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I asked this question just for curiosity! I guess that it's an unsolved problem, but I can't find any reference that mentions that.
You can see http://oeis.org/A001203 and How to find continued fraction of pi

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A_Sh
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  • Yeah, probably unknown.Highly likely, though. – Thomas Andrews Jan 14 '17 at 21:14
  • @ThomasAndrews Thanks, but have you any reference? – A_Sh Jan 14 '17 at 21:16
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    To what, "probably unknown?" If I had an answer, I'd post an answer. Don't tag people in comments to re-ask your question. – Thomas Andrews Jan 14 '17 at 21:17
  • @ThomasAndrews I thought that you had some evidence about that, or you read some related texts about that. – A_Sh Jan 14 '17 at 21:21
  • At least we know that there must be infinite many non-ones :) – Peter Jan 16 '17 at 13:34
  • Numerical analysis seems to indicate that the ratio $\frac{\ number\ of \ ones}{\ number\ of\ entries}$ is approximately $0.414$ , which is about $\sqrt{2}-1$. Details later, I am currently calculating the continued fraction of $\pi$ with $2\cdot 10^6$ digits precision. – Peter Jan 17 '17 at 10:02

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As a rule, problems of this kind are extremely difficult. We know for any reasonable method of expressing real numbers as sequences of integers, for almost every real number we'll see a given digit (or more broadly any finite string of digits) with a particular frequency. For instance, if I'm looking at a number's decimal representation, then for almost every real number we'll see the digit string $17357$ with frequency $10^{-5}$. A similar rule holds for continued fractions, i.e. there exists a (in this case positive) constant $C > 0$ such that for almost every real number, the digit $1$ will occur with frequency $C$.

However, although we know almost every real number has this property, it's generally fairly difficult to construct examples of such numbers, and (as of this time) pretty much impossible to determine if a given number has this property. We can give very trivial answers to the negative (e.g. rationals are easy to check), but when it comes to determining whether a number encountered "in real life", say $\sqrt[7]{2}, e, \pi$ has this property, we really don't have any kind of machinery to deal with this. We have conjectures of what we think will probably happen, but no techniques have been discovered to test them. It's just an area of math where we have no clue how to approach the problem.

EDIT: I would bet money this problem is still open.

AJY
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  • +1 Thanks for the explanation, but it's not really an answer. – A_Sh Jan 14 '17 at 21:38
  • Edited to be clearer as to my response. – AJY Jan 14 '17 at 21:39
  • How much?! :))) – A_Sh Jan 14 '17 at 21:42
  • That being said, my guess is it's generally accepted that yes, $\pi$ probably has infinitely many $1$s in its continued fraction. – AJY Jan 15 '17 at 00:57
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    @Ab_Sh I also have great doubts that this problem can be actually solved. But I am more careful with my prognoses than AJY because I have had some surprises about discovery's in mathematics, I would have assumed to be impossible. Nevertheless, it should be even extremely difficult to decide whether the continued fraction of $\pi$ eventually contains only of ones and twos. – Peter Jan 16 '17 at 13:28
  • If I would be forced to bet, I would bet that the problem is open as well. It seems much more likely to me than the opposite. – Peter Jan 16 '17 at 13:31
  • @Peter Thanks for your response. – A_Sh Jan 16 '17 at 15:31
  • I'm less familiar with the continued fraction question, but I know that for the analogous problem with $b$-ary expansions, the problem is unsolved. – AJY Jan 17 '17 at 14:33
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I calculated the first $3.8\cdot 10^6$ entries of the continued fraction of $\pi$ (This requires about $2\cdot 10^6$ digits precision) and noticed the following :

If $d(n)$ denotes the ratio of the number of ones to the total number of entries, if we consider the first $n$ entries (The initial "$3$" counts as an entry as well), the following holds :

For $68692<n\le 3.8\cdot 10^6$ , we have $0.414\le d(n)\le 0.416$

Of course, there is no guarantee that this goes on forever, but it seems that roughly $d\approx 0.415$ for sufficient large $n$.

This is a good evidence (of course no proof) that infinite many ones appear in the continued fraction of $\pi$

Peter
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  • Sidenotes : The longest sequence of consecutive ones appearing is $18$, the longest sequence of consecutive non-ones appearing is $23$ – Peter Jan 17 '17 at 12:38
  • And what is the biggest number?! – A_Sh Jan 19 '17 at 13:10
  • The $453\ 294$ th entry is $12\ 996\ 958$ – Peter Jan 19 '17 at 18:18
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    It is known that for almost all real numbers, the positive integer $a$ occurs in the continued fraction with relative frequency$$\log_2\Bigl(\frac{a+1}a!\Big/\frac{a+2}{a+1}\Bigr)\ .$$For $a=1$ this gives $0.415$, which is consistent with your calculation. Unfortunately, there is no guarantee that $\pi$ is not one of the infinitely rare exceptions to this rule :) – David Nov 16 '17 at 05:36