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I've been interested in counting how many binary trees there are with n leaves. I consider 2 trees to be the same if I can swap the children of nodes to get the other one.

I've started by figuring out the first terms: 1, 1, 1, 2, 3, 6, 11

I then quickly found a recursive formula for it: $$ u_1 = 1 \\ \forall\ n > 1,\ u_n = \sum_{i = 1}^{\text{floor}(n/2)}u_i\times u_{n-i} $$

It appears this sequence comes with the sweet name of the "Half-Catalan numbers" (found here https://oeis.org/A000992). I can see where this name comes from as this is defined as the Catalan numbers except we do the sum up to $\text{floor}(n/2)$ instead of $n$.

But the Catalan numbers also have a neat non-recursive formula : $$ u_n = \frac{1}{n+1}{2n \choose n} $$

So my question is, is there any non-recursive formula for the Half-Catalan numbers as well?

Regards.

QuentinJanuel
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1 Answers1

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For the Catalan Numbers there's copious detail of how to derive $C_n=\dfrac1{n+1}\dbinom{2n}n$ starting from the recurrence relation $$C_n=\sum_{k=0}^{n-1}C_kC_{n-1-k}\tag{0}$$ and the initial value $C_0$, via the ordinary generating function at the following link.

Simplifying Catalan number recurrence relation

Perhaps trying to mirror the arguement might result in what you ask for, although it's not for the faint hearted.

The "half catalans" seem to be starting from a less enticing recurrence relation; if that doesn't prove tractable, maybe jump in with the ordinary generating function for the "half catalans", which is given at the link you provided to Sloan's.

Martin Hansen
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