TITLE: Enumeration of lattice paths of a specific type
QUESTION [6 upvotes]: One of the approaches to "Special" meanders led (in particular) to the following question:

What is the number $a_{m,n}(\ell)$ of $\ell$-step paths from $(1,1)$ to $(m,n)$ using the following four kinds of steps: $(i,j)\mapsto$ $(i+1,j)$, $(i+1,j-1)$, $(i,j+1)$ or $(i-1,j+1)$ and with the restriction that $i$ and $j$ stay positive at each step?

The only thing I managed to find out is sort of a trivial reformulation of the definition: the polynomials
$$
P_\ell(x,y):=\sum_{m,n\geqslant1}a_{m,n}(\ell)x^my^n
$$ 
satisfy a recurrence: since $P_\ell(x,0)=P_\ell(0,y)=0$, each $F_\ell(x,y):=(x+y+x/y+y/x)P_\ell(x,y)$ is a polynomial, and we have
$$
P_{\ell+1}(x,y)=F_\ell(x,y)-F_\ell(x,0)-F_\ell(0,y).
$$
In a way of example - here is the table of the $a_{m,n}(7)$:
$$
\begin{array}{cccccccc}
 0 & 1 & 21 & 80 & 125 & 85 & 21 & 1 \\
 1 & 28 & 139 & 254 & 210 & 76 & 7 & 0 \\
 21 & 139 & 306 & 308 & 140 & 21 & 0 & 0 \\
 80 & 254 & 308 & 168 & 35 & 0 & 0 & 0 \\
 125 & 210 & 140 & 35 & 0 & 0 & 0 & 0 \\
 85 & 76 & 21 & 0 & 0 & 0 & 0 & 0 \\
 21 & 7 & 0 & 0 & 0 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
$$
As literature on the subject of lattice paths is way too vast for me to handle, this is mostly a reference request: I believe that working hardly enough I could find at least a generating function, but I also believe it must be done somewhere already.
In fact a more (or less?) specific question in this direction is whether there is some kind of searchable database of combinatorial objects where I could find such things.
There is a question Combinatorial Databases here on MO but I could not find anything about path enumeration in the answers. The closest I could get is the Dyck path enumeration on FindStat

REPLY [2 votes]: This is an extension of the accepted answer. Using it, and ideas from some other papers, I managed to obtain an alternative expression for the generating function which I find interesting.
As explained in that answer, the problem reduces to finding $Q(x,0,z):=\sum_{i,\ell}a_{n,1}(\ell)x^nz^\ell$.
Let $z=\frac1{q+q^{-1}}$ and $x=-\frac{(1-q) (1-q^2) t}{(1-t) (1-q t)q}$. Then,
$$
x^2zQ(x,0,z)=-x+\frac{1-4z^2}{z^2}\sum _{k=1}^\infty \frac{(-1)^kq^kt}{\left(1-q^kt\right)^2}.
$$
The series $\sum _{k=1}^\infty \frac{(-1)^kq^kt}{\left(1-q^kt\right)^2}$ seems to be related to elliptic functions, Jacobi forms and such stuff, but I cannot figure out how exactly. Maybe I will ask a separate question about it.
Few words about the proof. The expression for the generating function in the other answer can be equivalently described as follows. Suppose $x$, $y$ satisfy $(x^2+y^2+x^2y+xy^2)z=xy$. Then,
$$
(*)\qquad\qquad\qquad\qquad\qquad\qquad x^2zQ(x,0,z)=xy-y^2zQ(y,0,z).\qquad\qquad\qquad\qquad\qquad\qquad
$$
Now the curve $(x^2+y^2+x^2y+xy^2)z=xy$ admits rational parametrizations $x(t)$, $y(t)$; moreover it is possible to choose such parametrization that satisfies $y(t)=x(qt)$, with $z=\frac1{q+q^{-1}}$ and $x(t)=-\frac{(1-q) (1-q^2) t}{(1-t) (1-q t)q}$. Then, denoting $F(t,q)=\frac{x(t)^2}{q+q^{-1}}Q\left(x(t),0,\frac1{q+q^{-1}}\right)$, we get from $(*)$ above the following functional equation for $F$:
$$
F(t,q)=\frac{(1-q)^2 \left(1-q^2\right)^2 t^2}{(1-t) (1-q t)^2 \left(1-q^2 t\right)q}-F(qt,q).
$$
This leads more or less straightforwardly to the claimed equality.
Update
Expression in terms of "original" variables can be also derived from this. Although less pretty, it is, I believe, still more explicit than the one in terms of iterated radicals.
We have
$$
x^2zQ(x,0,z)=-x+\frac{1-4z^2}{z^2}\sum _{k=1}^\infty\frac{(-1)^k}{A(x,z)T_k\left(\frac{1}{2 z}\right)+B(x,z)U_{k-1}\left(\frac{1}{2 z}\right)-2},
$$
where $T$ and $U$ are the Chebyshev polynomials, while $A(x,z)$ and $B(x,z)$ are quadratic irrationalities determined by
$$
\begin{aligned}
A+\frac{2z}{1+2z}B&=2\left(1-\frac{1-2 z}{x z}\right),\\
A^2-\frac{4z^2}{1-4z^2}B^2&=4
\end{aligned}
$$