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Here is a generating-function proof of your conjectured identity (and an answer to question 2). The main ingredient is a formula for the appearing symmetric sums. Let $T(z)$ (the ``tree function'') …

Here is an argument for the leading coefficient (and more). We use (formal) generating functions. Let $f_{n,x}(t):= \sum_{m=0}^n {n-x \choose m } {n+x \choose n-m} e^{(x+2m-n)t}$, we are interested i …

Here is a proof using (formal) generating functions. The Lah number $L(n,m+1)$ $$L(n,m+1)=\frac{n!}{(m+1)!}[x^n] \bigg(\frac{x}{1-x}\bigg)^{m+1}$$ counts the number of unordered partitions of the se …

A generating function proof. As $\arcsin(z)=\sum_{k\geq 0} \frac{1}{2k+1} {2k \choose k} \frac{z^{2k+1}}{4^k}$ and $\frac{1}{\sqrt{1-z^2}}=\sum_{k\geq 0}{2k \choose k}\frac{z^{2k}}{4^k}$ we have that …