# 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?

• $$\Large\frac{1}{30^{\pi^e}}\approx h \tag{Planck's constant}$$ – Mr Pie Oct 8 '18 at 0:40

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

It's mine :)

Hope you like it

EDIT

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

• Nice! Quite a high degree of accuracy as well – TheSimpliFire Dec 2 '17 at 13:52

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

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

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.

• Thanks. Is there a proof for this? – TheSimpliFire Dec 14 '17 at 19:40
• @TheSimpliFire: This was found using an Integer relations algorithm, so no derivation yet from first principles – Tito Piezas III Dec 15 '17 at 2:57

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 $$\pi$$.