Is it possible to separate Pi from Eulers constant with rational constants?

Question

1. $\pi^2 \cdot e^2 \sim \frac{10000}{137}$, is quite accurate, although I have no idea what this "fine structure constant" $\frac{1}{137}$ from physics really is.
2. For any prime numbers $p_1<p_2<p_3<p_4$, $(p_1 * p_2 * p_3 * p_4)^9$ has always exactly $10000$ divisors and the powers of $2$ do relate to the powers of $10$ in some way dependent of their divisors, which also seems to relate in some form to $\pi$.

So my question is, is it possible to separate $\pi$, from $e$ in the first equation without using recursions or irrational numbers and how does $\pi$ relate to the $2^{10} = 1024, 10^3 = 1000$ coincidence?

Answer 1

This does not answer the question but relates the approximation in the question to another one.

Since $137\pi^2e^2 \approx 9991$, the accuracy of the following approximation is better.

$$\pi^2e^2 \approx \frac{100^2-3^2}{137}$$

Now if we bother to write $$\frac{100^2-3^2}{\pi^2e^2} \approx 136.999899991$$ this is closer to an integer than the nice

$$e^\pi-\pi \approx 19.999099979$$

And the ratio of the distance of both expressions from an integer is close to yet another integer.

$$\frac{e^\pi-\pi-20}{\frac{100^2-3^2}{\pi^2e^2}-137} \approx 9$$

I believe that the following expression is also worth mentioning. $$\frac{1}{\alpha}\approx137.035999 \approx \frac{34259}{250}=5^3+12+\frac{3^2}{2·5^3}=5^3+\frac{5^2}{2}-\frac{1}{2}+5^{-2}-\frac{5^{-3}}{2}$$

In other words, the polynomial $(2n+1)^3+3n^2$ takes the value $137$ for $n=2$. Equivalently, $(4n+1)^3+3(2n)^2$ gives $137$ for $n=1$. Here is an illustration of $137$ as a three-dimensional shape.

Only the faces seen have additional cubes, shown in red. Those extra cubes could be halved and shared among the seen and unseen faces for another symmetry.

Regarding the $2^{10} \approx 10^3$, since $10=2·5$ we can simplify it down a bit to the following.

$$2^{10} \approx 10^3 = 2^35^3$$ $$2^7 \approx 5^3$$

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Update

A simpler fourth power $(9^4)$ yields a better approximation: $$\pi \approx \frac{27}{e\sqrt{10}} \approx 3.1410097$$

compared to $$\pi \approx \frac{100}{e\sqrt{137}} \approx 3.143006$$

However, from your approximation the following almost-integer is obtained.

$\frac{10^4}{137\pi^2e^2} \approx 1.0009000 $

This happens to match the gap in $e^\pi-\pi$, so combining both gives

$$e^\pi-\pi+\frac{10^4}{137\pi^2e^2} \approx 21.000000058$$