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.

• Maybe it would be clearer if you formulated the problem as: is there a positive integer $n$ such that $\pi - 3 = a (10^{-n} + 10^{-2n}) + c$ for some positive integer $a < 10^{n+1}$ and some positive real $c < 10^{-2n}$? – A.P. Mar 30 '15 at 10:00
• Does $\pi$ start with two identical decimal sequences ? - No. It doesn't. – Lucian Mar 30 '15 at 13:26
• @Lucian based on...? – Hannes Petri Mar 31 '15 at 9:38
• In other words, using the language of stringology, "Is there a square prefix of the decimal digit-string of the fractional part of $\pi$?"; i.e., "Can the decimal numeral for $\pi$ be written in the form $3.xxy$, where $x$ and $y$ are strings of digits?" I suspect that a proof one way or the other is not humanly achievable. (One might also ask for the largest $k$ such that $\pi = 3.x^ky$.) – r.e.s. Apr 1 '15 at 0:58

This is a really cool question! And the answer is: almost certainly not, but maybe.

First, let me explain why it's possible. We have calculated about 13 trillion digits of pi so far, meaning we don't know anything about the digits after that. This means the digits immediately following the last one we've calculated could very well be 1415926... etc. The chances that the next 13 trillion digits are equivalent to the first 13 trillion are very slim: $(\frac{1}{10})^{\text{13 trillion}}$. Super unlikely, yet possible!

However, that's not the only case that would make pi start with two identical decimal sequences; the repetition could start anywhere, so we need to sum all possible cases. First let's assume we don't know anything about the decimal expansion of pi (For all we know, maybe it repeats after the first digit [i.e. 3.114...]). The probability of a repetition after the $n^{\text{th}}$ digit is $(\frac{1}{10})^{n}$, since the changes of each of the $n$ digits matching is $(1/10)$. So, the probability that this is true for some value of $n$ would be: $$\sum_{i=1}^{\infty}{\bigg(\frac{1}{10}\bigg)^i} = \frac{1}{9}$$

That seems like a pretty good chance! And that would be the chance that a number picked at random from a uniform distribution like $[0,1]$ would start with two identical sequences.

However, with pi, we KNOW that it doesn't start with 3.11, or 3.1414, or 3.141141, so that eliminates a big part of that probability. But then the question becomes, how many digits do we know this for? Do we know that the first million digits don't equal the second million? What about the first and second billion, or trillion, or (13 trillion)/2? I honestly don't know what kind of analyses the people who did the massive calculations were doing on the digits they found, but let's assume we know that in the first 13 trillion digits, there are no repeating sequences that start at the beginning. This would change our probability formula to $$\sum_{i=6.5 \text{ trillion}}^{\infty}{\bigg(\frac{1}{10}\bigg)^i}$$

Which is roughly equal to $\frac{1}{6.5 \text{ trillion}}$, or about $10^{-10^{12}}$.

Therefore, the answer to your question is that it is possible that pi starts with two identical decimal sequences, but it is extremely unlikely.

• "The chances that the next 13 trillion digits are equivalent to the first 13 trillion are very slim: $1/10^{10^{12}}$" -- it's implied here that the digits of Pi are random-uniform distributed. Is that the case? – OJFord Jun 6 '15 at 1:14
• @OllieFord, we don't know. For all we know, the next 13 trillion digits could all be 7, but to our knowledge, no sequence of digits is more likely than any other, so I assumed equal probability. Kinda how if I flip a weighted (biased) coin and have you guess the result, you have a fifty percent chance of getting it right even though there's not a 50-50 chance of heads vs tails – Christopher Shroba Jun 12 '15 at 15:57
• By "Boole's inequality", the probability that for some $n$ the second $n$ digits duplicate the first $n$ digits is upper-bounded by $\sum_{i=1}^{\infty}{\bigg(\frac{1}{10}\bigg)^i}= \frac{1}{9}$. Because the events are not mutually exclusive, the correct probability is slightly less than this (approximately $0.1099899$, as can be found by inclusion-exclusion). Taking this into account, your argument has even more force. – r.e.s. Dec 13 '15 at 5:45

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 $\pi$.

• +1, the connection to irrationality measure is nice. To be a stickler, though, are those finitely many rational approximants ($p/q$ where $|\pi - p/q| < q^{-8}$) known for sure? I can imagine a proof that there were finitely many of them that nevertheless failed to produce an effective bound. – mjqxxxx Jun 5 '15 at 19:17
• @mjqxxxx: Indeed. I don't have access to Salikhov's paper which shows that it's less than $8$, but Hata's paper, which is the previous-best technology, is available online here, and appears to be effective (though extracting the actual constant looks annoying). So the worst-case scenario might be that I have to change all the $8$s in my answer to $9$s to be completely honest. – Micah Jun 5 '15 at 21:48