\section{Tight gradings and protected matching}\label{sec:tight}

Fix a primitive pair $d,e>0$ and write
\[
 G_{d,e}(z)=1+\sum_{k\geq1}c_kz^k.
\]
For a formal series $J$ in named variables, brackets such as $[z^k]J$ denote the coefficient of the indicated monomial; absent monomials have coefficient zero. Thus $c_k=[z^k]G_{d,e}$.

For each $k\geq1$, choose positive integers $D_1,D_2$ satisfying
\begin{equation}\label{eq:ambient}
 eD_1-dD_2=1,\qquad D_1>kd+1,\qquad D_2>ke.
\end{equation}
B\'ezout's identity gives one integral solution, and adding a sufficiently large multiple of $(d,e)$ gives these inequalities. In particular $\gcd(D_1,D_2)=1$. All constructions involving $D_1,D_2$ in this section are for this fixed $k$.

\subsection{Paths, shadows, and the coefficient formula}

The maximal lower Dyck path $C(D_1,D_2)$ joins $(0,0)$ to $(D_1,D_2)$ by east and north unit edges and stays as close as possible to the segment from $(0,0)$ to $(D_1,D_2)$ while remaining below it. Its word is
\[
 \prod_{x=1}^{D_1} EN^{\lfloor xD_2/D_1\rfloor-\lfloor(x-1)D_2/D_1\rfloor},
\]
where concatenation reads from left to right. We also read the path cyclically, or extend it periodically by translation by $(D_1,D_2)$. Its edges retain their labels under cyclic reading.

A grading $\omega$ assigns a nonnegative integer to each edge. An edge is selected if its grade is positive. For an oriented interval $J$ of the path, let $J_E,J_N$ be its horizontal and vertical edges, respectively, and write $\omega(J_E)=\sum_{h\in J_E}\omega(h)$ and $\omega(J_N)=\sum_{v\in J_N}\omega(v)$. We consider gradings of totals
\[
 p=\omega(C_N)=kd,\qquad q=\omega(C_E)=ke.
\]
For a selected horizontal edge $h$, its horizontal shadow interval starts at $h$ and ends at the first vertical edge for which $|J_N|=\omega(J_E)$. Its shadow $\sh(h)$ is the set of vertical edges in this interval. For a selected vertical edge $v$, read backwards from $v$ to the first horizontal edge for which $|J_E|=\omega(J_N)$; the horizontal edges in that interval form $\sh(v)$. A zero-grade edge has empty shadow. Since $q<D_2$ and $p<D_1$, these first returns exist within one period. Shadows of collections of edges mean unions of their shadows.

Compatibility means the following. For every selected horizontal edge $h$ and selected vertical edge $v$, in the cyclic interval from $h$ to $v$ at least one of these conditions holds:
\begin{itemize}
\item a prefix ending at a vertical edge strictly before $v$ has as many vertical edges as total horizontal grade;
\item a suffix beginning at a horizontal edge strictly after $h$ has as many horizontal edges as total vertical grade.
\end{itemize}
These are the compatible-grading conditions of~\cite[Definition~3.14]{BLM}, with the endpoints made explicit. A compatible grading is \emph{shadowed} if all selected vertical edges lie in the horizontal shadows, or all selected horizontal edges lie in the vertical shadows. Under \eqref{eq:ambient} and the prescribed totals, these are exactly the tight gradings of~\cite[Definition~5.4]{BLM}, since
\[
 pD_2-qD_1=-k=-\gcd(p,q).
\]
Its weight is
\begin{equation}\label{eq:weight}
 \wt(\omega)=\prod_{h\in C_E}v_{\omega(h)}
              \prod_{v\in C_N}w_{\omega(v)}.
\end{equation}
Thus horizontal grades are bounded by $b$ and vertical grades by $a$ in every nonzero contribution.

We use the following coefficient formula, in precisely the determinant-one case of~\cite[Theorem~1.2]{BLM}:
\begin{equation}\label{eq:tight}
 c_k=\sum_{\omega\text{ tight on }C(D_1,D_2)}\wt(\omega).
\end{equation}
The hypotheses follow from \eqref{eq:ambient}. Equivalently, \cite[Corollary~5.14(2), Theorem~5.17]{BLM} applies with exponent $|dD_2-eD_1|=1$. Appendix~\ref{sec:coefficient} derives the formula in this strict determinant-one case.

