in search of a transformation between determinants

Question

Motivated by this MO question. Consider the two matrices $A_n$ and $B_n$ with entries $\binom{2j}i$ and $\binom{n+1}{2j-i}$, respectively; for $1\leq i, \,j\leq n$.

I can show $\det A_n=\det B_n=2^{\binom{n+1}2}$; it is not a problem (see, for instance, my answer here).

Question.

(1) Is there a transformation converting $A_n$ into $B_n$ (or vice-versa)?

(2) Is there a combinatorial interpretation/bijection revealing the counts by $\det A_n$ and $\det B_n$?

My asking for (2) is due to the fact that $2^{\binom{n+1}2}$ enumerates domino tilings of an Aztec diamond.

(3) Let $s(k)=$the number of $1$'s in the binary expansion of $k$. Then, $A_n$ and $B_n$ share the same Smith normal form (showing the diagonal vector) given by $$[2^{\max(4k-2n+s(n-k)-s(k),0)}:\, 1\leq k\leq n].$$ This claim is based on data from Noam Elkies' comments seen below. Any proof?

Remark. Let $\lceil x\rceil=$the smallest integer greater than or equal to $x$ (ceiling function). So, (3) implies $$\sum_{k=1}^n\max(4k-2n+s(n-k)-s(k),0)=\frac{n(n+1)}2;$$ Or, equivalently $\sum_{k=\lceil\frac{n}2\rceil}^n(s(k)-s(n-k))=\lceil\frac{n}2\rceil$.

Answer 1

There is such a transformation, of the form predicted in Linear transformation that preserves the determinant.

Denoting $R$ the involution matrix $e_i\mapsto e_{n+1-i}$, it turns out that the matrix $A$ has an $LU$-decomposition in which $U:=\left[{j\choose i}\right]_{{1\le i\le n}\atop{1\le j\le n}}$, and the lower triangular part is $L=RCR$, where $C$ is the upper triangular matrix in Suvrit's decomposition , $B=VCV^{-1}$ (warning: $B$ of this question is named "$A$" there). So $B= (VR) A(VRU)^{-1}$, with $\det(VR)=\det(VRU)^{-1}=1$.

$$*$$ [edit] The description becomes a bit more gracious if we include the indices $i=0$ and $j=0$. So, if we define the $n\times n$ matrices with integer entries $$A_n:=\left[{2j\choose i}\right]_{{0\le i< n}\atop{0\le j< n}}\qquad B_n:=\left[{n\choose 2j-i}\right]_{{0\le i< n}\atop{0\le j< n}}$$ $$U_n:=\left[{j\choose i}\right]_{{0\le i< n}\atop{0\le j< n}}\qquad L_n:=\left[2^{2j-i}{j\choose2j- i}\right]_{{0\le i< n}\atop{0\le j< n}}$$ $$N_n:=\Big[ \delta_{i+1,j}\Big]_{{0\le i< n}\atop{0\le j< n}}\qquad R_n:=\Big[ \delta_{n-i,j}\Big]_{{0\le i< n}\atop{0\le j< n}}$$ Then, (hiding the subscript $n$) $$A=LU$$ and $$B=VCV^{-1}$$ with $$V:=U^{T}R\qquad C:=(I+N)AU^{-1}\ .$$

Answer 2

This doesn't answer the original question but answers the later SNF question for the matrix $B_n$. Let $C_n$ be the $n\times n$ matrix whose $(i,j)$-entry ($1\leq i,j\leq n$) is $\binom{x+1}{2j-i}$. Up to row and column permutations that preserve the sign of the determinant, this is the dual Jacobi-Trudi matrix for the Schur function $s_{n,n-1,\dots,1}$, specialized by setting $x+1$ variables equal to 1 and the others to 0. I compute the SNF of this matrix over the field $\mathbb{Q}[x]$ in http://math.mit.edu/~rstan/papers/jtsnf.pdf. Now set $x=n$ and consider the SNF over $\mathbb{Z}_{(2)}$ (the integers localized at 2, i.e., invert all primes except 2). My proof technique can still be used since all hook lengths are odd and therefore units in $\mathbb{Z}_{(2)}$. Namely, one shows that the bottom-left $k\times k$ minor $M$ divides all $k\times k$ minors (with a special argument when $M=0$), etc. (Actually, we can deduce the SNF of $B_n$ directly from that of $C_n$ since for the partition $(n,n-1,\dots,1)$ we can work over $\mathbb{Z}_{(2)}[x]$ rather than $\mathbb{Q}[x]$.) We get that the $i$th largest diagonal element of the SNF is $\prod_u(n+1+c(u))$, where $u$ ranges over all squares of the $i$th diagonal hook of the partition $(n,n-1,\dots,1)$, and $c(u)$ is the content of the square $u$. This product is just $(2n-2i+2)!/(2i-1)!$, so over the integers the $i$th
largest diagonal element of the SNF of $B_n$ is the largest power of 2 dividing $(2n-2i+2)!/(2i-1)!$.