TITLE: Does every irrational number contain arbitrarily long sequences of some digit?
QUESTION [5 upvotes]: Suppose we have an irrational number represented in base $3$ such that there can only be a maximum of $n$ consecutive $1$'s or $2$'s in the ternary expansion. Does this imply there are arbitrarily long sequences of $0$'s in the expansion? Or does such a number even exist?

REPLY [2 votes]: Start with any irrational number (between $0$ and $1$ for convenience) in ternary expansion and translate the digits after the initial "$0.$" per $0\mapsto 1212$, $1\mapsto 1202$, $2\mapsto 1012$. The the resulting number is also irrational (because nonperiodic) and there are no runs of digits longer than $1$.