Mathematical Intuition Behind Schizophrenic Numbers?

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.

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1 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.

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Thanks, fixed the typo. How are you expanding that square root in the last step? –  ant11 Jul 23 '14 at 7:23
@ant11 Taylor expand $f(x)=\sqrt(1-x)$ around zero –  Arun Kumar Jul 23 '14 at 9:47