The formula is a polynomial identity in the initial coefficients and consequently holds over the given field with the given weights. One can implement specialization in the completed tropical vertex group by first substituting the universal coefficients as $p_{1,s}\mapsto w_s\tau^s$ and $p_{2,h}\mapsto v_h\tau^h$, with an auxiliary formal variable $\tau$. These substitutions are continuous for the coefficient-ideal and $\tau$-adic completions. Homogeneity identifies the result with $P_1(\tau x),P_2(\tau y)$, and identifies every coefficient of bidegree $(p,q)$ after removing its common factor $\tau^{p+q}$. This proves \eqref{eq:tight} for every specialization, including vanishing weights.

\subsection{The matching rule}

A positive horizontal grade $h$ opens a packet of $h$ blue tokens. A vertical edge removes one token from the most recently opened nonempty packet. The emptying of a packet is the first return defining its horizontal shadow interval: all packets opened inside it must empty before its last token is removed.

This description is valid also for the cyclic path. On the periodically extended path, give a horizontal edge increment equal to its grade and a vertical edge increment $-1$. A selected horizontal edge starts a positive excursion; the total increment over one period is $q-D_2<0$, so its first return exists before one full period. If another excursion starts inside it, the later excursion returns before the earlier one: otherwise at the earlier return its increment from its own starting height would already be negative, and a downward step of size one must have crossed zero first. Thus the intervals are nested or disjoint, including across the cyclic endpoint. Every selected horizontal edge lies in a maximal interval. Summing the first-return equalities over disjoint maximal intervals shows that their total number of vertical edges is $q$. Since $q<D_2$, a vertical edge outside them exists; cutting there makes all maximal intervals ordinary linear intervals. Reading backwards with vertical increments equal to their grades and horizontal increments $-1$ proves the same facts for vertical shadows, whose total horizontal mass is $p$.

Independently, let each horizontal edge create one red token, and let a vertical edge of grade $s$ remove the $s$ most recent unmatched red tokens whenever possible. The first-return interpretation of this second matching will encode compatibility.

\begin{lemma}\label{lem:red}
Suppose the red matching succeeds through a positive-grade vertical edge $v$. If $h$ is the earliest horizontal edge whose token is removed at $v$, then the interval from $h$ to $v$ is the shortest backwards interval satisfying $|J_E|=\omega(J_N)$.
\end{lemma}
\begin{proof}
Let the red height after a prefix be the number of its horizontal edges minus its total vertical grade. Write $H$ for its value just after $v$. The token created at $h$ was inserted from height $H$ to height $H+1$ and remained unmatched until $v$. Every horizontal edge strictly between $h$ and $v$ was therefore inserted from a height greater than $H$. The interval starting at $h$ is balanced, and no interval starting at a later horizontal edge is balanced.
\end{proof}

Consider a maximal blue interval. Its first horizontal edge is its root. A positive horizontal edge is called active until its blue packet empties. If it empties strictly before a given vertical edge, its closing edge supplies the horizontal compatibility witness. Otherwise its red token must be protected from the demand at that vertical edge. In particular the demand must stay above the most recent active root.

This gives a stack of records $(r,z)$. Here $r$ is the remaining number of blue tokens of an active root, and $z$ is the number of free red tokens above its protected token and below any more recent active root. Only the top record is changed by the next edge:
\begin{enumerate}
\item A zero-grade horizontal edge increases its $z$ by one.
\item A horizontal edge of grade $h>0$ pushes the record $(h,0)$.
\item A vertical edge of grade $s\geq0$ requires $z\geq s$ and replaces $(r,z)$ by $(r-1,z-s)$.
\item If $r$ becomes zero, pop the record. If a parent remains, add $z+1$ to its free balance; the extra token is the former root's token.
\end{enumerate}
The demand in step 3 is tested before step 4. For example, adjacent edges of grades $E_1N_1$ fail this test and are incompatible, whereas $E_1E_0N_1$ is accepted with final free balance zero.

\begin{lemma}\label{lem:stack}
On a maximal blue interval, compatibility for all forward selected pairs is equivalent to legality of the protected matching. If the interval has $m+1$ horizontal edges, its total vertical grade is at most $m$.
\end{lemma}
\begin{proof}
Assume all earlier steps are legal, and let $h$ be the most recent active root at a vertical edge $v$ of positive grade $s$. If $s\leq z$, Lemma~\ref{lem:red} gives a backwards witness strictly after $h$. It is also strictly after every older active root. Roots that have already closed have a strict horizontal witness.

If $s>z$, the relative red height after $v$, measured immediately after $h$, is negative. Before this step that relative height was nonnegative throughout the interval after $h$, since the token of $h$ remained protected. In particular the height just before every intervening horizontal edge was nonnegative. None can start a balanced suffix ending at $v$. The horizontal witness for $(h,v)$ also fails because $h$ is active at $v$. Thus compatibility fails at the first illegal step. Zero-grade vertical edges impose no selected-pair condition and satisfy the demand automatically.

The root remains protected through the last vertical edge of the interval. Its token is therefore never consumed inside the interval. At most the other $m$ red tokens can be consumed, proving the bound.
\end{proof}

Cyclic pairs are covered by the same witnesses. If a selected vertical edge is earlier than a selected horizontal edge after our cut, the cyclic interval passes the latter's blue closing edge before reaching its target. This is a strict horizontal witness. The same argument treats pairs in different maximal blue intervals.

\section{The single-root frame}\label{sec:frames}

We next determine the underlying words of all the tight gradings in \eqref{eq:tight}. This will replace the path of $k$-dependent size by a fixed finite automaton.

Assign to an edge starting at $(x,y)$ the rank
\[
 \rho(x,y)=D_2x-D_1y.
\]
The horizontal starts have ranks in $[0,D_1)$ and the vertical starts have ranks in $[D_1,D_1+D_2)$. Indeed, a horizontal edge starts at $(x,\lfloor D_2x/D_1\rfloor)$, and a vertical edge in column $x$ starts at a height between $\lfloor D_2(x-1)/D_1\rfloor$ and $\lfloor D_2x/D_1\rfloor-1$. Along each step the rank increases by $D_2$ modulo $D_1+D_2$. Coprimality implies that the ranks of all edge starts are exactly $0,1,\ldots,D_1+D_2-1$.

\begin{lemma}\label{lem:orientation}
Every tight grading of totals $(p,q)=(kd,ke)$ has all its selected vertical edges covered by horizontal shadows.
\end{lemma}
\begin{proof}
We show that the other shadow inclusion would force $pD_2-qD_1>0$. A maximal vertical shadow interval begins with a horizontal edge and ends with its root vertical edge. Write its numbers of edges as $n_i$ horizontal and $m_i+1$ vertical. First-return balance says that its total vertical grade is $n_i$.

For each selected horizontal edge in this interval, vertical compatibility with the root fails: the root's first backwards return reaches the beginning of the interval. Horizontal compatibility consequently supplies a first-return interval ending strictly before the root. All these horizontal shadows lie inside the maximal vertical interval and avoid its last edge. Restricting the horizontal grading to this interval preserves those shadows. Their union has size equal to its total horizontal grade, by the same blue matching as above. This grade is at most $m_i$.

Let $r_i$ be the rank of the first horizontal edge. The start of the last vertical edge has rank $r_i+n_iD_2-m_iD_1$. Since the former rank is below $D_1$ and the latter is at least $D_1$, we have $n_iD_2-m_iD_1>0$. All selected horizontal edges are in these maximal intervals under the assumed inclusion. Summing, and using the vertical shadow mass $\sum_i n_i=p$, gives
\[
 pD_2-qD_1
 =\sum_i(n_iD_2-m_iD_1)+(\sum_i m_i-q)D_1>0.
\]
But \eqref{eq:ambient} gives $pD_2-qD_1=-k$. Thus only the stated shadow inclusion is possible.
\end{proof}

