Are the first $n$ digits of $\pi$ equal to the second $n$ digits for some $n\ge1$?

Question

Are the first $n$ digits of $\pi$ equal to the second $n$ digits for some $n\ge1$?

$$\pi \stackrel{?}{=}\underbrace{3.1415926\ldots}_{\text{the first }n\text{ digits}}~\underbrace{31415926\ldots}_{\begin{array}{c}\text{the same }n\text{ digits}\\\text{in the same order}\end{array}}~\underbrace{\ldots\ldots\ldots}_{\text{more digits}}$$

If so, is the smallest such $n$ known?

Answer 1

It is very unlikely. If we take the digits of $\pi$ to be "random", the chance of a repeat after $n$ digits is $10^{-n}$. We can exclude however many digits we know do not repeat. Say we know it doesn't repeat by one million digits. Then the chance we have a repeat is less than $$\sum_{i=10^6}^\infty 10^{-i}=\frac {10^{-10^6}}{1-.1}=\frac 1{9\cdot 10^{10^6-1}} $$ which is extremely small.

Answer 2

Take a look at Khinchin's constant. The continued fraction coefficients of most real numbers have a have a finite geometric mean that equals Khinchin's constant.

If Pi or one of these other real numbers had a big repeat as described, it would happen after 10 trillion digits (since we know Pi that far), and would introduce a truly huge continued fraction coefficient, enough to skew away from Khinchin. So far, there are only a handful of transcendental numbers that are not Khinchin numbers.

It would be better to look for Khinchin violation elsewhere, since numbers like $log(2)+log(7)+1/e$ can be checked in a split second. You could check sextillions of real numbers with the same amount of effort that it would take to extend Pi another 100 trillion digits.