# On conjectured continued fractions and $e$

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.

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 $$\tanh(\alpha):=\frac{e^{2\alpha} -1}{e^{2\alpha}+1}.$$

• See here for more continued fractions involving $e$. – Dietrich Burde Feb 24 '20 at 9:19
• @downvoter may you please explain why you downvoted? Is there something I could fix/improve? Thanks. – Mr Pie Feb 24 '20 at 10:44
• It appears that $\frac{e^{1/n}+1}{e^{1/n}-1}$ has the continued fraction $[2n;6n,10n,14n,\ldots]$ – Jaycob Coleman Feb 24 '20 at 10:51

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.
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.
• Thank you for your answer, and congratulations! $(+1)$ $\color{green}{\checkmark}$ – Mr Pie Feb 24 '20 at 21:35