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As we learn in high school $\pi$ has infinitely many digits, if we consider any string of digits, it will of course repeat infinitely many times throughout $\pi$.

I would like to consider the first $n$ digits of $\pi$ and ask when will they repeat themselves for the first time. One could make the question of the following kind, for each $n$ (the number of first digits of $\pi$ we are considering) there is an $m$ such that the string of the first $n$ digits reappears for the first time (beginning) at the $m$'th position. Assuming there is no redundancy we have the bound that, at least, $n< m$.

I am interested if there is any well know theorem that can produce an $m$ for any $n$. Since I suppose no such theorem exists, is there at least a bound on $m$ that is known? Something like $m \leq n!+n^2$ (or some other algebraic expression in $n$).

Furthermore, can anything be said about the nature of the function $m(n)$, it is for sure monotony increasing, but is it rising exponentially, polynomially, with some other more exotic function, if so I would really be interested to know what that function is.

Also, would this function be the same for the hexadecimal expression of $\pi$ for example (or any other basis)?

I have $\epsilon$ knowledge of number theory (I mostly do geometry and analysis) so feel free to redirect me to some standard texts if this question seems trivial.

A second more philosophical question, since $\pi$ infinitely many digits we can find a string of numbers somewhere in $\pi$ with any characterisation we like, thus it is often said that somewhere in $\pi$ there is the whole of Shakespeare written in binary. With this said, what prevents us from looking for a string of digits with the characteristic that if it starts at the point $x$ the sting is of length $x$ and is equal to the first $x$ digits of $\pi$. This seems a bit contradictory since it would mean that there is an $x$ after which the digits that occurred up to that point repeat once more before they continue their random descent into the depths of $\pi$.

Anyhow this simply confuses me, and if somebody could elucidate what is the contradiction in the reasoning in this I also think this would be helpful for me.

El Ruño
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    Very little is known for certain about the digits of $\pi$. It's an irrational number so we know they are not periodic, even allowing for an initial stub block. For all we know, it's possible that, beyond a certain point, only two digits appear. Of course, this would surprise everyone, but, really, very little is known for sure. – lulu Dec 29 '22 at 02:03
  • Cf. this question and this MathOverflow question – J. W. Tanner Dec 29 '22 at 02:03
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    Your claim that "since $\pi$ has infinitely many digits we can find a string of numbers somewhere in $\pi$ with any characterization we like" is simply false. $\frac 13=.33333\cdots$ has infinitely many digits, for instance. So does $.101001000100001\cdots$ and the latter is irrational, though it only contains $0$ and $1$. Again, it is conjectured that $\pi$ is normal which would imply what you said (roughly speaking), but nobody knows if the conjecture is true or not. – lulu Dec 29 '22 at 02:05
  • @lulu, correct, and on the other way, "almost all" real numbers contains all possible finite digits sequences. Sure "almost all" depends slightly on the measure used. – z100 Dec 29 '22 at 02:11
  • There is some tabulation at https://oeis.org/A159345 "$a(n)$ is the number of digits in the decimal expansion of $\pi$ needed to contain a repeated $n$-digit substring." – Gerry Myerson Dec 29 '22 at 03:16
  • had a look at that link, El R? – Gerry Myerson Dec 30 '22 at 04:09
  • This has been asked many times here (perhaps not exactly with the same words). In fact, we know almost nothing about the long term frequency of the digits in $\pi$. We cannot even rule out that there eventually are only two digits , say the tail is $0110001000\cdots$ , only zeros and ones following. – Peter Dec 30 '22 at 12:06
  • Yes, most mathematicians "believe" that $\pi$ is even normal , but the only evidence is that within the incredibly many digits of $\pi$ that have been calculated , nothing else happened. – Peter Dec 30 '22 at 12:07
  • That $\pi$ has no period (since it is known to be irrational) does not mean that long strings cannot repeat (In fact , if $\pi$ is actually normal, every finite string appears even infinite many times!) . An irrational number need not contain every finite string at all , as pointed out , and that the complete work of Shakespeare appears somewhere in $\pi$ (after a suitable modification , of course) , is purely speculative, but nevertheless widely believed. – Peter Dec 30 '22 at 12:11
  • @z100 But $\pi$ is not a "random" number, it is a concrete real number, so the argument that almost all real numbers are normal cannot be used. In fact, this way , we could claim that every number for which we cannot disprove its normality , is normal. Of course this is not true. – Peter Dec 30 '22 at 12:16

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When you write "it will of course repeat infinitely many times throughout $\pi$", this essentially refers to a property called being a normal number. It is conjectured that $\pi$ is normal, but we are far from having a proof for this. As far as I know, it has not even been proven that the digit $7$, say, appears infinitely often in the decimal expansion.

Then again, from what numerical results suggest, the (decimal or other) digits of $\pi$ appear pretty much like random. With that in mind, any $n$-digit string would be expected to appear with probability $10^{-n}$ at any given place, so it is expected to occur about once per $10^n$ places, so with probability $\approx1-\frac1e$ within $10^n$ candidate places, or if you want to go on the safe side, with probability $\approx1-\frac1{e^k}$ within $k\cdot 10^n$ candidate places. But nothing is for sure (and would not even be for sure with truly random digits).

Hagen von Eitzen
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  • To add some detail : A truly random infinite sequence of digits would be normal with probability $1$ , which does however not mean that a not-normal number cannot be chosen. – Peter Dec 30 '22 at 12:19