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.

  • 4
    $\begingroup$ There are no short answers to your questions. Recommend the inexpensive and recent cambridge.org/us/academic/subjects/mathematics/number-theory/… $\endgroup$
    – Will Jagy
    Mar 3, 2015 at 19:07
  • 3
    $\begingroup$ No such representation is known, it doesn't mean no such representation exists. $\endgroup$ Mar 3, 2015 at 19:15
  • $\begingroup$ The question deals with the structure of continued fractions. $\endgroup$
    – kryomaxim
    Mar 3, 2015 at 19:15
  • $\begingroup$ I have asked a similar question before, which you might find related. See here. $\endgroup$ Mar 3, 2015 at 19:18

1 Answer 1


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! :)

  • $\begingroup$ and are there some mathematical constants with irregular continued fraction behavior? how regularity/irregularity can be explained in this case? $\endgroup$
    – kryomaxim
    Mar 3, 2015 at 19:23
  • $\begingroup$ I've added more detail to my answer. Hopefully it helps! $\endgroup$
    – Sam OT
    Mar 3, 2015 at 20:54
  • $\begingroup$ Does this en.wikipedia.org/wiki/Benford%27s_law apply to why there are so many $1$'s in the continued fraction? $\endgroup$
    – user141592
    Mar 3, 2015 at 23:24
  • $\begingroup$ Ah, no, that's actually something completely different. It's do to with decimal expansions. Consider this. Pick a number randomly in from 0 to 199 (inclusive). We observe that all the number at least 100 start with a 1, as do all the numbers 10 to 19, as well as the number 1. This gives 111 numbers starting with a 1, out of a total 200 numbers. So if you pick an integer uniformly in $\{0, ..., 199\}$, then there is a probability $111/200 > 1/2$ of it's starting with a 1. $\endgroup$
    – Sam OT
    Mar 4, 2015 at 9:15
  • $\begingroup$ This fact is applied to detecting fraud in accounting. It's possible to see what a 'standard' number for a form is (say, £3.49), and then work out how likely the first digit is (obviously, my above case is contrived as a stark example). If people have randomly filled in the form, then this can be predicted. $\endgroup$
    – Sam OT
    Mar 4, 2015 at 9:16

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