# Specific strings that cannot be found in PI

Considering that the decimals of PI aren't like prime numbers (random), they are pseudo-random (can be calculated through a formula or an infinite series, yet it is calculable). Keeping this in mind, I've been seeing a lot lately, these pictures that say that If you convert the digits into letters you will find every book ever and you date of birth and death every conversation you will ever have and had etc.. etc.... But what I was really curious about. Are there any strings of N numbers where N > 3 (because I've tested them all with 1 and 2) which can be mathematically proven to NOT be found in pi, or the contrary, to prove that there are no such strings for any N

• Nobody knows how to prove statements of this form. As far as I am aware, it's possible that from some point on the only digits in the expansion are $0,1$ (not very probable, mind you). And it is possible (and more probable) that every finite string occurs somewhere in the expansion. But, it is not known. – lulu Mar 21 '18 at 21:27
• You're asking if $\pi$ forms a disjunctive sequence. As far as I know this is still an open question. – krirkrirk Mar 21 '18 at 21:30
• If you convert the digits into letters you will find every book ever and you date of birth and death every conversation you will ever have and had etc.. etc... This is not known to be true. It is known to be overwhelmingly likely, though (as in the probability is $1$). – Arthur Mar 21 '18 at 21:31
• If the probability is one doesn't it mean it is sure to happen? Or is the probability more like 0.999....999999? – Mattia Marziali Mar 21 '18 at 21:38
• @MattiaMarziali No, the probability is exactly $1$, but we aren't sure it will happen. We are almost sure. Compare to throwing a dart at a mathematical dartboard, and asking about the probability of missing the exact center. That also has probability $1$, but it's not a certainty. – Arthur Mar 22 '18 at 11:09

## 2 Answers

Those pictures you've been reading are wrong - well, not wrong, but they're jumping the gun. They're making a claim about $\pi$ that has not been proven.

Likewise, your question is an open question. One of the major open questions about $\pi$ is whether or not it is a normal number: a number whose digits are "essentially random", which basically means that every string of digits can be found in it, in roughly the proportions you'd expect of a random number. For example, in a normal number, the string "123" should show up about $0.1\%$ of the time (because there's a $0.1\%$ chance of a random sequence of three digits being 123). It is known that almost all numbers are normal (in a rare feat of accurate nomenclature on the part of mathematicians), but it is not known whether $\pi$ itself is normal. It's also interesting to note that it is really hard to find normal numbers - only a handful have been identified, and they were specifically "cooked up" to be normal numbers. To my knowledge, no "natural" constant is known to be normal.

A positive answer to your question ("yes, such an $N$ exists") would mean that $\pi$ is not normal. A negative answer would not prove that $\pi$ is normal (it wouldn't necessarily meet the "proportion" requirement) but it would be significant progress.

Yes, statistically speaking you will find all strings if you indeed buy a ticket to infinity and if you can prove the digits of $\pi$ are uniformly distributed.

See if the length of your string is $3$ you have only $10^3=1000$ such strings.

Assuming randomness (uniform distribution ) if you continue with the digits of $\pi$
you expect to hit your string in the first $1002$ digits.

That is, the expected value of finding your string is 1.

Thus if you pick 1,000,000 digits, the expected value of finding your string is about 1000. Of course we are assuming randomness.

All it take is about 10,000,000 digits to be almost sure that you will find many copies of your string.

• This is a common error - this is only true if the digits of $\pi$ are uniformly distributed. For example, the number $0.101001000100001\ldots$ is irrational, but nowhere in its decimal expansion does the string $2$ appear. – Reese Mar 21 '18 at 21:55
• Thanks, that clarifies the point that digits of prime numbers are not necessarily arranged at random. I have assumed randomness without proof. – Mohammad Riazi-Kermani Mar 21 '18 at 22:21