# Is there a combinatorial reason that the (-1)st Catalan number is -1/2?

The $n$th Catalan number can be written in terms of factorials as $$C_n = {(2n)! \over (n+1)! n!}.$$ We can rewrite this in terms of gamma functions to define the Catalan numbers for complex $z$: $$C(z) = {\Gamma(2z+1) \over \Gamma(z+2) \Gamma(z+1)}.$$ This function is analytic except where $2n+1, n+2$, or $n+1$ is a nonpositive integer -- that is, at $n = -1/2, -1, -3/2, -2, \ldots$.

At $z = -2, -3, -4, \ldots$, the numerator of the expression for $C(z)$ has a pole of order 1, but the denominator has a pole of order $2$, so $\lim_{z \to n} C(z) = 0$.

At $z = -1/2, -3/2, -5/2, \ldots$, the denominator is just some real number and the numerator has a pole of order 1, so $C(z)$ has a pole of order $1$.

But at $z = -1$: - $\Gamma(2z+1)$ has a pole of order $1$ with residue $1/2$; - $\Gamma(z+2) = 1$; - $\Gamma(z+1)$ has a pole of order $1$ with residue $1$. Therefore $\lim_{z \to -1} C(z) = 1/2$, so we might say that the $-1$st Catalan number is $-1/2$.

Is there an interpretation of this fact in terms of any of the countless combinatorial objects counted by the Catalan numbers?

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Have you read any of these links? math.ucr.edu/home/baez/counting – Qiaochu Yuan Feb 17 '10 at 21:49
Some of them, but not recently. – Michael Lugo Feb 17 '10 at 21:50
Michael, I think you are making things a bit more complicated than they really are: write the explicit formula as a recurrence relation: $$C_{n+1} = \frac{4n+2}{n+2} C_n$$ From here you immediately conclude that $C_{-1}=-\frac12 C_0 =-\frac12$, and that the formula makes no sense for $n=-2$, so you cannot extend the sequence in the negative direction any further. – Igor Pak Feb 18 '10 at 4:22
Well, Igor, I think you're right. – Michael Lugo Feb 18 '10 at 13:00
One of the papers by Propp discusses how the well-known expression for negative binomial coefficients can be given a concrete combinatorial interpretation following ideas of Schanuel; it seems potentially relevant. – Qiaochu Yuan Feb 19 '10 at 3:41

Whatever combinatorial interpretation someone comes up with has to somehow count the number of ordered trees on 0 vertices as $-1/2$. Using counting techniques like the ones advocated by Baez allows you to make sense of generalized counts, but negative counts stretch 'combinatorics' pretty far!