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Using Pietro Majer's bisection formula we find by a straightforward computation (I did it with Maple, but I'm sure it could be done without too much difficulty by hand) that the OP's formula for $$\su …

I'm not completely sure what the problem is, but $$(1-x-y) c(x,y) = 1 - y C(xy) = 1 - \frac{1-\sqrt{1-4xy}}{2x},$$ where $C(z)$ is the Catalan number generating function, $$C(z) =\sum_{n=0}^\infty C_ …

The coefficient of $z^n$ is $$\sum_{0\le k\le n/2} (-1)^k \frac{k+1}{2n-3k+1}\binom{2n-3k+1}{n-2k}.$$ To see this, let $C(z)$ be the Catalan number generating function, $$C(z) = \frac{1-\sqrt{1-4z}} …

As David noted, since the summands aren't in general integers, it's difficult to give a combinatorial interpretation to the formula. However, if we multiply by $a$ or $b$ we get integers and we can gi …

For Bender and Goldman's theory of prefabs, see http://www.iumj.indiana.edu/docs/20060/20060.asp. Michael Henle's theory of dissects is introduced in Dissection of generating functions, Studies in App …

Here's a sketch of a derivation of (3) from (1). It's fairly straightforward to compute $$\sum_{n=0}^\infty F_n(x) z^n = \frac{1-2z-xz}{(1-xz)^2 -4z}.$$ If you expand this in powers of $z$ you get (3) …

If you multiply both sides by $e^{x+y}$ there's a simple bijective proof: In a nutshell, every 0-1 matrix consists of a matrix with a 1 in every row and column together with some all-zero rows and col …

These numbers count balanced signed graphs (without loops). A signed graph is a graph in which every edge has a sign, either positive or negative. It is balanced if every cycle has an even number of n …

Here's a sketch of a proof, using Lagrange inversion, of the equation $(27x^2/4-1)F^3+3F-2=0$, where $$F(x) = \sum_{n=0}^\infty \binom{3n/2}{n}x^n.$$ One form of Lagrange inversion says that if $h(x) …

Formal power series with radius of convergence 0 often arise in counting labeled graphs. For example, the exponential generating function for labeled connected graphs is $\log G(x)$, where $$G(x) = \s …

See Gilbert Labelle, On asymmetric structures, Discrete Math. 99 (1992), 141–164.

I started writing this before Richard's answer appeared, with which it overlaps a lot, but I still have something to add. Let us look at a more general problem: Suppose that $G(x) = 1+g_1x+g_2x^2+\cd …

No, there is no name for this kind of generating function, except in the case $k=0$, when they are called “doubly exponential generating functions”. I do not know of any applications for $k>0$.

The exponential generating function for permutations of order dividing $k$ is $$\exp\biggl(\sum_{d\mid k} \frac{x^d}{d}\biggr).$$ See, e.g., L. Moser and M. Wyman, On solutions of $x^d = 1$ in symmetr …

For a simple but noncombinatorial proof of the unimodality of the coefficients of $[n]!_q$, see George E. Andrews, A theorem on reciprocal polynomials with applications to permutations and compositi …