Why $e^{\pi}-\pi \approx 20$, and $e^{2\pi}-24 \approx 2^9$?

This was inspired by this post. Let $q = e^{2\pi\,i\tau}$. Then,

$$\alpha(\tau) = \left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24} = \frac{1}{q} - 24 + 276q - 2048q^2 + 11202q^3 - 49152q^4+ \cdots\tag1$$

where $\eta(\tau)$ is the Dedekind eta function.

For example, let $\tau =\sqrt{-1}$ so $\alpha(\tau) = 2^9 =512 $ and $(1)$ "explains" why,

$$512 \approx e^{2\pi\sqrt1}-24$$ $$512(1+\sqrt2)^3 \approx e^{2\pi\sqrt2}-24$$ and so on.

Q: Let $q = e^{-\pi}$. Can we find a relation, $$\pi = \frac{1}{q} - 20 +c_1 q + c_2 q^2 +c_3 q^3 +\cdots\tag2$$ where the $c_i$ are well-defined integers or rationals such that $(2)$ "explains" why $\pi \approx e^{\pi}-20$?

Update: For example, we have the rather curious,

$$\beta_1 := \frac{1}{q} - 20 +\tfrac{1}{48}q - \tfrac{1}{300}q^3 -\tfrac{1}{972}q^5 +\tfrac{1}{2160}q^7+\tfrac{1}{\color{brown}{2841}}q^9-\tfrac{1}{\color{brown}{2369}}q^{11}-\cdots\tag3$$

$$\beta_2 := \frac{1}{q} - 20 +\tfrac{1}{48}q - \tfrac{1}{300}q^3 -\tfrac{1}{972}q^5 +\tfrac{1}{2160}q^7+\tfrac{1}{\color{brown}{2842}}q^9-\tfrac{1}{\color{brown}{2810}}q^{11}-\cdots\tag4$$

With $q = e^{-\pi}$,

$$\pi \approx \beta_1 = e^\pi -20 +\tfrac{1}{48}q-\dots\; (\text{differ:}\; {-4}\times 10^{-22})\\ \pi \approx \beta_2 = e^\pi -20 +\tfrac{1}{48}q-\dots\; (\text{differ:}\; {-3}\times 10^{-22})$$

However, there seems to be an indefinite number of formulas, where the choice of a coefficient (say, $2841$ or $2842$) determines an ever-branching tree of formulas. But there might be a subset where the coefficients have a nice closed-form.

This does not follow your proposal exactly but it is built on series with rational terms only.

From expansions

$$ e^\pi=\sum_{k=0}^\infty\frac{3\left(e^{3\pi}-\left(-1\right)^ke^{-3\pi}\right)\Gamma\left(\frac{k}{2}+3i\right)\Gamma\left(\frac{k}{2}-3i\right)}{2 \pi k!}=\sum_{k=0}^\infty a(k) $$ http://oeis.org/A166748

and

$ \pi=3+\sum_{k=1}^\infty\frac{1}{4·16^k}\left(-\frac{40}{8k+1}+\frac{56}{8k+2}+\frac{28}{8k+3}+\frac{48}{8k+4}+\frac{10}{8k+5}+\frac{10}{8k+6}-\frac{7}{8k+7}\right)=3+\sum_{k=1}^\infty b(k) $

https://oeis.org/wiki/User:Jaume_Oliver_Lafont/Constants#Series_involving_convergents_to_Pi

the following representation is obtained

$$e^\pi-\pi-20 = \frac{1201757159}{10580215726080}+\sum_{k=16}^\infty a(k) -\sum_{k=2}^\infty b(k) \approx -0.00090002 \approx -\left(\frac{3}{100}\right)^2$$

Cancellation comes from the first three decimal digits: $$ \sum_{k=0}^{15} a(k) = \frac{991388824265291953}{42849873690624000}\approx 23.136(2) $$ $$ b(1) = \frac{49087}{360360} \approx .136(1) $$ [EDIT] Three digit cancellation may also be obtained by taking 14 terms from the series for $e^\pi$ and 3 terms from the simpler $$\pi-3=\sum_{k=1}^{\infty}\frac{3}{(1+k)(1+2k)(1+4k)}$$

(Lehmer, http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf, pag 139-140)

[EDIT] Another expression with a "wrong digit" that leads to higher precision when corrected is given by $$ e - \gamma-log\left(\frac{17}{2}\right) \approx 0.0010000000612416$$ $$ e \approx \gamma+log\left(\frac{17}{2}\right) +\frac{1}{10^3} +6.12416·10^{-11}$$

Why is $e$ close to $H_8$, closer to $H_8\left(1+\frac{1}{80^2}\right)$ and even closer to $\gamma+log\left(\frac{17}{2}\right) +\frac{1}{10^3}$?

[EDIT] $$\sum_{k=0}^\infty \frac{2400}{(4 k+9) (4 k+15)} = 100 \pi-\frac{58480}{231} \approx 60.99909$$