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 $\sum_{k=\lceil\frac{n}2\rceil}^n(s(k)-s(n-k))=\lceil\frac{n}2\rceil$.

gpcode:A(n) = matrix(n,n,i,j,binomial(2*j,i))$\phantom{000000000000000000000}$B(n)=matrix(n,n,i,j,binomial(n+1,2*j-i))$\phantom{000000000000000000000}$S(M)=round(log(matsnf(M,4))/log(2))$$ $$ and then the vector in my first comment is bothvector(16,n,S(A(n)))andvector(16,n,S(B(n)))$\endgroup$ – Noam D. Elkies Dec 31 '16 at 21:17