# Evaluate the determinant $\det\left[ \binom{2n}{n+i-j} \right]_{i,j=0}^{n-1}$

I am trying to show that: $$\det\left[ \binom{2n}{n+i-j} \right]_{i,j=0}^{n-1}=\prod_{i=0}^{n-1} \frac{\binom{2n+i}{n}}{\binom{n+i}{n}}$$

I have tried playing with the algebra for some time. For example, if we fix $i$ and consider a particular row vector, we have: $$\left[ \begin{array}{c} \binom{2n}{n+i} & \binom{2n}{n+i-1} & \binom{2n}{n+i-2} & \dots & \binom{2n}{i+1} \end{array} \right]$$ Which equals $$\left[ \begin{array}{c} \frac{(2n)!}{(n+i)!(n-i)!} & \frac{(2n)!}{(n+i-1)!(n-i+1)!} & \frac{(2n)!}{(n+i-2)!(n-i+2)!} & \dots & \frac{(2n)!}{(i+1)!(2n-i-1)!} \end{array} \right]$$ It seems that our goal should be to factor out $\binom{2n+i}{n} / \binom{n+i}{n}$ and leave a matrix whose determinant evaluates to 1. Clearly: $$\binom{2n+i}{n} / \binom{n+i}{n} = \frac{(2n+i)!}{n!(n+i)!} \cdot \frac{n!(i!)}{(n+i)!} = \frac{(2n+i)!}{(n+i)!} \cdot \frac{i!}{(n+i)!}$$ However, I am unsure of how to proceed.

For those interested, the determinant given enumerates plane partitions contained within an $n\times n \times n$ cube, or equivalently, rhombic tilings of a regular hexagon with side length $n$.

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I don't have a complete answer, but I can point out that your matrix is of a special type (symmetric Toeplitz). Maybe that will help find some general algorithm which is applicable. – Sidious Lord Aug 19 '12 at 10:21

It seems that our goal should be to factor out ... and leave a matrix whose determinant evaluates to 1.

Actually, one can factor out something to leave a (generalized) Vandermonde matrix.

Using column operations one can see that $$\det\binom{2n}{n+i-j}=\det\binom{2n+j}{n+i}.$$ Now $\binom{n}{k}=\frac{n^{\downarrow k}}{k!}$, where $n^{\downarrow k}:=n(n-1)\ldots(n-k+1)$, so the determinant we want to compute is just $$\det\frac{(2n+j)^{\downarrow n+i}}{(n+i)!}= \prod_i\frac1{(n+i)!}\cdot\prod_j(2n+j)^{\downarrow n}\cdot \det(n+j)^{\downarrow i}$$ (here we used that $a^{\downarrow k+l}=a^{\downarrow k}\cdot(a-k)^{\downarrow l}$).

Observe that $$\det(n+j)^{\downarrow i}=\det(n+j)^i=\prod j!,$$ so the answer is $$\prod_j\frac{(2n+j)!j!}{((n+j)!)^2}= \prod_j\frac{\binom{2n+j}n}{\binom{n+j}n}.$$

(Perhaps, all this is also contained in Krattenthaler's paper quoted above. But anyway.)

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Krattenthaler, in this article, proves a more general formula, of which the OP's determinant is a special case. Given the $n\times n$ matrix $\mathbf A$ with elements

$$\mathbf a_{j,k}=\binom{p+q}{p+j-k}, \quad 1\leq j,k\leq n$$

then

\begin{align*} \det\mathbf A&=\prod_{j=1}^n \prod_{k=1}^p \prod_{\ell=1}^q \frac{j+k+\ell-1}{j+k+\ell-2}\\ &=\prod_{j=1}^n \frac{(p+q+j-1)!(j-1)!}{(p+j-1)!(q+j-1)!} \end{align*}

where the triple product formula is attributed to MacMahon. A number of proofs for this determinantal identity are given in the linked article. For your particular special case,

$$\prod_{j=0}^{n-1} \frac{(2n+j)!j!}{((n+j)!)^2}=\prod_{j=0}^{n-1} \frac{\frac{(2n+j)!}{(n+j)!n!}}{\frac{(n+j)!}{n!j!}}=\prod_{j=0}^{n-1} \frac{\binom{2n+j}{n}}{\binom{n+j}{n}}$$