\begin{proposition}\label{prop:single}
A tight grading of totals $(kd,ke)$ has exactly one maximal horizontal shadow interval. It has $kd+1$ horizontal edges and $ke$ vertical edges, begins at a horizontal edge of rank $\ell$ with $0\leq\ell<k$, and ends with free red balance zero. All selected edges belong to this interval.
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:orientation}, all selected vertical edges belong to the maximal blue intervals. Write their edge counts as $(m_i+1,n_i)$, their root ranks as $r_i$, and set
\[
 \delta_i=n_iD_1-m_iD_2.
\]
The last vertical edge has starting rank $r_i+D_1+D_2-\delta_i$, which is less than $D_1+D_2$. Hence $0\leq r_i<\delta_i$. Blue balance and Lemma~\ref{lem:stack} give $\sum_i n_i=q$ and $\sum_i m_i\geq p$. Therefore
\begin{equation}\label{eq:mass}
 k=qD_1-pD_2
  =\sum_i\delta_i+(\sum_i m_i-p)D_2.
\end{equation}
All $\delta_i$ are positive, and $D_2>ke\geq k$. The nonnegative integer $\sum_i m_i-p$ must be zero. Consequently $\sum_i m_i=p$ and $\sum_i\delta_i=k$.

Using $eD_1-dD_2=1$ gives
\[
 d\delta_i-m_i=(dn_i-em_i)D_1.
\]
Both $d\delta_i$ and $m_i$ lie in $[0,kd]$, whose length is less than $D_1$. Their difference is a multiple of $D_1$, so
\begin{equation}\label{eq:proportional}
 m_i=d\delta_i,\qquad n_i=e\delta_i.
\end{equation}
Starting at rank $r_i$, the horizontal edge reached after $r_id$ horizontal steps has relative height
\[
 \left\lfloor\frac{r_i+r_idD_2}{D_1}\right\rfloor=r_ie,
\]
and hence rank zero. It belongs to the interval because $r_i<\delta_i$ and $r_id\leq m_i$. Every maximal blue interval thus contains the same rank-zero horizontal edge. Since these intervals are disjoint, there is exactly one.

Its counts follow from \eqref{eq:proportional} and \eqref{eq:mass}, and its root rank satisfies $0\leq\ell<k$. All positive horizontal grades belong to a blue interval by construction. The $kd$ units of vertical grade consume all red tokens other than the root token, so its final free balance is zero.
\end{proof}

Let $A$ be the maximal lower Dyck word from $(0,0)$ to $(d,e)$. It starts with $E$ and ends with $N$; write
\begin{equation}\label{eq:words}
 A=EMN,\qquad U=EM,\qquad B=ENM,
\end{equation}
where $M$ is the intervening word, possibly empty. These words encode the two periodic phases of a frame. For example, $(d,e)=(2,3)$ gives $A=ENENN$, $M=NEN$, $U=ENEN$, and $B=ENNEN$.

\begin{lemma}\label{lem:word}
For every $0\leq\ell<k$, the interval starting at the horizontal edge of rank $\ell$ and ending at local displacement $(kd+1,ke)$ exists in $C(D_1,D_2)$ and has word
\begin{equation}\label{eq:frameword}
 W_{k,\ell}=A^\ell U B^{k-\ell-1}EN,\qquad 0\leq\ell<k.
\end{equation}
In particular this describes the interval in Proposition~\ref{prop:single}.
\end{lemma}
\begin{proof}
After $x$ horizontal steps from the root, the local height of the next horizontal edge is
\begin{align}
 y_\ell(x)
 &=\left\lfloor\frac{\ell+xD_2}{D_1}\right\rfloor\notag\\
 &=\left\lfloor\frac{ex}{d}\right\rfloor
   -\mathbf{1}_{\{d\mid x,\ x/d>\ell\}},
   \qquad 1\leq x\leq kd.\label{eq:height}
\end{align}
Here $\mathbf{1}$ is $1$ if its subscripted condition holds and $0$ otherwise. To verify the second equality, substitute $D_2=(eD_1-1)/d$. If $d\nmid x$, the distance of $ex/d$ to an integer is at least $1/d$, whereas the perturbation $(\ell-x/d)/D_1$ has absolute value less than $k/D_1<1/d$. For $x=jd$, the sign of $\ell-j$ gives the asserted correction.

