Why is $e^{1/e}$ close to $\frac{13}{9}$? $$e^\frac{1}{e} = \frac{13.002010749 \cdots}{9}$$
Is there a reason why it is close to $\frac{13}{9}$?
Or is this just a mathematical coincidence?
 A: Although not a ''coincidence'' in the sense that $\frac{13}{9}$ is closely related to the continued fraction expansion of $e^{1/e}$, surprising ''coincidences'' can be generated rather easily. I recommend the following excellent article on the subject:
Seemingly Remarkable Mathematical Coincidences Are Easy to Generate
A: By using the theory of continued fractions, you get a systematic method for generating as many similar "coincidences" as you like. Here's how.
In general, every irrational number $x > 0$ has a unique convergent continued fraction expansion of the form
$$x = n_0 + \frac{1}{n_1 + \frac{1}{n_2 + \frac{1}{n_3 + \frac{1}{...}}}}
$$
for unique integers $n_0 \ge 0$ and $n_1,n_2,n_3,\ldots \ge 1$. Truncating this infinite continued fraction to form a sequence of finite continued fractions gives you a sequence of very good rational approximations, starting with
$$x \approx n_0
$$
$$x \approx n_0 + \frac{1}{n_1}
$$
$$x \approx n_0 + \frac{1}{n_1 + \frac{1}{n_2}}
$$
$$x \approx n_0 + \frac{1}{n_1 + \frac{1}{n_2 + \frac{1}{n_3}}}
$$
If a large value of $n_i$ pops up, then the approximation is particularly close when you truncate the expression just before $n_i$.
So, for example, using
$$\pi = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \frac{1}{292 + ...}}}}
$$
if we truncate just before $292$ then we should get a particularly good approximation, namely
$$\pi \approx 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1}}} = \frac{355}{113}
$$
which we can compare to the decimal approximation
$$\pi \approx 3.14159265358979... \approx \frac{354.999969856...}{113}
$$
What a coincidence!
A: $13/9$ is a convergent of the continued fraction for $e^{1/e}$, so it is expected to be unusually close to $e^{1/e}$.
The continued fraction looks like
$$
e^{1/e} = 1+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 4+\frac{1}{55+\ddots}}}
$$
and since $55$ is exceptionally large, the approximation
$$
1+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 4}} = \frac{13}{9}
$$
is exceptionally close.
