# 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.

$$\det B_n^1=n,$$ $$\det B_n^2=\frac{n(n^2-1)}{3},$$ $$\det B_n^3=\frac{n^2(n^2-1)(n^2-2^2)}{3^2\cdot5},$$ $$\det B_n^4=\frac{n^2(n^2-1)^2(n^2-2^2)(n^2-3^2)}{3^3\cdot5^2\cdot7},$$ $$\det B_n^5=\frac{n^3(n^2-1)^2(n^2-2^2)^2(n^2-3^2)(n^2-4^2)}{3^4\cdot5^3\cdot7^2\cdot9},$$ $$\dots$$
$$\det B_n^l=n^{\lceil l/2\rceil}\prod_{k=1}^{l-1}\frac{(n^2-k^2)^{\lfloor (n-k+1)/2\rfloor}}{(2k+1)^{n-k}}.$$
The fact that $$\det B_n^n=2^{n(n-1)/2}$$ follows from Rational fraction expression for triangular powers of 2