On the near-integer $163/\ln(163)$

Question

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.

Answer 1

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 $2$.

1st test: Unfortunately, it seems 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).

2nd test: 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,

$$\frac{427}{\ln(427)} \approx \tfrac{1}{2}(52415-6693\sqrt{\color{blue}{61}}) = \color{brown}{70.499459}57\dots$$

To recall, $427 = 7\times61$ so the sudden appearance of $61$ seems a wee bit curious.

P.S. I've collected $163$-related stuff in this list.

Answer 2

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.

Answer 3

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.

Answer 4

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!