$G(l,m,n_1,n_2)$ as defined by (2) is not a polynomial.

@PerAlexandersson In cases like this it's often useful to apply Lagrange inversion to an equation like $f = R(f)$ rather than $f=xR(f)$. Thus if we want to solve $F =x (1+F^p)$ we can set $F(x) = xG(x^p)$ and $y=x^p$; then $F=x(1+F^p)$ reduces to $G(y) = 1+ yG(y)^p$, the well-known Fuss-Catalan functional equation.

@PerAlexandersson But in applying Lagrange inversion there is usually no problem in finding these powers; Lagrange inversion gives a formula for the coefficients of arbitrary powers of the power series to be solved for.

There is no problem in taking the $p$th root of a formal power series. If $G(x)$ is any formal power series we can write $G(x) = ax^b H(x)$ where $H(x)$ is a formal power series with constant term 1. Then $G(x)^{1/p}=a^{1/p} x^{b/p} H(x)^{1/p}$.

Incidentally, this other interpretation of $\frac{j}{2(j+k)}\binom {2j}{j}\binom{2k}{k}$ has been published. It appears in my paper Super Ballot Numbers, sciencedirect.com/science/article/pii/0747717192900342, section 7.

The original sum (with $x$ and $y$, falling factorials) is also indefinitely summable, according to Maple.

Whenever a sum $\sum_{m=1}^n t(n,m)$ with weird looking parameters has a nice closed form, it's a good bet that the indefinite sum $\sum_{m=1}^k t(n,m)$ also has a closed form. (This is almost always the case with Putnam problems!) Tewodros's solution shows that this is the case here; we don't need creative telescoping, just indefinite summation, which Maple (for example) can do easily. This also shows that the identity can easily be greatly generalized, though it would take a bit of work to find which specializations of the most general form are nice.

@FedorPetrov This isn't quite true. The WZ method, $$𝐹(𝑛+1,𝑘)−𝐹(𝑛,𝑘)=𝐺(𝑛,𝑘+1)−𝐺(𝑛,𝑘),$$ usually works, but not always. What always works is Zeilberger's algorithm, which is a little more complicated. See Theorem 6.2.1 of $A=B$. (www2.math.upenn.edu/~wilf/Downld.html)

According to this paper sciencedirect.com/science/article/pii/S0095895606000682, a dart is the same as an arc. A similar (but not exactly the same) use of dart can be found in planarity.org/Klein_basic_graph_definitions.pdf.

You might look at Edelman's paper core.ac.uk/download/pdf/82751622.pdf, which discusses the joint distribution of cycles and inversions in permutations.

It might be noted that this integral is, up to normalization, a special case of the beta integral. The general beta integral corresponds to Jacobi polynomials (though with a different normalization than usual).

The integral representation $$C_n = \frac{1}{2\pi} \int_0^4 t^n \sqrt{\frac{4-t}{t}}\, dt$$ can be found in cs.uwaterloo.ca/journals/JIS/VOL4/SIXDENIERS/Catalan.html, though I am sure it is much older.