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Martin Rubey
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The number of lattice paths below y=n/m x for gcd(m,n) = 1

The motivation of my question is the recent preprent of Armstrong, Rhoades and Williams http://arxiv.org/abs/1305.7286 on rational Catalan combinatorics.

An important starting point of this paper is the fact that the number of lattice paths with steps $(1,0)$ and $(0,1)$ from the origin to $(m,n)$, where $gcd(m,n)=1$ weakly below the diagonal $y=\frac{n}{m} x$ equals $$ \frac{1}{m+n}\binom{m+n}{m}. $$

This can be proved (as done by Bizley, http://www.jstor.org/discover/10.2307/41139633?uid=3737528&uid=2&uid=4&sid=21102085756863) by considering all cyclic permutations of each path from the origin to $(m,n)$ and demonstrating that among these there is exactly one path below the diagonal.

I would like to know of any other proof of this fact, for example using generatingfunctionology.

One approach I looked at (from the paper by Gessel and Ree http://people.brandeis.edu/~gessel/homepage/papers/faber.pdf) extends the recurrence $B(m,n)=B(m,n-1)+B(m-1,n)$ to all of $\mathbb N^2$ and then uses suitable initial values $B(m,0)$ and $B(0,n)$ to obtain generating functions for paths below $y=\frac{1}{m} x$. But it seems that the initial values for the general case are unpleasant.

Martin Rubey
  • 5.8k
  • 1
  • 24
  • 39