On conjectured continued fractions and $e$

Question

Playing around with numbers, I conjectured three incredibly interesting things:

$$9+\cfrac{1}{18+0\times 12\cfrac{1}{18+1\times 12+\cfrac{1}{18+2\times 12+\cfrac{1}{18+3\times 12+\ddots}}}}=\frac{4e^{1/3}-2}{e^{1/3}-1}$$

$$6+\cfrac{1}{9+0\times 6+\cfrac{1}{9+1\times 6+\cfrac{1}{9+2\times 6+\cfrac{1}{9+3\times 6+\ddots}}}}=\frac{4e^{2/3}-2}{e^{2/3}-1}$$

$$5+\cfrac{1}{6+0\times 4+\cfrac{1}{6+1\times 4+\cfrac{1}{6+2\times 4+\cfrac{1}{6+3\times 4+\ddots}}}}=\frac{4e-2}{e-1}$$

So, what in God's name is going on behind the scenes? Why does this seem to be true, why does it involve $e$, so many questions! All I did was play around on a calculator with some continued fractions, went to Wolfram Alpha and asked for the result to be written in terms of $e$ and then I noticed some patterns. But what is really going on? Well, apart from a bit of luck, I have no idea.

Any ideas? Thanks.

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Edit:

This question may be of help, as it reveals general continued fractions regarding the hyperbolic tangent, which to those who don't know, is a function with respect to some value $\alpha$ defined as $\tanh(\alpha):=\frac{e^{2\alpha} -1}{e^{2\alpha}+1}.$

Answer 1

First of all, the terms on the right are of the form $$ \frac{4e^z-2}{e^z-1} = 4 + \frac2{e^z-1} $$

Let's examine the last of your cases which is for $z=1$. It is known that $$ e = 1 + \frac2{[1;6\;10\;14\;18\cdots]} $$ hence $$ 4+\frac2{e-1} = 4+[1;6\;10\;14\;18\cdots] = [5;6\;10\;14\;18\cdots] $$ which is your 3rd fraction. The Wikipedia page linked above has also fractions for $e^{x/y}$, you can use it to see your other equations just as easy: $$e^{x/y} = 1+ \cfrac{2x}{2y-x+}\; \cfrac{x^2}{6y+}\; \cfrac{x^2}{10y+}\; \cfrac{x^2}{14y+}\; \cdots$$ thus $$ \frac{2x}{e^{x/y}-1} = 2y-x+\; \cfrac{x^2}{6y+}\; \cfrac{x^2}{10y+}\; \cfrac{x^2}{14y+}\; \cdots $$ and with $x=1$ and $y=3$ we have: $$\begin{align} 4+\frac2{e^{1/3}-1} &= 4+2\cdot3-1+\; \cfrac{1}{6\cdot3\,+}\; \cfrac{1}{10\cdot3\,+}\; \cfrac{1}{14\cdot3\,+}\; \cdots\\ &= [9;18\;30\;42\;\cdots]\\ \end{align}$$ which is your 1st fraction. And finally, for the 2nd case, we rewrite the above to $$\begin{align} \frac{2x}{e^{x/y}-1} &= 2y-x+\; \cfrac{x}{\frac6xy+}\; \cfrac{1}{\frac{10}x y+}\; \cfrac{1}{\frac{14}x y+}\; \cdots\\ &\stackrel{x=2}= 2y-2+ \cfrac{2}{3y+}\; \cfrac{1}{5y+}\; \cfrac{1}{7y+}\; \cfrac{1}{9y+}\; \cdots\\ \end{align}$$ dividing by $x=2$:

$$\begin{align} \frac{2}{e^{2/y}-1} &= y-1+\; \cfrac{1}{3y+}\; \cfrac{1}{5y+}\; \cfrac{1}{7y+}\; \cdots\\ &\stackrel{y=3}= 2+ \cfrac{1}{9+}\; \cfrac{1}{15+}\; \cfrac{1}{21+}\; \cfrac{1}{27+}\; \cdots\\ \end{align}$$ Adding 4 we arrive at the 2nd equation.

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p.s.: If we didn't evaluate for $x$ and just continued, we'd get $$\begin{align} \frac{2}{e^{x/y}-1} &= 2\frac yx - 1+\; \cfrac{1}{\frac6xy+}\; \cfrac{1}{\frac{10}x y+}\; \cfrac{1}{\frac{14}x y+}\; \cdots\\ \end{align}$$ which is after dividing the equation by $x$ and moving the $x$'s to the denominators in all fractions. This can be rewritten as, now with $z=2y/x$: $$\begin{align} 4+\frac{2}{e^{2/z}-1} &= 4 + z - 1+\; \cfrac{1}{3 z+}\; \cfrac{1}{5 z+}\; \cfrac{1}{7 z+}\; \cdots\\ &= [z+3;3z\;5z\;7z\;\cdots] \end{align}$$ We then get your equations for $z=6$, $z=3$ and $z=2$, respectively.