How do we prove that the simple continued fraction for $e^{2/n}=[1;\frac{n-1}{2},6n,\frac{5n-1}{2},1,1,...]$? Motivation: remarkably, the simple continued fraction - which is unique - for $e^{1/n},e^{2/n}$ is known for every $n\in\Bbb N$. The expansions for $e^{3/n}$, or even $e^{p/q}$, are not known in general, and it is a strange miracle that we can figure out any of them at all... Euler first derived the continued fraction for $e$ and $e^{1/n}$. The expansion for $e^{2/n}$ is viewable on the web, but I have not been able to find a proof of this beautiful fact.
This original text from Euler proves that, for real $s$: $$\begin{align}\tag{1}e^{2/s}&=1+\frac{2}{s-1+}\frac{1}{3s+}\frac{1}{5s+}\cdots\\e^{1/s}&=1+\frac{1}{s-1+}\frac{1}{1+}\frac{1}{1+}\frac{1}{3s-1+}\cdots\end{align}$$It derives the expansion for $e^{1/s}$ from that for $e^{2/s}$ by means of his "interpolation" formulae: $$\begin{align}a+\frac{1}{m+}\frac{1}{n+}\frac{1}{b+}\frac{1}{m+}\frac{1}{n+}\frac{1}{c+}\cdots&=\frac{1}{mn+1}\left((mn+1)a+n+\frac{1}{(mn+1)b+m+n+}\frac{1}{(mn+1)c+m+n+}\cdots\right)\\A+\frac{1}{a+}\frac{1}{m+}\frac{1}{n+}\frac{1}{b+}\cdots&=A+\frac{mn+1}{(mn+1)a+n+}\frac{1}{(mn+1)b+m+n+}\cdots\\a+\frac{1}{b+}\frac{1}{c+}\cdots&=a-n+\frac{mn+1}{m+}\frac{1}{n+}\frac{1}{\frac{b-m-n}{mn+1}+}\frac{1}{m+}\frac{1}{n+}\frac{1}{\frac{c-m-n}{mn+1}+}\cdots\end{align}$$Which all have nice special cases for $m=n=1$.
Wikipedia claims that in fact: $$\tag{$\ast$}e^{2/s}=1+\frac{1}{\frac{s-1}{2}+}\frac{1}{6s+}\frac{1}{\frac{5s-1}{2}+}\frac{1}{1+}\frac{1}{1+}\frac{1}{\frac{7s-1}{2}+}\frac{1}{18s+}\frac{1}{\frac{11s-1}{2}+}\frac{1}{1+}\cdots$$
Which I'd like to understand. First note that, by equivalence transformation (multiply by $2,1/2,2,1/2,\cdots$) the a first try might be to convert $(\ast)$ into: $$e^{2/s}=1+\frac{2}{s-1+}\frac{1}{3s+}\frac{1}{5s-1+}\frac{1}{\frac{1}{2}+}\frac{1}{2+}\frac{1}{\frac{7s-1}{4}+}\cdots$$Which unfortunately doesn't work out. Also, although two $1s$ have clearly been interpolated somehow to get $(\ast)$, we can't directly employ any of the standard results shown above, since, instead of $a,1,1,b,1,1,...$ it is of the form $a,b,c,1,1,d,e,f,1,1,...$.
If we try to realise $(1)$ as an interpolation, using the middle formula, we need to solve $2a+1=s-1\implies a=\frac{s-1}{2}$, $2b+2=3s\implies b=\frac{3s-2}{2}$, etc., to get: $$e^{2/s}=1+\frac{1}{\frac{s-1}{2}+}\frac{1}{1+}\frac{1}{1+}\frac{1}{\frac{3s-2}{2}+}\frac{1}{1+}\frac{1}{1+}\frac{1}{\frac{5s-2}{2}+}\cdots$$Which seems better, but is still not quite right.
I'm convinced the answer is going to be a simple transformation: the denominators of $(\ast)$ are of the form $\frac{x-1}{2},2x$ for $x$ the denominator in $(1)$. The presence of a $2$-multiplier and a $1/2$-multiplier suggests some simple cancellation. However, the two interpolating $1s$ are causing me a headache.
Does anyone know how we get $(\ast)$?
