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\title{An exact plateau for three binary summands modulo a Mersenne
number, with finite-layer carry balancing}
\author{Anonymous\thanks{\textbf{AI disclosure.}  The mathematics in this
manuscript was produced principally by AI systems, and the human author
contributed no mathematical content.  Section~\ref{sec:ai} states which system
did what.}}
\date{15 August 2026; revised 27 August 2026}

\begin{document}
\maketitle

\begin{abstract}
Put $N=2^k-1$ and let $F_k(t)$ count ordered triples $(a,b,c)\in
\{0,\ldots,N-1\}^3$ such that $a+b+c\equiv t\pmod N$ and
$\wt(a)+\wt(b)+\wt(c)<k$.  We give a three-state polynomial transfer matrix for
this count.  Our terminal result is an exact all-width evaluation: if $t$ has
exactly two nonadjacent zero digits, then
\[
 F_k(t)=(k+23)3^{k-4}\qquad(k\ge4),
\]
with no error term and no hypothesis on $k$.  The proof is structural rather than
computational.  The cyclic digit equations imply that total input weight equals
$\wt(t)$ plus total carry mass, so a target containing $z$ zero digits requires
only the first $z$ carry layers; a two-zero target therefore admits carry masses
zero and one alone, and those two layers are evaluated in closed form.  The rest
of the paper studies what happens when $z$ grows.  In the chamber in which the
zero digits of $t$ are isolated, we derive the layers of carry mass at most five
and prove that their cumulative sum is nondecreasing under Robin--Hood balancing
of the intervening one-runs; the analogous secondary balancing operation is also
nondecreasing through mass five.  For every fixed carry mass we prove that the
corresponding coefficient is a universal polynomial in bounded-length cyclic
pattern counts.  We emphasise what these do \emph{not} give: the balancing
results are finite-layer, the extension from five layers to all layers remains
open as Conjecture~\ref{conj:all}, and we classify no global maximizer of $F_k$.  For the reverse-oriented receiver boundary, we additionally prove a
uniform compensation theorem for all six lengths at which the unanchored degree-two prefix
is negative, and coefficientwise prefix positivity for every positive residual length.
\end{abstract}

\section{Introduction}

The Tu--Deng problem asks for a sharp upper bound on a Hamming-weight-truncated
two-summand representation count modulo $2^k-1$ \cite{TuDeng,SpiegelhoferWallner,LiuLuoXie}.
Its recent complete solution uses cyclic carries and a subsequence-language
factorization \cite{LiuLuoXie}.  We study the first multisummand case.

For $0\le t<N$, define
\[
 F_k(t)=\#\{(a_1,a_2,a_3)\in\{0,\ldots,N-1\}^3:
 a_1+a_2+a_3\equiv t\pmod N,\ \sum_i\wt(a_i)<k\}.
\]
The conventions at the residue zero do not affect any statement below; each
residue is represented by its $k$-bit word.

Our terminal result is the following exact evaluation, proved in
Section~\ref{sec:plateau}.

\begin{theorem}[two-zero plateau]\label{thm:plateau}
Let $k\ge4$ and let $t$ be a $k$-bit cyclic word with exactly two zero digits,
no two of them adjacent.  Then
\[
 F_k(t)=(k+23)3^{k-4}.
\]
\end{theorem}

The value depends only on $k$, not on the spacing of the two zeros: the whole
chamber is flat, which is why we call it a plateau.  The proof is short once the
carry grading of Section~\ref{sec:transfer} is in place, and it is exact at every
width rather than asymptotic.

The remainder of the paper asks what survives as the number $z$ of zero digits
grows, and the honest answer is: less than one would like.  Our results there are
finite-layer, and we state at the outset what we do not prove.  We do not
classify the global maximizers of $F_k$.  Theorem~\ref{thm:plateau} is not a
statement about maximizers and the plateau value is not maximal: at $k=12$ the
plateau reads $35\cdot3^8=229{,}635$, while direct evaluation gives
$F_{12}(110101101010)=293{,}499$, a target with five zero digits.  Nor do we
prove the extremal conjecture that motivated the balancing analysis; that is
Conjecture~\ref{conj:all}, and Section~\ref{sec:open} explains why
Theorem~\ref{thm:main} does not imply it.

The balancing analysis concerns words whose zeros are isolated.  Such a cyclic
word can be written
\[
 t=0,1^{L_1}0,1^{L_2}\cdots0,1^{L_z},\qquad L_i\ge1,
\]
up to rotation.  We call the $L_i$ the run lengths and put $w=\sum_iL_i=k-z$.

\begin{definition}
A \emph{primary balancing move} replaces two run lengths $p,q$, with
$p\ge q+2$, by $p-1,q+1$, without changing their cyclic positions.  After all
run lengths differ by at most one, a \emph{secondary balancing move} applies the
same operation to two cyclic spacings between the longer runs.
\end{definition}

For a polynomial $P$, write $[x^{\le m}]P=\sum_{j=0}^m[x^j]P$.

\begin{theorem}[finite-layer balancing]\label{thm:main}
Let $t$ lie in the isolated-zero chamber and let $t'$ be obtained from $t$ by a
primary balancing move.  If $H_t$ is the carry polynomial defined in
Section~\ref{sec:transfer}, then
\[
 [x^{\le C}]H_{t'}\ge[x^{\le C}]H_t\qquad(0\le C\le5).
\]
After primary balancing, the corresponding inequality through $C=5$ also holds
for every secondary balancing move.
\end{theorem}

The theorem is deliberately finite-layer.  When $z\le6$ it controls the complete
truncated count whenever all admitted carry layers lie below five; for arbitrary
$z$ it is structural information rather than a proof of the global extremal
conjecture.  The cumulative form of the statement is not a convenience.  Single
layers are genuinely not monotone: the audit of Section~\ref{sec:verify} finds,
through width twelve, that thirteen of the thirty balancing moves reaching carry
mass three strictly decrease that layer, while no move decreases the cumulative
sum through mass three.  Any argument that treats the layers one at a time
therefore fails already at mass three.

