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Consider the "Chacon words": $B_0=0$ and $B_{n+1} = B_nB_n1B_n$. The word $B_n$ has $\ell_n := \frac{3^{n+1}-1}{2}$ digits and the number of $1$'s in $B_n$ is $\ell_n - 3^n = \ell_{n-1} \sim \ell_n/3$...