TITLE: Numerical coincidence involving the number 1663
QUESTION [5 upvotes]: The numerical coincidence
$\displaystyle \frac{1663e^2}{3} \approx 2^{12}$.
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.

REPLY [3 votes]: It can be related to the simpler
$$\frac{13e^2}{3}\approx 2^5$$
because
$$ \frac{3}{e^2} \approx \frac{1663}{2^{12}} = \frac{13·2^7-1}{2^{12}}= \frac{13}{2^5}-\frac{1}{2^{12}}$$
Further decomposition into alternating sign bits is
$$\frac{1663}{2^{12}} = \frac{1}{2}-\frac{1}{2^3}+\frac{1}{2^5}-\frac{1}{2^{12}}$$