\section{The cyclic carry transfer matrix}\label{sec:transfer}

Write $t_j\in\{0,1\}$ for the target digit, $q_j\in\{0,1,2,3\}$ for the number
of input summands having a one in position $j$, and $c_j\in\{0,1,2\}$ for the
carry entering that position.  Cyclic addition is equivalent to
\begin{equation}\label{eq:digit}
 q_j+c_j=t_j+2c_{j+1},\qquad c_k=c_0.
\end{equation}
There are $\binom3{q_j}$ choices of the three input digits.  Hence, with rows and
columns indexed by $0,1,2$,
\begin{equation}\label{eq:M}
 M_b(z)_{c,c'}=\binom3{b+2c'-c}z^{b+2c'-c},
\end{equation}
where an out-of-range binomial coefficient is zero, and
\begin{equation}\label{eq:E}
 E_t(z)=\tr\prod_{j=0}^{k-1}M_{t_j}(z).
\end{equation}
Thus $[z^d]E_t(z)$ counts triples of total input weight $d$.

Summing \eqref{eq:digit} around the cycle gives the key regrading.

\begin{lemma}[carry grading]\label{lem:grading}
Every cyclic solution satisfies
\[
 \sum_jq_j=\wt(t)+\sum_jc_j.
\]
Consequently, if $z=k-\wt(t)$ and
\[
 N_0(x)=\begin{pmatrix}1&3&0\\0&3x&x\\0&x^2&3x^2\end{pmatrix},\qquad
 N_1(x)=\begin{pmatrix}3&1&0\\x&3x&0\\0&3x^2&x^2\end{pmatrix},
\]
then
\begin{equation}\label{eq:H}
 H_t(x)=\tr\prod_jN_{t_j}(x),\qquad
 E_t(x)=x^{\wt(t)}H_t(x),\qquad
 F_k(t)=[x^{\le z-1}]H_t(x).
\end{equation}
\end{lemma}

\begin{proof}
Sum \eqref{eq:digit}; cyclicity makes $\sum_jc_{j+1}=\sum_jc_j$.
Weighting a transition by its incoming carry gives the displayed matrices and
the trace imposes cyclic closure.  The strict budget is equivalent to total
carry mass $<z$.
\end{proof}

\section{The first two layers and an exact plateau}\label{sec:plateau}

Let $r$ be the number of cyclic one-runs.  Direct classification of carry supports
gives the following formulas.

\begin{proposition}\label{prop:firstlayers}
For $w=\wt(t)\ge2$,
\[
 [x^0]H_t=3^w,\qquad [x^1]H_t=3^{w-2}(w+8r).
\]
\end{proposition}

\begin{proof}
With no carry, each target-one digit chooses one of the three summands.  With
carry mass one, its unique support position either begins a one-run or is one of
the $w-r$ internal one positions.  A run start contributes $3^w$, whereas an
internal position contributes $3^{w-2}$.  Summing gives the formula.
\end{proof}

\begin{corollary}[two-zero plateau; this is Theorem~\ref{thm:plateau}]
\label{cor:plateau}
If $k\ge4$ and $t$ has exactly two nonadjacent zero digits, then
\[
 F_k(t)=(k+23)3^{k-4},
\]
independently of the distance between the zeros.
\end{corollary}

\begin{proof}
Here $z=2$, $w=k-2$, and $r=2$.  Only carry masses zero and one occur in
\eqref{eq:H}; substitute Proposition~\ref{prop:firstlayers}.
\end{proof}

The isolation hypothesis is essential.  At $k=6$, balancing zero gaps
$(1,2,3)$ to $(2,2,2)$ changes the admitted layer vector from
$(27,57,147)$ to $(27,81,108)$ and decreases the total from $231$ to $216$.

