One of the approaches to ["Special" meanders](https://mathoverflow.net/questions/146802/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](https://mathoverflow.net/q/187001/41291) 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](http://www.findstat.org/DyckPaths?highlight=%28lattice%29%7C%28paths%29) on [FindStat](http://www.findstat.org)



Enumeration of lattice paths of a specific type
It seems the first solution to this problem appeared in Theorem 4 of


*Raschel, Kilian*, [**Counting walks in a quadrant: a unified approach via boundary value problems**](http://dx.doi.org/10.4171/JEMS/317), J. Eur. Math. Soc. (JEMS) 14, No. 3, 749-777 (2012). [ZBL1238.05014](https://zbmath.org/?q=an:1238.05014).


Notice that the walk corresponds to the fourth singular walk of Figure 2. The kernel $K(x,y;z)$ of the walk is given by $$K(x,y;z) = x y z(x +x/y +y +y/x -1/z).$$
Let $X\_0(y) = X\_0(y;z)$ (resp. $Y\_0(x)=Y\_0(x;z)$) be one of the solutions $x$ (resp. $y$) to $K(x,y;z) = 0$. Then Theorem 4 states that
$$Q(x,y;z) = \sum\_{m,n\geq 1} x^{m-1}y^{n-1} \sum\_{\ell\geq 0} z^l a\_{m,n}(\ell) $$
satisfies
$$ Q(x,y;z) = \frac{1}{K(x,y;z)}\left(z x^2Q(x,0;z)+z y^2 Q(0,y;z)-xy\right)$$
with
$$ Q(x,0;z) = \frac{1}{zx^2} \sum\_{p\geq 0} Y\_0 \circ (X\_0\circ Y\_0)^{\circ p}(x;z)\,\left[ (X\_0\circ Y\_0)^{\circ p}(x;z) - (X\_0\circ Y\_0)^{\circ (p+1)}(x;z) \right].$$
Here $f^{\circ p}$ means $f\circ \cdots \circ f$ with $p$ occurrences of $f$. Notice that by symmetry $Q(0,y;z) = Q(y,0;z)$.


A quick check with Mathematica reproduces the table for $\ell=7$:



```
k = x y z (x + x/y + y + y/x - 1/z);
x0[y_] = x /. Solve[k == 0, x][[1]] // FullSimplify[#, y > 0 && x > 0] &;
y0[x_] = y /. Solve[k == 0, y][[1]] // FullSimplify[#, y > 0 && x > 0] &;
Series[SeriesCoefficient[1/k (Sum[y0[xp] (xp - x0[y0[xp]]) /. 
   xp -> Nest[x0[y0[#]] &, x, p], {p, 0, 2}] 
   // -x y + # + (# /. x -> y) &), {z, 0, 7}], {x, 0, 7}, {y, 0, 7}]

```

yields the output


$$
\begin{aligned}
&\ \ \ \left(y+21 y^2+80 y^3+125 y^4+85 y^5+21 y^6+y^7+O\left(y^8\right)\right)\\+&x\ 
 \left(1+28 y+139 y^2+254 y^3+210 y^4+76 y^5+7 y^6+O\left(y^8\right)\right)\\+&x^2
 \left(21+139 y+306 y^2+308 y^3+140 y^4+21 y^5+O\left(y^8\right)\right)\\+&x^3
 \left(80+254 y+308 y^2+168 y^3+35 y^4+O\left(y^8\right)\right)\\+&x^4
 \left(125+210 y+140 y^2+35 y^3+O\left(y^8\right)\right)\\+&x^5 \left(85+76 y+21
 y^2+O\left(y^8\right)\right)\\+&x^6 \left(21+7
 y+O\left(y^8\right)\right)\\+&x^7+O\left(x^8\right)
\end{aligned}
$$


