Why Is $\ln(23) + 1/(\color{red}{163} + 1/(1+1/\color{red}{41}))\approx\pi$

Question

I know from reading that the Heegner number 163 yields the prime generating or Euler Lucky Number 41. Now apparently $\ln23<\pi$ and this can be shown without calculators. I noticed that $$ \pi-\ln23= \cfrac{1}{\color{red}{163} + \cfrac{1}{1 + \cfrac{1}{\color{red}{41} + \cfrac{1}{2 + \cdots}}}} $$

Question: Are there any "good" mathematical reasons why the largest Heegner and largest Euler Lucky number occur within the first three (-four ?) terms of the expansion? Or is it purely coincidence ?

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Indeed the finite c.f. $$\cfrac{1}{\color{red}{163} + \cfrac{1}{1 + \cfrac{1}{\color{red}{41}}}}:=\frac{42}{6887}\approx 0.00609843\ldots.$$ In turn this yields the crude approximation $$\ln 23+\frac{42}{6887}\approx\pi;$$ which I believe gives the first 8 digits of $\pi$ correctly.

Answer 1

I believe there is no good reason as to why the largest Heegner number $163$ or Euler lucky number $41$ appears in the simple continued fraction for $\pi - \log(23)$; perhaps it is simply a coincidence. I do not think there is any real connection between Euler lucky numbers and Heegner numbers :)

For example, consider the simple continued fraction representation of $\pi$ is $[\color{green}{3};\color{red}{7},\color{blue}{15},1,292,1,1,1,2,1,3,1,...]$, no pattern has ever been found in this representation. Is there any special reason for why $\color{green}{p_2}$ (the second prime) and $\color{red}{p_4}$ (the fourth prime) and $\color{blue}{B_4}$ (the fifth bell number) occur within the first three terms of this continued fraction? Probably not.