A problem concerning repetitions in decimal expansions of $\pi$ and other irrational numbers

Question

I am a logician with a degree in philosophy, not in mathematics. I came up with this problem while studying Brouwer's intuitionism and his weak counterexamples to the Law of Excluded Middle. This being said, I am interested in knowing whether this has a solution in everyday standard mathematics, not in the intuitionistic one.

Let $X$ be a sequence of digits. I call initial segment of $X$ a sequence $Y$ such that $X = YZ$ for some other (possibly empty) sequence $Z$. I call the shortest initial segment of $X$ that validates a property $\varphi$ a sequence $Y$ such that:

1. $Y$ is an initial segment of $X$
2. $Y$ validates the property $\varphi$
3. If $Z$ is an initial segment of $Y$ that validates the property $\varphi$, then $Z=Y$

Given these definitions, we may say that the decimal expansion of a real number $r$ has always the form $AAB$: if $A$ is empty, we say that the decimal expansion of $r$ "has no repeating initial segment"; if $A$ is non-empty, the decimal expansion of $r$ "has a repeating initial segment".

Now call $I(r)$ the shortest repeating initial segment of the decimal expansion of $r$: that is, the shortest $A$ for which $r = n.AAB$ holds.

The function $f$ from real numbers to positive integers takes the number of elements in $I(r)$ plus 1.

In other words, $f(r)$ is the place in the sequence where the first repetition of the whole initial segment of the sequence occurs. If there is no such repetition, then $f(r)=1$, which is equivalent to say that the sequence starts with two empty sequences, and then the rest of the numbers.

So, for example, $f(1) = 2$, because the shortest initial segment of the decimal expansion of $1$ ($1.0000...$) repeats for the first time in position $2$: $1.[0][0]0000...$ (I am well aware that $1$ is also equal to $0.99999...$ But in this case also $f(1)=2$. If there are other numbers with more than one possible decimal expansion, I am pretty confident that there is a boring way of chosing a canonical one so that the function is well defined.)

For a more illustrative example, consider the number $0.\overline{1367}$. In this case $f(0.\overline{1367})=5$, because at position $5$ the initial segment of four digits repeats for the first time. Also, this is the shortest initial segment to repeat. (There are others, for instance: $\langle 13671367 \rangle$, etc.) And so on.

There are irrational numbers in which this function gives a definite answer. For instance, the number $0.10100100010000\dots$ has $f(0.10100100010000\dots)=3$: $0.[10][10]0100010000\dots$

Now, for my question, consider $f(\pi)$. Can we prove that $f(\pi)>1$ is contradictory? In principle, I would say no: $\pi$ being irrational means that an initial segment of some length will not repeat "endlessly", but it may well happen that it only repeats once (or twice, or... a finite number of times). As indeed is the case with $0.10100100010000\dots$ and many other irrational numbers. And if $f(\pi)>1$ is not contradictory, is there a way to know the value of $f(\pi)$?

The interesting fact, at least for me, is that, intuitively, I would say the the larger the initial segment is, the less "likely" it becomes that it will ever repeat. So, the less likely it becomes that $f(\pi)$ will come up to be greater than 1. Is there a rigorous mathematical fact that reflects this idea, or not?

(I am assuming that $f(\pi)$ is not knowable "as of today", given the available lists of initial decimals of $\pi$. If this is not the case, the question may be restated in full generality: is there a way to find the value of $f(r)$ for an arbitrary irrational number $r$, or to prove that $f(r)>1$ is contradictory?)

I am thankful for any input you may give me about this problem. (Also, thanks for the comments, that helped me improve the question overall.)

Answer 1

It is almost certainly hopeless to prove anything one way or the other about $f(\pi)$ or about $f(\alpha)$ for any "typical" irrational number $\alpha$ such as $\sqrt{2}, e$, anything that hasn't been explicitly defined to start with a repeated segment. Heuristically if we don't have $f(x) = 2$ or $3$ it's quite unlikely to be anything bigger, and we can make this precise.

The interesting fact, at least for me, is that, intuitively, I would say the the larger the initial segment is, the less "likely" it becomes that it will ever repeat. So, the less likely it becomes that $f(\pi)$ will come up to be greater than 1. Is there a rigorous mathematical fact that reflects this idea, or not?

Yes, we can consider the probability that a random real number $x \in [0, 1]$ satisfies $f(x) = n$. Choosing such a random real number is equivalent to choosing its digits independently and uniformly from $\{ 0, \dots 9 \}$ so this is an easy process to reason about, and in particular the probability $\mathbb{P}(f(x) = n + 1)$ is approximately

$$\mathbb{P}(f(x) = n + 1) \approx \frac{1}{10^n}$$

(this isn't exact because the condition that the first repetition has length $n$ rules out some patterns but the error term doesn't really matter for our purposes), because by independence, if we condition on the first $n$ digits the distribution over the next $n$ digits is unchanged, so we just compute the probability that the second batch of $n$ digits matches the first batch. So the probability of $n = 1$ is not so bad but if we don't have repetitions for $n = 1, 2, 3$ they get exponentially less likely from there. The closest things I know to such a repetition among "famous" constants are the near-misses

$$\sqrt{2} = \color{red}{1.4}\color{blue}{14}2135 \dots $$ $$e = 2.7\color{red}{1828}\color{blue}{1828}459 \dots $$

which almost repeat segments of length $2$ and $4$ (honestly, this one is quite strange and surprising) respectively but not starting from the first decimal.

$\pi$ itself has been computed to $200$ trillion digits, and I bet no one has looked but I would be shocked if those digits contained a repetition of the desired form (in any case I once memorized the first $100$ digits of $\pi$ so I can confirm there are no repetitions of length $\le 98$). Given that, any even longer repetitions are ludicrously unlikely.