 A: If you have prior knowledge of the answer, you can figure it out. I don’t know how you might derive it “from first principles” though! I’m a little surprised by the lack of publicly available information about continued fractions - you’d think ones concerning notable values, such as $e$, would have well-documented explanations.
If $\alpha_1:=\frac{1}{e^{2/s}-1}-\frac{s-1}{2}$ then $e^{2/s}=1+\frac{1}{\frac{s-1}{2}+\alpha_1}$.  A simple series calculation finds: $$\alpha_1=\frac{\sum_{n\ge2}\frac{2^{n-1}\cdot(n-1)}{s^n\cdot(n+1)!}}{\sum_{n\ge1}\frac{2^n}{s^n\cdot n!}}$$If $\alpha_2:=\frac{1}{\alpha_1}-6s$, then $e^{2/s}=1+\frac{1}{\frac{s-1}{2}+}\frac{1}{6s+\alpha_2}$ and one calculates: $$\alpha_2=\frac{\sum_{n\ge3}\frac{2^n\cdot(n-1)(n-2)}{s^n\cdot(n+2)!}}{\sum_{n\ge2}\frac{2^{n-1}\cdot(n-1)}{s^n\cdot(n+1)!}}$$Continuing, let $\alpha_3:=\frac{1}{\alpha_2}-\frac{5s-1}{2}$: $$\alpha_3=\frac{\sum_{n\ge3}\frac{2^n\cdot n(n-1)(n-2)}{s^n\cdot(n+3)!}}{\sum_{n\ge3}\frac{2^n\cdot(n-1)(n-2)}{s^n\cdot(n+2)!}}$$Then, if (deliberately skipping $4$) $\alpha_5:=\frac{2\alpha_3-1}{1-\alpha_3}$, we have $e^{2/s}=1+\frac{1}{\frac{s-1}{2}+}\frac{1}{6s+}\frac{1}{\frac{5s-1}{2}+}\frac{1}{1+}\frac{1}{1+\alpha_5}$ and: $$\alpha_5=\frac{\sum_{n\ge4}\frac{2^n\cdot(n-1)(n-2)(n-3)}{s^n\cdot(n+3)!}}{3\sum_{n\ge3}\frac{2^n\cdot(n-1)(n-2)}{s^n\cdot(n+3)!}}$$Et cetera.
Inductively these values follow the progression (performing both $\frac{1}{1+}\frac{1}{1+}$ steps together, as done above when skipping from $\alpha_3\to\alpha_5$): $$\frac{\sum_{n\ge m+1}\frac{2^{n-1}\cdot\prod_{k=1}^m(n-k)}{s^n\cdot(n+m)!}}{\sum_{n\ge m}\frac{2^n\cdot\prod_{k=1}^{m-1}(n-k)}{s^n\cdot(n+m-1)!}},\frac{\sum_{n\ge m+2}\frac{2^n\cdot\prod_{k=1}^{m+1}(n-k)}{s^n\cdot(n+m+1)!}}{\sum_{n\ge m+1}\frac{2^{n-1}\cdot\prod_{k=1}^m(n-k)}{s^n\cdot(n+m)!}},\frac{\sum_{n\ge m+2}\frac{2^n\cdot n\prod_{k=1}^{m+1}(n-k)}{s^n\cdot(n+m+2)!}}{\sum_{n\ge m+2}\frac{2^n\cdot\prod_{k=1}^{m+1}(n-k)}{s^n\cdot(n+m+1)!}},\frac{\sum_{n\ge m+3}\frac{2^n\cdot\prod_{k=1}^{m+2}(n-k)}{s^n\cdot(n+m+2)!}}{(m+2)\sum_{n\ge m+2}\frac{2^n\cdot\prod_{k=1}^{m+1}(n-k)}{s^n\cdot(n+m+2)!}}$$With $m=1,4,7,10,...$.
Using some convergence results we can see that, since $\alpha_n\to0$ (empirically: I haven’t proven that yet, and will edit it in when I do) the continued fraction converges to $e^{2/s}$ for nonzero positive $s$. Certainly all terms are positive (eventually) when $s>0$ so convergence “to the right thing” is automatic (see the linked post).
Empirically it also converges for negative $s$ but that’s more annoying to deal with, due to the negative continued fraction coefficients. Anyway, if $s$ is negative you can just take a reciprocal as $e^{-2/s}=\frac{1}{e^{2/s}}$.
