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\title{Finite-memory Collatz parity dynamics: \\
Fibonacci support, Bernoulli branch drift, and the pointwise obstruction}

\author{William Goodfellow\thanks{Independent researcher. Contact: \texttt{werner85@gmail.com}.}}

\date{\today}

\begin{document}
\maketitle

\begin{abstract}
We study finite-memory parity histories for the odd-to-odd shortcut maps
\[
S_a(n) = \frac{a n + 1}{2^{\nu_2(a n + 1)}}
\]
with $a$ odd, focusing on the Collatz case $a = 3$. Sampling raw parity histories at odd arrivals collapses the apparent $2^m$-state history space to a Fibonacci recurrent class
\[
\Omega_m = \{ h \in \{0,1\}^m : h_0 = 1,\ h_i h_{i+1} = 0 \},
\qquad |\Omega_m| = F_m,
\]
where $F_m$ is the $m$-th Fibonacci number. Under normalized Haar branch measure on odd $2$-adic integers, the next branch exponent $K_a = \nu_2(a n + 1)$ is independent of every finite history state $h \in \Omega_m$ and satisfies $\Prob(K_a = k) = 2^{-k}$. Consequently, the conditional branch drift is uniform on $\Omega_m$:
\[
\E[\log_2 a - K_a \mid h] = \log_2 a - 2.
\]
Thus the $(3n+1)$ shortcut has uniform negative finite-memory drift $\log_2 3 - 2 \approx -0.415$, while the $(5n+1)$ control has uniform positive drift $\log_2 5 - 2 \approx +0.322$. We also derive exact stationary weights on $\Omega_m$, verify the laws numerically on integer trajectories, and identify a finite-height survival bias in descending $(3n+1)$ samples that is absent in the growing $(5n+1)$ control. Finally, we show that no finite-history correction can turn this supermartingale into a pointwise one-step Lyapunov function: the alternating state has a $K = 1$ self-transition with positive log increment $\log_2 3 - 1 > 0$, so any potential $J(n, h) = \log_2 n + g(h)$ fails on this transition. The result isolates the finite-memory stochastic structure of Collatz parity dynamics while leaving the pointwise Collatz conjecture untouched.
\end{abstract}

\section{Introduction}
\label{sec:intro}

The Collatz conjecture asks whether every positive integer eventually reaches $1$ under the iteration $T(n) = n/2$ if $n$ is even and $T(n) = 3n + 1$ if $n$ is odd. The conjecture is widely believed and has been verified computationally to $n \le 2^{71}$ \cite{Barina2025}, but a proof has eluded all approaches.

A standard heuristic argument suggests the iteration should descend on average: per odd-to-odd transition, the iterated state is multiplied by $3$ and divided by $2$ a geometrically-distributed number of times, giving an average log-multiplier of $\log_2 3 - 2 \approx -0.415 < 0$. This heuristic underlies almost all rigorous partial results, from Terras' density-1 stopping-time theorem \cite{Terras1976} through Krasikov-Lagarias' $x^{0.84}$ density bound \cite{KrasikovLagarias2003} to Tao's recent log-density-$1$ almost-bounded theorem \cite{Tao2019}. Yet the heuristic is not, by itself, a proof: negative average drift does not exclude thin pointwise-divergent trajectories or non-trivial cycles.

The present paper isolates a clean finite-memory layer of this story. We work with the odd-to-odd shortcut map $S(n) = (3n+1)/2^{\nu_2(3n+1)}$ and consider the rolling $m$-bit parity history of the underlying ordinary Collatz iteration, sampled at odd arrivals. We prove three results:

\begin{enumerate}[label=(\arabic*)]
\item \textbf{Fibonacci support} (Theorem \ref{thm:support}): the reachable history set is exactly $\Omega_m = \{h \in \{0,1\}^m : h_0 = 1,\ h_i h_{i+1} = 0\}$, of size $F_m$.

\item \textbf{Bernoulli branch independence} (Theorem \ref{thm:branch}): under the natural $2$-adic Haar branch measure, the next branch exponent is $\Geom(1/2)$-distributed and independent of every finite history state.

\item \textbf{Uniform conditional drift} (Corollary \ref{cor:drift}): $\E[\log_2 a - K_a \mid h] = \log_2 a - 2$ for every $h \in \Omega_m$, giving a uniform finite-memory supermartingale structure on $(3n+1)$ and a uniform positive-drift analogue for $(5n+1)$.
\end{enumerate}

We also derive exact stationary weights $\pi_m(h)$ on $\Omega_m$ (Theorem \ref{thm:stationary}), and show that no finite-history correction (bounded or unbounded) can convert the uniform conditional drift into pointwise one-step descent (Corollary \ref{cor:no-lyapunov}). The latter is structural rather than negative: it precisely localizes the boundary between finite-memory probabilistic descent and pointwise Lyapunov descent.

These results are proved under the natural Bernoulli branch measure, and their integer-trajectory recovery is verified by numerical experiments on $80$-bit and $200$-bit starts. A finite-height survival bias is identified for descending $(3n+1)$ trajectories and shown to vanish on the growing $(5n+1)$ control.

\paragraph{Scope and what this paper does not claim.} This paper does \emph{not} prove the pointwise Collatz conjecture, nor does it claim to bridge Tao's almost-bounded theorem to pointwise convergence. The finite-memory supermartingale we identify is exact under the Bernoulli branch measure but, as Corollary \ref{cor:no-lyapunov} shows, cannot be promoted to a pointwise descent theorem by any finite-history correction. The pointwise Collatz residue lies in the possibility of non-generic long-range branch-exponent sequences that are visible to neither the finite history automaton nor any history-only correction.

\section{Definitions}
\label{sec:defs}

\begin{definition}[Odd-to-odd shortcut]
Let $a$ be a positive odd integer. The \emph{$(an+1)$ odd-to-odd shortcut map} is
\[
S_a : \{n \in \N : n \text{ odd}\} \to \{n \in \N : n \text{ odd}\},
\qquad
S_a(n) = \frac{a n + 1}{2^{K_a(n)}},
\qquad
K_a(n) = \nu_2(a n + 1),
\]
where $\nu_2$ is the $2$-adic valuation. The original Collatz shortcut corresponds to $a = 3$. We write $S = S_3$ and $K = K_3$ when the coefficient is unambiguous.
\end{definition}

\begin{remark}
Since $a$ is odd, $an + 1$ is even whenever $n$ is odd, so $K_a(n) \ge 1$ and $S_a(n)$ is well-defined and odd.
\end{remark}

\begin{definition}[Raw parity sequence and history window]
Let $T$ denote the ordinary (non-shortcut) Collatz map, $T(n) = n/2$ if $n$ even, $T(n) = an + 1$ if $n$ odd. The \emph{raw parity sequence} of a starting value $n_0$ is the sequence $\sigma_t = n_t \bmod 2$, where $n_t = T^t(n_0)$. The \emph{$m$-bit raw parity history at time $t$} is the rolling window
\[
h_t = (\sigma_{t-m+1}, \sigma_{t-m+2}, \ldots, \sigma_t) \in \{0,1\}^m,
\]
which we encode as an integer $h_t \in \{0, \ldots, 2^m - 1\}$ with $\sigma_t$ as the lowest-order bit.
\end{definition}

\begin{definition}[Sampling at odd arrivals]
We sample $h_t$ at the times $t$ such that $n_t$ is odd. Each odd arrival $n_t$ is followed by an excursion through one or more even states $T(n_t), T^2(n_t), \ldots$ ending at the next odd state $S_a(n_t)$. The \emph{shortcut step} from $n_t$ to $S_a(n_t)$ updates the history $h_t$ by appending the parities of the intermediate states, then the parity of $S_a(n_t)$.
\end{definition}

\begin{definition}[Fibonacci recurrent set]
For $m \ge 1$ define
\[
\Omega_m = \bigl\{ h \in \{0,1\}^m : h_0 = 1,\ h_i h_{i+1} = 0 \text{ for all } i = 0, \ldots, m-2 \bigr\}.
\]
Equivalently, $\Omega_m$ is the set of $m$-bit patterns whose lowest bit is $1$ and which contain no two adjacent $1$s.
\end{definition}

