I have the numbers $f(j, a, b)$ for $j=0,1,\dots$ and $b=0,1, \dots, j$ and $a=0,1,2,3$. They satisfy the recursion
$$ f(j, a, b) = \begin{cases} 0, & j=0, (a,b)\neq (0,0) \\ 1, & j=0, (a,b) = (0,0) \\ 0, & j>0, (a,b) = (0,0) \\ f(j-1, a-1, b+1), & j>0, a>0 \\ \sum_{k=0}^3 f(j-1, k, b-1), & j>0, a=0 \end{cases} $$
I'm trying to find the generating function
$$F(z, u, v) = \sum_{j=0}^\infty \sum_{a=0}^3 \sum_{b=0}^j f(j,a,b)z^ju^av^b.$$
And then especially $G(z)=F(z,1,1)$
If the sum in the last case goes to $2$ instead of $3$ (let's call it $r$), we get (for $f(n)=\sum_{a=0}^2 \sum_{b=0}^n f(n,a,b) $) the sequence A191519 which has the generating function
$$G_{r=2}(z)=\frac{2}{1-2x-x^2+\sqrt{1-2x^2-3x^4}}$$
And for $r=1$, we get the Fibonacci numbers.
I was wondering if we get something similar for $r=3$ (and higher)?
My try:
$$F(z, u, v) = \sum_{j=0}^\infty \sum_{a=0}^3 \sum_{b=0}^j f(j,a,b)z^ju^av^b \\ = 1 + z\sum_{j=0}^\infty z^j \sum_{b=0}^j \left ( \sum_{k=0}^3 f(j,k,b-1)v^b + \sum_{a=1}^3 f(j,a-1,b+1)u^av^b \right ) \\ =1 + z\sum_{j=0}^\infty z^j \left ( \sum_{k=0}^3 \sum_{b=1}^j f(j,k,b-1)v^b + \sum_{a=1}^3 \sum_{b=0}^{j-1} f(j,a-1,b+1)u^av^b \right ) \\ = 1 + z\sum_{j=0}^\infty z^j \left ( \sum_{k=0}^3 \sum_{b=0}^{j-1} f(j,k,b)v^b + v\sum_{a=1}^3 \sum_{b=0}^{j-1} f(j,a-1,b)u^av^b \right ) \\ = 1 + z\sum_{j=0}^\infty z^j \sum_{b=0}^{j-1}v^b \left( \sum_{k=0}^3 f(j,k,b) + vu\sum_{a=0}^2 f(j,a,b)u^a \right ) \\ = 1 + zF(z,1,v) + z\sum_{j=0}^\infty \sum_{b=0}^{j-1} vu\sum_{a=0}^2 f(j,a,b)u^a z^j v^b $$
Because $f(j,3,b) = f(j-1, 2, b+1) = f(j-2, 1, b+2) = f(j-3, 0, b+3)$, we have
$$ \sum_{a=0}^2 f(j,a,b)u^a = \frac{3}{4} \sum_{a=0}^2 f(j,a,b)u^a + \frac{1}{4}\left(f(j+3,3,b-3)+f(j+2,3,b-2)+f(j+1,3,b-1)\right) \\ = \frac{1}{4} \left (\sum_{m=1}^3 \left(\sum_{a=0}^2 f(j,a,b)u^a + f(j+m,3,b-m) \right) \right) $$
I was trying to make the "missing" $a=3$ term appear, but they are of the wrong sort: if we put the $z$'s and the $v$'s in there, they don't fit together, I think. Anyway, in order to get squareroots in the formula for $F$, we should have $F^2$ in a formula where we solve it from. How is the case $r=2$ generating function found?