Mathematical coincidences concerning the numbers $\pi$, $e$ and $163$ 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?
 A: 
The diagram shows $n$ circles and $n-1$ equilateral triangles.
We have the following elegant result:

There exists a constant $C$ such that, if the total area is $2^n C$ then
(product of the areas of all the enclosed regions) $=2C$.

Numerical investigation shows that $C=2.71858...$
So $C=e$, right?
Wrong!

$C=\dfrac{3\left(2^{1/6}\right)}{\sqrt{\dfrac{15\sqrt3}{\pi}-\dfrac{27}{\pi^2}-4}}\approx 2.718586969\approx 1.000112e$.

A mathematical imposter.
A: $$\frac{163}{e}+\frac{e}{163}+\frac{\pi }{163}\approx 60$$
It's mine :)
Hope you like it
EDIT
$163 (\pi -e)\approx 69$
A: The number $e^{\pi\sqrt{163}}$ is very close to the integer $262537412640768744$ the difference is about $7.5\times 10^{-13}$
A: 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.
A: 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 $\pi$.
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