# Sequences of consecutive digits of arbitrary length in irrational ternary numbers (take two).

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. Furthermore, suppose the only digit immediately before or after a $1$ is a $0$ or $1$, and likewise, the only digit immediately before or after a $2$ is a $0$ or $2$. Does this imply there are arbitrarily long sequences of $0$'s in the expansion? Or does such a number even exist?

For example, numbers of this form would look something like:

$0.111111100002220200000000202001010102010100000000200102000202022222220201020\ldots$

Let $b$ be the sequence $1^n0$ consisting of $n$ ones followed by a zero. The number whose ternary expansion is
$$0.b0b^20b^30\ldots$$
Added: E.g., if $n=3$ this is
$$0.\underbrace{1110}_b0\underbrace{1110111}_{b^2}0\underbrace{0111011101110}_{b^3}0\ldots$$