Why is $e^\pi - \pi$ so close to $20$? $e^\pi-\pi\approx 19.99909998$
Why is this so close to $20$?
 A: If you think that's good agreement, try Ramanujan's constant $e^{\pi \sqrt{163}}.$ There is a deep explanation for the proximity in that case though.
A: You may find :

This curious near-identity was apparently noticed almost simultaneously around 1988 by N. J. A. Sloane, J. H. Conway, and S. Plouffe, but no satisfying explanation as to "why" $e^\pi-\pi \simeq 20$ is true has yet been discovered.

after some googling.
A: Partial Answer
If $e^\pi-\pi\approx20$ then $e^x-x-20$ must have a root around $x=\pi$. Hence the function defined by $z=e^{f(z)}-f(z)-20$ should pass $f(0)\approx\pi$. So with the Lagrange Inversion Theorem, we can find a series representation for $f(z)$. If there's any special reason why $f(0)\approx\pi$, it would explain why $e^\pi-\pi\approx20$. We can find $f(z)$ by inverting $e^x-x-20$ at $x=3$, as follows. *
$$f(z)=3+\frac{[z-(e^3-23)]}{e^3-1}-\frac{e^3[z-(e^3-23)]^2}{(e^3-1)^3\cdot 2!}+\frac{(e^3+2e^6)[z-(e^3-23)]^3}{(e^3-1)^5\cdot 3!}\ldots$$
If we substitute $z=0$ and simplify the series, our question is equivalent to asking why: $$3+\sum_{n=1}^{+\infty}\frac{(23-e^3)^np_n(e^3)}{(e^3-1)^{2n-1}(2n-1)!}\approx\pi$$
Where $p_n(e^3)$ are polynomials in $e^3$. Indeed we see that the left hand side is approximately $3.1416$. This is expected since solving $e^x-x-20=0$ gives us $x\approx 3.141\,633\,302\,801\,037$. 
Figure: $\color{red}{y=e^x-x-20}$ in red, $\color{blue}{y=x}$ in blue, $\color{purple}{y=f(x)}$ in purple and the partial sum for $\color{green}{y=f(x)}$ up to $n=3$ in green. The purple curve represents the reflection of the red curve over the line $y=x$. The purple curve crosses the $y$ axis around $(0,\pi)$, implying $e^\pi-\pi\approx20$. Geogebra link: https://www.desmos.com/calculator/nhbpehuuds


Potential Next Steps
The coefficients of $e^{3k}$ for each $p_n(e^3)$ are shown in the table below. For example, when $n=3$, $p_3(e^3)=e^3+2e^6$.
$$\begin{array}{|c|c|c|c|c|}
\hline
n&\text{Coefficient of }e^3&e^6&e^9\\
\hline
1&0&0&0\\
\hline
2&-1&0&0\\
\hline
3&1&2&0\\
\hline
4&-1&-8&-6\\
\hline
\end{array}$$
If you think you can see any particular pattern in the table, or would like to calculate more entries, it could poetntially lead to an explanation.
We could also attempt to compare the terms in the series to a series for $\pi$. Namely $f(0)=3+0.153-0.0123+0.00135-0.000169\ldots$. We could narrow our search to the series for $\pi$ with the same rate of convergence. I think this series has first order convergence.

* We can't invert $e^x-x-20$ around $x=0$ because the derivative is $0$ here. I tried at $x=1$ but couldn't get it to work. 
A: The proximity to $\pi$ is purely coincidental, as is often the case. Take for instance the root of the equation $x^4+x^5=e^6$, which is $x=\pi+0.000 000 029$. Many other far more accurate yet equally coincidental approximations are shown on this page: http://www.contestcen.com/pi.htm.
For each of these coincidences, one could be surprised and wonder: "Why is it so close? ". This is a naive question because a particular relationship is especially selected to be close to an exact value and chosen among an infinity of relationships more or less approximated.
One can find very easily as many coincidences as we want, with $\pi$ or/and with other usual constants, thanks to computer automatic search. The general principle of the method is described with examples in the paper "Mathématiques expérimentales" published on Scribd (in French): http://www.scribd.com/JJacquelin/documents.
A: The right question is why 
$$e^\pi-\pi\approx 20-\frac{1}{1111+\frac{1}{11+\frac{1}{\sqrt{2}}}} $$
A: As much as I wish there was some deep connection here, all (lack of) evidence points to the fact that this is nothing but mere, brilliant mathematical coincidence.
