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The identity in my recent answer can be stated in a particularly neat form: $${}_2F_0\left({-n, n+1\atop{}};\frac{x}{2}\right) ~\cdot~ {}_2F_0\left({-n, n+1\atop{}};-\frac{x}{2}\right) ~=~ {}_3F_0\...

I have reduced solving this question to proving the following identity, for $n, \ell \ge 0$: $$ (n-2\ell+1)^{n-1} \binom{n}{\ell-1} = \\ \frac{1}{2} \sum_{n_1+n_2=n-1}\left[ (n_1+1)^{n_1-1} (n_2-2\...

In process of doing some computations on Hilbert schemes, I stumbled across the following identity, for $k \ge 2$: $$ k^{k-3} = \frac{1}{2} \sum_{i=1}^{k-1} \binom{k-2}{i-1} i^{i-2} (k-i)^{k-i-2} $$ ...

Incidentally I've obtained a hypergeometric identity that I've not seen before: $${}_3F_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2$$ So, I wonder if it is well-known and ...