Catalan generating function

c[x_]=((1-(Sqrt[1-4(x^2)]))/(2(x^2)))

Chech this looks right:

Series[c[x],{x,0,20}]

Generating function for dimensions of ATL irreps, found on the Online Encylopedia of Integer Sequences

v[n_][x_]=((((1-(Sqrt[1-4(x^2)])))^(2n))/((((2x))^(2n))(Sqrt[1-4(x^2)])))

Chech this looks right:

Table[Series[v[a][x],{x,0,24}],{a,1,4}]

Table[Binomial[2b,b-a],{a,1,4},{b,0,12}]//MatrixForm

Testing the recursion relation on v[n+2] and v[n]:

Table[{Series[v[n+2][x],{x,0,24}],Series[ (((x^2)(c[x]-1))/(1-(x^2)-(x^2)c[x]))v[n][x],{x,0,24}]},{n,1,5}]

Now, let's see if the graph generating function g(x) = ( 1-x( f(x) ) ) / ( 1 - x( f(x)+1 ) ) works for f=c[x]+v[k1][x] + ... + v[kn][x]

Simple example:  k1=4

LHS2[x]=Simplify[c[x]+v[4+2][x]]
RHS2[x]=Simplify[((1-(x^2)(c[x]+v[4][x]))/(1-(x^2)(1+c[x]+v[4][x])))]

{Series[LHS2[x],{x,0,50}],Series[RHS2[x],{x,0,50}]}

Notice above the different coefficients at x^22.  boo.

LHS2[x]=Simplify[c[x]+v[5+2][x]]
RHS2[x]=Simplify[((1-(x^2)(c[x]+v[5][x]))/(1-(x^2)(1+c[x]+v[5][x])))]

{Series[LHS2[x],{x,0,50}],Series[RHS2[x],{x,0,50}]}

Series[Simplify[(((x^2)(c[x]-1))/(1-(x^2)-(x^2)c[x]))],{x,0,20}]