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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
$$
P_ell(x,y):=sum_m,ngeqslant1a_m,n(ell)x^my^n
$$
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 $a_m,n(7)$:
$$
beginarraycccccccc
0 & 1 & 21 & 80 & 125 & 85 & 21 & 1 \
1 & 28 & 139 & 254 & 210 & 76 & 7 & 0 \
21 & 139 & 306 & 308 & 140 & 21 & 0 & 0 \
80 & 254 & 308 & 168 & 35 & 0 & 0 & 0 \
125 & 210 & 140 & 35 & 0 & 0 & 0 & 0 \
85 & 76 & 21 & 0 & 0 & 0 & 0 & 0 \
21 & 7 & 0 & 0 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
endarray
$$

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

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, J. Eur. Math. Soc. (JEMS) 14, No. 3, 749-777 (2012). ZBL1238.05014.

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
$$Q(x,y;z) = sum_m,ngeq 1 x^m-1y^n-1 sum_ellgeq 0 z^l a_m,n(ell) $$
satisfies
$$ Q(x,y;z) = frac1K(x,y;z)left(z x^2Q(x,0;z)+z y^2 Q(0,y;z)-xyright)$$
with
$$ Q(x,0;z) = frac1zx^2 sum_pgeq 0 Y_0 circ (X_0circ Y_0)^circ p(x;z),left[ (X_0circ Y_0)^circ p(x;z) - (X_0circ Y_0)^circ (p+1)(x;z) right].$$
Here $f^circ p$ means $fcirc cdots circ f$ 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

$$left(y+21 y^2+80 y^3+125 y^4+85 y^5+21 y^6+y^7+Oleft(y^8right)right)+x
left(1+28 y+139 y^2+254 y^3+210 y^4+76 y^5+7 y^6+Oleft(y^8right)right)+x^2
left(21+139 y+306 y^2+308 y^3+140 y^4+21 y^5+Oleft(y^8right)right)+x^3
left(80+254 y+308 y^2+168 y^3+35 y^4+Oleft(y^8right)right)+x^4
left(125+210 y+140 y^2+35 y^3+Oleft(y^8right)right)+x^5 left(85+76 y+21
y^2+Oleft(y^8right)right)+x^6 left(21+7
y+Oleft(y^8right)right)+x^7+Oleft(x^8right)$$

Bousquet-Melou's work might be pertinent. See
https://arxiv.org/abs/1708.06192
Figure 4 there seems to be your lattice walk.