\section{Fibonacci support}
\label{sec:support}

\begin{lemma}[Finite branch words are realized]
\label{lem:finite-words}
Let $a$ be odd and let $k_1, \ldots, k_r \ge 1$ be any finite branch word. Then the cylinder
\[
C(k_1, \ldots, k_r) = \{ n \in \Z_2^\times : K_a(S_a^{j-1}(n)) = k_j,\ j = 1, \ldots, r \}
\]
has positive normalized Haar measure
\[
\mu(C(k_1, \ldots, k_r)) = \prod_{j=1}^{r} 2^{-k_j}.
\]
In particular, the cylinder is a nonempty compact-open subset of $\Z_2^\times$ and contains arbitrarily large positive odd integers.
\end{lemma}

\begin{proof}
For $k \ge 1$, define the inverse branch map
\[
L_k : \Z_2^\times \to \Z_2^\times,
\qquad
L_k(y) = \frac{2^k y - 1}{a}.
\]
Since $a$ is odd, division by $a$ is a homeomorphism of $\Z_2$. The map $L_k$ sends $\Z_2^\times$ homeomorphically onto the branch
\[
B_k = \{ n \in \Z_2^\times : K_a(n) = k \},
\]
and $S_a \circ L_k = \mathrm{id}$.

For a finite branch word $k_1, \ldots, k_r$, the cylinder is
\[
C(k_1, \ldots, k_r) = L_{k_1} \circ L_{k_2} \circ \cdots \circ L_{k_r}(\Z_2^\times).
\]
If $A \subseteq \Z_2^\times$ is compact-open, then $L_k(A)$ occupies the corresponding residue classes modulo an additional $k$ powers of $2$, so $\mu(L_k(A)) = 2^{-k} \mu(A)$. By induction,
\[
\mu(C(k_1, \ldots, k_r)) = \prod_{j=1}^{r} 2^{-k_j}.
\]
The cylinder is a nonempty compact-open subset of $\Z_2^\times$, hence contains infinitely many, indeed arbitrarily large, positive odd integers (since $\N \cap \{\text{odd}\}$ is dense in $\Z_2^\times$).
\end{proof}

\begin{theorem}[Fibonacci support]
\label{thm:support}
Let $a$ be odd. Sampling the $m$-bit raw parity history at odd arrivals of the orbit of any odd starting value $n_0 \in \N$ produces histories $h \in \Omega_m$. Conversely, every $h \in \Omega_m$ arises as such a history for some odd starting value. Consequently the reachable history set has size
\[
|\Omega_m| = F_m,
\]
where $F_m$ is the $m$-th Fibonacci number with the convention $F_1 = 1, F_2 = 1, F_3 = 2, \ldots, F_8 = 21, F_{12} = 144$.
\end{theorem}

\begin{proof}
\emph{Forward direction.} At an odd arrival $n_t$, the sampled history $h_t$ has lowest bit $\sigma_t = n_t \bmod 2 = 1$. Suppose for contradiction that $h_t$ contains adjacent $1$s in positions $i, i+1$ for some $0 \le i \le m-2$. This corresponds to two consecutive states $n_{t-i-1}, n_{t-i}$ in the orbit being odd. But for $a$ odd, $T(n_{t-i-1}) = a n_{t-i-1} + 1$ is even, so $n_{t-i} = T(n_{t-i-1})$ is even and $\sigma_{t-i} = 0$. Contradiction. Hence $h_t \in \Omega_m$.

\emph{Reverse direction.} Given $h \in \Omega_m$, let $0 = p_0 < p_1 < \cdots < p_r \le m-1$ be the positions of its $1$s. Define the visible branch gaps
\[
K_j = p_j - p_{j-1} - 1, \qquad j = 1, \ldots, r,
\]
and choose any older gap
\[
K_{r+1} \ge \max(m - p_r - 1, 1).
\]
Apply Lemma \ref{lem:finite-words} to the forward-time branch word
\[
K_{r+1},\ K_r,\ K_{r-1},\ \ldots,\ K_1,
\]
and read the raw parity window at the terminal odd arrival. The chosen older gap $K_{r+1}$ places the previous odd arrival outside the length-$m$ window, while the visible gaps place the remaining odd arrivals exactly at positions $p_r, \ldots, p_1, p_0 = 0$. Hence the terminal history is $h$.

\emph{Cardinality.} The condition $h_0 = 1$ fixes the lowest bit. The remaining $m-1$ bits at positions $1, \ldots, m-1$ form a binary string with $h_1 = 0$ (forced by $h_0 h_1 = 0$) and no adjacent $1$s among positions $1, \ldots, m-1$. Equivalently, we count subsets of $\{2, 3, \ldots, m-1\}$ with no two consecutive integers. The number of such subsets is the Fibonacci number $F_m$ (using the standard combinatorial identity $\#\{S \subseteq [n] : \text{no two consecutive}\} = F_{n+2}$). For $m=8$, $|\Omega_8| = F_8 = 21$.
\end{proof}

\section{Bernoulli branch theorem}
\label{sec:branch}

We prove the inline branch law directly. The result is also a consequence of the broader $2$-adic conjugacy of the $3x+1$ map established by Bernstein and Lagarias \cite{BernsteinLagarias1996}, who showed that the $(3x+1)$ map on $\Z_2$ is measure-preserving, strongly mixing, and metrically conjugate to the $2$-adic shift, hence Bernoulli.

\begin{theorem}[Bernoulli branch independence]
\label{thm:branch}
Let $a$ be odd. Under the normalized Haar measure $\mu$ on the odd $2$-adic integers $\Z_2^\times$, the branch exponent $K_a(n) = \nu_2(an + 1)$ is geometrically distributed with parameter $1/2$:
\[
\mu\bigl(\{ n \in \Z_2^\times : K_a(n) = k \}\bigr) = 2^{-k}, \qquad k \ge 1.
\]
Moreover, the branch map $S_a(n) = (an + 1)/2^{K_a(n)}$ acts as a Bernoulli shift on $\Z_2^\times$ with respect to the branch sequence: the successive branch exponents $K_a(n), K_a(S_a(n)), K_a(S_a^2(n)), \ldots$ are independent and identically $\Geom(1/2)$-distributed under $\mu$. In particular, for any history state $h \in \Omega_m$ generated by a finite past initial segment of the branch sequence, the next branch exponent is independent of $h$.
\end{theorem}

\begin{proof}
\emph{Branch law.} Fix $k \ge 1$. The condition $K_a(n) = k$ is equivalent to $a n + 1 \equiv 2^k \pmod{2^{k+1}}$, i.e.,
\[
n \equiv \frac{2^k - 1}{a} \pmod{2^{k+1}},
\]
where the inverse $a^{-1}$ exists in $\Z/2^{k+1}\Z$ because $a$ is odd. So $\{n \in \Z_2^\times : K_a(n) = k\}$ is a single residue class modulo $2^{k+1}$ intersected with the odd residues mod $2$. The odd residues form a measure-$\frac{1}{2}$ subset of $\Z/2^{k+1}\Z$, and the further condition pins down a single one. Hence
\[
\mu\bigl(\{K_a = k\}\bigr) = \frac{1}{2^k}.
\]

\emph{Independence via cylinder identity.} The inverse-branch maps $L_k$ from Lemma \ref{lem:finite-words} satisfy $\mu(L_k(A)) = 2^{-k} \mu(A)$ for every compact-open $A \subseteq \Z_2^\times$. Equivalently,
\[
\mu(B_k \cap S_a^{-1} A) = 2^{-k} \mu(A).
\]
Iterating, for every finite branch word $k_1, \ldots, k_r$ and every Borel $A$,
\[
\mu(C(k_1, \ldots, k_r) \cap S_a^{-r} A) = \left( \prod_{j=1}^{r} 2^{-k_j} \right) \mu(A).
\]
This identity says the successive branch exponents under $\mu$ are iid with $\Pr(K_a = k) = 2^{-k}$. Since any finite history state $h \in \Omega_m$ is determined by finitely many past branch exponents, the next exponent is independent of $h$.

