Numerical coincidence involving the number 1663 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.
 A: 12288/1663 is the 7th convergent in the infinite continued fraction representation of $e^2$, so it naturally will be a very good rational approximation.
For completeness, here's the list of the first 10 convergents (via Mathematica):
{7, 15/2, 22/3, 37/5, 133/18, 2431/329, 12288/1663, 14719/1992, 27007/3655, 176761/23922}
A: Let's rewrite this as $12288 \approx 1663 e^2$. The coincidence, then, is that $1663e^2$ is close to an integer. In fact $1663e^2 \approx 12288.0002925$, or $1663e^2 - 12288 \approx 0.0002925$. That is, $1663e^2 - 12288 \approx 1/3448$.
Since we'd be equally impressed if $1663e^2$ were just below an integer, that means that we're impressed because $1663e^2$ is in an interval of width about $2/3448$ or $1/1724$; call this $\epsilon$.
But we'd expect $1663\epsilon $ integers $n \le 1663$ to have the property that $ne^2$ is within $\epsilon/2$ of an integer. And $1663\epsilon \approx 1$. So this is really not impressive.
(Except that $12288$ has such a simple prime factorization.)
A: 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
$$\frac{1663}{2^{12}} = \frac{1}{2}-\frac{1}{2^3}+\frac{1}{2^5}-\frac{1}{2^{12}}$$
A: 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.
