Schizophrenic numbers (A014824) are numbers whose square roots "look" like rational numbers. They were first discussed in 2004 by Darling in the Universal Book of Mathematics (page 282), and I personally read about them from Pickover. They are defined as follows:
$$S_n=10S_{n-1}+n$$
$$S_0=0$$
So they start $1,12,123,1234\ldots$ and larger schizophrenic numbers look like $12345679012345679012\ldots$
The interesting thing about schizophrenic numbers is that the square root of the odd ones display long runs in their decimal representation. From the wikipedia page,
$$\begin{gather*}\sqrt{S_{49}}=\\
1111111111111111111111111.\\
1111111111111111111111\\
0860\\
555555555555555555555555555555555555555555555\\
2730541\\
66666666666666666666666666666666666666666\\
0296260347\\
2222222222222222222222222222222222222\\
0426563940928819\\
4444444444444444444444444444444\\
38775551250401171874\\
9999999999999999999999999999\\
808249687711486305338541\\
66666666666666666666666\\
5987185738621440638655598958\\
33333333333333333333\\
0843460407627608206940277099609374\\
99999999999999\\
0642227587555983066639430321587456597\\
222222222\\
1863492016791180833081844\\
\cdots\end{gather*}
$$
after which the pattern disintegrates into nothingness. This sequence of repeating digits $1,5,6,2,4,\ldots$ is A060011.
Nowhere in the wikipedia page, nor Darling's or Pickover's writing, is justification given for this behavior. I noticed the relation between
$$\sqrt{123456790}\approx11111.1111$$
$$11111.1111^2=123456789.87654321\approx123456790$$
but that doesn't explain the bizarre pattern shown. I was wondering if anyone has or could point me to an explanation.
Best Answer
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.