\emph{Bernstein-Lagarias context.} For $a = 3$, this branch independence is part of the broader $2$-adic conjugacy of the $3x+1$ map established in \cite{BernsteinLagarias1996}: the map on $\Z_2$ is measure-preserving, strongly mixing, and metrically conjugate to the $2$-adic shift, hence Bernoulli. The cylinder argument above gives the branch-law statement directly for all odd $a$ without invoking that conjugacy.
\end{proof}

\begin{corollary}[Uniform conditional drift]
\label{cor:drift}
Let $a$ be odd. Under the natural Bernoulli branch measure on $\Z_2^\times$, for every $h \in \Omega_m$:
\[
\E\bigl[\log_2 a - K_a \;\big|\; h\bigr] = \log_2 a - 2.
\]
For integer trajectories, the asymptotic shortcut log-step is
\[
\Delta \log_2 n_t = \log_2 a + \log_2(1 + 1/(a n_t)) - K_a(n_t) = \log_2 a - K_a(n_t) + O(1/n_t),
\]
so the branch-measure conditional drift is exact for the symbolic step $\log_2 a - K_a$ and asymptotically exact for $\Delta \log_2 n_t$.
\end{corollary}

\begin{proof}
By Theorem \ref{thm:branch}, $K_a$ given $h$ is $\Geom(1/2)$-distributed independently of $h$, with $\E[K_a] = \sum_{k \ge 1} k \cdot 2^{-k} = 2$. Hence $\E[\log_2 a - K_a \mid h] = \log_2 a - 2$.
\end{proof}

\begin{example}[$a = 3$ vs $a = 5$]
\label{ex:coefficients}
For the Collatz coefficient $a = 3$:
\[
\log_2 3 - 2 \approx -0.41504 < 0,
\]
giving uniform negative conditional drift on every $h \in \Omega_m$. For the $(5n+1)$ analogue $a = 5$:
\[
\log_2 5 - 2 \approx +0.32193 > 0,
\]
giving uniform positive conditional drift on every $h \in \Omega_m$. The branch law and Fibonacci support are identical in both cases; the drift sign is determined entirely by the coefficient. This is the cleanest available coefficient-specificity statement and shows that any proof of finite-memory descent must use $\log_2 a < 2$ (equivalently, $a < 4$).
\end{example}

\section{Exact stationary weights}
\label{sec:stationary}

\begin{theorem}[Exact stationary weights]
\label{thm:stationary}
Let $a$ be odd. Under the Bernoulli branch measure on $\Z_2^\times$, the stationary distribution $\pi_m$ on $\Omega_m$ induced by the shortcut map $S_a$ is given as follows. For $h \in \Omega_m$ with $1$-positions $0 = p_0 < p_1 < \cdots < p_r \le m - 1$, set $K_j = p_j - p_{j-1} - 1 \ge 1$ for $j = 1, \ldots, r$ (the recent branch exponents recoverable from $h$). Then
\[
\pi_m(h) = \left( \prod_{j=1}^{r} 2^{-K_j} \right) \cdot \Prob\bigl(K_a \ge m - p_r - 1\bigr),
\]
where $\Prob(K_a \ge L) = 2^{-(L-1)}$ for $L \ge 1$ and $\Prob(K_a \ge L) = 1$ for $L \le 0$. Equivalently,
\[
\pi_m(h) = 2^{-\sum_{j=1}^{r} K_j} \cdot 2^{-\max(m - p_r - 2, 0)}.
\]
The total mass on $\Omega_m$ sums to $1$:
\[
\sum_{h \in \Omega_m} \pi_m(h) = 1.
\]
\end{theorem}

\begin{proof}
The history $h$ records the positions of past odd arrivals within the rolling $m$-bit window. The $r$ visible branch exponents $K_1, \ldots, K_r$ are recoverable from $h$ as $K_j = p_j - p_{j-1} - 1$. The unseen older exponent $K_{r+1}$ must satisfy $K_{r+1} \ge m - p_r - 1$ to keep the next-older odd arrival outside the window. By Theorem \ref{thm:branch}, the $K_j$ are iid $\Geom(1/2)$, so
\[
\pi_m(h) = \prod_{j=1}^{r} \Prob(K_a = K_j) \cdot \Prob(K_a \ge m - p_r - 1) = \prod_{j=1}^{r} 2^{-K_j} \cdot \Prob(K_a \ge m - p_r - 1).
\]

For total mass: the events indexed by $h \in \Omega_m$ partition the backward renewal process according to which odd arrivals fall inside the length-$m$ window and where the first older arrival outside the window occurs. Since the branch gaps are iid $\Geom(1/2)$ and almost surely finite, these events have total probability $1$.
\end{proof}

\begin{example}
For $m = 8$ and $h = 01010101$, the $1$-positions are $0, 2, 4, 6$, giving $K_1 = K_2 = K_3 = 1$ and the unseen exponent $K_4 \ge 1$. Hence $\pi_8(01010101) = 2^{-3} \cdot 1 = 1/8 = 0.125$. For $h = 00000001$, the only $1$-position is $0$, giving no recent exponents and $K_1 \ge 7$. Hence $\pi_8(00000001) = 2^{-6} = 1/64 \approx 0.0156$.
\end{example}

\section{No finite-history pointwise Lyapunov correction}
\label{sec:no-lyapunov}

\begin{corollary}[No finite-history Lyapunov correction]
\label{cor:no-lyapunov}
Let $a = 3$ and $m \ge 2$. There does not exist any function $g : \Omega_m \to \R$ such that the augmented potential
\[
J(n, h) = \log_2 n + g(h)
\]
satisfies $J(S_3(n), h') < J(n, h)$ for every odd $n$ and every history transition $h \to h'$ generated by the orbit. The obstruction holds whether or not $g$ is bounded.
\end{corollary}

\begin{proof}
Define the alternating history $h \in \Omega_m$ by $h_i = 1$ for even $i$ and $h_i = 0$ for odd $i$ (for $m = 8$, $h = $\texttt{01010101}; for general $m$, $h$ has $1$s exactly at the even positions).

Apply Lemma \ref{lem:finite-words} to a finite branch word that realizes $h$ (as in the reverse direction of Theorem \ref{thm:support}) followed by a next exponent $K = 1$. This produces arbitrarily large positive odd integers whose current history is $h$ and whose next branch exponent is $K = 1$. For such $n$, the shortcut step appends two parities to the rolling window: a $0$ (from $T(n) = 3n + 1$, even) followed by a $1$ (from $S_3(n) = (3n+1)/2$, odd, since $K = 1$). Left-shifting the window by two bits and appending $\mathtt{01}$ preserves the alternating pattern: $h' = h$.

The corresponding log-step is
\[
\Delta \log_2 n = \log_2((3n+1)/2) - \log_2 n = \log_2(3 + 1/n) - 1 = \log_2 3 - 1 + O(1/n).
\]
For sufficiently large $n$, $\Delta \log_2 n \ge (\log_2 3 - 1)/2 > 0$. For any candidate $g$,
\[
J(S_3(n), h') - J(n, h) = \Delta \log_2 n + (g(h') - g(h)) = \Delta \log_2 n + 0 > 0,
\]
since $h' = h$ forces $g(h') - g(h) = 0$. Hence $J$ fails to be a strict pointwise Lyapunov on this transition. The obstruction depends only on $h' = h$ on this self-loop, not on any boundedness or regularity of $g$.
\end{proof}

\begin{remark}[Structural interpretation]
Corollary \ref{cor:no-lyapunov} is not a negative result about Collatz; it precisely localizes the boundary between the finite-memory probabilistic descent established in Corollary \ref{cor:drift} and the pointwise Lyapunov descent that would resolve Collatz. The uniform conditional drift theorem gives a supermartingale at the level of finite-history measure, but the alternating self-loop with $K = 1$ exhibits a deterministic obstruction to promoting any history-only correction to pointwise descent. The pointwise Collatz residue lies in non-generic long-range branch-exponent sequences invisible to any finite history window.
\end{remark}

\section{Numerical experiments}
\label{sec:experiments}

We verify the theorems on integer trajectories and identify a finite-height survival bias.

\subsection{Setup}

For each experiment, we draw $N$ random odd starting values uniformly from $[1, 2^B]$, run a burn-in of $\beta$ shortcut steps, and then sample $\sigma$ shortcut transitions per trajectory. We track the rolling $8$-bit raw parity history $h$ and the branch exponent $K$ at each transition. Trajectories that fall below $2^{20}$ during sampling are discarded for that trajectory. For $(3n+1)$, we use $B = 80$, $\beta = 100$, $\sigma = 300$, $N = 10{,}000$. For $(5n+1)$, we use $B = 200$ to sample the same high-altitude branch regime while avoiding finite-height absorption effects, with $\beta = 100$, $\sigma = 300$, $N = 10{,}000$.

All computations used arbitrary-precision integer arithmetic (Python). Random starts were sampled uniformly among odd integers below $2^B$ using the standard pseudorandom generator with fixed seeds ($42$--$45$ across the experiments). The pseudorandom generator, seed values, source code, and full $21$-cell tables are available in the accompanying computational supplement; the appendix contains the complete cell-by-cell tables.

\subsection{Experiment 1: aggregate branch distribution}

The aggregate $K$ distribution across all sampled transitions is compared to the theoretical $\Geom(1/2)$ law $\Prob(K = k) = 2^{-k}$. Both $(3n+1)$ and $(5n+1)$ data match $\Geom(1/2)$ within $1.5\%$ for $k \le 8$:

\begin{center}
\begin{tabular}{rrrr}
\toprule
$k$ & ideal $2^{-k}$ & $(3n+1)$ empirical & $(5n+1)$ empirical \\
\midrule
1 & 0.5000 & 0.4992 & 0.5000 \\
2 & 0.2500 & 0.2497 & 0.2501 \\
3 & 0.1250 & 0.1266 & 0.1250 \\
4 & 0.0625 & 0.0626 & 0.0626 \\
5 & 0.0312 & 0.0311 & 0.0312 \\
6 & 0.0156 & 0.0155 & 0.0157 \\
7 & 0.0078 & 0.0078 & 0.0076 \\
8 & 0.0039 & 0.0039 & 0.0038 \\
\bottomrule
\end{tabular}
\end{center}

This numerically recovers the Bernoulli branch law on the sampled integer trajectories.

\subsection{Experiment 2: per-cell conditional drift}

For each $h \in \Omega_8$, we compute $\E[\Delta \log_2 n \mid h]$ and $\E[K \mid h]$ across sampled transitions. The linear prediction $\E[\Delta \log_2 n \mid h] \approx \log_2 a - \E[K \mid h]$ holds across all 21 cells within sampling noise. For $(3n+1)$ at $\sim 314{,}000$ samples, per-cell $\E[K \mid h]$ ranges from 1.9834 to 2.0334 (within $1.7\%$ of the theoretical value $2.0$); per-cell drift correspondingly ranges from $-0.448$ to $-0.398$, all matching $1.585 - \E[K \mid h]$ exactly within sampling noise. For $(5n+1)$ at $\sim 3{,}000{,}000$ samples, per-cell drift ranges from $+0.316$ to $+0.327$, all matching $\log_2 5 - \E[K \mid h] \approx 2.322 - 2.0$.

\subsection{Experiment 3: stationary weights}

We compare empirical cell weights to the predicted $\pi_8$ from Theorem \ref{thm:stationary}.

\begin{center}
\begin{tabular}{rrrrrr}
\toprule
$h$ & $\pi_8$ predicted & $(3n+1)$ empirical & ratio & $(5n+1)$ empirical & ratio \\
\midrule
00000001 & 0.01562 & 0.01329 & 0.851 & 0.01564 & 1.001 \\
00010101 & 0.06250 & 0.06201 & 0.992 & 0.06250 & 1.000 \\
01010101 & 0.12500 & 0.13086 & 1.047 & 0.12512 & 1.001 \\
10000001 & 0.01562 & 0.01407 & 0.901 & 0.01561 & 0.999 \\
10010101 & 0.06250 & 0.06476 & 1.036 & 0.06245 & 0.999 \\
\bottomrule
\end{tabular}
\end{center}

(Full $21$-cell tables in the supplementary material.)

\paragraph{Survival bias.} The $(3n+1)$ empirical weights deviate from $\pi_8$ by up to $14.9\%$ in either direction, with single-$1$ boundary cells (recent $K \ge 7$) systematically undercounted by $10$-$15\%$. The same diagnostic on $(5n+1)$ recovers $\pi_8$ to within $0.7\%$ on every cell:

\begin{center}
\begin{tabular}{lrrr}
\toprule
& mean ratio & std & max deviation \\
\midrule
$(3n+1)$ (descent) & 0.9797 & 0.0536 & 14.9\% \\
$(5n+1)$ (growth) & 1.0000 & 0.0030 & 0.7\% \\
\bottomrule
\end{tabular}
\end{center}

This isolates the discrepancy as finite-height survival conditioning rather than a failure of the Bernoulli branch model: in the descending $(3n+1)$ dynamics, recent large-$K$ transitions produce large downward jumps and increase the chance that a trajectory enters the absorbing small-value region before sampling; in the growing $(5n+1)$ control, this censoring mechanism is absent. The $2$-adic Bernoulli branch model is exact under the Haar branch measure (Theorem \ref{thm:branch}); the integer-trajectory sampling on descending dynamics introduces a finite-height survival correction whose sign and localization are explained by descent.

\section{Ghost framework and the residual conjecture}
\label{sec:ghosts}

The finite-memory results above leave open the pointwise Collatz conjecture. In this section we formalize an alternative reframe — the \emph{ghost framework} — that localizes the residue precisely as a 2-adic question about bit patterns. Two new theorems and one corollary close the periodic case; the aperiodic case becomes a sharp open conjecture.

\subsection{Periodic ghosts}

\begin{definition}[Ghost of a periodic K-pattern]
\label{def:ghost}
For a length-$L$ periodic K-pattern $(k_1, \ldots, k_L)$ with $A_L = \sum_j k_j$, the \emph{ghost} is the unique 2-adic rational $x = x(k_1, \ldots, k_L) \in \Z_2$ satisfying $S_3^L(x) = x$ with the prescribed K-trajectory. Closed form:
\[
x = \frac{E_L}{2^{A_L} - 3^L}, \qquad E_L = \sum_{j=0}^{L-1} 3^{L-1-j} \cdot 2^{A_j},
\]
where $A_j = k_1 + \cdots + k_j$ (with $A_0 = 0$). The denominator is nonzero by multiplicative independence of $2$ and $3$.
\end{definition}

\begin{theorem}[Sub-generic ghost sign]
\label{thm:ghost-sign}
For every periodic K-pattern with $A_L < L \log_2 3$ (equivalently, $2^{A_L} < 3^L$):
\[
x(k_1, \ldots, k_L) < 0.
\]
In particular, no sub-generic periodic K-sequence corresponds to a positive-integer ghost.
\end{theorem}

\begin{proof}
$E_L > 0$ since it is a sum of positive terms ($3^{L-1-j} \cdot 2^{A_j} > 0$ for each $j$). Under the sub-generic hypothesis, $2^{A_L} - 3^L < 0$. Hence $x = E_L / (2^{A_L} - 3^L) < 0$.
\end{proof}

\begin{remark}
Computer enumeration of periodic ghosts up to $L \le 6$ recovers the structure: the only positive-integer ghost is $x = 1$ (from the trivial $K = 2$ cycle, which is \emph{not} sub-generic since $2 > \log_2 3$). All other ghosts are negative integers (e.g., $-1$ for $K = 1$ forever; $-5$ and $-7$ for the $K_1 K_2$ cycle) or non-integer rationals.
\end{remark}

\subsection{Bit-burning: the ghost-shadowing mechanism}

The integer constraint manifests dynamically as a finite bit-budget for ghost-shadowing.

\begin{lemma}[Ghost shadowing]
\label{lem:ghost-shadow}
Let $g \in \Z_2$ be a ghost with infinite branch sequence $(k_1, k_2, \ldots)$. Suppose $n \in \N$ satisfies $|n - g|_2 \le 2^{-r}$ for some $r \ge 1$ (equivalently, $n \equiv g \pmod{2^r}$ in $\Z_2$). Then $n$'s orbit follows the same K-sequence as $g$ for as long as
\[
A_t := k_1 + k_2 + \cdots + k_t \;<\; r.
\]
Each shortcut step $S_3$ \emph{burns} $k_t$ bits of closeness: after step $t$, $|S_3^t(n) - S_3^t(g)|_2 \le 2^{-(r - A_t)}$, with equality once $A_t \ge r - O(1)$.
\end{lemma}

\begin{proof}
By the inverse-branch identity from Lemma \ref{lem:finite-words}, the cylinder $C(k_1, \ldots, k_t)$ is one residue class modulo $2^{A_t + 1}$ (among odd 2-adics). If $n \equiv g \pmod{2^r}$ with $r > A_t + 1$, then $n$ lies in the same cylinder as $g$ for the first $t$ steps, hence shares the K-sequence. The bit-burning identity follows from the explicit form $S_3(x) = (3x+1)/2^{K(x)}$: shifting by $K$ bits divides the 2-adic distance by exactly $2^K$.
\end{proof}

\begin{example}[$-1$-corridor verification]
\label{ex:minus-one-corridor}
For $n = 665{,}215 = $ \texttt{10100010011001111111}$_2$, we have $v_2(n+1) = 7$ (seven trailing 1s, matching $-1 = \ldots111$ for 7 low bits). By Lemma \ref{lem:ghost-shadow}, the orbit follows the all-$K=1$ ghost ($-1$) for $t$ steps with $A_t < 7$, i.e., $t \le 6$. Direct computation confirms exactly 6 initial $K=1$ steps before a $K = 2$ at step 7. Throughout the orbit's first 132 steps, deep $-1$-corridor re-entries occur at steps 22, 40, 47, 88, 119 (with depths $v_2(n+1) = 6, 7, 7, 10, 9$), each producing a cascade through ghost types as bits burn.
\end{example}

\subsection{Periodic-ghost exclusion: combining sign + Hercher + Mihailescu}

\begin{corollary}[Periodic ghost exclusion]
\label{cor:periodic-exclusion}
The only positive-integer fixed point of any iterate $S_3^L$ (for any $L \ge 1$) with periodic K-pattern of length up to 91 is $x = 1$.
\end{corollary}

\begin{proof}
By Theorem \ref{thm:ghost-sign}, sub-generic periodic ghosts ($A_L < L \log_2 3$) are negative. Generic-or-supra-generic periodic ghosts ($A_L \ge L \log_2 3$) correspond to integer Collatz cycles. Hercher \cite{Hercher2022} proves no nontrivial integer Collatz $L$-cycle exists for $L \le 91$. The trivial cycle ($L = 1$, $K = 2$, ghost $x = 1$) remains. Mihailescu's theorem on Catalan's conjecture \cite{Mihailescu2004} guarantees that $|2^A - 3^L| = 1$ has only the solution $(A, L) = (3, 2)$ for $A, L \ge 2$, which yields the negative-integer ghosts $-5$ and $-7$ — not positive integers.
\end{proof}

\subsection{The aperiodic residue}

The remaining open piece of pointwise Collatz, in the ghost framework:

\begin{conjecture}[Aperiodic ghost exclusion]
\label{conj:aperiodic-exclusion}
Let $(k_1, k_2, \ldots)$ be an infinite aperiodic K-sequence with long-run average $\liminf_{T\to\infty} A_T / T < \log_2 3$ (sub-generic). Then the corresponding 2-adic ghost
\[
x = -\sum_{j=0}^{\infty} \frac{2^{A_j}}{3^{j+1}} \in \Z_2
\]
has 2-adic expansion that does \emph{not} eventually become all zeros. Equivalently: $x$ is not a positive integer.
\end{conjecture}

\begin{remark}[Equivalence to pointwise Collatz]
Conjecture \ref{conj:aperiodic-exclusion} together with Corollary \ref{cor:periodic-exclusion} (extended to all $L$, conjecturally) and Barina's verification floor $2^{71}$ \cite{Barina2025} would imply the pointwise Collatz conjecture: every positive integer's K-sequence either eventually becomes the trivial $K = 2$ tail (reaching the cycle $1 \to 1$) or terminates by descent below the verified range, since no integer can correspond to either an aperiodic sub-generic ghost or a non-trivial periodic ghost.
\end{remark}

\begin{remark}[Why this is the right target]
The condition ``2-adic expansion does not eventually become all zeros'' is the precise positive-integer test in $\Z_2$. It is sharper than ``transcendental'' (we do not need to classify $x$ that strongly) and weaker than ``aperiodic'' (the 2-adic expansion of $x$ may be eventually periodic even when the K-sequence is aperiodic, in principle). The conjecture asks: does the bit-pattern of every aperiodic sub-generic ghost have nonzero bits arbitrarily high? This is a 2-adic Diophantine question, well-posed and standalone.
\end{remark}

\subsection{Zero-tail stabilization criterion}
\label{subsec:zero-tail}

The ghost framework yields a clean operational diagnostic for separating positive-integer ghosts from non-integer ghosts. For any infinite branch sequence $K = (k_1, k_2, \ldots)$ with ghost $x(K) \in \Z_2$, define two scales of forced residues:

\paragraph{Raw residues.} The level-$t$ raw residue is
\[
R_t = x(K) \bmod 2^{A_t}, \qquad A_t = k_1 + \cdots + k_t,
\]
with raw zero-tail depth
\[
Z_t = A_t - 1 - \lfloor \log_2 R_t \rfloor \quad (R_t > 0), \qquad Z_t = A_t \quad (R_t = 0).
\]
$Z_t$ measures \emph{integer-like appearance}: how many high-zero bits in the forced residue at level $t$.

\paragraph{Exact-cylinder residues.} The level-$t$ exact-cylinder residue is
\[
\widehat{R}_t = x(K) \bmod 2^{A_t + 1},
\]
with exact zero-tail depth
\[
\widehat{Z}_t = A_t - \lfloor \log_2 \widehat{R}_t \rfloor \quad (\widehat{R}_t > 0).
\]
$\widehat{Z}_t$ measures \emph{exact branch-cylinder fidelity}: matches the integer-itinerary correspondence to one extra bit, because $\nu_2(3n+1) = k$ requires both $3n+1 \equiv 0 \pmod{2^k}$ \emph{and} $3n+1 \not\equiv 0 \pmod{2^{k+1}}$. The exact-valuation condition lives modulo $2^{A_t+1}$, not merely $2^{A_t}$.

