Okay, so Greg Egan simplified the formula as I was slowly attempting to:

Focusing on the most interesting case, where \\(d=2e, n=2m\\), then we can write

\[
\text{moment}(2e,2m) =\displaystyle{ \frac{(e-1)! (2e+2m-2)!}{ (e+m-1)! (2e+m-2)!} }
\]

or in other words

\[ \text{moment}(2e,2m) = \displaystyle{\frac{ 2(e+m-1) \choose m}{e+m-1 \choose m} }
\]

or in other words

\[ \text{moment}(2e,2m) = \displaystyle{\frac{(2e+m-1) \cdots (2e+2m-2)}{e \cdots (e+m-1)} }\]

So, our job is to explain why this is often but not always an integer... but _always_ an integer if \\(e=1,2\\).