25
$\begingroup$

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.

$\endgroup$

1 Answer 1

27
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks, fixed the typo. How are you expanding that square root in the last step? $\endgroup$
    – ant11
    Commented Jul 23, 2014 at 7:23
  • 3
    $\begingroup$ @ant11 Taylor expand $f(x)=\sqrt(1-x)$ around zero $\endgroup$
    – Arun Kumar
    Commented Jul 23, 2014 at 9:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .