I'm working on a rational approximation of the square root function by continued fractions in the complex plane. The following kind binomial coefficients Hurwitz matrices (for $n=4$ and $6$) play important role: $ \begin{pmatrix} 4&4&0&0\\ 1&6&1&0\\ 0&4&4&0\\ 0&1&6&1 \end{pmatrix} $ or $ \begin{pmatrix} 6&20&6&0&0&0\\ 1&15&15&1&0&0\\ 0&6&20&6&0&0\\ 0&1&15&15&1&0\\ 0&0&6&20&6&0\\ 0&0&1&15&15&1 \end{pmatrix}. $ The computer tells me that the eigenvalues of these matrices turn out to be consecutive powers of $2$, $(1,2,4,8,\dots)$ and as a consequence the determinants are powers of $2$ as well. I'd like to know the reason for that. (I see only that $(1,1,\dots,1)$ is an eigenvector with eigenvalue $2^{n-1}$ and $(0,0,\dots,1)$ is the eigenvector with eigenvalue $1$).
Also, the coefficients of the corresponding continued fractions are ratios of determinants of the principal minors of the matrices consisting of intersections of the first rows and columns. The factorizations of the determinants of the minors (e.g. $ \det\begin{pmatrix} 6 \end{pmatrix} $, $\det \begin{pmatrix} 6&20\\ 1&15 \end{pmatrix}, $ $\det \begin{pmatrix} 6&20&6\\ 1&15&15\\ 0&6&20 \end{pmatrix}, \dots $) don't contain primes larger than $2n$, even though they grow exponentially. For the above examples these are: ($4$,$20$ and $64$) and ($6$, $70$, $896$, $8064$ and $32726$). (for the $n=6$ case the factorizations are $2 * 3$, $2 * 5 * 7$, $2^7 * 7$, $2^7 * 3^2 * 7$ and $2^15$). This implies that the continued fractions have coefficients that are products of small fractions, and I'd like to know the reason for that too.
