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?