A friend of mine came across this rather odd combinatorial identity. We've spent a while but haven't been able to prove it. Any ideas?
The following holds exactly for even integers $n$, and is approximately true for odd integers $n$: $$n = \dfrac{n+1}{n^n - 1} \sum_{k=1}^{n/2} \dbinom{n-k}{k-1} n^k (n - 1)^{n+1-2k}$$
So we want to prove that for even $n$ $$ \frac{n^n-1}{n+1}=\sum_{l\ge0}\binom{n-1-l}l n^l(n-1)^{n-1-2l}. $$ The RHS is the coefficient of $t^{n-1}$ in $\frac1{1-(n-1)t-nt^2}$. But $$ \frac1{1-(n-1)t-nt^2}=\frac1{(1+t)(1-nt)}=(1-t+t^2-\ldots)(1-nt+n^2t^2-\ldots), $$ so this coefficient is a sum of a geometric progression and it's equal to $\frac{n^n+(-1)^{n-1}}{n+1}$ and we're done (as a bonus we get the correct version of the identity for odd $n$).
