Mathematical coincidences concerning the numbers $\pi$, $e$ and $163$

Question

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?

Answer 1

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

It's mine :)

Hope you like it

EDIT

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

Answer 2

The number $e^{\pi\sqrt{163}}$ is very close to the integer $262537412640768744$ the difference is about $7.5\times 10^{-13}$

Answer 3

This is not an approximation but an equality,

$$\big(12\sqrt2\big)^2 \sum_{k=1}^\infty\frac1{k^6\,\binom{2k}{k}}+\big(12\sqrt3\big)^2\sum_{k=1}^\infty\frac1{k^2\,\binom{2k}{k}}\sum_{n=1}^{k-1}\frac1{n^4}=\color{blue}{163}\,\zeta(6)$$

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.

Answer 4

The diagram shows $n$ circles and $n-1$ equilateral triangles (and $6n-5$ cells).

We have the following elegant result:

There exists a constant $C$ such that, if the total area is $2^nC$ then
{product of the areas of the cells} $=2C$.

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.

Calculation

$r_k=\{\text{radius of $k$th circle from outside}\}=\sqrt{\dfrac{2^nC}{\pi}}\left(\frac{1}{2}\right)^{k-1}$

$A_k=\{\text{area of $k$th circle from outside}\}=\pi{r_k}^2$

$B_k=\{\text{area of $k$th triangle from outside}\}=\frac{3\sqrt3}{4}{r_k}^2$

$\text{product areas of cells}=A_n\prod\limits_{k=1}^{n-1}\left(\dfrac{A_k-B_k}{3}\right)^3\left(\dfrac{B_k-A_{k+1}}{3}\right)^3=2C$

This leads to $C=\dfrac{3\left(2^{1/6}\right)}{\sqrt{\dfrac{15\sqrt3}{\pi}-\dfrac{27}{\pi^2}-4}}$