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

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?

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Just head breaking question! –  Milingona Ana Jun 28 '13 at 9:45
I don't think this is known or even conjectured. –  Chris Eagle Jun 28 '13 at 9:45
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. –  Chris Eagle Jun 28 '13 at 9:52

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.

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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? –  Glen The Udderboat Jun 28 '13 at 20:47
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. –  Ross Millikan Jun 28 '13 at 20:55
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 $.9$. :) –  Glen The Udderboat Jun 28 '13 at 20:58
I fixed that in the exponent. Thanks –  Ross Millikan Jun 28 '13 at 21:53

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.

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How could any finite repetition effect the value of the Kinchin liimt? –  David Speyer Jun 28 '13 at 15:15
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. –  Erick Wong Jun 28 '13 at 16:57

Yes. This interesting and useful link shows that the sequence of digits $314$ repeats at the 2,120 decimal digit of Pi.

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Sorry, I wrongly understood the question. –  user64494 Jun 28 '13 at 9:47
.. but a nice link... –  draks ... Aug 30 '13 at 21:37