\section{Carry masses two and three}

In the isolated-zero chamber put
\[
 u_j=\#\{i:L_i\ge j\}\quad(j\ge2).
\]
Thus $u_2$ counts nonsingleton one-runs and $u_3$ counts runs of length at
least three.

\begin{proposition}[mass two]\label{prop:mass2}
At fixed $(w,z)$, the carry-two coefficient has the form
\[
 [x^2]H_t=C_2(w,z)+64\,3^{w-4}u_2,
\]
where $C_2(w,z)$ is independent of the run arrangement.
\end{proposition}

\begin{proof}
A mass-two carry sequence consists of two unit carries.  Separating adjacent and
nonadjacent supports gives
\begin{align*}
[x^2]H_t={}&n_{001}3^{w+1}+{\tbinom r2+n_{011}}3^w+n_{101}3^{w-1}\\
&+(r(w-r)-n_{011}+n_{111})3^{w-2}
 +(\tbinom{w-r}2-n_{111})3^{w-4}.
\end{align*}
In the isolated-zero chamber $r=n_{101}=z$, $n_{001}=0$,
$n_{011}=u_2$, and $n_{111}=w-2z-u_2$.  Substitution leaves precisely the
displayed arrangement-dependent term.
\end{proof}

For mass three the finite component classification yields the following closed
form; it is included because its only negative exchange explains why the theorem
must be cumulative.

\begin{proposition}[mass three]\label{prop:mass3}
For $w\ge6$,
\[
 [x^3]H_t=\frac{3^{w-6}}6P(w,z,u_2,u_3),
\]
where
\begin{align*}
P={}&w^3+24w^2z+45w^2+192wz^2+234wz+290w\\
&+512z^3-1008z^2+11374z\\
&+u_2(384w+3072z-13680)+3072u_3.
\end{align*}
For every primary move, if $\delta_C=[x^C](H_{t'}-H_t)$, then
\[
 \delta_3\ge-\frac58\delta_2,qquad \delta_2+delta_3>0
\]
whenever the move changes either coefficient.
\end{proposition}

\begin{proof}
A mass-three support is either three unit carries or an oriented adjacent
$(1,2)$ component.  Splitting the former into component types $1+1+1$, $2+1$,
and $3$, multiplying the forced local binomial weights, and collecting by whether
a run has length at least two or three gives $P$.

Under a balancing move all cases are nonnegative except
$(L_D,L_R)=(3,1)\mapsto(2,2)$.  In that case
\[
 \delta_3=3^{w-6}(64w+512z-2792),\qquad
 \delta_2=576\,3^{w-6}.
\]
The layer exists only for $z\ge4$, and the move requires $w\ge6$; the first
expression is minimized at $(w,z)=(6,4)$ and equals $-360$.  The assertions
follow.
\end{proof}

\section{Masses four and five}\label{sec:45}

Let
\[
 a_{22}=\#\{i:L_i\ge2,\ L_{i+1}\ge2\},\qquad
 a_{23}=\#\{i:L_i\ge2,L_{i+1}\ge3\}+\#\{i:L_i\ge3,L_{i+1}\ge2\}.
\]
Indices are cyclic.  Carry mass four is the first layer that detects cyclic order.

\begin{proposition}[mass four]\label{prop:mass4}
For $w\ge8$,
\[
 [x^4]H_t=\frac{3^{w-8}}{24}Q(w,z,u_2,u_3,u_4,a_{22}),
\]
where
\begin{align*}
Q={}&w^4+32w^3z+90w^3+384w^2z^2+1188w^2z+1835w^2\\
&+2048wz^3+1728wz^2+50996wz+5946w\\
&+4096z^4-16128z^3+369260z^2-701412z+49152u_2^2\\
&+u_2(768w^2+12288wz-43200w+49152z^2-470016z+1402584)\\
&+u_3(12288w+98304z-462336)+98304u_4-331776a_{22}.
\end{align*}
Moreover $\sum_{C=0}^4\delta_C\ge0$ for every primary move.
\end{proposition}

\begin{proof}[Proof sketch with complete reduction]
Classify supports as four unit carries or one $(1,2)$ component plus two unit
carries.  Inclusion--exclusion for overlapping placements gives $Q$.  Subtraction
under a move has seven threshold cases according to the receiver length
$1,2,3,4,$ or at least $5$.  Cases not crossing the $u_2$ threshold are immediate.
When the receiver crosses it, put $d=\delta a_{22}$.  Then $0\le d\le2$ and
$u_2\ge d$.  After multiplication by the positive normalization, the three
possible donor thresholds give
\begin{align*}
R_4={}&-13824d+4096u_2+32w^2+512wz-1736w\\
&+2048z^2-19072z+59809,\\
R_5={}&-13824d+4096u_2+32w^2+512wz-1224w\\
&+2048z^2-14976z+41057,\qquad R_6=R_5+4096.
\end{align*}
They increase on $w\ge8,z\ge5$ and their minima, at $d=u_2=2$ and
$(w,z)=(8,5)$, are $4833,10657,14753$.  The six boundary pairs with
$5\le z\le w<8$ are checked directly from \eqref{eq:H}; this is a finite
verification of rational polynomial identities, not an asymptotic assumption.
\end{proof}

The mass-five formula is lengthy but records the complete statistic needed at
this layer.

