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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.

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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

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