In the decimal expansion of the following infinite sum (number of $9$s is doubled in each term) we get a beautiful pattern: $$x=\sum_{n=0}^\infty \frac{1}{10^{2^n}-1}=\frac{1}{9}+\frac{1}{99}+\frac{1}{9999}+\dots$$
$$x=0.1213121412131215121312141213121\dots$$
$$x=0.\color{blue}{1}2\color{blue}{1}\dots$$
$$x=0.\color{blue}{121}3\color{blue}{121}\dots$$
$$x=0.\color{blue}{1213121}4\color{blue}{1213121}\dots$$
$$x=0.\color{blue}{121312141213121}5\color{blue}{121312141213121}\dots$$
This number shows a perfect pattern until it's $2^9=512$th digit, when $10$ should appear for the first time, and of course it messes the pattern up because it contains two digits instead of $1$.
I get why this happens intuitively, but how exactly does this pattern appear?
Will we get a similar pattern in any base $b$ for the sum $x=\sum_{n=0}^\infty \frac{1}{b^{2^n}-1}$?
Observe that
$$\begin{align} 1/9=&\ 0.111111111...,\\ 1/99=&\ 0.010101010...,\\ 1/9999=&\ 0.000100010.... \end{align}$$
Sum them. At every decimal place you get a $1$ from $1/9$; at every $2n$ th decimal you add an extra $1$ because of $1/99$; and at every $4n$ th decimal you add even one more extra $1$ because of $1/9999$. You hence get
$$1/9+1/99+1/9999=0.121312131...$$
Now adding more terms will produce the pattern you saw.