Coefficients of products of binomials

Question

What is the coefficient of the term $\prod_{k=1}^N{z_k^{N-1}}$ in the expansion of $\prod_{1\leq i<j\leq N}{(z_i+z_j)^2}$? From the symmetry of the problem all that I can argue is that this coefficient shall be the largest of all the other coefficients.

Answer 1

This is more of a comment but slightly too large to post as such. If you are merely interested in how your coefficient grows with $N$, here are two crude bounds:

Let $C_N^*$ denote the coefficient of $\prod_{k=1}^N{z_k^{N-1}}$ in the expansion of $\prod_{1\leq i<j\leq N}{(z_i+z_j)^2}$. First note that

\begin{equation} C_N^* \geq 2^{\frac{N (N-1)}{2}} \, . \end{equation}

This can be easily seen by observing that from the expansion of

\begin{equation} \prod_{1\leq i<j\leq N}{(z_i+z_j)^2} = \prod_{1\leq i<j\leq N}{(z_i^2 + 2 z_i z_j +z_j^2)} \end{equation}

we can extract

\begin{equation} \prod_{1\leq i<j\leq N}{2 z_i z_j} = 2^{\frac{N (N-1)}{2}} \prod_{k=1}^N{z_k^{N-1}} \, . \end{equation}

An upper bound can be obtained by considering the expression $\prod_{1\leq i<j\leq N}{(2 z_i^2 + 2 z_i z_j + 2 z_j^2)}$; certainly any coefficient in the expansion of the preceding product will be larger than its counterpart for the original expression. An upper bound on $C_N^*$ is thus

\begin{equation} C_N^* < 2^{\frac{N (N-1)}{2}} M_N\, . \end{equation}

The variable $M_N$ is the amount of how many ways there are to combine the terms in the expansion of $\prod_{1\leq i<j\leq N}{(z_i^2 + z_i z_j + z_j^2)}$ to yield $\prod_{k=1}^N{z_k^{N-1}}$.

For the sake of argument, let us assume that $N$ is odd. An upper bound on $M_N$ can be obtained by noticing that this quantity certainly lies below the amount of ways to obtain any term involving $z_N^{N-1}$. To that end, we can pick $\frac{N-1}{2} - k$ out of $N-1$ terms to provide $z_N^2$. From the remaining $\frac{N-1}{2} - k$ terms involving $z_N$ we pick $2 k$ to provide $z_i z_N$ (for some $i$). This leaves us with $\frac{(N-1)^2}{2} - k$ terms to pick any of at most 3 items not involving $z_N$. Putting the above together we arrive at

\begin{align} M_N &\leq \sum_{k=0}^{\frac{N-1}{2}} \binom{N-1}{\frac{N-1}{2}-k} \binom{\frac{N-1}{2}+k}{2k} 3^{(N-1)^2-k} \\ &\leq \sum_{k=0}^{\frac{N-1}{2}} 2^{N-1} 2^{\frac{N-1}{2}+k} 3^{(N-1)^2-k} \\ &< \sum_{k=0}^{\frac{N-1}{2}} 2^{N-1} 2^{\frac{N-1}{2}+k} 2^{2 (N-1)^2-2k} \\ &= 2^{\frac{3}{2}(N-1) + 2(N-1)^2} \sum_{k=0}^{\frac{N-1}{2}} 2^{-k} \\ &< 2^{\frac{3}{2}(N-1) + 2(N-1)^2 + 1} \, . \end{align}

Thus

\begin{equation} C_N^* < 2^{\frac{N (N-1)}{2} + \frac{3}{2}(N-1) + 2(N-1)^2 + 1} \, . \end{equation}

The two bounds tell you that

\begin{equation} \frac{1}{2} \leq \lim_{N \to \infty} \frac{1}{N^2} \log{C_N^*} < 2 + \frac{1}{2} \end{equation}