13
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Just head breaking question! $\endgroup$ Jun 28, 2013 at 9:45
  • 2
    $\begingroup$ I don't think this is known or even conjectured. $\endgroup$ Jun 28, 2013 at 9:45
  • 5
    $\begingroup$ Using a crude probabilistic heuristic: if this doesn't happen for $n \le m$, then the chance it happens at all is about $1/9\cdot 10^m$. So if it doesn't happen early it's very unlikely to happen at all. $\endgroup$ Jun 28, 2013 at 9:52

2 Answers 2

9
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Thanks. However, you seem to treat the event of a repeat at $i$ to be independent of, e.g., the event of a repeat at $i+1$. Is that justifiable? $\endgroup$
    – Řídící
    Jun 28, 2013 at 20:47
  • $\begingroup$ They are not strictly independent, but very close. Given that there is no repeat starting with digit $1,000,000$ it is very slightly more likely that there is a repeat starting at digit $1,000,001$, but it is a tiny effect. Even if it made it ten times more likely, that doesn't matter here at all. $\endgroup$ Jun 28, 2013 at 20:55
  • $\begingroup$ OK, I'll think a bit about that, because I don't immediately see that. Also, you might actually make it ten times more likely by correcting the $9$ to $.9$. :) $\endgroup$
    – Řídící
    Jun 28, 2013 at 20:58
  • $\begingroup$ I fixed that in the exponent. Thanks $\endgroup$ Jun 28, 2013 at 21:53
1
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ How could any finite repetition effect the value of the Kinchin liimt? $\endgroup$ Jun 28, 2013 at 15:15
  • $\begingroup$ I don't see why a single repetition (rather than two or more) would cause a huge CF coefficient. If $q=10^n - 1$, then we'd have a rational approximation $p/q$ with accuracy about $1/10q^2$, only slightly better than a typical convergent. The coincidence seems to me not to be in the goodness of fit, but in the denominator being a repunit. $\endgroup$
    – Erick Wong
    Jun 28, 2013 at 16:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .