This series was introduced in [KY15], where this model was first solved, using heavy computer algebra.

By setting $y = 0$ in (A.6) (with $t$ and $A_1$ expressed in terms of $Z$), we find that $Q(0,0)$ lies in $\mathbb{Q}(\lambda, \sqrt{1+4Z})$, and has degree 6 over $\mathbb{Q}(t, \lambda)$. More generally, $Q(0, y)$ is quadratic over $\mathbb{Q}(\lambda, y, Z)$, and we recover its expression in terms of $Z$ given in [KY15, Sec. 5.3] (note that the model we consider here differs by a diagonal symmetry from the one of [KY15]). Moreover, $Q(0, y)$ has a rational expression in terms of $\lambda, Z$ and $V$, where $V$ is the unique series in $t$, with coefficients in $\mathbb{Q}(\lambda, y)$ and constant term 0, satisfying $V = Z(Zy - 1)(1 + V)^2$.

It remains to get back to $Q(x, 0)$, which can be done using the equation $R(x) + S(Y_0) = xY_0 + R(0)$, where $Y_0$ is the root of the kernel that is a power series in $t$. It is found that $Q(x, 0)$ has degree 4 over $\mathbb{Q}(\lambda, x, Z)$, degree 12 over $\mathbb{Q}(\lambda, t, x)$, and one recovers the expression given in [KY15, Sec. 5.3]. Moreover, $Q(x, 0)$ admits a rational expression in terms of $\lambda, \sqrt{1+4Z}$ and $U$, where $U$ is the unique series in $t$ with coefficients in $\mathbb{Q}(\lambda)$ and constant term 0 satisfying $U = Z((2 + \lambda)xZ + x - 1)(1 + U)^2$.

An alternative solution is described in [BM16a, Sec. 4].

## A.6. SECOND WEIGHTED MODEL

In this example, $L(y)$ has simple poles at 0 and at $-1$, while $J(y)$ has a double pole at both points. Fortunately, eliminating the pole at 0 also eliminates the pole at $-1$, and Lemma 4.18, applied with $\rho = 1$, yields

$$J(y) = C_2L(y)^2 + C_1L(y) + C_0,$$

with

$$C_2 = t^2, \quad C_1 = -t - 4t^2, \quad C_0 = -2t - 4t^2 + 2t^2S'(0).$$

Returning to $Q(0, y) \equiv Q(y)$ gives a quadratic equation of the form (A.1), involving a single auxiliary series $A_1 = Q(0, 0)$:

$$y^2 t^2 (y+1)^2 Q(y)^2 + (2ty^3 - 2ty^2 - y^2 - 2t + y)Q(y) + 2tA_1 + y^2 - y = 0. \quad (A.7)$$

Equation (A.2) has two solutions (one more than needed to determine $A_1$!). One of them reads $2t + O(t^2)$, the other is $1 - 2t + O(t^3)$. Both are double roots of the discriminant of $Pol$ with respect to its first variable. This gives for $A_1$ a cubic equation over $\mathbb{Q}(t)$, and $A_1$ admits a rational expression in terms of the unique series $Z \equiv Z(t)$, with constant term 0, satisfying

$$Z = t(2 + 2Z + 4Z^2 + Z^3). \quad (A.8)$$

More precisely,

$$Q(0, 0) = \frac{Z(4 - 4Z + Z^3)}{8t}. \quad (A.9)$$

(In fact, $Z$ is one of the series $Y$ satisfying (A.2).) The series $Q(0, y)$ is quadratic over $\mathbb{Q}(Z, y)$, as follows from (A.7), and admits a rational expression in $Z$ and $\sqrt{1 - yZ(2 + Z)}$. Since the model is $x/y$-symmetric, this completes its solution.

As mentioned in [BM16a, Sec. 4], this model can also be solved using the “half-orbit” approach of [BMM10, Sec. 6].

## A.7. THIRD WEIGHTED MODEL

This model is obtained by reversing steps of the previous one. In particular, its $y$-invariant is obtained by replacing $y$ by $1/y$ in the invariant of the previous model. It has poles at 0 and $-1$, while $L(y)$ has a simple pole at 0 only. We first eliminate the (double) pole of $J(y)$ at $-1$, by considering $(L(y) - L(1))J(y)$: this indeed suffices, as $L(y) - L(1)$ has a double root at $-1$. Then we eliminate the resulting double pole at 0, and apply Lemma 4.18 with $\rho = 1$ to obtain:

$$(L(y) - L(-1))J(y) = C_2L(y)^2 + C_1L(y) + C_0,$$

where

$$C_2 = t^2, \quad C_1 = -t(2+5t+tS(0)), \quad C_0 = t(5t+2)S(0)+t(1+3t)S'(0)+\frac{(1+3t)(13t^2+7t+1)}{t}.$$
Returning to $Q(0, y)$ gives a quadratic equation of the form (A.1), involving two auxiliary series $A_1 = Q(0, 0)$ and $A_2 = Q'_y(0, 0)$:

