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Is there an irrational number that the digits never repeat anywhere and have all 10 digits appear everywhere?

let's look at one that doesn't work like $$\pi=3.141592653589793238462643383...$$ starting at the 23rd digit you get 33 so it fails another example of one that fails is $0.10102101023135791...$ even tho no digit ever repeats twice a pair of digits do $10,10$ and and here 5 digits in a row do $10102,10102$.

my question is there an irrational number such that all digits are used equally and no sequence of the digits repeat twice like this. $123547123547,8989,0909,182182,99,...$

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    I don't understand the question. What do you mean by "have all 10 digits appear everywhere" ? – tristan Aug 03 '20 at 16:12
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    Here is a closely related question. – pregunton Aug 03 '20 at 16:15
  • so is there a irrational number where no sequence of digits repeat twice if you look at 0.1001011001101001... wouldn't work because not all digits are there and 00,11,0110 repeats @tristan –  Aug 03 '20 at 16:15
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    thanks @pregunton –  Aug 03 '20 at 16:16
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    The OP probably means that for every $n\ge2$, every word containing $2n$ consecutive digits is not a run of two identical words of length $n$. – user1551 Aug 03 '20 at 16:18
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    While I don't have a proof at hand, it's very likely that such numbers are "rare" in the sense of having Lebesgue measure zero (equivalently, having 0 probability of being chosen at random in a uniform distribution on a bounded interval). – R.. GitHub STOP HELPING ICE Aug 04 '20 at 00:19
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    For $x$ chosen with uniform density from the interval $[0,1]$, for each positive integer $n$ the probability that digits number $2n$ and $2n+1$ are equal is $1/10$, and these are independent, so with probability $1$ there are infinitely many repeated digits. Similarly, with probability $1$ there are infinitely many repeated $k$-tuples of digits for every $k$. – Robert Israel Aug 04 '20 at 01:15

1 Answers1

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The sequence of digits you want is an infinite square-free word on the alphabet 0123456789.

EDIT: Consider an infinite square-free word on the alphabet 012, which we know exists by Thule's construction. Let the positions of $2$ in this word be $i_1, i_2, \ldots$: there must be infinitely many, otherwise after some point we would have an infinite square-free word on alphabet 01, which is impossible. For each $k$, change the letter in position $i_k$ to $3$, $4$, \ldots $9$ or leave it as $2$ if $k \equiv 0, 1, \ldots, 7 \mod 8$ respectively. The resulting infinite word is still square-free, and now has infinitely many of each of $0,1, \ldots, 9$.

Robert Israel
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    This answer really doesn't go into the existence of a number whose decimal expansion matches such a word. It's not hard to show, but based on how OP's question is worded I suspect it might be nontrivial to OP. – R.. GitHub STOP HELPING ICE Aug 04 '20 at 00:22
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    Link-only answers are discouraged. What happens when that link stops working? It would be well to pull the most relevant parts of that page into your answer. – Stig Hemmer Aug 04 '20 at 07:54
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    Besides the other criticisms in comments, nothing in “infinite square-free word on the alphabet 0123456789” requires that all digits appear infinitely often. E.g. an infinite square-free word from ${0,2,1}$ (which exists by Thule’s construction) can be viewed as over ${0,…,9}$, but doesn’t satisfy what the question asks for. – Peter LeFanu Lumsdaine Aug 04 '20 at 10:55
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    @PeterLeFanuLumsdaine That's true. However, you can take such a word and change some of the infinitely many $2$'s to $3$'s, $4$'s, ..., $9$'s. Well, maybe I should edit my answer to go into detail on that. – Robert Israel Aug 04 '20 at 14:01