Why are there no continued fraction representation for $\pi$ obeying mathematical rules? There are several irrational numbers that can be represented with continued fraction such that a mathematical rule arises in this continued fraction. For example, the Euler number $e$ can be represented as follows:
$$e=[2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,\ldots]= 2+ \cfrac{1}{1+ \cfrac{1}{2 + \cfrac 1 \ddots}}$$
It's very fascinating that in the case where the number sequence after the $,$ is chaotic, it does exists a well-ordered scheme when considering continued fractions. Every three steps in the continued fraction, the parameter increases by $2$ and elsewhere the parameter is $1$. Why it is so? Another popular mathematical constant $\pi$ however, does not have a regular structure in the continued fraction:
$$\pi = [3,7,15,1,292,1,1,1,\ldots].$$
But I can see many $1$-s in this series of parameters. May it be that $\pi$ has an irregular continued fraction, but not a very intensive irregularity? Why these structural differences in continued fractions arise between $\pi$ and $e$?
And what is with other mathematical constant like the Euler-Mascheroni-constant $\Gamma$; is there also a regularity in the continued fractions?
I would be happy for any comments and answers.
 A: If you allow generalised continued fractions, then you can actually. This one is courtesy of Lord Brouncker is 1654:
$$
\frac{1}{4}\pi=\cfrac{1}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\dotsb}}}}
$$
This has very slow convergence. Apparently people didn't believe it initially. It takes nearly 50 terms for five decimal places of accuracy and nearly 120 for six! (It convergence in an oscillatory manner: above, below, above, below, etc.)

If you have a decimal expansion $x=0.x_1x_2x_3...$, then, given that $x$ was chosen uniformly randomly in $[0,1]$, we can see that each of the $x_i$ is uniform on $\{0, 1, ..., 9 \}$. This is not the case, however, with continued fractions. In essence, we can see this best with an example. If you simply start to calculate, say $101/37$, then you'll see that to get a large number in the fraction (in the standard type as you have in your question) you need to have the number that you're inverting close to $0$. Let me explain:
$${101 \over 37} = 2 + {27 \over 37} \ \ \text{giving a 2},$$
$${37 \over 27}  = 1 + {10 \over 27} \ \ \text{giving a 1},$$
and so we continue; however, suppose that at some point we have
$${1003 \over 1000} = 1 + {3 \over 1000}  \ \ \text{giving a 1},$$
$${1000 \over 3}    = 333 + {1 \over 3}   \ \ \text{giving a 333}.$$
We can actually show, with quite a bit of work, what the distribution is: it has density
$$f(x) = {1 \over {\log 2}}{1 \over {1+x}}.$$
(You can either try to prove this yourself or I'm sure you can look it up - in fact, there's probably a related SE question.)

Hopefully this helps! :)
