A non-crossing pairing of size $n$ is commonly defined by linearly arranging $n$ objects and pairing them up with "brackets" that don't cross. Equivalently, if $a<b<c<d$, then we can't pair as $(a,c), (b,d)$.
I'm curious about the following generalization: the linearly arranged objects can be thought of as forming a linear graph with edges connecting neighboring elements as in $\cdots i-1 \leftrightarrow i \leftrightarrow i+1\cdots $. Now what if the objects are connected in some other graph type, say for example, a tree? Now the condition would be: take the unique path connecting $b-d$. If $c$ is on this path and $a$ is not, then we can't pair as $(a,c), (b,d)$.
Are there any results on counting this type of structure? The holy grail for me would be the generating function $f(z)=\sum_n a_n z^n$ where $a_n$ counts the number of non-crossing pairings on any rooted plane tree with $n$ edges. Even knowing the generating function up to a few unknown constants would be fine.
I've managed to come up with a counting argument and write down a horrible-looking functional equation involving $f'$ and terms like $f(x)$ and $f\left(\frac{1}{1-\sqrt{x}}\right)$, which I can describe if anyone's interested (it would take a bit to write out and I'd prefer to do it only if it'd help). I'm worried that this might be as good as it gets, but I still had some hope that I had overlooked a simpler counting argument.