\begin{proposition}[mass five]\label{prop:mass5}
For $w\ge10$,
\[
 [x^5]H_t=\frac{3^{w-10}}{120}R,
\]
where
\begin{align*}
R={}&w^5+40w^4z+150w^4+640w^3z^2+3180w^3z+6275w^3\\
&+5120w^2z^3+18720w^2z^2+168240w^2z+73230w^2\\
&+20480wz^4-3840wz^3+1880700wz^2-1684920wz+88344w\\
&+32768z^5-215040z^4+7491040z^3-32833560z^2+83997432z\\
&+u_2(1280w^3+30720w^2z-79200w^2+245760wz^2-1889280wz\\
&\hspace{18mm}+5332120w+655360z^3-10045440z^2+76407680z-197059800)\\
&+u_3(30720w^2+491520wz-1850880w+1966080z^2-19783680z+57716160)\\
&+u_4(491520w+3932160z-19476480)+3932160u_5\\
&+u_2^2(245760w+1966080z-17510400)+3932160u_2u_3\\
&-a_{22}(1658880w+13271040z-96629760)-13271040a_{23}.
\end{align*}
Every primary move satisfies $\delta_5\ge0$.
\end{proposition}

\begin{proof}
The formula follows by the same connected-component expansion.  For receiver
length at least two, direct subtraction and the constraints forced by the crossed
thresholds give a nonnegative polynomial in unused run length and unused runs.
For receiver length one, only the two neighbors of the receiver and the two
neighbors of the donor affect $(a_{22},a_{23})$.  Marking those neighbors by
membership in $\{L\ge2\}$ and $\{L\ge3\}$ gives a finite local basis.  Minimizing
the four donor-threshold expressions on that basis leaves, respectively, the
positive normalized constants
\[
 240765,qquad946896,qquad1237200,qquad1455123.
\]
All remaining variables are nonnegative slack.  Hence $\delta_5\ge0$.
The accompanying exact verifier reconstructs the component formula and every
local-basis case directly from \eqref{eq:H}.
\end{proof}

Theorem~\ref{thm:main} for primary moves now follows from
Propositions~\ref{prop:firstlayers}--\ref{prop:mass5}: mass one and mass two are
nondecreasing, the only mass-three loss is compensated by mass two, the possible
mass-four losses are compensated cumulatively, and mass five is nondecreasing.

\section{Secondary balancing}

After primary balancing the zero gaps are $s$ and $s+1$; equivalently the run
lengths are $s-1$ and $s$.  Let $n$ be the number of longer gaps, so $k=sz+n$.
If $s\ge3$, every statistic in Propositions~\ref{prop:firstlayers}--
\ref{prop:mass5} is invariant under rearrangement.  If $s=2$, only $a_{22}$ varies:
it counts adjacent longer gaps.  A secondary move that separates such a pair has
$\delta a_{22}=-1$.  The mass-four gain is $13824\,3^{w-8}$, and the mass-five
change is
\[
 13824\,3^{w-10}(9z+n-233/4).
\]
Their sum is
\[
 13824\,3^{w-10}(9z+n-197/4)>0
\]
whenever mass five is admitted; if only mass four is admitted, its gain is already
positive.  This proves the secondary assertion of Theorem~\ref{thm:main}.

\section{Why only bounded correlations occur}

\begin{theorem}[bounded-correlation principle]\label{thm:bounded}
For fixed $C$, the coefficient $[x^C]H_t(x)$ is a universal polynomial in the
numbers of cyclic factors of $t$ of lengths at most $C+1$.
\end{theorem}

\begin{proof}
Decompose a carry sequence of mass $C$ into connected components of nonzero
states.  Each component has length at most $C$, and its feasibility and weight are
determined by it and the preceding target bit, hence by a window of length at most
$C+1$.  Begin with products of placement counts for the finitely many component
types.  Inclusion--exclusion removes overlaps.  Each overlap merges windows whose
total carry mass is still at most $C$, so every correction is again a count of a
factor of length at most $C+1$.  There are finitely many component types, and the
resulting polynomial depends only on $C$.
\end{proof}

This theorem explains the successive appearance of $u_2,u_3$, then $a_{22}$, and
then $a_{23}$; it does not by itself prove an all-layer inequality.

\section{Open all-mass problem}\label{sec:open}

The computations and finite-layer theorem suggest the following statement.

\begin{conjecture}\label{conj:all}
In the isolated-zero chamber, every primary balancing move satisfies
\[
 [x^{\le z-1}]H_{t'}\ge[x^{\le z-1}]H_t.
\]
After primary balancing, secondary balancing has the same property.  Consequently
the maxima of $F_k(t)$ in this chamber occur at cyclically balanced run words.
\end{conjecture}

Conjecture~\ref{conj:all} is not a consequence of
Theorem~\ref{thm:main}.  A prospective proof can exploit the zero-anchor fact:
every carry cycle of mass $C<z$ enters at least one zero block in state zero.
Cutting there reduces the cyclic three-state process to anchored excursions through
states $1,2$.  The primitive transfer matrices are coefficientwise totally
nonnegative, so Cauchy--Binet supplies positive two-path minors.  Turning the full
exchange difference into a prefix sum of such minors is the outstanding step.

\subsection{A uniform receiver-boundary compensation}

The reverse-oriented one-excursion sector admits a useful exact boundary theorem.  Put
$A_j=N_0N_1^j$ and write $a_j=(A_j)_{00}$, $b_j=(A_j)_{01}$, and
$c_j=(A_j)_{10}$.  For the shortest receiver, the residual-$d$ kernel after clearing its
positive monomial factor is
\[
 U_{d,r}=\{64xa_d+24x^2b_{d+1}+3x^3b_{d+2}\}a_r
          -64\operatorname{rev}(b_d)c_r,                 \tag{15}
\]
where $\operatorname{rev}$ reverses the coefficient vector on its natural support.  The
general receiver kernel is coefficientwise no smaller after normalization, so (15) is the
boundary case of this sector.