$$ 
\begin{aligned}
& yt^3(y+1)^4 Q(y)^2 + Q(y)\times \\
& (t^2y^5 + 2t^2y^4 - t^2y^3 - ty^4 - t^2y^2 - 3ty^3 - 11t^2y + ty^2 - 3t^2 - 4ty + y^2 - t - yt^3(y+1)^2A_1) \\
& - t(ty^3 - ty^2 - 11ty - y^2 - 3t - 4y - 1)A_1 + ty(1+3t)A_2 + y^2(ty^2 + 2ty - 2t - 1) = 0.
\end{aligned} 
\quad (A.10) 
$$

Two series (in $\sqrt{t}$) satisfy (A.2). Hence the discriminant of $Pol$ with respect to its first variable admit two double roots. This gives a pair of equations satisfied by $A_1$ and $A_2$, and finally both series turn out to be cubic over $Q(t)$. Moreover, they have rational expressions in terms of the series $Z$ defined by (A.8). Of course, $A_1 = Q(0, 0)$ is still given by (A.9), since reversing steps does not change the excursion generating function. For $A_2$, we find

$$ A_2 = \frac{Z^2(Z+2)(Z^5 + 28Z^4 + 4Z^3 - 56Z^2 + 32)}{256t(1+Z)^2}. $$

Returning to (A.10) shows that $Q(0, y)$ is quadratic over $Q(y, Z)$, and can be expressed rationally in terms of $y, Z$ and $\sqrt{4 - 4(y-2)Z + (4 - 8y + y^2)Z^2 - 6yZ^3 - yZ^4}$.

As mentioned in [BM16a, Sec. 4], this model can also be solved using the “half-orbit” approach of [BMM10, Sec. 6].

### A.8. FOURTH (AND LAST) WEIGHTED MODEL

This model, which has never been solved so far, differs from the one of Section A.6 by a reflection in a vertical line. Hence it has the same $y$-invariant, but the $x/y$-symmetry is lost. The $y$-invariant has double poles at $0$ and $-1$, while $L(y)$ only has a simple pole at $0$. We first eliminate the pole of $J(y)$ at $-1$ by considering $(L(y) - L(-1))J(y)$ (again, this is sufficient since $-1$ is a double root of $L(y) - L(-1)$). Then we eliminate the pole at $0$, apply Lemma 4.18 with $\rho = 1$, and obtain:

$$ (L(y) - L(-1))J(y) = C_3L(y)^3 + C_2L(y)^2 + C_1L(y) + C_0, \quad (A.11) $$

where

$$ C_3 = t^2, \quad C_2 = -t - 2t^2 S(0), \quad C_1 = -3t(1 + 3t) + t(1 - 2t)S(0) + t^2 S(0)^2 - 2t^2 S'(0), $$

and

$$ C_0 = -1 - 7t - 17t^2 + 2t(1 + 2t)S(0) + 2t^2 S(0)^2 + t(1 - 2t)S'(0) - t^2 S''(0). $$

Returning to $Q(0, y) \equiv Q(y)$, this gives a cubic equation of the form (A.1), involving no less than three auxiliary series, namely $A_1 = Q(0)$, $A_2 = Q'(0)$ and $A_3 = Q''(0)$, the derivatives being still taken with respect to $y$.

As in Gessel's case, we find that three series $Y$ cancel (A.2). One of them is a series in $t$, namely $Y_0 = 2t + 4t^2 + 24t^3 + O(t^4)$, and the other two are series in $s := \sqrt{t}$, namely $Y_+ = s + s^2 + 7s^5/2 + O(s^6)$ and $Y_- = -s + s^2 - 7s^5/2 + O(s^6)$. Here there is no obvious guess for their exact values. However, upon eliminating $A_2$ and $A_3$ between (A.1), (A.2) and (A.3), we find that each series $Y$ must satisfy:

$$ Y(Y+1)(3tY^3 - 3tY^2 + Y^3 + tY - Y^2 + t)(tY^3 + 4tY^2 + 2tY + 2t - Y)(tYQ(Y) + 1) = 0. $$

Using the first few coefficients of the three series $Y$, we conclude that $P_0(Y_0) = 0$ and $P_1(Y_+) = P_1(Y_-) = 0$, with

$$ P_0(y) = ty^3 + 4ty^2 + 2ty + 2t - y \quad \text{and} \quad P_1(y) = 3ty^3 - 3ty^2 + y^3 + ty - y^2 + t. $$

In particular, the three series $Y$ are cubic. Following the approach of [BMJ06, Sec. 7], we conclude that the discriminant of $Pol$ with respect to its first variable, evaluated at $A_1, A_2, A_3, t, y$, has three double roots in $y$, namely $Y_0, Y_+$ and $Y_-$.
To get a clearer view of what happens, we first note that $P_1(y) = y^3 P_0(-1+1/y)$. Hence the three roots of $P_0$ are $Y_0, -1+1/Y_+$ and $-1+1/Y_-$. Equivalently, $P_0(y) = (1+y)^3 P_1(1/(1+y))$, and the three roots of $P_1$ are $Y_+, Y_-$ and $1/(1+Y_0)$.

