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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:

$0.111111100002220200000000202001010102010100000000200102000202022222220201020\ldots$

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Let $b$ be the sequence $1^n0$ consisting of $n$ ones followed by a zero. The number whose ternary expansion is

$$0.b0b^20b^30\ldots$$

is irrational, satisfies your condition, and has no subsequence of more than two consecutive zeroes.

Added: E.g., if $n=3$ this is

$$0.\underbrace{1110}_b0\underbrace{1110111}_{b^2}0\underbrace{0111011101110}_{b^3}0\ldots$$

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  • $\begingroup$ Okay, sorry. I misunderstood your notation. $\endgroup$ – Paul Sinclair Sep 5 '15 at 20:06

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