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 $$Q(x,y;z) = \sum_{m,n\geq 1} x^{m-1}y^{n-1} \sum_{\ell\geq 0} z^l a_{m,n}(\ell) $$ 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

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

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 $$Q(x,y;z) = \sum_{m,n\geq 1} x^{m-1}y^{n-1} \sum_{\ell\geq 0} z^l a_{m,n}(\ell) $$ 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

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

It seems the first solution to this problem appeared 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 third 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

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

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 $$Q(x,y;z) = \sum_{m,n\geq 1} x^{m-1}y^{n-1} \sum_{\ell\geq 0} z^l a_{m,n}(\ell) $$ 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

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