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