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i.e. can we write

$\pi = 3.14159\dots X\dots$

where $X$ consists of (say) $10^{100}$ consecutive zeroes?

[Originally asked on reddit without response :-( ]

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9  
If pi is normal, then yes. But this is unknown! –  Dan Brumleve May 17 '11 at 3:36
7  
That would be hard to memorize for the pi decimal-counting geeks isn't it. 3.1415926....0000000000000000000000000000 (twenty hours later) 0000... Damn it! I messed up my count. –  Patrick Da Silva May 17 '11 at 3:39
    
Here is an amusing website that allows you to search for the first occurrence of a digit string in pi. angio.net/pi –  paw88789 Sep 4 at 20:29
    
The exact outcome of the 2016 US presidential election is coded in there somewhere. But there's no way for us to find it, or recognize it. Every possible false result is in there too! (assuming pi is normal) –  user4894 Sep 4 at 20:31

1 Answer 1

up vote 10 down vote accepted

To expand on Dan Brumleve's comment; it is widely believed, but not proved, that, given any finite string of digits, that finite string appears somewhere in the decimal for $\pi$, in fact, occurs infinitely often, in fact, occurs about once every $10^n$ digits, where $n$ is the length of the string. The same is believed to hold true for $e$, $\sqrt2$, in fact, for pretty much any number known to be irrational and not constructed specifically to falsify the belief, but, again, nothing has been proved. For all we know, all the digits of $\pi$ from some point on are sixes and sevens.

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Just to check: my gut says that "any finite string of digits appears somewhere in the decimal expansion of $x$" is weaker than "$x$ is normal (or 10-normal)." Is that so? –  Adam Saltz May 17 '11 at 3:54
    
@IAMBrian: Yes: "normal in base 10" implies that any finite string of $n$ digits appears with density $1/10^n$; in particular, it occurs infinitely many times. "Any finite string of digits appears somewhere" only guarantees at least one occurrence. –  Arturo Magidin May 17 '11 at 4:03
    
Yes. "Normal" means "appears with the right frequency" - that's the "once every $10^n$ digits" in my answer. It's easy to make a number which isn't normal even though every finite string appears; just start with a number that is normal, and then put really long strings of threes in it from time to time. Well, you do have to be a little careful where you put those strings, but it can be made to work. –  Gerry Myerson May 17 '11 at 4:07
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@Arturo, actually, if every finite string appears once, then every finite string appears infinitely often, since every finite string is a prefix to infinitely many different finite strings. But of course you are correct that there's no density implication. –  Gerry Myerson May 17 '11 at 4:09
    
D'oh. Good point. –  Arturo Magidin May 17 '11 at 4:09

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