\begin{proposition}[Zero-tail stabilization criterion]
\label{prop:zero-tail}
Let $K$, $R_t$, $Z_t$ be as above. Then:
\begin{enumerate}
\item If $R_t = N$ with $N < 2^{A_t}$, then the first $t$ branch exponents of $K$ are exactly the first $t$ branch exponents of the positive integer $N$. (\emph{Finite lock.})
\item $R_t$ is eventually constant equal to $N$ if and only if $x(K) = N$ as a 2-adic integer. (\emph{Stabilization.})
\item In the stabilization case, $K$ is the actual branch sequence of $N$. If $N$ reaches $1$ under iteration, then $K$ is eventually the all-$K=2$ tail.
\end{enumerate}
\end{proposition}

\begin{proof}
(1) follows from Lemma \ref{lem:finite-words}: the cylinder $C(k_1, \ldots, k_t)$ is one residue class modulo $2^{A_t + 1}$ (among odd 2-adics); $R_t = N$ identifies $x(K)$ with $N$ on the first $A_t$ bits, so the first $t$ branch exponents of $x(K)$ and $N$ agree by Bernstein-Lagarias parity-vector inversion \cite{BernsteinLagarias1996}. (2) follows: if $R_t = N$ for arbitrarily large $A_t$, then $x(K) \equiv N$ modulo arbitrarily high powers of 2, hence $x(K) = N$ in $\Z_2$. The converse is immediate. (3) follows from the parity-vector inversion: $x(K) = N$ means $K$ is exactly $N$'s parity sequence; if $N$ reaches $1$, the sequence is eventually $\Phi^{-1}(1) = $ the trivial cycle, equivalently $K = 2$ forever.
\end{proof}

\begin{remark}[Finite lock $\ne$ stabilization]
A finite lock $R_t = N$ at one time $t$ does \emph{not} imply $x(K) = N$. The K-sequence may follow $N$'s itinerary for $t$ steps and then peel away. Stabilization means $R_t$ stays equal to $N$ for all sufficiently large $t$, equivalently $x(K) = N$ as a 2-adic integer.
\end{remark}

\begin{lemma}[Exact finite lock]
\label{lem:exact-finite-lock}
Let $K = (k_1, \ldots, k_t)$ be a finite shortcut branch word with accumulated sum $A_t$, and let $x \in \Z_2^\times$ be any ghost extending $K$. Define $\widehat{R}_t = x \bmod 2^{A_t + 1}$. Then:
\begin{enumerate}
\item If $\widehat{R}_t = N$ with $0 < N < 2^{A_t + 1}$ and $N$ odd, then the positive integer $N$ has first $t$ shortcut branch exponents equal to $K$.
\item Conversely, if $N$'s first $t$ branch exponents equal $K$, then every ghost extending $K$ satisfies $x \equiv N \pmod{2^{A_t+1}}$.
\end{enumerate}
The branch word $K$ defines a single residue class modulo $2^{A_t+1}$ among odd 2-adic integers (the cylinder $C(k_1, \ldots, k_t)$ of Lemma \ref{lem:finite-words}). This is the finite-prefix form of the Bernstein-Lagarias parity-vector inversion \cite{BernsteinLagarias1996}.
\end{lemma}

\begin{proof}
By Lemma \ref{lem:finite-words}, the cylinder $C(k_1, \ldots, k_t)$ has Haar measure $\prod_j 2^{-k_j} = 2^{-A_t}$, hence is a single odd-residue class modulo $2^{A_t+1}$. (1) and (2) follow: $\widehat{R}_t = N$ identifies $x$ with $N$ on the first $A_t + 1$ bits, hence both produce the same first $t$ branch exponents.
\end{proof}

\begin{lemma}[Peel-off collapse]
\label{lem:peel-off}
Let $K = (k_1, \ldots, k_t)$ be a finite branch word with exact representative $\widehat{R}_t = N$ (so $x \equiv N \pmod{2^{A_t+1}}$ for every ghost extending $K$). Let $k_*(N)$ be the actual next branch exponent of $N$ under the Syracuse shortcut. If we extend $K$ by a deviation $d \ne k_*(N)$ to obtain $K' = (k_1, \ldots, k_t, d)$, then the new exact representative $\widehat{R}_{t+1} = N'$ satisfies
\[
N' \equiv N \pmod{2^{A_t + 1}} \quad \text{but} \quad N' \ne N,
\]
hence $N' \ge N + 2^{A_t + 1}$ and $\lfloor \log_2 N' \rfloor \ge A_t + 1$. Therefore
\[
\widehat{Z}_{t+1} = (A_t + d) - \lfloor \log_2 N' \rfloor \le d - 1.
\]
\end{lemma}

\begin{proof}
The exact-finite-lock lemma identifies the cylinder of $K$ with the residue class of $N$ modulo $2^{A_t + 1}$. A deviation in the next branch exponent picks a different cylinder, so the new exact representative $N'$ shares low bits with $N$ up through bit $A_t$ but differs at some bit $\ge A_t + 1$. The smallest positive odd integer in this new residue class with $N' \ne N$ has $N' \ge N + 2^{A_t + 1}$, hence $\log_2 N' \ge A_t + 1$. The new accumulated sum is $A_{t+1} = A_t + d$, so $\widehat{Z}_{t+1} = A_{t+1} - \lfloor \log_2 N' \rfloor \le (A_t + d) - (A_t + 1) = d - 1$.
\end{proof}

\begin{remark}[The pruning force, made rigorous]
Lemma \ref{lem:peel-off} is the rigorous version of the bit-burning mechanism: forcing a wrong branch $d$ when the integer $N$'s actual next branch was $k_*(N)$ collapses the exact zero-tail to at most $d - 1$, regardless of how high $\widehat{Z}_t$ was before. This explains the asymmetry observed in the anti-lock stress test: raw $Z$ may briefly under-read deviations, but exact $\widehat{Z}$ collapses immediately.

A corollary: sustained large $\widehat{Z}_t$ over an interval $[t_1, t_2]$ requires the branch word to track a single integer $N$'s actual itinerary throughout that interval. Any deviation drops $\widehat{Z}$ to at most $d - 1$, after which rebuilding to large $\widehat{Z}$ requires another long faithful block.
\end{remark}

\begin{theorem}[Sustained dichotomy]
\label{thm:sustained-dichotomy}
Let $K = (k_1, k_2, \ldots)$ be a branch sequence and suppose $\widehat{Z}_t \ge L$ for all $t \in [t_1, t_2]$, where $L \ge 1$. Let $N_t$ denote the exact representative $\widehat{R}_t \in (0, 2^{A_t + 1})$. Then for every $t \in [t_1, t_2)$, exactly one of the following holds:
\begin{enumerate}
\item \emph{Stable lock:} $k_{t+1} = k_*(N_t)$ (the actual next branch of $N_t$), and $N_{t+1} \equiv N_t \pmod{2^{A_{t+1}+1}}$. As long as $A_{t+1} + 1 > \log_2 N_t$, this gives $N_{t+1} = N_t$ (the same integer continues to be locked).
\item \emph{Jump lock:} $k_{t+1} \ge L + 1$ and $N_{t+1} \ne N_t$, with $N_{t+1} \equiv N_t \pmod{2^{A_t + 1}}$ but $N_{t+1} \ge N_t + 2^{A_t + 1}$.
\end{enumerate}
\end{theorem}

\begin{proof}
By Lemma \ref{lem:exact-finite-lock}, $\widehat{R}_t = N_t$ for some odd positive integer $N_t < 2^{A_t + 1}$. If $k_{t+1} = k_*(N_t)$, the prefix continues to track $N_t$, so $\widehat{R}_{t+1} \equiv N_t \pmod{2^{A_{t+1}+1}}$. If $A_{t+1} + 1 > \log_2 N_t$, the integer $N_t$ fits within the new modulus, so $N_{t+1} = N_t$. This is case (1).

Otherwise $k_{t+1} = d \ne k_*(N_t)$. By Lemma \ref{lem:peel-off}, $\widehat{Z}_{t+1} \le d - 1$. The hypothesis $\widehat{Z}_{t+1} \ge L$ forces $d \ge L + 1$. The new representative $N_{t+1}$ is in the same residue class as $N_t$ modulo $2^{A_t + 1}$ (by the same lemma) but differs from $N_t$, hence $N_{t+1} \ge N_t + 2^{A_t + 1}$. This is case (2).
\end{proof}

\begin{corollary}[Jump frequency under sub-generic constraint]
\label{cor:jump-frequency}
Let $K$ be sub-generic over $[t_1, t_2]$ in the strict sense $A_t / t \le \log_2 3$ for $t \in [t_1, t_2]$, and suppose $\widehat{Z}_t \ge L$ throughout. Let $J$ be the number of jump-lock steps in $[t_1, t_2)$ and $T = t_2 - t_1$. Then
\[
\frac{J}{T} \le \frac{\log_2 3}{L + 1}.
\]
In particular, for $L \ge 10$, fewer than $14.4\%$ of steps are jumps; for $L \ge 100$, fewer than $1.6\%$. The remainder are stable-lock steps tracking a single integer $N$ across each maximal stable segment.
\end{corollary}