At $x=kd$ the height is $ke-1$. The horizontal edge there has rank $\ell+D_1-k<D_1$; its successor has rank $\ell+D_1+D_2-k\geq D_1$, so the next edge is vertical. The frame stops at this edge. Formula~\eqref{eq:height} leaves the first $\ell$ copies of $A$ unchanged and moves the last $N$ of every later primitive block past the next $E$. This gives \eqref{eq:frameword}.
\end{proof}

An \emph{accepted decoration} of a word $W_{k,\ell}$ is a grading with horizontal grades at most $b$ and vertical grades at most $a$, whose first horizontal grade is positive, for which the protected matching is legal, whose root closes exactly at the last vertical edge, and whose final free balance is zero. Let $\operatorname{Accept}(W_{k,\ell})\in\FF$ be the sum of its weights as in \eqref{eq:weight}.

\begin{proposition}\label{prop:bijection}
For every $k\geq1$,
\begin{equation}\label{eq:accepted}
 c_k=\sum_{\ell=0}^{k-1}\operatorname{Accept}(W_{k,\ell}).
\end{equation}
\end{proposition}
\begin{proof}
Proposition~\ref{prop:single} and Lemmas~\ref{lem:stack} and~\ref{lem:word} send each nonzero-weight tight grading to an accepted decoration with its root rank. Conversely, start an accepted decoration at the unique horizontal edge of rank $\ell$ in $C(D_1,D_2)$ and set all outside grades to zero. Lemma~\ref{lem:word} identifies its word with the actual interval. The total length $k(d+e)+1$ is less than $D_1+D_2$, so no edge is repeated. Protected matching and the cyclic-pair argument prove compatibility. The root shadow covers every selected vertical edge.

Blue and red balance respectively give total horizontal grade $ke$ and total vertical grade $kd$, so this is a tight grading of the prescribed totals. The constructions are inverse and preserve weights. Distinct root ranks give distinct labelled gradings even when the words coincide. Equation~\eqref{eq:tight} now gives \eqref{eq:accepted}.
\end{proof}

\section{A finite system with one catalytic variable}\label{sec:system}

\subsection{The frame automaton}

For the fixed direction $(d,e)$, the family of words in \eqref{eq:frameword}, counted with its root-rank multiplicities, is the noncommutative series
\[
 \sum_{\ell,r\geq0}A^\ell U B^rEN.
\]
Construct a directed labelled graph with three distinguished vertices: the start, a second-stage vertex, and the finish. Attach an $A$-labelled cycle at the start, a $U$-labelled path from the start to the second stage, a $B$-labelled cycle at the second stage, and an $EN$-labelled path from the second stage to the finish. Replace each word-labelled path by its individual $E,N$ edges, choosing pairwise distinct internal vertices for the four pieces. Write $|W|$ for the number of letters of a word $W$. There are no empty edges, since $|U|=d+e-1\geq1$.

The graph has $m=3(d+e)$ vertices: the four pieces contribute respectively $d+e-1,d+e-2,d+e-1,1$ internal vertices in addition to the three distinguished vertices. An accepting path is uniquely specified by the numbers $\ell,r$ of traversals of the two cycles, so it reads $W_{k,\ell}$ with $k=\ell+r+1$.

Number the vertices from $1$ to $m$. Let $\EE,\NN$ be the $m\times m$ matrices whose $(i,j)$ entries count, respectively, the $E$-edges and $N$-edges from $i$ to $j$. Let $\alpha$ be the row vector with a $1$ at the start and zeros elsewhere, and let $\beta$ be the corresponding column vector for the finish. For a word $W$, $W(\EE,\NN)$ means the matrix product obtained by replacing each letter by its matrix in the same order; the empty word gives the identity matrix $I$. Then $\alpha W(\EE,\NN)\beta$ counts accepting paths reading $W$. This counts paths with their multiplicities, which is precisely the convention required by \eqref{eq:accepted}.

\subsection{Subtrees and free red balance}

Introduce two independent formal variables: $t$ marks word length and $u$ marks free red balance. All matrix series in this section have entries in $\FF[u][[t]]$, the ring of formal $t$-series with polynomial coefficients in $u$. For any fixed word length, the number of red tokens is bounded by that length, so each such coefficient is a polynomial.

A complete positive-root subtree starts at a positive horizontal edge, includes its nested packets, and stops when that root's blue packet empties. Its free balance at closure excludes the root's own token. Let $F$ be the matrix series counting all such legal subtrees, weighted by \eqref{eq:weight}, by $t^{|W|}$, by $u$ to the free balance, and by $W(\EE,\NN)$. Thus the $(i,j)$ entry also counts all paths from vertex $i$ to vertex $j$ reading the decorated word. The balance at closure may be any nonnegative integer.

