$\displaystyle \frac{1663e^2}{3} \approx 2^{12}$.
showed up in a comment of this mostly-unrelated question.
Numerically, it's not a surprise that $e^2$ is close to a rational number whose numerator and denominator are in this range — similarly good approximations to most numbers can be obtained by truncating the continued fraction at the desired level of accuracy. What's much more of a surprise to me is the appearance of a 12th power in this expression. Is there a good explanation for having such a smooth number here? E.g. see the j-invariant explanation for Ramanujan's observation that
$e^{\pi\sqrt{163}}$
is close to an integer, or the Pisot number explanation for the fact that even powers of the golden ratio are close to integers.
Best Answer
The explanation is that $1663/12288$ is a convergent in the continued fraction expansion of $e^{-2}$. In other words, the continued fraction expansion of $e^{-2}$ starts out as $$[0;7,2,1,1,3,18,5,1,...].$$ If you truncate this expansion after the $5$ then you end up with the rational number $$[0;7,2,1,1,3,18,5]=\frac{1663}{12288}.$$ By basic properties of continued fractions it follows that $$\left|e^{-2}-\frac{1663}{12288}\right|\le\frac{1}{12288^2},$$ which explains the behaviour you are observing. Furthermore there is a theorem which says that all `best approximations' to a real number come from truncating the continued fraction expansion.