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?
 A: Here's an explanation for why, as far as I can tell, the answer to your question should be "no", and -1/2 isn't really a Catalan number.
The generating function $C(x)=\sum_{n=0}^\infty C_n x^n$ 
for the Catalan numbers (with the usual interpretation starting from $C_0=1$) satisfies the functional relation
$$C(x) = 1 + xC(x)^2,$$
and from this we can derive the closed-form expression $C(x)=\frac{1-\sqrt{1-4x}}{2x}$.
The formula you wrote above for the Catalan number $C_n$ in terms of factorials, or the gamma function, is really expressing the coefficients of the Taylor expansion of $F(x)= \frac{-\sqrt{1-4x}}{2x}$.
This just so happens to match up with the coefficients of $C(x)$ for non-negative powers of $x$, but not for the $x^{-1}$ term, since
$ F(x) = -\frac12 x^{-1} + C(x).$
So now the question is, are we really sure that $C(x)$ is the "right" Catalan generating function rather than $F(x)$? Well yes, because $C(x)$ is the one that satisfies the functional relation above, which leads to all the nice combinatorial interpretations. The function $F(x)$, on the other hand, doesn't satisfy any particularly nice equation---there is $F(x)^2 = \frac14 x^{-2} - x^{-1}$, which I don't see a natural interpretation for combinatorially.
A: 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!