While a parent is active, the atoms between two successive vertical edges that consume its own blue tokens are zero-grade horizontal edges and complete child subtrees. The former contributes $tu\EE$; a child counted by $F$ contributes $uF$, because its formerly protected root token becomes free. If $K$ counts arbitrary sequences of these atoms, including the empty sequence, then
\begin{equation}\label{eq:forest}
 K=I+(tu\EE+uF)K.
\end{equation}
The child cannot consume any of the parent's existing free tokens: its own root token separates them from its internal demand. Consequently its legality is independent of the balance with which the parent enters this atom. Every atom has positive length, so the geometric expansion $K=\sum_{j\geq0}(tu\EE+uF)^j$ is well defined coefficient by coefficient in $t$.

For a matrix series $J(t,u)$ over $\FF[u][[t]]$, define the divided-difference operator and its weighted sum entrywise by
\begin{equation}\label{eq:delta}
 \Delta J(t,u)=\frac{J(t,u)-J(t,0)}{u},\qquad
 \Phi(J)=\sum_{s=0}^{a}w_s\Delta^sJ,
\end{equation}
where $\Delta^0$ is the identity and $\Delta^s$ is $s$ successive applications. In particular $\Delta^s(u^j)=u^{j-s}$ for $j\geq s$ and is zero for $j<s$. Thus $\Phi$ implements a vertical demand with its weight and discards insufficient balances.

For $0\leq j\leq b$, let $P_j$ count prefixes after $j$ vertical edges assigned directly to a root, with the root edge itself omitted and with no final atom sequence appended. The tree decomposition gives
\begin{equation}\label{eq:prefix}
 P_0=I,\qquad
 P_j=t\Phi(P_{j-1}K)\NN\quad(1\leq j\leq b),\qquad
 F=t\EE\sum_{h=1}^{b}v_hP_h.
\end{equation}
The product $P_{j-1}K$ combines the preceding free balance with that returned by the next sequence of atoms before the demand is applied. Blue matching uniquely determines the child intervals and the directly assigned vertical edges. Hence this decomposition counts each decoration once for each graph path reading it. Matrix multiplication sums over the intermediate graph vertex at each concatenation.

\subsection{Polynomial fixed-point equations}

To put \eqref{eq:forest}--\eqref{eq:prefix} into the form required by a discrete-differential theorem, keep $K,F$ as unknowns and introduce the matrices
\[
 R_0=K,\qquad R_j=P_jK\quad(1\leq j<b).
\]
The matrices $R_j$ represent a prefix followed by the next atom sequence. The equations become
\begin{equation}\label{eq:dde}
\begin{aligned}
 R_j&=t\Phi(R_{j-1})\NN K &&(1\leq j<b),\\
 K&=I+tu\EE K+t^2u\EE
       \left(\sum_{h=1}^{b}v_h\Phi(R_{h-1})\right)\NN K,\\
 F&=t^2\EE\left(\sum_{h=1}^{b}v_h\Phi(R_{h-1})\right)\NN.
\end{aligned}
\end{equation}
When $b=1$, there are just the unknown matrices $K,F$. These equations are equivalent to the tree equations: from a solution of \eqref{eq:dde}, define $P_j$ by \eqref{eq:prefix}. Starting with $P_0K=R_0$, induction gives $P_jK=R_j$ for $j<b$; the formula for $F$ and then \eqref{eq:forest} follow. The forward implication follows by substitution.

List all the matrix entries as scalar unknowns $Z_i$, $1\leq i\leq(b+1)m^2$, in any fixed order. Let $f_i\in\{0,1\}$ be their respective initial values. Expanding the products in \eqref{eq:dde} gives polynomials $\mathcal Q_i$ over $\FF$ such that every equation has the form
\begin{equation}\label{eq:proper}
 Z_i=f_i+t\mathcal Q_i
 \bigl((\Delta^sZ_r)_{1\leq r\leq(b+1)m^2,\ 0\leq s\leq a},t,u\bigr),
\end{equation}
For example,
\[
 [\Phi(R_{j-1})\NN K]_{i\ell}
 =\sum_{r=1}^{m}\sum_{s=1}^{m}\sum_{h=0}^{a}
       w_h\Delta^h(R_{j-1})_{ir}\NN_{rs}K_{s\ell}.
\]
Every other matrix product expands in the same way into a finite sum of products of scalar entries. Thus each $\mathcal Q_i$ is a polynomial in commuting scalar unknowns and their listed divided differences.