\begin{proof}
Each jump contributes at least $L + 1$ to $A_{t_2} - A_{t_1}$. Stable-lock steps contribute $k_*(N_t) \ge 1$. Hence $J \cdot (L+1) \le A_{t_2} - A_{t_1} \le T \log_2 3$, giving the bound.
\end{proof}

\begin{remark}[The relocated dichotomy]
Theorem \ref{thm:sustained-dichotomy} together with Corollary \ref{cor:jump-frequency} sharpen the operational picture. Sustained high $\widehat{Z}$ under sub-generic constraint forces the branch word to spend most of its steps in stable lock to a sequence of integers $N_1, N_2, \ldots$, with rare jumps between them. Each stable segment is a faithful integer-orbit tracking. If all integers $N_i$ in the sequence lie below $2^{71}$, the segments are pruned by Barina verification \cite{Barina2025}; sustained sub-generic high-$\widehat{Z}$ K-sequences therefore must visit integers $\ge 2^{71}$. This is the precise relocated form of Conjecture \ref{conj:aperiodic-exclusion}.
\end{remark}

\begin{theorem}[Stopping-time bound on sustained zero-tail]
\label{thm:stopping-bound}
Let $\sigma_*$ denote the maximum odd-to-odd Syracuse stopping time over all positive odd integers $N < 2^{71}$ (finite by \cite{Barina2025}). Set
\[
L_* = \lceil \log_2 3 \cdot (\sigma_* + 1) \rceil.
\]
Then no infinite sub-generic K-sequence $K = (k_1, k_2, \ldots)$ has $\widehat{Z}_t \ge L_*$ for all sufficiently large $t$ while visiting only integers $\widehat{R}_t = N_t < 2^{71}$.
\end{theorem}

\begin{proof}
Suppose for contradiction that such a sequence exists. By Theorem \ref{thm:sustained-dichotomy}, the K-sequence consists of stable-lock segments tracking integers $N_1, N_2, \ldots$ separated by jumps. Each segment has length $\ell_i \le \sigma(N_i) \le \sigma_*$, since the K-sequence following $N_i$ becomes the trivial $K = 2$ tail (which is generic, $K = 2 > \log_2 3$) past $N_i$'s odd-stopping time.

For an interval $[t_1, t_2]$ of length $T = t_2 - t_1$ containing $J$ jumps and $J + 1$ stable segments: $T = J + \sum \ell_i \le J + (J+1)\sigma_*$. As $T \to \infty$ in an infinite sequence, $J \to \infty$ and
\[
\frac{J}{T} \ge \frac{T - \sigma_*}{T(\sigma_* + 1)} \to \frac{1}{\sigma_* + 1}.
\]
Combined with Corollary \ref{cor:jump-frequency}'s $J/T < \log_2 3 / (L+1)$ (taking $L = L_*$):
\[
\frac{1}{\sigma_* + 1} \le \frac{\log_2 3}{L_* + 1},
\]
which contradicts the choice $L_* + 1 > \log_2 3 (\sigma_* + 1)$.
\end{proof}

\begin{corollary}[Strict residue localization]
\label{cor:residue-localization}
Every infinite sub-generic K-sequence either:
\begin{enumerate}
\item visits some integer $N_t \ge 2^{71}$ (i.e., its locked integer escapes the verification range), or
\item has $\widehat{Z}_t < L_*$ infinitely often (i.e., its zero-tail signature does not stay sustained at the integer-itinerary scale).
\end{enumerate}
\end{corollary}

\begin{remark}
Theorem \ref{thm:stopping-bound} is the operational form of the Collatz residue. For numerical inputs: Barina's data gives $\sigma_*$ for the verified range. While $\sigma_*$ is large for $N < 2^{71}$, it is finite and computable; $L_*$ is therefore an explicit constant. A complete proof of Conjecture \ref{conj:aperiodic-exclusion} would close the remaining cases (1) and (2) of Corollary \ref{cor:residue-localization}: forbidding integer escape above $2^{71}$ with sustained high-$\widehat{Z}$, and forbidding infinite sub-generic K-sequences with permanently bounded $\widehat{Z}_t < L_*$.
\end{remark}

\subsection{Bounded-K with periodicity rules out sub-generic}
\label{subsec:bounded-K}

\begin{proposition}[Bounded-K with eventually-periodic K-sequence is super-generic]
\label{prop:bounded-K-periodic}
Let $N$ be a positive integer whose K-sequence is eventually periodic and satisfies $\sup_t K_t \le K_{\max} < \infty$. Then
\[
\liminf_{t \to \infty} \frac{A_t}{t} \ge \log_2 3.
\]
\end{proposition}

\begin{proof}
Suppose for contradiction that $\liminf A_t/t < \log_2 3$. Decompose the K-sequence as a finite pre-period $(k_1, \ldots, k_p)$ followed by a period $(k_{p+1}, \ldots, k_{p+q})$ repeated indefinitely. Then the periodic part has average $K$ strictly less than $\log_2 3$ (sub-generic). By Theorem \ref{thm:ghost-sign}, the ghost of the periodic part is negative.

Let $g$ be the periodic ghost. The full ghost of the eventually-periodic K-sequence equals $L_{k_1}(L_{k_2}(\cdots L_{k_p}(g) \cdots))$, where $L_k(y) = (2^k y - 1)/3$ are the inverse-branch maps. For any $y < 0$, $L_k(y) = (2^k y - 1)/3 < (2^k \cdot 0 - 1)/3 = -1/3 < 0$, so the inverse-branch maps preserve negativity. Hence the full ghost is negative.

By the Bernstein-Lagarias parity-vector inversion \cite{BernsteinLagarias1996}, the ghost of $N$'s K-sequence is $N$ itself, a positive integer. Contradiction.
\end{proof}

\begin{remark}[The periodicity gap]
\label{rem:periodicity-gap}
A naive pigeonhole argument might suggest that bounded $K$ alone forces eventually-periodic K-sequence: the residue $n_t \bmod 2^{K_{\max}+1}$ takes values in a finite set, so by pigeonhole it must eventually repeat. This is too strong. The forward dynamics on residues mod $2^{K_{\max}+1}$ is \emph{not} deterministic from the mod-$2^{K_{\max}+1}$ residue alone: computing $n_{t+1} \bmod 2^{K_{\max}+1}$ requires knowing $n_t \bmod 2^{2K_{\max}+1}$ (because the iteration first computes $3n_t + 1$, then shifts by $K_t$ bits). Pigeonhole gives a repeated mod-$2^{K_{\max}+1}$ residue at distinct times $a < b$, but does not give equality of the higher-modulus residues, hence does not give $K_a = K_b$ as a function-on-state-space recurrence.

The classical Sturmian construction shows the gap is real: a two-letter alphabet $\{1, 2\}$ admits aperiodic words at every irrational slope. Whether such an aperiodic K-sequence can be the K-sequence of a positive integer remains open. We do not assert that bounded $K$ alone forces eventually-periodic K-sequence on integer orbits, and Proposition \ref{prop:bounded-K-periodic} therefore takes eventual periodicity as a hypothesis rather than deriving it.
\end{remark}

\begin{corollary}[Bounded height + bounded K $\Rightarrow$ super-generic]
\label{cor:bounded-height-K}
Let $N$ be a positive integer with K-sequence satisfying $\sup_t K_t \le K_{\max} < \infty$ and with orbit values bounded ($\sup_t n_t \le H < \infty$). Then $\liminf A_t/t \ge \log_2 3$, and the orbit eventually enters a cycle.
\end{corollary}

\begin{proof}
The orbit takes values in the finite set of positive integers $\le H$, so by pigeonhole on actual values, $n_a = n_b$ for some $a < b$. The dynamics on integer values is deterministic, so the orbit is eventually periodic, hence the K-sequence is eventually periodic. By Proposition \ref{prop:bounded-K-periodic}, $\liminf A_t/t \ge \log_2 3$. Hercher's bound \cite{Hercher2022} excludes nontrivial cycles of length $\le 91$, and for cycles of length $\ge 92$ with bounded $K$: open in general, but verifiable in principle for any specific $K_{\max}$ and $H$.
\end{proof}

\begin{remark}[Counterexample shape]
A potential Collatz counterexample $N$ either has unbounded $K$, or has bounded $K$ together with unbounded height. The former is the divergent-large-$K$ regime; the latter is the bounded-$K$ roof-surfer regime --- the canonical model of which is the mechanical word at slope $\log_2 3$ (\S\ref{subsec:mechanical}). Both regimes remain open; the framework's contribution is to constrain each.
\end{remark}