\begin{proposition}[six-case receiver boundary]\label{prop:receiverboundary}
For every $\ell\ge1$ and $1\le r\le6$, every coefficient of
\[
 \frac{U_{0,r}(x)a_\ell(x)}{1-x}
\]
is nonnegative.  Moreover, for $r\ge2$,
\[
 [x^{\le2}]U_{0,r}=8\,3^{r-2}(8r-53),                   \tag{16}
\]
while the value at $r=1$ is $-120$.  Thus the unanchored degree-two prefix is negative
exactly for $1\le r\le6$ and positive for $r\ge7$.  In fact, every coefficient of
$U_{0,r}/(1-x)$ is nonnegative for every $r\ge7$.
\end{proposition}

\begin{proof}
The active block of the anchor is
\[
 a_\ell(x)=(1,3)
 \begin{pmatrix}3&1\\x&3x\end{pmatrix}^{\!\ell}
 \binom10.                                                \tag{17}
\]
After division by $4^{\ell+1}$, its coefficient of $x^q$ is the joint mass that a
two-state Markov chain, initially distributed as $(1/4,3/4)$ and switching with probability
$1/4$, visits state one $q$ times before ending in state zero.  The endpoint-conditioned
partial mass sequence is log-concave by Dekking--Kong \cite{DekkingKong}, hence the
coefficients of $a_\ell$ are unimodal.

Direct multiplication in (15) shows that $P_r=U_{0,r}/(1-x)$ has only one negative
coefficient for $1\le r\le6$.  The three coefficients around it are
\[
\begin{array}{c|rrr}
r&[x]P_r&-[x^2]P_r&[x^3]P_r\\\hline
1&192&120&627\\2&576&296&817\\3&1728&696&2523\\
4&5184&1512&7849\\5&15552&2808&24579\\6&46656&3240&77409
\end{array}
\]
and all other coefficients are nonnegative.  Both neighbors dominate the negative
magnitude.  In a coefficient of $P_ra_\ell$, pair the negative summand with the right
neighbor on the increasing side of the unimodal anchor sequence and with the left neighbor
on its decreasing side.  This proves coefficientwise nonnegativity.  Finally, extracting
the first two coefficients of (15) gives
$[x^0]a_r=3^r$, $[x^1]a_r=3^{r-2}(r+8)$, and
$[x^2]c_r=3^r$; simplification yields (16), including its stated sign threshold.

It remains to prove the final assertion uniformly.  Write $P_r=U_{0,r}/(1-x)$ and
$Q_r=P_{r+1}-3P_r$.  The characteristic recurrence of the active two-state block, together
with the single state-two path in $c_r$, gives
\[
 P_{r+2}=3(1+x)P_{r+1}-8xP_r-\frac{576x^{2r+4}}{1-x},
 \qquad
 Q_{r+1}=x(P_r+3Q_r)-\frac{576x^{2r+4}}{1-x}.            \tag{18}
\]
Exact multiplication gives $P_7,Q_7\ge0$ coefficientwise.  Moreover $P_r$ is constant from
degree $2r$ onward, with constant value
\[
 T_r=164\,2^r+440\,4^r-192.                              \tag{19}
\]
Assume $P_r,Q_r\ge0$.  Before degree $2r+4$, the second recurrence in (18) has no negative
correction.  From degree $2r+4$ onward, the referenced coefficients have reached their
constant tails, and the available coefficient is
\[
 T_r+3(T_{r+1}-3T_r)
   =-328\,2^r+1760\,4^r+960\ge576.                       \tag{20}
\]
It absorbs the correction in (18), proving $Q_{r+1}\ge0$.  Since
$P_{r+1}=3P_r+Q_r$, induction proves $P_r\ge0$ for every $r\ge7$.
\end{proof}

\begin{theorem}[complete receiver-opening kernel]\label{thm:receiverkernel}
For every $d\ge1$ and $r\ge1$, every coefficient of $U_{d,r}/(1-x)$ is nonnegative.
Together with Proposition~\ref{prop:receiverboundary}, this settles the zero-residual cases
after one outside anchor, and the result persists at every receiver length after the positive
normalization of the receiver-length reduction.
\end{theorem}

\begin{proof}
Write $P_{d,r}=U_{d,r}/(1-x)$.  Coefficient extraction from the residual generating function
gives
\[
 P_{d+2,r}=3(1+x)P_{d+1,r}-8xP_{d,r}.                  \tag{21}
\]
The coefficientwise cone
\[
 P_{d,r}\ge0,\qquad 3P_{d+1,r}\ge8P_{d,r}              \tag{22}
\]
is invariant under (21): the next term is
$3P_{d+1,r}+x(3P_{d+1,r}-8P_{d,r})$, and three times this expression minus
$8P_{d+1,r}$ is again nonnegative.

