<s>[INST] Determinants of products of binary matrices and binomial coefficients Consider two binary semi-infinite matrices with obvious patterns:
$$
C=
\begin{bmatrix}
1 &0 &0 &0 &0 &0 &0 &\cdots\\
1 &0 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &0 &1 &0 &0 &0 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
and
$$
T=
\begin{bmatrix}
1 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &1 &0 &0 &0 &\cdots\\
0 &0 &0 &1 &1 &0 &0 &\cdots\\
0 &0 &0 &0 &1 &1 &0 &\cdots\\
0 &0 &0 &0 &0 &1 &1 &\cdots\\
0 &0 &0 &0 &0 &0 &1 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
Let $A_n=T^n C$, then the non-zero entries of $A_n$ are the ${n+1}\choose{m}$, $0\le m \le n+1$ binomial coefficients. For example, 
$$
A_3=
\begin{bmatrix}
4 &4 &0 &0 &0 &0 &\cdots\\
1 &6 &1 &0 &0 &0 &\cdots\\
0 &4 &4 &0 &0 &0 &\cdots\\
0 &1 &6 &1 &0 &0 &\cdots\\
0 &0 &4 &4 &0 &0 &\cdots\\
0 &0 &1 &6 &1 &0 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
Is there a simple formula for the determinants of the $k\times k$ upper left blocks of the matrices $A_n$? That is: what is $\det(B_n^k)$, where $$B_n^k(ij)=A_n(ij),$$ $1\le i,j \le k\le n+1$?
NOTES:


*

*The computer factorisation of $\det(B_n^k)$ shows that
$\det(B_n^{(n+1)}=2^{n(n+1)/2}$ and the determinants factorisations
have only factors less than $2(n+1)$, which suggests that the
determinants are products of binomial coefficients of the form
${l}\choose{k}$, $0\le k,l\le 2(n+1)$.

*This question is motivated by the continued fraction approximation of
the square root function, the matrices $A_n$ being the Hurwitz
matrices of the continued fractions.
 [/INST] It follows from the answer to Coefficients of binomial continued fractions, since the matrices in the question are Hurwitz matrices of the continued fractions that:
$$ \det B_n^1=n,$$
$$ \det B_n^2=\frac{n(n^2-1)}{3},$$
$$ \det B_n^3=\frac{n^2(n^2-1)(n^2-2^2)}{3^2\cdot5},$$
$$ \det B_n^4=\frac{n^2(n^2-1)^2(n^2-2^2)(n^2-3^2)}{3^3\cdot5^2\cdot7},$$
$$ \det B_n^5=\frac{n^3(n^2-1)^2(n^2-2^2)^2(n^2-3^2)(n^2-4^2)}{3^4\cdot5^3\cdot7^2\cdot9},$$
$$ \dots$$
$$ \det B_n^l=n^{\lceil l/2\rceil}\prod_{k=1}^{l-1}\frac{(n^2-k^2)^{\lfloor (n-k+1)/2\rfloor}}{(2k+1)^{n-k}}.$$
The fact that $$\det B_n^n=2^{n(n-1)/2}$$ follows from Rational fraction expression for triangular powers of 2
</s>