Mathematical Intuition Behind Schizophrenic Numbers?
That recursion should be $S_n = 10 S_{n-1} + n$.
Solving the recursion, we get $$S_n = \dfrac{10^{n+1}}{81} - \dfrac{9n + 10}{81}$$
If $n$ is odd, say $n=2k-1$, write this as
$$ S_n = \left(\dfrac{10^{k}}{9}\right)^2 \left(1 - \dfrac{9n+10}{10^{2k}}\right)$$
so that
$$\sqrt{S_n} = \dfrac{10^k}{9} \left(1 - \dfrac{9n+10}{2 \times 10^{2k}} - \dfrac{(9n+10)^2}{8\times 10^{4k}} - \ldots\right)$$
We get one block of nearly $10^{2k}$ digits from the $10^k/9 = 1\ldots1.1\ldots$, then another of nearly $10^{2k}$ digits where the $(9n+10)/(2\times 10^{2k})$ is included, etc.