It is therefore enough to prove, uniformly in $r$,
\[
 P_{1,r}\ge0,\qquad P_{2,r}\ge0,qquad
 3P_{3,r}-8P_{2,r}\ge0.                                \tag{23}
\]
For any of these three families $S_r$, closing-length extraction gives
\[
 S_{r+2}=3(1+x)S_{r+1}-8xS_r-\frac{x^{2r+4}C_S(x)}{1-x},\tag{24}
\]
where the coefficient vectors of $C_S$ are respectively
\[
 (1728,192),\quad(5184,1152,576),\quad
 (5184,6336,6336,5184).                                 \tag{25}
\]
Put $Q_r=S_{r+1}-3S_r$.  Exact multiplication supplies the six initial rows and gives
$S_r\ge0$ there and $Q_6\ge0$.  On each diagonal touched by the correction in (24), the
margin in $x(S_r+3Q_r)$ is $A2^r+B4^r+C_j$, with
\[
\begin{array}{c|rr|l}
S&A&B&C_j\\\hline
P_1&-528&8128&1472,1280\\
P_2&-544&34688&8064,5184,4608\\
3P_3-8P_2&7232&151808&114432,42432,20544,15360.
\end{array}                                             \tag{26}
\]
The last constant repeats beyond the displayed diagonals.  These expressions are
nonnegative for $r\ge6$: use $4^r=(2^r)^2$ for the first two rows, while the third is
immediate.  Thus (24) propagates $S_r,Q_r\ge0$, proving (23).  The invariant cone (22)
then proves the theorem for all $d\ge1$.
\end{proof}

\begin{remark}[why donor closing must be paired]
The oppositely oriented donor-closing Wronskian does not become prefix-positive merely by
adding anchor factors.  In the sharp family $p=q+2$, its signed expansion gives
\[
 [x^{\le q+4}]\,W_{\rm close}(x)a_1(x)^m
   =-3^{m+1}8^q<0\qquad(q\ge1,\ m\ge0).                 \tag{27}
\]
Thus the remaining first-row argument must pair donor closing with the other incident dimers;
an arbitrarily large standalone positive padding reservoir is insufficient.

The minimal candidate pairing, after clearing endpoint denominators, is
\[
 J_{d,q,u,v}=a_ua_vW(a,a)+a_ub_vW(c,a)+b_ua_vW(a,c),     \tag{28}
\]
where $p=q+2+d$.  Both endpoint variables obey the homogeneous recurrence with
characteristic coefficients $3(1+x),-8x$; the residual variable obeys the same recurrence
plus a coefficientwise nonnegative correction.  The invariant cone used in
Theorem~\ref{thm:receiverkernel} therefore reduces prefix positivity of (28) to twelve
one-parameter receiver-length families.  These families pass the exact finite audit through
$q=12$.  They share a universal order-five recurrence whose modal roots are
$\alpha^2,8x,\beta^2,x^2\alpha,x^2\beta$, where
$\alpha+\beta=3(1+x)$ and $\alpha\beta=8x$.  Applying the quadratic factor
$1-3x^2(1+x)E+8x^5E^2$ leaves three nonnegative initial profiles and thereafter the exact
law $R_{q+1}=8xR_q$.  The inverse quadratic has coefficients $x^{2n}a_n(x)\ge0$, so positive
convolution reconstructs every family.  Consequently (28) is prefix-positive for all
$d\ge0$ and $q,u,v\ge1$.

For the full five-Wronskian expression, the same quadratic filter leaves three fixed
receiver profiles and then obeys $R_{q+1}=8xR_q$ exactly.  Receiver length can therefore be
eliminated without splitting the monomer reservoir.  The remaining five variables admit a
finite positive-reconstruction frontier.  Put
$A(E)=E^2-3(1+x)E+8x$.  The two active endpoint sequences are annihilated by $A$, while the
residual and two closing-endpoint sequences are annihilated by $(E-x^2)A$.  For the latter
write $G_n=F_{n+1}-x^2F_n$, and for the former write $D_n=F_{n+1}-F_n$.  In the variable
order $(d,u_b,u_c,v_f,v_g)$ the required frontier sizes are respectively
$4,5,4,3,5$: retain the initial $F$ row, the necessary initial $G$ or $D$ rows, and then the
first nonnegative margin $3G_{n+1}-8G_n$ or $3D_{n+1}-8D_n$.  The exceptional cones are
delayed through $D_3$ for $u_b$ and through $G_3$ for $v_g$.

The resulting $4\cdot5\cdot4\cdot3\cdot5=1200$ tensor-frontier polynomials are
coefficientwise nonnegative for each of the three receiver profiles.  These are finite exact
certificates: their unprefixed degree is at most $105$, and the distributed verifier checks
the stabilized prefix tail through degree $119$.  The invariant active cone reconstructs
each $G$ or $D$ sequence positively, the identities
$F_{n+1}=x^2F_n+G_n$ and $F_{n+1}=F_n+D_n$ reconstruct the five length variables, and the
positive inverse receiver filter reconstructs $q$.  Thus the arbitrary-context
five-Wronskian core is prefix-positive.  An earlier undelayed tensor frontier had 192 signed
cells out of 576, all arising from the $u_b$ margin; the delayed increment frontier is the
essential repair rather than a larger numerical audit.
\end{remark}

\begin{proposition}[Boundary mass two for five and six zero blocks]
\label{prop:boundary-two-bounded}
Let a primary balancing exchange move one unit from a donor run of length $p$ to a
nonadjacent receiver run of length $q$, with $p\ge q+2$.  If the cyclic target has five or
six zero blocks, then the exchange difference accumulated through zero-boundary carry mass
two is prefix-positive in the total-carry variable, for all positive run lengths.
\end{proposition}

\begin{proof}
Weight a zero-block boundary state by $y$ and truncate at $y^2$.  Since
$(K_L)_{0,2}=0$, boundary mass two consists only of two state-one boundaries.  Adjacent
ones give a three-block weight $b_{i-1}(K_{L_i})_{1,1}c_{i+1}$; separated ones give two
disjoint $bc$ dimers.  Multiplying the two arc contexts through $y^2$ reduces the exchange
to eleven Wronskians.  Five are the boundary-mass-one orientations above.  The six new
orientations are
\[
 W(a,d),\quad W(b,b),\quad W(b,c),\quad W(c,b),\quad W(c,c),\quad W(d,a),
 \qquad d_L=(K_L)_{1,1}.
\]
The first two terms and $W(c,b)$ can be signed, so the combination must be retained.

