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. $123547123547,8989,0909,182182,99,…$
Best Answer
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 $0,1, \ldots, 9$.