeter $n+2$ (but shifted by 2, that is, there are 3 classes of triangulations of a 6-gon, and 3 4-cell polyiamonds with perimeter 6). –  Gerry Myerson Jul 24 '13 at 7:39

If I understand correctly, these are A001683 if turning over the $n$-gon is not allowed as an isomorphism, and A000207 if it is. In both articles there are formulae.
You might also be interested that plantri can compute these isomorphism classes very quickly, several hundred thousand per second. The argument lists for triangulating an $n$-gon are "plantri -P# -c2m2 -o #" and "plantri -P# -c2m2 #" respectively, where # is the value of $n$.
ADDED: Every abstract isomorphism between two triangulations preserves the $n$-gon, for the following reason. The $n$-gon is a hamiltonian cycle in the graph. Every chord is a separating edge (removing its two vertices disconnects the graph) and such edges cannot lie on a hamiltonian cycle. I.e., the $n$-gon is the unique hamiltonian cycle.
I asked for clarification from the OP, but then I thought: any abstract isomorphism maps triangles to triangles, so it may be a bit hard to avoid mapping the $n$-gon to itself. Up to $n=20$, they are all non-isomorphic as graphs (i.e 12 for an 8-gon, not 11). How can we resolve this? –  Brendan McKay Jul 24 '13 at 12:44