# Can we find every finite sequence of $\pi$ within $\pi$?

Inspired by one of the top posts I was wondering if $\pi = 3.14159\dots$ were normal--as in you could find every finite string of numbers within $\pi$'s digits--would that mean we could find every finite sequence of the digits of $\pi$ in $\pi$'s digits. Would we find

$$3141592653...\left(\text{n-th digit of \pi} \right)$$

For any $n$? Also is there an example of a number that is normal that is easy to prove to be normal? I'm unfamiliar with the property.

• you can read about normal numbers here. It is not known whether or not $\pi$ is normal. Indeed, it is not known whether every digit occurs infinitely often in the decimal expansion of $\pi$.. The wiki article linked to provides some fairly natural examples. – lulu Jun 20 '18 at 20:12
• @lulu: Yes. Note, however, that the question is about what it would mean if $\pi$ were normal, not about whether it's normal. – joriki Jun 20 '18 at 20:25
• @joriki Ah, you are correct. Read too quickly. I'll leave the comment up for the sake of the link. – lulu Jun 20 '18 at 20:33
• You can, of course, find any finite sequence of digits of $\pi$ somewhere, at least once, among the digits of $\pi$, namely where the string first occurs; that's true of any number, normal or not. Normality would imply that you can find any such string infinitely often, with asymptotic density equal to $1/10^n$, where $n$ is the length of the string. – Barry Cipra Jun 20 '18 at 20:33
• What do you mean by "one of the top posts"? – Rob Arthan Jun 20 '18 at 20:52

The answer to your question is yes: the first billion digits of $\pi$ forms a finite block of base-$10$ digits, and every finite block of base-$10$ digits must occur infinitely often in any normal-to-base-$10$ number. (I wonder if you asked exactly what you wanted to?)

A quick comment at this point: it looks from your question like you've gotten the definition of "normal" incorrectly. Normality and its relatives don't just assert that lots of variety occurs, they assert that it occurs statistically: e.g. simple normality to base $10$ implies that asymptotically $1\over 10$th of the number's digits in base $10$ are $2$, not just that infinitely many digits are $2$. Precisely:

• A number is simply normal to base $b$ if each digit in base $b$ occupies $1\over b$th of the places of the base-$b$ expansion of the number (asymptotically speaking). Note that the number whose binary expansion is $0.01001000100001..._2$ is not simply normal to base $2$, even though both $0$ and $1$ occur infinitely often in its expansion, since their distribution isn't right: the asymptotic probability of a digit being $0$ is $1$, not ${1\over 2}$.

• A number is normal to base $b$ if it is simply normal to base $b^n$ for all $n$. Intuitively, this means that not only are the individual base-$b$ digits stochastically distributed, but so are the finite blocks of digits; e.g. $1\over 3$ is simply normal to base $2$ but not normal to base $2$ (since e.g. the block "$00$" doesn't appear anywhere in the binary expansion of $1\over 3$, let alone the required "$1\over 4$th of the time").

• A number is normal (or absolutely normal) if it is simply normal to all bases.

Despite their objective commonality (in a precise sense, "most" real numbers are normal), verified examples of naturally-occurring normal numbers are hard to come by. Champernowne's constant is easy to prove "normal to base $10$," but isn't exactly "naturally occurring" and it is not known to be normal to all bases. Examples of absolutely normal numbers can be explicitly constructed; however, these examples are even less naturally-occurring than Champernowne's constant.