Apply the same quadratic receiver filter used above.  It leaves three profiles and then
satisfies $R_{q+1}=8xR_q$.  In every other length variable each profile is annihilated by
\[
 (E-x^2)\bigl(E^2-3(1+x)E+8x\bigr).
\]
After setting $G_n=R_{n+1}-x^2R_n$, the active quadratic cone reconstructs the sequence from
a delayed finite frontier.  A five-block context has two arc layouts and $5^4=625$ frontier
cells per layout.  A six-block context has the three layouts $(1,3),(2,2),(3,1)$; the two
lengths adjacent to the receiver require one additional seed, giving
$5^3 6^2=4500$ cells per layout.  Every cell is coefficientwise nonnegative for all three
receiver profiles.  These are finite certificates: the maximum filtered degree is below
$95$, and the audit checks the stabilized prefix tail through degree $104$.  Positive
reconstruction in the remaining lengths and then in receiver length proves the claim.
The exact certificate is reproduced by
\texttt{inquiries/33d-boundary-mass-two-frontier.py}.
\end{proof}

\begin{proposition}[Five-to-six insertion monotonicity]
\label{prop:boundary-two-insertion}
In the setting of Proposition~\ref{prop:boundary-two-bounded}, begin with five zero blocks
and insert one positive-length zero block at any position on either arc between the donor and
receiver.  The boundary-mass-two exchange difference after insertion minus that before
insertion is coefficientwise prefix-positive in every total-carry degree.
\end{proposition}

\begin{proof}
There are ten ordered insertion types among the three six-block arc layouts.  The same
quadratic receiver filter leaves three profiles and then the geometric mode $8x$.  All
ordinary length parameters obey the cubic recurrence used in
Proposition~\ref{prop:boundary-two-bounded}.  If $H_w$ denotes the insertion difference as a
function of the inserted length, the pre-insertion summand contributes an additional root
one.  Set
\[
 D_w=H_{w+1}-H_w,\qquad J_w=D_{w+1}-x^2D_w.
\]
Then $J$ obeys the active quadratic recurrence.  Its positive frontier retains the base
$H$ row, the first $D$ row, three successive $J$ rows, and the next active-cone margin.
Together with the ordinary delayed frontiers this gives either $4500$ or $5400$ cells for
each insertion type.  All $50400$ cells are coefficientwise nonnegative for each of the
three receiver profiles, for $151200$ sign certificates.  Positive reconstruction in
$J,D,H$ and receiver length proves the result.  The exact recurrence, tail, and frontier
audit is \texttt{inquiries/33e-boundary-mass-two-insertion.py}.
\end{proof}

\begin{proposition}[Six-to-seven insertion monotonicity]
\label{prop:boundary-two-insertion-seven}
In the setting of Proposition~\ref{prop:boundary-two-bounded}, begin with six zero blocks
and insert one positive-length zero block at any position on either arc between the donor and
receiver.  The boundary-mass-two exchange difference after insertion minus that before
insertion is coefficientwise prefix-positive in every total-carry degree.
\end{proposition}

\begin{proof}
There are eighteen ordered insertion types among the four seven-block arc layouts.  The
receiver and length recurrences are unchanged from
Proposition~\ref{prop:boundary-two-insertion}.  Six receiver-adjacent insertion types have
$22500$ frontier cells each, and the other twelve have $27000$ each, for $459000$ cells and
$1377000$ three-profile sign certificates.

The finite contraction need not materialize a sixth tensor variable.  The truncated cyclic
row is multilinear in its zero-block matrices.  Apply each frontier functional directly to
the corresponding matrix and rotate the cyclic trace so that the inserted block is last.
Up to the final closure the six- and seven-block contractions are identical.  At closure the
latter passes through the transformed inserted block, whereas the former is multiplied by
the sum of that functional's coefficients.  This realizes the root-one difference used in
the preceding proof.  Every resulting polynomial is coefficientwise nonnegative, and its
prefix coefficients are constant from degrees $95$ through $104$.  Recurrence reconstruction
proves the result for arbitrary positive lengths.  The exact bounded-memory audit is
\texttt{inquiries/33f-boundary-mass-two-six-to-seven.py}.
\end{proof}

\section{Verification and reproducibility}\label{sec:verify}

The exact Python verifier distributed with this manuscript performs integer
polynomial arithmetic.  It checks \eqref{eq:E} against direct triple enumeration,
the regrading \eqref{eq:H}, all displayed layer formulas, the local basis used in
Proposition~\ref{prop:mass5}, and exhaustive low-width exchange audits.  Run
\begin{verbatim}
python3 inquiries/33c-orbit-exchange.py --kmax 12
python3 inquiries/33d-boundary-mass-two-frontier.py
python3 inquiries/33e-boundary-mass-two-insertion.py
python3 inquiries/33f-boundary-mass-two-six-to-seven.py
\end{verbatim}
from the project root.  The script is supplementary evidence and a check on
algebra; the only finite boundary checks invoked in the proof are explicitly
identified above.

