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I want to find a closed form to the bivariate generating function $$ G(x, y) = \sum\limits_{i, j} \binom{i+j}{i}^2 x^i y^j. $$ Ideally, I would prefer a direct approach that is based on the definition above.

I know that there is a closed form here, as one can reduce the summation to Legendre polynomials: $$ G(x, y) = \sum\limits_n y^n \sum\limits_k \binom{n}{k}^2 \left(\frac{x}{y}\right)^k = \sum\limits_n (y-x)^n P_n \left(\frac{y+x}{y-x}\right), $$ where $P_n(x)$ is the $n$-th Legendre polynomial. From this, using the generating function formula $$ \sum\limits_n P_n(x) t^n = \frac{1}{\sqrt{1-2xt+t^2}}, $$ we get the closed-form expression for $G(x, y)$ as $$ \boxed{G(x, y) = \frac{1}{\sqrt{1-2(y+x)+(y-x)^2}}} $$ But I totally fail to see any meaningful way to derive it in a more direct and self-contained way. Any hints? And while we're at it, are there similar closed-form expressions for higher powers of $\binom{n}{k}$?

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  • $\begingroup$ Some possibly useful fact: $$ 1-2(y^2+x^2)+(y^2-x^2)^2 = (1-x-y)(1+x-y)(1-x+y)(1+x+y). $$ Meaning that $G(x^2, y^2)$ can be represented as $$ G(x^2, y^2) = \frac{1}{\sqrt{(1-x-y)(1+x-y)(1-x+y)(1+x+y)}}. $$ Not sure if it helps though. $\endgroup$ Jun 11, 2023 at 15:52

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After some thoughts, I couldn't find any reasonable way to do it directly, but at least now I understand how to fill the gap in the derivation of the genfunc of Legendre polynomials. First, let's expand the genfunc.

$$ G(t, x) = \frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_k \frac{t^k(2x-t)^k}{4^k}\binom{2k}{k}. $$ If we want to extract the coefficient near $t^n$, we get $$ [t^n] G(t, x) = \frac{1}{2^n}\sum\limits_k \binom{2k}{k} \binom{k}{n-k} (-1)^{n-k} x^{2k-n}, $$ where $k$ goes from $\lceil n/2 \rceil$ to $n$. Changing $k \to n-k$, we get $$ [t^n] G(t, x) = \frac{1}{2^n} \sum\limits_k (-1)^k \binom{2n-2k}{n-k} \binom{n-k}{k} x^{n-2k}. $$ On the other hand, $$ \binom{2n-2k}{n-k}\binom{n-k}{k}=\binom{2n-2k}{n-k,k,n-2k}=\binom{2n-2k}{n}\binom{n}{k}, $$ which gives one of the standard expressions $$ [t^n] G(t, x) =\boxed{ \frac{1}{2^n} \sum\limits_k (-1)^k \binom{n}{k}\binom{2n-2k}{n} x^{n-2k}} $$

Now, we need to connect the dots. To make it relevant to $\binom{n}{k}^2$, we define Legendre polynomials as $$ P_n(x) = \frac{1}{2^n} \sum\limits_{k=0}^n \binom{n}{k}^2(x-1)^{n-k} (x+1)^k. $$ It's easy to check that it compacts into $$ P_n(x) = [t^n] \frac{(t+x-1)^n(t+x+1)^n}{2^n} = [t^n] \frac{((t+x)^2-1)^n}{2^n}. $$

We can now expand it back to get the same expression: $$ [t^n]\frac{((x+t)^2-1)^n}{2^n} = [t^n]\frac{1}{2^n}\sum\limits_k (-1)^k \binom{n}{k}(x+t)^{2n-2k} =\boxed{ \frac{1}{2^n} \sum\limits_k (-1)^k \binom{n}{k}\binom{2n-2k}{n} x^{n-2k}} $$ I still don't like this approach, as I needed to know $G(t, x)$ in advance to justify it, and it's quite far from being direct in terms of the genfunc for $\binom{n}{k}^2$, but I suppose it's still nice to have proofs that are directly based on binomial identities, rather than physics...

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Ok, after some more thought I figured out how to work with $\binom{n}{k}^2$ directly. Consider $$ Q_n(x, y) = \sum\limits_{k=0}^n \binom{n}{k}^2 x^k y^{n-k} = [t^n] (x+t)^n (y+t)^n. $$ We need to sum it up over all $n$, and we already know that $Q_n(x) = (y-x)^n P_n(\frac{y+x}{y-x})$.

But let's analyze $Q_n(x, y)$ on its own: $$ [t^n](x+t)^n (y+t)^n = [t^n](t(t+x+y)+xy)^n $$ This expands into $$ \sum\limits_k \binom{n}{k} x^k y^k [t^k] (t+x+y)^{n-k} = \sum\limits_k \binom{n}{k} \binom{n-k}{k} x^k y^k (x+y)^{n-2k}. $$ Note that $$ \binom{n}{k} \binom{n-k}{k} = \binom{n}{k, k, n-2k} = \binom{n}{2k} \binom{2k}{k}, $$ hence we want to compute the sum $$ \sum\limits_{n, k} \binom{n}{2k} \binom{2k}{k} x^k y^k (x+y)^{n-2k}. $$ Let's sum up over $n$ for each fixed $k$: $$ \sum\limits_{n} \binom{n}{2k} (x+y)^{n-2k} = \sum\limits_{t} \binom{2k+t}{2k} (x+y)^{t} = \frac{1}{(1-x-y)^{2k+1}}. $$ Thus, we want to compute $$ \sum\limits_k \binom{2k}{k} \frac{x^k y^k}{(1-x-y)^{2k+1}}. $$ On the other hand we know that $$ \sum\limits_k \frac{x^k}{4^k} \binom{2k}{k} = \frac{1}{\sqrt{1-x}}, $$ thus the sum above compacts into $$ \frac{1}{1-x-y} \frac{1}{\sqrt{1-4\frac{xy}{(1-x-y)^2}}}= \frac{1}{\sqrt{(1-x-y)^2-4xy}}, $$ which then expands into the same expression in the denominator.

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