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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    $\begingroup$ Have you read any of these links? math.ucr.edu/home/baez/counting $\endgroup$ Feb 17, 2010 at 21:49
  • $\begingroup$ Some of them, but not recently. $\endgroup$ Feb 17, 2010 at 21:50
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    $\begingroup$ 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. $\endgroup$
    – Igor Pak
    Feb 18, 2010 at 4:22
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    $\begingroup$ Well, Igor, I think you're right. $\endgroup$ Feb 18, 2010 at 13:00
  • $\begingroup$ 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. $\endgroup$ Feb 19, 2010 at 3:41

2 Answers 2


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.


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!

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    $\begingroup$ There are, of course, hundreds of combinatorial interpretations of the Catalan numbers, and it may be that only some of them "degenerate" in the correct way. Stanley knows a lot about both the Catalan numbers and combinatorial reciprocity; I hope he takes a look at this thread. $\endgroup$ Feb 19, 2010 at 3:43
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    $\begingroup$ Right - I looked at about 20 of them, and all of them either made no sense at all for n=-1, or they were essentially equivalent to the one above. So I figured I would post at least that much information. $\endgroup$ Feb 19, 2010 at 12:57
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    $\begingroup$ One of the problems in Stanley says that C_n is the dimension of the space of invariants of SL(2, C) acting on the (2n)th tensor power of the defining 2-dimensional representation. Does anyone know how to take negative tensor powers? Can you get an object with a "generalized dimension" of -1/2? $\endgroup$ Mar 4, 2010 at 19:38

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