# Certain matrices of interesting determinant

Let $$M_n$$ be the $$n\times n$$ matrix with entries $$\binom{i}{2j}+\binom{j}{2i}, \qquad \text{for 1\leq i,j\leq n}.$$

QUESTION. Is this true? There is some evidence. The determinant $$\det(M_{2n+1})=0$$ and $$\det(M_{2n})=(-1)^n\binom{2n}n^22^{n(n-3)}.$$

• The odd case is easy because $M_{2n-1}$ has an $n \times n$ block of zeros at bottom right. In the even case, $M_{2n}$ still has that $n \times n$ block of zeros, so its determinant is $(-1)^n$ times the product of the determinants of the $n \times n$ blocks at top right and bottom left, which are equal because they are each other's transpose. So it comes down to a determinant of $i \choose 2j$ for $i \in [n-1,2n]$ and $j \in [1,n]$, which must be known. – Noam D. Elkies Dec 17 '18 at 5:29
• @NoamD.Elkies typo: $i\in [n+1,2n]$ – Fedor Petrov Dec 17 '18 at 6:31
• Right, thanks. Alas it is long after the 5-minute window for editing comments. – Noam D. Elkies Dec 17 '18 at 15:35
• @NoamD.Elkies: Good observation. Thanks! – T. Amdeberhan Dec 17 '18 at 18:56

Noam Elkies in the comments reduces the problem to the identity $$\det\left(\binom{(n+1)+i}{2j+2}\right)_{i,j=0}^{n-1}=\binom{2n}n2^{n(n-3)/2}.$$ In general the determinant $$\binom{N+i}{c_j+j}$$, $$i,j=0,\dots,n-1$$ for integers $$0\leqslant c_0\leqslant c_1\leqslant \dots \leqslant c_{n-1}\leqslant N$$ may be calculated by the following trick.

Consider the Young diagrams $$\lambda,\mu$$ with rows length $$c_0\leqslant c_1\leqslant \dots \leqslant c_{n-1}$$ and $$N,N,\dots,N$$ ($$n$$ rows in total) respectively. Count the skew Young tableaux of the shape $$\mu\setminus \lambda$$. On one hand, the number of such tableaux is known (see Okounkov, A., Olshanski, G.: Shifted Schur functions. St. Petersburg Math. J. 9(2), 73–146 (1997), Theorem 8.1) to be equal to $$\frac{(N\cdot n-\sum c_i)!}{\prod_{i=0}^{n-1} (N+i)!}\cdot \prod_{i=0}^{n-1} (c_i+i)!\cdot \det\left(\binom{N+i}{c_j+j}\right)_{i,j=0}^{n-1}.$$ On the other hand, it is the same as the number of usual Young tableaux of the shape $$(N-c_0,\dots,N-c_{n-1})$$, which may be evaluated by Hook Length Formula as $$\frac{(N\cdot n-\sum c_i)!}{\prod_{i=0}^{n-1} (N-c_{n-i-1}+i)!}\cdot \prod_{i Therefore $$\det\left(\binom{N+i}{c_j+j}\right)_{i,j=0}^{n-1}=\\ \prod_{i=0}^{n-1}\frac{(N+i)!}{(c_i+i)!(N-c_{n-i-1}+i)!}\cdot \prod_{i It remains to substitute $$N=n+1$$ and $$c_i=i+2$$. We get $$\prod_{i and $$(*)$$ rewrites as $$\det\left(\binom{(n+1)+i}{2j+2}\right)_{i,j=0}^{n-1}= \frac{1!2!\dots (n-1)! (n+1)!\dots (2n)!} {2!4!\dots (2n)! 2! 4!\dots (2n-2)!}2^{n(n-1)/2}.$$ Replace $$2!4!\dots (2n)!$$ to $$2\cdot 1!\cdot 4\cdot 3!\cdot 6\cdot 5!\cdot \ldots\cdot (2n)\cdot(2n-1)!=\\=2^nn! 1!3!\dots (2n-1)!.$$ We get $$\frac{(2n)!}{n!n!} 2^{n(n-3)/2}$$ as supposed.

Remark: of course the above determinant is a polynomial in $$N$$, and if we replace $$(N+i)!/(N-c_{n-i-1}+i)!$$ in $$(*)$$ by $$c_{n-i-1}! \binom{N+i}{c_{n-i-1}}$$, the formula becomes true for arbitrary $$N$$, not necessarily positive integer.

As a follow-up on Fedor's wonderful proof, I would like to pen down just a comment.

Let's replace $$c_j+j$$ by $$u_j$$ and treat $$N$$ as an indeterminate. Then, for non-negative integers $$a$$ and $$b$$, we have the determinantal evaluation $$\det\left(\binom{N+i+a}{u_{j+b}}\right)_{i,j=0}^{n-1} =\prod_{i=0}^{n-1}\frac{(N+i+a)!}{(N+n-a-1)!}\binom{N+n+a-1}{u_{i+b}}\prod_{i The proof follows from the method of condensation; see Desnanot–Jacobi identity.

Remark. The introduction (or generalization) of the parameters $$a$$ and $$b$$ is what makes above-mentioned method work, effectively.

• are not there typos? This $u_{j+b}$ looks strange. – Fedor Petrov Dec 17 '18 at 19:03
• Not typo, I believe. I am just shifting by $b$. You would see that is not strange if you try to use condensation (as you move up and down in the infinite matrix). – T. Amdeberhan Dec 17 '18 at 19:05
• also $N+a$ are always together except the denominator where $N-a$ appears. Is it ok? – Fedor Petrov Dec 17 '18 at 19:57
• Good point. That was some annoying part that did not allow me to get further generalization of the matrices, such that $N+a+i=x_i$. On the positive side, it allows the reader to see the roles of the "shifting parameters" $a$ and $b$ with the proof. – T. Amdeberhan Dec 17 '18 at 21:31
• Tewodros, is it still ok? If we replace $N$ by $N+5$ and $a$ by $a-5$ it looks like nothing except denominator changes... – Fedor Petrov Dec 17 '18 at 23:58