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Something similar to this has probably been posted, but since I can't find any at the moment I will post it here.

There are many numerical expressions to do with $\pi$, $e$ and $163$ (Wikipedia has many of these). The following are some of the approximations I have discovered when trying out different operations using the three numbers on my calculator:

$$e^\pi - \pi^{1-e} \approx 23$$ $$\sqrt[e]{\pi} \approx \dfrac{\pi+1}e$$ $$\sqrt{\pi+e+163} \approx 13$$ $$\sqrt[3]{163}-\sqrt[3]{\pi} \approx 4 $$ $$\sqrt{163}-\sqrt{\pi}\approx11$$ $$\dfrac{\sqrt{163}}{\sqrt[3]e} \approx 6+\pi $$ $$\dfrac{\pi}{2e} \approx \dfrac1{\sqrt3}$$ $$\sqrt[3]{\dfrac{\pi^3}{\sqrt[3]e}+\dfrac{e^3}{\sqrt[3]{\pi}}}\approx 3.3 \,\text{(my favourite)}$$ $$ e^\pi-2(4\pi-1)\approx0$$ $$ \dfrac{\pi}e\left(e^{\sqrt[3]{\pi}}\right)\approx5$$

EDIT: Inspired by @Raffaele's approximation I find that if $$x=\frac{163}{e}+\frac{e}{163}+\frac{\pi}{163}-e^{\pi}$$ then $\sin x \approx 0.6$, $\cos x \approx 0.8$ and $\tan x \approx 0.75$.

Do you have any others?

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    $\begingroup$ $$\Large\frac{1}{30^{\pi^e}}\approx h \tag{Planck's constant}$$ $\endgroup$
    – Mr Pie
    Oct 8, 2018 at 0:40

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$$\frac{163}{e}+\frac{e}{163}+\frac{\pi }{163}\approx 60$$

It's mine :)

Hope you like it

EDIT

$163 (\pi -e)\approx 69$

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  • $\begingroup$ Nice! Quite a high degree of accuracy as well $\endgroup$
    – TheSimpliFire
    Dec 2, 2017 at 13:52
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The number $e^{\pi\sqrt{163}}$ is very close to the integer $262537412640768744$ the difference is about $7.5\times 10^{-13}$

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  • $\begingroup$ How did you find this coincidence? $\endgroup$ Dec 2, 2017 at 12:19
  • $\begingroup$ It's fairly well known, that number is sometimes called the "Ramanujan constant". $\endgroup$
    – Ben P.
    Dec 2, 2017 at 12:45
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    $\begingroup$ This is not a coincidence. $\endgroup$
    – MJD
    Dec 2, 2017 at 14:26
  • $\begingroup$ @MJD Ah, you got me... still it'll look like a coincidence for "most people". $\endgroup$
    – Ben P.
    Dec 2, 2017 at 15:04
  • $\begingroup$ $e^{\pi\sqrt{163}}=2.62537412640768743999999999999250\times 10^{17}$ $\endgroup$ Dec 3, 2017 at 8:31
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This is not an approximation but an equality,

$$288\sum_{k=1}^\infty\frac1{k^6\,\binom{2k}{k}}+432\sum_{k=1}^\infty\frac1{k^2\,\binom{2k}{k}}\sum_{n=1}^{k-1}\frac1{n^4}=\color{blue}{163}\,\zeta(6)$$

where $\displaystyle\zeta(6) = \frac{\pi^6}{945}$, but the appearance of $163$ here is just a coincidence, and not related to its being a Heegner number.

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  • $\begingroup$ Thanks. Is there a proof for this? $\endgroup$
    – TheSimpliFire
    Dec 14, 2017 at 19:40
  • $\begingroup$ @TheSimpliFire: This was found using an Integer relations algorithm, so no derivation yet from first principles $\endgroup$ Dec 15, 2017 at 2:57
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Just sharing one of mine. [Sorry: It is only for $e$ and $\pi$]

$$(1+9^{-4^{6\times 7}})^{3^{2^{85}}}\approx e$$

The left-hand side is equal to first 18,457,734,525,36,901,453,873,570 (18 trillion trillion) digits of $e$ after decimal.

See that the left-hand side consists of all the numbers from $1$ to $9$.

A similar thing can be formed for $\pi$ also, $$\pi\approx 2^{5^{.4}}-.6-(\frac{.3^9}{7})^{.8^{.1}}$$

Correct up to 10 digits of $\pi$.

Video_Link

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