Up to powers of $t, y$ and $y+1$, the discriminant of $P_1$ with respect to its first variable, evaluated at $A_1, A_2, A_3, t, y$, is a polynomial in $t, A_1, A_2, A_3, y$, of degree 15 in $y$, with dominant coefficient $-4t^3(1+3t)^3$. We denote it by $\Delta(y)$. Then the polynomial

$$ (y-1)^6 \Delta(y) - y^{18} (y+1)^6 \Delta(-1+1/y) \qquad (A.12) $$

has a factor $P_1(y)^2$. Specializing this identity at $y = Y_+$ shows that $-1+1/Y_+$ is also a root of $\Delta(y)$. Of course, the same holds for $-1+1/Y_-$, so that finally all roots of $P_0(y)$ cancel $\Delta(y)$. Moreover, since the above polynomial (A.12) admits $P_1(y)^2$ as a factor, $-1+1/Y_+$ and $-1+1/Y_-$ are in fact double roots of $\Delta(y)$. This means that $\Delta(y)$ has a factor $P_0(y)^2$.

Replacing $y$ by $1/(1+y)$ in (A.12) shows that

$$ y^6 (y+1)^{18} \Delta(1/(1+y)) - (y+2)^6 \Delta(y) $$

admits $P_0(y)^2$ as a factor. Using the same argument as before, we conclude that $\Delta(y)$ is also divisible by $P_1(y)^2$.

We have now proved that $\Delta(y)$ is divisible by $P_0(y)^2 P_1(y)^2$. Performing the Euclidean division of $\Delta(y)$ by $P_0(y)^2 P_1(y)^2$ yields a remainder of degree 11 in $y$, and all its coefficients (which are polynomials in $t$ and the $A_i$'s) must vanish. By performing eliminations between three of them (we have chosen the coefficients of $y^{11}, y^1$ and $y^0$), we find that the three series $A_i$ have degree 8, and in fact belong to the same extension of degree 8 of $\mathbb{Q}(t)$. In particular, $A_1 = \mathbb{Q}(0,0)$ reads

$$ A_1 = \frac{-1 - 6t + \sqrt{Z}}{2t^2}, $$

where $Z = 1 + 12t + 40t^2 + O(t^3)$ satisfies a quartic equation:

$$ \begin{aligned} & 27 Z^4 - 18 (10000 t^4 + 9000 t^3 + 2600 t^2 + 240 t + 1) Z^2 \\ & \quad + 8 (10 t^2 + 6 t + 1) (102500 t^4 + 73500 t^3 + 14650 t^2 + 510 t - 1) Z \\ & = (10000 t^4 + 9000 t^3 + 2600 t^2 + 240 t + 1)^2. \end{aligned} $$

This equation has genus 1, so there will not be any rational parametrization in this case. The Galois group of the above polynomial is the symmetric group on four elements, hence there is no extension of order 2 between $\mathbb{Q}(t)$ and $\mathbb{Q}(t, Z)$.

Once $A_1, A_2$ and $A_3$ are determined, one returns to (A.11). Expressing the series $A_i$ as rational functions in $t$ and $A_1$ shows that $Q(0, y)$ is cubic over $\mathbb{Q}(t, y, A_1)$, and by eliminating $A_1$, one finds that it has degree 24 over $\mathbb{Q}(t, y)$.

It remains to get back to $Q(x, 0)$, which can be done using the equation $R(x) + S(Y_0) = xY_0 + R(0)$, where $Y_0$ is the root of the kernel that is a power series in $t$. It is found that $Q(x, 0)$ has degree 3 over $\mathbb{Q}(t, x, A_1)$, and degree 24 over $\mathbb{Q}(t, x)$.

OB: BRANDEIS UNIVERSITY, DEPARTMENT OF MATHEMATICS, 415 SOUTH STREET, WALTHAM, MA 02453, USA, MBM: CNRS, LABRI, UNIVERSITÉ DE BORDEAUX, 351 COURS DE LA LIBÉRATION, 33405 TALENCE CEDEX, FRANCE, KR: CNRS, INSTITUT DENIS POISSON, UNIVERSITÉ DE TOURS ET UNIVERSITÉ D'ORLÉANS, PARC DE GRANDMONT, 37200 TOURS, FRANCE