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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 $$\det(M_{2n})=(-1)^n\binom{2n}n^22^{n(n-3)}.$$

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    $\begingroup$ 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. $\endgroup$ – Noam D. Elkies Dec 17 '18 at 5:29
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    $\begingroup$ @NoamD.Elkies typo: $i\in [n+1,2n]$ $\endgroup$ – Fedor Petrov Dec 17 '18 at 6:31
  • $\begingroup$ Right, thanks. Alas it is long after the 5-minute window for editing comments. $\endgroup$ – Noam D. Elkies Dec 17 '18 at 15:35
  • $\begingroup$ @NoamD.Elkies: Good observation. Thanks! $\endgroup$ – T. Amdeberhan Dec 17 '18 at 18:56
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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 $$ \frac{(N\cdot n-\sum c_i)!}{\prod_{i=0}^{n-1} (N-c_{n-i-1}+i)!}\cdot \prod_{i<j} (c_j+j-c_i-i). $$ Therefore $$ \det\left(\binom{N+i}{c_j+j}\right)_{i,j=0}^{n-1}=\\ \prod_{i=0}^{n-1}\frac{(N+i)!}{(c_i+i)!(N-c_{n-i-1}+i)!}\cdot \prod_{i<j} (c_j+j-c_i-i).\quad (*) $$ It remains to substitute $N=n+1$ and $c_i=i+2$. We get $$\prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}\prod_{i<j}(j-i)=\prod_{i<j} (c_j+j-c_i-i)=2^{n(n-1)/2}\prod_{i=0}^{n-1} i!$$ 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.

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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 $$\det\left(\binom{N+i+a}{u_{j+b}}\right)_{i,j=0}^{n-1} =\prod_{i=0}^{n-1}\frac{(N+i+a)!}{(N+n-a-1)!}\binom{N+n+a-1}{u_{i+b}}\prod_{i<j}^{0,n-1}(u_{j+b}-u_{i+b}).$$ 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.

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  • $\begingroup$ are not there typos? This $u_{j+b}$ looks strange. $\endgroup$ – Fedor Petrov Dec 17 '18 at 19:03
  • $\begingroup$ 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). $\endgroup$ – T. Amdeberhan Dec 17 '18 at 19:05
  • $\begingroup$ also $N+a$ are always together except the denominator where $N-a$ appears. Is it ok? $\endgroup$ – Fedor Petrov Dec 17 '18 at 19:57
  • $\begingroup$ 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. $\endgroup$ – T. Amdeberhan Dec 17 '18 at 21:31
  • $\begingroup$ Tewodros, is it still ok? If we replace $N$ by $N+5$ and $a$ by $a-5$ it looks like nothing except denominator changes... $\endgroup$ – Fedor Petrov Dec 17 '18 at 23:58

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