On the near-integer $163/\ln(163)$ This question, concerning the approximation $\frac{163}{\ln(163)}\approx 2^5$, was posted on MO 5 years ago: Why Is 163/ln(163) a Near-Integer?.
It was concluded that it had nothing to do with 163 being a Heegner number, and that it is most likely just a mathematical coincidence.
Playing with my calculator, I noticed that $163\pi\approx2^9$, and $\ln(163)\pi\approx 2^4$, so I thought maybe $\pi$ has something to do with this? I proceeded to press more buttons on my calculator, 
and came up with $\pi\approx\frac{2^9}{163}+\frac1{2^{11}}\approx\frac{2^4}{\ln(163)}+\frac1{2^{11}}$. What's going on here? 
I noticed also that $67$ exhibits somthing similar: $\frac{67}{\ln(67)}\approx2^4-\frac{67}{2^{10}}$.
I haven't found such relations with other Heegner numbers, but I still remain unsatisfied. 
Maybe it is the start of some Ramanujan-type infinite series for $\frac1{\pi}$, or..? I am not convinced that these relations are just meaningless numerology. Can someone explain what's going on? And what does $\pi$ has to do with this? I post this hoping that someone who knows more than I do could shed some light on it, and am sorry in advance if this is not the appropriate place to do so.
 A: What makes the Heegner numbers $d$ special is that, since they have class number $h(-d) = 1$, then the j-function $j(\tau)=j \Big( \tfrac{1+\sqrt{-d}}{2} \Big)$ at those points is an algebraic integer of degree $1$.
However, there are $d$ such that $j(\tau)$ is an algebraic integer of degree $2$. The largest of which is $d = 7\times61=427$ and explains why,
$$e^{\pi\sqrt{427}} \approx 5280^3(236674+30303\sqrt{\color{blue}{61}})^3+743.999999999999999999999987\dots$$
Thus, if your (and many others) observation about $d=163$ has to do with its "Heegner-ness" and class number, then it is reasonable to ask if something analogous happens with, 
$$x = \frac{427}{\ln(427)} = \color{brown}{70.499459}62\dots$$
this time assuming that $x$ is close to an algebraic integer of degree $2$. 
Unfortunately, Mathematica doesn't recognize this as near the root of a quadratic equation $ax^2+bx+c = 0$ with small coefficients and $a=1$ (since we require that $x$ be near an algebraic integer). 
Update: Expanding the search radius, most of the results were like,
$$\tfrac{1}{2}(52417-\sqrt{2732780289}) = \color{brown}{70.499459}59\dots$$
with hundreds of discriminants a huge meaningless number. Within this radius, and with an accuracy above $10^{-7}$, the single discriminant that was small, of all numbers, turned out to be,
$$\tfrac{1}{2}(52415-6693\sqrt{\color{blue}{61}}) = \color{brown}{70.499459}57\dots$$
Sigh. (End update.)
The observation about $\frac{163}{\ln(163)} \approx 31.9999987$ seems to be nothing more than mathematical coincidence, as interesting as it is. But there are other aspects of $163$ that are not coincidence. For example,
$$1^2+40^2 = 42^2-163$$
is a consequence of the well-known prime generating polynomial $F(n) = n^2+n+41$. I've collected some of these gems in this list.
A: Seem to be 5 numbers between 1 and $10^6$ and 45 numbers between 1 and $16 \cdot 10^6$ which are closer to an integer than 163/log(163). I guess what is special is that 163 is such a small number. The next smallest which is closer to an integer is 53453.
A: Let $k$ be the number of bits in your LaTeX expression for the approximation. We expect to be able to find an approximation to within $2^{-k}+O(1)$ as $k \to \infty$, because we already have one, namely $2^{-k} \approx 0$. So your examples are rather mundane because they hardly get close to $2^{-k}$ precision.
A: Another funny observation related to approximations of $2^5 = 32$: $\pi^3+1 =/almost/= 32 = 2^5$, actually it is $32.00627...$, so we can almost factor $2^5$ as $$\pi^3+1 = \pi^3-(-1) = \pi^3-(-1)^3 = (\pi+1)(\pi^2-\pi+1) = (\pi+1)((\pi-1)^2+\pi) = (\pi+1)((\pi-1)^2-(-\pi)) = (\pi+1)(\pi-1+i\sqrt\pi)(\pi-1-i\sqrt\pi)$$
You can pick your favorite one!
