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I was recently introduced to the set of Catalan numbers, which are of the form $\prod_{i = 2}^n \frac{n + i}{n}$. The first few Catalan numbers (which I might as well mention are all integers) are: 1, 1, 2, 5, 14, 42, 132, 429, 1430.

I've recently learned that Catalan numbers represent the number of paths that, in creating paths of lattice points from the origin to $(n, n)$, only intersect the line $y = x$ at the points $(0, 0)$ and $(n, n)$, where $n \in \mathbb{Z}$.

I've also learned that the number of such paths from $(0, 0)$ to $(n, n)$ that intersect $y = x$ at the points $(0, 0)$, $(n, n)$, and exactly one other point in between, is $C_{n-1}.$

How would I go about proving these two statements? They are undoubtedly related, in both their content and their similar structure. Any steps in the right direction would be appreciated.

bof
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Jacob Schiller
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1 Answers1

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Let me start with your second question, since this is probably going to be harder to Google.

The Catalan numbers are well-known to satisfy the following recurrence: $$ C_n = \sum_{k=1}^{n} C_{k-1}C_{n-k}$$ For instance, your first lattice path interpretation of the Catalan numbers $C_n$ satisfies this by conditioning on the first place after the origin that the paths touch the diagonal. You have to think about that for a minute, but the point is the $k-1$ shows up because to ensure that $k$ is the first time you hit the diagonal, you have to "step out one square" for the steps between the origin and $(k,k)$.

So you can see where that argument takes you. Let $D_n$ be the number of lattice paths that stay strictly above the diagonal except at $(0,0)$ and $(n,n)$ and exactly one other point. So if you want $k$ to be the only time you hit the diagonal, then you have to step out one square on both sides of $(k,k)$:

$$D_n=\sum_{k=1}^{n-1} C_{k-1}C_{n-k-1}$$

By comparing the two RHS, we can clearly see that $D_n=C_{n-1}$.

When you understand this argument, it's straightforward to unwind it into an explicit bijection between the $D_n$-type paths and the Catalan paths of length $2(n-1)$, but I'll leave that to you :)

Your first question has numerous answers on e.g. the Wikipedia page.

The recurrence above can be turned into the product formula that you mention; it is a standard exercise in generating functions and indeed is usually the first solution to a nonlinear recurrence relation that is solved in this way. You can find the details in any decent book on combinatorics, or the first proof on Wikipedia, or on this website: e.g. Simplifying Catalan number recurrence relation.

There are also direct proofs. They are all at least a little tricky, but you can get a sense of what they entail by looking at the other Wikipedia proofs. The way I understand this trickiness is that for your product formula to be an integer, it must be the case that $n+1$ is a divisor of $\binom{2n+1}{n}$. This in turn can be understood as the freeness of a natural cyclic group action; this and a number of similar/related phenomenon in algebraic combinatorics go by the name of "The Cycle Lemma".

In any case, there are many ways in which I think Catalan objects are more naturally quotients of the set of (appropriate) lattice paths, rather than subsets— and so I view the weirdness in these proofs as the same weirdness inherent anytime we work with "distinguished representatives" of equivalence classes, rather than the classes themselves.

(On a brief look I wasn't able to find direct proofs on MSE. Surely they're here; vets feel free to tell us where in the comments ^.^)

Eric Nathan Stucky
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