\subsection{What the Lean development does and does not establish}
\label{sec:leanscope}

A Lean~4 development accompanies this manuscript.  It contains no \texttt{sorry},
no \texttt{admit}, no \texttt{native\_decide} and no declared axiom.  Because
formalization claims are easy to overstate, we separate its genuine content from
its bookkeeping.

The substantive theorems quantify over arbitrary inputs.  They are
\texttt{summed\_digit\_equation} and
\texttt{inputWeight\_eq\_targetWeight\_add\_carryMass}, which prove the grading
law of Lemma~\ref{lem:grading};
\texttt{inputWeight\_lt\_width\_iff\_carryMass\_lt\_targetZeros}, the budget
equivalence; \texttt{zero\_anchor}, the pigeonhole behind the zero-anchor fact
used in Section~\ref{sec:open}; \texttt{active\_coefficient\_step} with
\texttt{active\_cone\_reconstruction}, the coefficientwise cone propagation;
\texttt{dominating\_neighbor\_absorbs}; \texttt{receiver\_tail\_margin} with
\texttt{delayed\_receiver\_step\_preserves\_growth}; and
\texttt{paired\_receiver\_order\_five\_factorization}.

Four items are weaker than a reader might assume from their names, and we state
this plainly.  \texttt{zeroLayer w := 3\^{}w} is a \emph{definition}, so the
identity $[x^0]H_t=3^w$ is closed in Lean by \texttt{rfl} and carries no
combinatorial content; the classification in
Proposition~\ref{prop:firstlayers} is what does the work, and it is on paper.
Likewise \texttt{oneLayerBySupport} is the support partition supplied by hand,
and the Lean theorem is the algebra taking it to closed form.
\texttt{twoZeroCount} builds in the fact that only carry masses zero and one are
admitted, so the formalized plateau reduces to the arithmetic
$9+(k+14)=k+23$ after factoring $3^{k-4}$; the step that only masses zero and one
occur is Lemma~\ref{lem:grading} applied by hand, not a Lean proof about
$F_k$.  Finally \texttt{first\_row\_frontier\_cardinality} is the arithmetic
identity $4\cdot5\cdot4\cdot3\cdot5=1200$ discharged by \texttt{decide}; it
records the shape of the frontier index set and does not construct or check the
frontier polynomials, which live in the Python certificate.

Most importantly, the bridge from $F_k$ to the formal objects is \emph{assumed,
not formalized}.  The Lean structure \texttt{CarryConfig} takes the digit
equation \eqref{eq:digit} as a hypothesis, and its field \texttt{next} is an
arbitrary \texttt{Equiv.Perm (Fin k)} rather than the cyclic successor.  Nothing
in the development ties \texttt{CarryConfig} to the counting problem defining
$F_k$; a reader who wants that link must take it from
Section~\ref{sec:transfer}.  The development should therefore be read as a
machine-checked account of the carry algebra, not as a formalization of
Theorem~\ref{thm:plateau}.

\section{AI disclosure}\label{sec:ai}

The mathematics in this manuscript was produced principally by AI systems.  The
human author framed no argument, supplied no proof step, and contributed no
mathematical content; the role was to run the systems, collect the output, and
publish it.

GPT-5.6 Sol, at high reasoning effort, did the majority of the mathematics: the
three-state cyclic-carry transfer matrix of Section~\ref{sec:transfer}, the
carry-mass regrading, the exact carry-layer coefficients through mass five, the
primary and secondary balancing exchange arguments of Theorem~\ref{thm:main},
the bounded-correlation principle of Theorem~\ref{thm:bounded}, the
receiver-boundary compensation and receiver-opening kernel theorems, the exact
integer-polynomial certificate programs of Section~\ref{sec:verify}, and the
Lean~4 development of Section~\ref{sec:leanscope}.

Claude Opus 5, at high reasoning effort, supplied direction rather than
derivations: framing the problem as the first multisummand case after Tu--Deng,
selecting which sub-questions to attack, enforcing the distinction between what
is proved and what is conjectured, and carrying out the prior-art search
underlying the novelty statement below.

No step in this manuscript has been checked by hand by a human mathematician.
The reader is asked to weight the certificates and the Lean development
accordingly, and to read the scope limits of Section~\ref{sec:leanscope} as part
of the claim rather than as a caveat appended to it.

\section*{Novelty statement}

To the knowledge of the author and of the systems named in
Section~\ref{sec:ai}, whose search is described there, the three-summand transfer
matrix, carry-layer formulas, two-zero plateau, finite-layer balancing theorems,
and receiver-opening kernel theorem above have not appeared previously.  The
nearest direct predecessor is the two-summand Tu--Deng literature.  Balanced-word
matrix-product extremality and total positivity provide neighboring techniques but
do not imply Theorem~\ref{thm:main}.  This statement should be rechecked against
current bibliographic databases immediately before submission.

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\end{document}