TITLE: Does π start with two identical decimal sequences?
QUESTION [11 upvotes]: Let d$(x,y)$ be the sequence of decimals of π from the x:th one to the y:th one.
My question: is there a number $n$ such that d$(1,n)$ = d$(n+1, 2n)$? I.e., does π start with two identical decimal sequences, given a high enough $n$?
For example, suppose π = 3.1415914159243234... then $n$ = 5 would satisfy the above condition.

REPLY [6 votes]: It's worth noting that this kind of problem is not necessarily totally intractable, even if the answer is negative. Given the current state of mathematical knowledge, I believe we couldn't know if $\pi$ starts with $2$ identical decimal sequences, but we certainly could know if it starts with $8$.
Why is this? Because the irrationality measure of $\pi$ is less than $8$: that is, there are only finitely many $a$ and $b$ such that
$$
\left|\pi - \frac{a}{b}\right| < \frac{1}{b^8}\tag{1}
$$
However, if $\pi$'s decimal expansion starts with eight identical sequences, say $\pi=3.xxxxxxxx\dots$, then
$$
\left|\pi - \left(3 + \frac{x}{10^{|x|}-1}\right)\right|<\frac{1}{10^{8|x|}} < \frac{1}{(10^{|x|}-1)^8}
$$
because they both share the first same $8|x|$ decimal digits. So this could only happen if $\left(3 + \frac{x}{10^{|x|}-1}\right)$ is one of those finitely many $\frac{a}{b}$ that satisfy $(1)$ — and presumably you could go in the literature to see how big those are allowed to be, and check each one individually. Almost certainly, we know enough digits of $\pi$ to rule this possibility out completely.

In principle, this kind of approach can show that an irrational number does not begin with $3$ identical decimal sequences, as long as you can prove that its irrationality measure is strictly less than $3$. For example, all algebraic numbers as well as $e$ have irrationality measure $2$, so any algebraic number (or $e$) can only start with $3$ identical decimal sequences in finitely many ways, and you could in principle check all of them. If someone managed to significantly improve the known upper bound on the irrationality measure of $\pi$, it might become possible to know whether $\pi$ starts with $3$ identical decimal sequences in this way.
However, this approach isn't quite good enough for your original question. In fact any irrational number $\tau$ has infinitely many rational approximations $\frac{a}{b}$ such that
$$
\left|\tau-\frac{a}{b}\right|<\frac{1}{b^2}
$$
and, if there was some such good approximation with $b$ of the form $10^k-1$, this could potentially lead to $\tau$ starting with $2$ identical decimal sequences. So a full answer to your question would require a more involved argument, about which kinds of fractions $\frac{a}{b}$ were allowed to be close rational approximations of $\pi$.