Motivated by [this MO question](https://mathoverflow.net/questions/258284/is-the-matrix-left2m-choose-2j-i-right-i-j-12m-1-nonsingular/258291#258291). 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](https://mathoverflow.net/questions/258284/is-the-matrix-left2m-choose-2j-i-right-i-j-12m-1-nonsingular/258291#258291)).



> 
> **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](https://www.mfo.de/math-in-public/snapshots/files/domino-tilings-of-the-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](https://en.wikipedia.org/wiki/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 $\sum\_{k=\lceil\frac{n}2\rceil}^n(s(k)-s(n-k))=\lceil\frac{n}2\rceil$.



in search of a transformation between determinants
There is such a transformation, of the form predicted in [Linear transformation that preserves the determinant](https://mathoverflow.net/questions/522/linear-transformation-that-preserves-the-determinant?rq=1).


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](https://mathoverflow.net/questions/258284/is-the-matrix-left2m-choose-2j-i-right-i-j-12m-1-nonsingular/258609#258609) , 
$B=VCV^{-1}$ (*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
$$V:=U^{T}R\qquad C:=(I+N)AU^{-1}\ .$$


