Coefficients of binomial continued fractions
For a natural number $n$, let $$ \begin{equation} \beta_n(z)=\frac{(1+z)^n+(1-z)^n}{(1+z)^n-(1-z)^n}. \end{equation} $$ Then the coefficients of the numerator and denominator of $\beta_n$ are binomial. For example: $$\begin{equation} \beta_4(z)=\frac{z^4+6z^2+1}{4z^3+4z}=\frac{1}{4} z+\cfrac{1}{\frac{4}{5}z+\cfrac{1}{\frac{25}{16}z+\cfrac{1}{\frac{16}{5} z}}}. \end{equation} $$ Is there a simple formula for the coefficients of the continued fraction for any $n$?
NOTES:
the fact that all the coefficients are positive numbers follows from complex analysis, since $\Re\beta_n(z)>0$ for $\Re z >0$
computer calculation shows that the factorizations of the coefficients consist of small primes, less than $2(n+1)$
the motivation for the problem is the rational approximation of the square root function, since $(z\beta_n)(z^2)\approx z$ for $\Re z >0$
a relevant discussion: Binary eigenvalues matrices and continued fractions
Solution 1:
Recall by De Moivre's Theorem
\begin{equation} \tan n\theta = \frac{n \tan\theta - C(n,3) \tan^3 \theta + \cdots}{1 - C(n,2)\tan^2\theta+ C(n,4)\tan^4 \theta-\cdots} \end{equation}
Let $z = i\tan w$. Substituting and applying this result $$ \beta_n(z) = \beta_n(i\tan w) = -i \cot nw $$ By 1 or 2 which clearly hold for a complex domain
$\displaystyle \tan(nx) = \cfrac{n\tan x}{1 -\cfrac{(n^{2} - 1^{2})\tan^{2}x}{3 -\cfrac{(n^{2} - 2^{2})\tan^{2}x}{5 -\cfrac{(n^{2} - 3^{2})\tan^{2}x}{7 -\cdots}}}}$
with final term $\dfrac{(n^{2} - (n - 1)^{2})\tan^{2}x}{(2n - 1)}$. Simple operations show
\begin{align} \beta_n(z) = -i \cot nw &= {\frac{1}{ni\tan w} +\cfrac{n^{-1}(n^{2} - 1^{2})i \tan w}{3 -\cfrac{(n^{2} - 2^{2})\tan^{2}w}{5 -\cfrac{(n^{2} - 3^{2})\tan^{2}w}{7 -\cdots}}}} \\ &= {\frac{1}{nz} +\cfrac{n^{-1}(n^{2} - 1^{2})z}{3 +\cfrac{(n^{2} - 2^{2})z^2}{5 + \cfrac{(n^{2} - 3^{2})z^2}{7 +\cdots}}}} \\ &= {\frac{z^{-1}}{n} +\cfrac{n^{-1}(n^{2} - 1^{2})}{3z^{-1} +\cfrac{(n^{2} - 2^{2})}{5z^{-1} + \cfrac{(n^{2} - 3^{2})}{7z^{-1} +\cdots}}}} \\ &= {a_1z^{-1} +\cfrac{1}{a_2z^{-1}+\cfrac{1}{a_2z^{-1} + \cfrac{1}{a_3z^{-1} +\cdots}}}} \end{align}
Where $a_1 = \frac{1}{n}$, $a_2 = 3\frac{n}{n^2-1},a_3 = 5\frac{n^2-1}{n(n^2-2^2)},a_4 = 7\frac{n(n^2-2^2)}{(n^2-1)(n^2-3^2)},\cdots$ to $n$ terms. In general
\begin{equation} a_k = (2k-1)\Big(\frac{n}{n^2-1}\frac{n^2-2^2}{n^2-3^2}\frac{n^2-4^2}{n^2-5^2}\cdots\Big)^{(-1)^k} \end{equation}
to $[k/2]$ fractional products, $[x]$ being the ceiling function, where if $k$ is odd the last term has denominator $1$.
Substituting $1/z$ for $z$ now gives an explicit solution of the problem by using $\beta_{2m}(z) = \beta_{2m}(z^{-1})$ and $\beta_{2m+1}(z)= \beta_{2m+1}(z^{-1})^{-1}$.
To verify your case substituting $n=4$ returns $a_1 = 1/4, a_2 = 4/5$, $a_3 = 25/16$, and $a_4 = 16/5$.
Edit: The form of $a_k$ can be simplified. For even $n$ and even $k$ \begin{equation} a_k = (2k-1) \frac{(n-(k-2))(n-(k-4))\cdots n \cdots (n+k-4)(n+k-2)}{(n-(k-1))(n-(k-3))\cdots(n+k-3)(n+k-1)} = (2k-1)\Bigg(\frac{2^{k-1}((n+k)/2-1)!}{((n-k)/2)!}\Bigg)^2 \frac{(n-k)!}{(n+k-1)!} = \frac{2^{2(k-1)}(2k-1)}{n+k-1}{n-k \choose (n-k)/2}{n+k-2 \choose (n+k-2)/2}^{-1} \end{equation} Setting $k=n$, $a_n = 2^{2(n-1)}((n-1)!)^2/(2n-2)!$ giving a power of $2$ in the numerator. For even $n$ and odd $k$ take the reciprocal of $a_{k+1}$ and multiply by $(2k-1)(2k+1)/(n^2-k^2)$
\begin{equation} a_k = \frac{2k-1}{2^{2k}(n-k)}{n-k-1 \choose (n-k-1)/2}^{-1}{n+k-1 \choose (n+k-1)/2} \end{equation}
Treating odd $n$ and odd $k$ separately because of $n=k=1$ \begin{align} a_k &= (2k-1)\frac{(n-(k-2))(n-(k-4))\cdots(n+k-4)(n+k-2)}{(n-(k-1))(n-(k-3)\cdots n \cdots (n+k-3)(n+k-1)} \\ &= (2k-1)\Bigg(\frac{2^{k-1}((n+k-2)/2)!}{((n-k)/2)!}\Bigg)^2 \frac{(n-k)!}{(n+k-1)!}\\ &= \frac{2^{2(k-1)}(2k-1)}{n+k-1}{n-k \choose (n-k)/2}{n+k-2 \choose (n+k-2)/2}^{-1} \end{align} For odd $n$ even $k$ take the reciprocal of $a_{k+1}$ and multiply by $(2k-1)(2k+1)/(n^2-k^2)$ \begin{equation} a_k = \frac{2k-1}{2^{2k}(n-k)}{n-k-1 \choose (n-k-1)/2}^{-1}{n+k-1 \choose (n+k-1)/2} \end{equation}