# Choosing points in fractions of the unit interval

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.

-
Mathworld calls this the "18-point problem": mathworld.wolfram.com/18-PointProblem.html –  joriki Jun 5 '11 at 2:06
add comment

## 1 Answer

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

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

-
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 5 '11 at 1:59
Fascinating. Interestingly, the solution roughly matches Ross' up to $5/8-$. –  joriki Jun 5 '11 at 1:59
See also mathoverflow.net/questions/19896/… and the freely available original proof at citeseerx.ist.psu.edu/viewdoc/… –  joriki Jun 5 '11 at 2:04
add comment