I am very aware that the equation in the sandbox might just be a pure coincidence...

...but I want to know for sure.

I mean, it was first assumed that $\pi^3$ was so close to $31$ because it was just a coincidence, but there exists an explanation as to why both the values are close (go here). However, it seems like the fact that $e^\pi - \pi$ is so close to $20$ does not have any known slick proofs, and may just be a beautiful coincidence (go here).

So here is the equation I found:

$$2\ln(662+\pi) = 12.99999987854\ldots\simeq 13.$$

Why is the $LHS$ so close to $13$? Is it just a coincidence, or is there some kind of explanation to it?

Thank you in advance.

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    $\begingroup$ "does not have a thorough explanation" is misinterpreting the fact. It is more like "all its known proofs are not as slick". Any of its proof is a thorough explanation. $\endgroup$ – user545497 Mar 27 '18 at 10:53
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    $\begingroup$ @acetone ok thank you. I just didn't know a better word :) $\endgroup$ – Mr Pie Mar 27 '18 at 10:55
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    $\begingroup$ Almost integers $\endgroup$ – Xander Henderson Mar 30 '18 at 20:01

Note that

$$\log 100 x=x \implies x\approx 6.5$$


$$\log (662+\pi) \approx \log 650 =\log (6.5\cdot 100) \approx 6.5$$


$$2\log (662+\pi)\approx 13$$

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    $\begingroup$ Ahhh I see. How did you figure that out so quickly? $\endgroup$ – Mr Pie Mar 27 '18 at 10:26
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    $\begingroup$ @user477343 form the fact that $665\approx 650$. Note that your best approximation is due to $\log 102 x=x \implies x\approx 6.49619...$ and $102\cdot 6.5=663$. $\endgroup$ – gimusi Mar 27 '18 at 10:29
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    $\begingroup$ @user477343 You are welcome, thanks for your appreciation! Bye $\endgroup$ – gimusi Mar 27 '18 at 10:34
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    $\begingroup$ This doesn't explain why $2\ln(662 + \pi)$ is so much better an approximation than $2 \ln(665) \simeq 12.999574$. $\endgroup$ – gj255 Mar 27 '18 at 13:29
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    $\begingroup$ @gj255 As I wrote in the comment above a better approximation is $\log 102 x=x \implies x\approx 6.49619...$ and $102\cdot 6.5=663\approx 662+\pi$. $\endgroup$ – gimusi Mar 27 '18 at 13:33

For $a,b \in \mathbb N$, let $E(a,b)$ be the distance of $a\ln(b+\pi)$ to the nearest integer.

Then $a=2$ and $b=662$ are special because

$E(2,662)\approx 1.21\cdot 10^{-7}$ is the minimum for $a+b \le 5000$.

The next best values are $E(3961,3726)\approx 1.16\cdot 10^{-7}$ and $E(7768,2134)\approx 1.04\cdot 10^{-7}$, for $a+b \le 10000$.

$E(2,662)$ is the second value at most $10^{-6}$ for $a+b \le 5000$.

The first value is $E(147,495)\approx 8.03\cdot 10^{-7}$, which gives $a\ln(b+\pi)\approx 913.0000008039$.

Although $147+495$ is slightly less than $2+662$, $\ (2,662)$ looks nicer than $(147,495)$.


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