\begin{lemma}\label{lem:unique}
System \eqref{eq:dde} has a unique solution in matrices over $\FF[u][[t]]$. It is the solution counted by the protected tree decomposition.
\end{lemma}
\begin{proof}
Its constant terms are $K(0,u)=I$, $R_j(0,u)=0$ for $j>0$, and $F(0,u)=0$. The operator $\Delta$ preserves divisibility by every power of $t$. Thus inputs agreeing modulo $t^n$ give right-hand sides agreeing modulo $t^{n+1}$ in \eqref{eq:proper}. Successive substitution from the constant terms stabilizes each coefficient and gives a unique solution. Each stabilization involves finitely many polynomial operations in $u$ and finitely many divided differences, so the coefficients remain polynomial in $u$. The tree decomposition satisfies the equations and the same constant terms, which identifies the solution.
\end{proof}

\section{Algebraicity and extraction}\label{sec:algebraicity}

We use the following form of the system theorem of Notarantonio and Yurkevich~\cite[Theorem~1.3]{NY}. Over any field of characteristic zero, a finite system of the form \eqref{eq:proper}, with polynomial initial terms, polynomial right-hand sides, one catalytic variable, and a fixed finite maximum order of divided differences, has a unique solution in $\FF[u][[t]]$, and every component is algebraic over $\FF(t,u)$. The difference operator in~\cite{NY} is taken at an arbitrary field element; here that element is zero.

All these hypotheses hold for \eqref{eq:dde}. In particular the state number $m=3(d+e)$ and the difference order $a$ are independent of $k$. It follows that every entry of $F$ is algebraic over $\FF(t,u)$.

\begin{lemma}\label{lem:specialize}
If $J(t,u)\in\FF[u][[t]]$ is algebraic over $\FF(t,u)$, then $J(t,0)$ is algebraic over $\FF(t)$.
\end{lemma}
\begin{proof}
Choose a nonzero polynomial $P(t,u,Z)\in\FF[t,u,Z]$ annihilating $J$. Divide its coefficients by their greatest common power of $u$. Since multiplication by $u$ is injective on $\FF[u][[t]]$, the resulting polynomial still annihilates $J$ and its specialization at $u=0$ is nonzero. Evaluating the identity at $u=0$ gives the desired relation.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
The graph selects the words $W_{k,\ell}$ with their root-rank multiplicities. Specializing $u=0$ selects subtrees whose final free balance is zero. Proposition~\ref{prop:bijection} therefore gives
\begin{equation}\label{eq:extract}
 \alpha F(t,0)\beta
   =\sum_{k\geq1}c_kt^{k(d+e)+1}
   =t\bigl(G_{d,e}(t^{d+e})-1\bigr).
\end{equation}
The extra unit of length is the root horizontal edge. By the system theorem and Lemma~\ref{lem:specialize}, the left side is algebraic over $\FF(t)$. Hence so is $G_{d,e}(t^{d+e})$. The extension $\FF(t)/\FF(t^{d+e})$ is finite algebraic, so transitivity makes this series algebraic over $\FF(t^{d+e})$. Pulling back a nonzero relation through the injective substitution $z\mapsto t^{d+e}$ proves that $G_{d,e}(z)$ is algebraic over $\FF(z)$.
\end{proof}

\begin{example}
For $P_1(X)=1+w_1X$ and $P_2(Y)=1+v_1Y$, the direction $(1,1)$ has wall function $G_{1,1}(z)=1+w_1v_1z$.
\end{example}
\begin{proof}
At $k=1$ the only accepted decoration is $EEN$ with grades $(1,0,1)$, of weight $v_1w_1$. For $k>1$ the frames are $(EN)^\ell E(EN)^{k-\ell-1}EN$. If $\ell>0$, the root closes at the first $N$, too early. If $\ell=0$, every $EN$ before the final $EN$ must be a child with grades $(1,0)$: a zero horizontal grade would close the global root, and a positive vertical grade would consume the child's protected token. The last vertical edge can have grade at most one, whereas the required total is $k$. Thus no accepted decoration occurs for $k>1$.
\end{proof}