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In the paper "Playing pool with $\pi$ (the number $\pi$ from a billiard point of view)" http://www.maths.tcd.ie/~lebed/Galperin.%20Playing%20pool%20with%20pi.pdf George Galperin provides the following interesting "experimental" method to find out digits of $\pi$:

  1. Take two billiard balls with the ratio of their masses $M/m=100^N$.

  2. Put the small ball, $m$, between the wall at the origin and the big ball, $M$.

  3. Push the big ball towards the small ball.

  4. Calculate the total number of hits in the system: the number of collisions between the balls plus the number of reflections of the small ball from the wall.

It turns out that this number of hits gives the first $N+1$ digits of $\pi$ ignoring the decimal point. The proof of this result involves a conjecture that $$\left [\frac{\pi}{\arctan{(10^{-N})}}\right]=\left [\frac{\pi}{10^{-N}}\right]$$ is true for any big enough natural number $N$. Galperin writes that the modern mathematics is far from proving this conjecture. Galperin's article was in 2003. Has anything changed as far as the proof of this conjecture is concerned?

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  • $\begingroup$ Is the issue that there may be an absurdly large series of 9's in the decimal expansion of $\pi$? Even if there is, that should make the difference between the two values at most 1. Also, some irrationality indices have been computed that may resolve this. Gerhard "Is The Conjecture Really Rational?" Paseman, 2017.03.13. $\endgroup$ – Gerhard Paseman Mar 14 '17 at 6:44
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    $\begingroup$ Happy π-day to everyone! $\endgroup$ – gsa Mar 14 '17 at 6:49
  • $\begingroup$ Yes, the issue is to prove that the following situation $\pi=3.1415\ldots a_N 99\ldots 99 \ldots$, where $a_N<9$ and the number of nines equals to $N-1$, never happens in the $2N$ first consecutive digits of $\pi$. $\endgroup$ – Zurab Silagadze Mar 14 '17 at 7:02
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    $\begingroup$ There is another exposition of Galperin's result at cut-the-knot.org/blue/nicollier_galperin_ctk.shtml $\endgroup$ – Gerry Myerson Mar 14 '17 at 11:31

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