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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.

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    $\begingroup$ that $\ln(163) \pi$ is approximately $2^4$ is just a result from the preceding two.. so it has nothing special $\endgroup$ – George Aug 16 '15 at 11:51
  • $\begingroup$ @George You are right, and i'm afraid i'm going in circles. $\endgroup$ – nospoon Aug 16 '15 at 11:55
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    $\begingroup$ $2^9/163$ is actually not a very good approximation to $\pi$; it gives only 4 correct digits for the 5 we put in... $\endgroup$ – Henning Makholm Aug 16 '15 at 12:00
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    $\begingroup$ Anyway I think it makes sense to ask such a question, so I don't know why someone downvoted. It is partly related to en.wikipedia.org/wiki/Strong_Law_of_Small_Numbers and math.stackexchange.com/questions/111440/…. $\endgroup$ – user21820 Aug 16 '15 at 12:03
  • $\begingroup$ Are $$\frac{90}{log(90)}\approx 20.0008$$ and $$\frac{3\times43\times67}{log(43\times67)}\approx 1085.0008$$ similar? $\endgroup$ – Jaume Oliver Lafont Apr 8 '16 at 21:10
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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.

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  • $\begingroup$ I appreciate your answer. $\endgroup$ – nospoon Aug 16 '15 at 15:45
  • $\begingroup$ @nospoon: I must confess that I have also investigated why this quotient is so curiously close to an integer. :) In fact, Steve Heston (the OP in the MO link), emailed me before he posted that asking if there was any other possible reason. $\endgroup$ – Tito Piezas III Aug 16 '15 at 15:55
  • $\begingroup$ I understand that is just a coincidence. It is still a little hard for to accept that for some divine reason $7 \frac{58}{\ln(58)}\approx100$ ..:). Anyway, i must confess that i'm a fan of your website. $\endgroup$ – nospoon Aug 16 '15 at 16:11
  • $\begingroup$ @nospoon: Thanks. By the way, I've added an update to my answer. I guess that's still a coincidence. :) $\endgroup$ – Tito Piezas III Aug 16 '15 at 17:37
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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.

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    $\begingroup$ Specifically, $$163/\log(163) = 31.9999987\dots$$ $$53453/\log(53453) = 4910.0000012\dots$$ $\endgroup$ – Tito Piezas III Aug 16 '15 at 13:09
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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.

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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!

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