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How long a series of points in (0,1) can be chosen such that the first two are in different halves, the first three are in different thirds, ... the first $n$ are in different $n^{\text{th}}$s? My first try of $(0+,1-,\frac{1}{2}-,\frac{3}{4}-, \frac{1}{5}+,\frac{5}{8}-,\frac{1}{3}-,\frac{7}{8}-,\frac{1}{3}+)$ works through $9$, but there are two points in $(0.3,0.4)$. The plus and minus signs indicate a shift of some distance away from the given point small enough not to move over any fraction of interest.

Ross Millikan
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    Mathworld calls this the "18-point problem": http://mathworld.wolfram.com/18-PointProblem.html – joriki Jun 05 '11 at 02:06

1 Answers1

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An answer is asserted but not proved at the following URL:

http://en.wikipedia.org/wiki/Irregularity_of_distributions

Michael Hardy
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    That is very surprising. I'm glad I didn't spend more than a few minutes on this! I never would have reached anything close to 17. – Sputnik Jun 05 '11 at 01:59
  • Fascinating. Interestingly, the solution roughly matches Ross' up to $5/8-$. – joriki Jun 05 '11 at 01:59
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    See also http://mathoverflow.net/questions/19896/slick-proof-related-to-choosing-points-from-an-interval-in-order and the freely available original proof at http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.113.29&rep=rep1&type=pdf – joriki Jun 05 '11 at 02:04