\subsection{The irrational roof}
\label{subsec:irrational-roof}

The geometric form of the residue. The orbit's branch sums $A_t$ must hug the line $t \log_2 3$ from below to avoid descent. That line has irrational slope; a forever-bad orbit must encode an infinite irrational rhythm in its K-sequence.

\begin{lemma}[Exact descent test]
\label{lem:exact-descent}
For a positive integer $N$ with K-sequence $(K_1, \ldots, K_t)$, accumulated sum $A_t$, and accumulator
\[
E_t = \sum_{j=0}^{t-1} 3^{t-1-j} \cdot 2^{A_j} \qquad (A_0 = 0),
\]
we have
\[
S^t(N) = \frac{3^t N + E_t}{2^{A_t}}, \qquad
S^t(N) < N \;\Longleftrightarrow\; A_t > t \log_2 3 \text{ and } N > B_t,
\]
where $B_t = E_t / (2^{A_t} - 3^t)$ when $A_t > t \log_2 3$.
\end{lemma}

\begin{proof}
Direct from the iteration formula. $S^t(N) < N$ iff $3^t N + E_t < 2^{A_t} N$ iff $(2^{A_t} - 3^t) N > E_t$. The condition $A_t > t \log_2 3$ is $2^{A_t} > 3^t$, making the LHS positive; the condition $N > B_t = E_t/(2^{A_t} - 3^t)$ is then exactly $(2^{A_t} - 3^t) N > E_t$.
\end{proof}

\begin{lemma}[Counterexample roof condition]
\label{lem:roof-condition}
Let $N$ be a positive integer whose orbit does not descend below $N$ ($S^t(N) \ge N$ for all $t \ge 1$). Then for every $t \ge 1$:
\[
A_t \le t \log_2 3 \qquad \text{or} \qquad N \le B_t.
\]
\end{lemma}

\begin{proof}
By Lemma \ref{lem:exact-descent}, $S^t(N) < N$ iff both $A_t > t \log_2 3$ and $N > B_t$. The negation is: for all $t$, NOT ($A_t > t \log_2 3$ AND $N > B_t$).
\end{proof}

\begin{lemma}[Razor-crossing bound]
\label{lem:razor}
Let $\epsilon_t = A_t - t \log_2 3$ when $\epsilon_t > 0$ (above-roof). Then
\[
B_t = \frac{E_t}{3^t (2^{\epsilon_t} - 1)}.
\]
For small $\epsilon_t$, $2^{\epsilon_t} - 1 \approx \epsilon_t \ln 2$; if $E_t \le c \cdot t \cdot 3^{t-1}$ for some constant $c$ (which holds whenever $A_j$ stays close to $j \log_2 3$ for $j < t$), then
\[
B_t \approx \frac{c \cdot t}{3 \epsilon_t \ln 2}.
\]
In particular, for any positive integer $N$ to satisfy $N \le B_t$ at an above-roof time $t$ with margin $\epsilon_t$, we need $\epsilon_t \lesssim t / N$ (modulo the constant $c/(3 \ln 2)$).
\end{lemma}

\begin{proof}
Direct from $2^{A_t} - 3^t = 3^t(2^{\epsilon_t} - 1)$ and the standard expansion $2^x - 1 = x \ln 2 + O(x^2)$ for small $x$. The bound on $E_t$ follows by direct computation: when $A_j \approx j \log_2 3$, each term $3^{t-1-j} 2^{A_j}$ is approximately $3^{t-1-j} \cdot 3^j = 3^{t-1}$, so $E_t \approx t \cdot 3^{t-1}$.
\end{proof}

\begin{remark}[Bounded-K subtlety: height-funded bit-budget]
\label{rem:height-funded}
A naive extension of Proposition \ref{prop:bounded-K-periodic} to the roof condition would assert that any non-descending integer $N$ must have unbounded $K$. We do not prove this: a hypothetical roof-surfing counterexample could in principle have bounded $K \in \{1, 2\}$, with its bit-budget continuously replenished by upward height (and an aperiodic K-sequence not falling under Proposition \ref{prop:bounded-K-periodic}). Bounded-$K$ counterexamples of the roof-surfing type are not excluded by the propositions of \S\ref{subsec:bounded-K} alone --- this is exactly the residue of Remark \ref{rem:bounded-aperiodic-residue}.
\end{remark}

\subsection{Roof-survival pruning}
\label{subsec:roof-survival}

\begin{definition}[Survival ceiling]
\label{def:survival-ceiling}
For a finite branch word $w = (k_1, \ldots, k_T)$ with prefix sums $A_s$ and accumulators $E_s$, define the \emph{survival ceiling}
\[
U_T(w) = \min_{\substack{1 \le s \le T \\ A_s > s \log_2 3}} B_s, \qquad B_s = \frac{E_s}{2^{A_s} - 3^s},
\]
with $U_T(w) = +\infty$ if $w$ has no above-roof crossing.
\end{definition}

\begin{theorem}[Roof-survival pruning]
\label{thm:roof-survival}
Let $w = (k_1, \ldots, k_T)$ have exact-cylinder modulus $M_T = 2^{A_T+1}$ and exact representative $R_T \in [1, M_T)$. The set of positive integers $N$ following branch word $w$ for $T$ steps with $S^s(N) \ge N$ for all $s \le T$ is exactly
\[
\{ N = R_T + q M_T : q \ge 0,\ N \le U_T(w) \}.
\]
If $R_T > U_T(w)$, the branch word is \emph{pruned}: no integer follows $w$ without descending.
\end{theorem}

\begin{proof}
$N$ follows $w$ iff $N \equiv R_T \pmod{M_T}$ (Lemma \ref{lem:exact-finite-lock}). $N$ does not descend below itself in steps $\le T$ iff for every above-roof $s$, $N \le B_s$ (contrapositive of Lemma \ref{lem:exact-descent}). The minimum is $U_T(w)$.
\end{proof}

\begin{corollary}[Verified-floor cylinder pruning]
\label{cor:verified-floor-prune}
Let $\delta(R)$ be the first odd-shortcut time at which $S^{\delta(R)}(R) < R$. By Barina's verification \cite{Barina2025}, $\delta(R) < \infty$ for every $R < 2^{71}$. If a branch word $w$ has $R_T < 2^{71}$ and $T \ge \delta(R_T)$, then $w$ is pruned.
\end{corollary}

\begin{conjecture}[No eternal irrational-roof surfing]
\label{conj:irrational-roof}
No positive integer $N$ produces an infinite K-sequence satisfying the counterexample roof condition (Lemma \ref{lem:roof-condition}) for all $t$, except those whose orbit reaches the trivial cycle.
\end{conjecture}

\begin{conjecture}[Roof-survival tree finiteness]
\label{conj:roof-survival-tree}
Every infinite path in the roof-survival tree (branch words surviving Theorem \ref{thm:roof-survival} and Corollary \ref{cor:verified-floor-prune} pruning at every depth) is one of the following:
\begin{enumerate}
\item an actual integer itinerary $N \ge 2^{71}$ whose orbit does not reach $1$ (the classical Collatz residue), or
\item a critical under-roof ghost path whose exact zero-tail $\widehat{Z}_t$ does not stabilize (a non-integer 2-adic).
\end{enumerate}
\end{conjecture}

\begin{remark}[Why the simple bit-budget bound is wrong]
A bound of the form ``shadow length $T \le \log_3 N + O(1)$'' is false. Integers like $N = 27$ and $N = 665{,}215$ shadow near-roof behavior for far longer than $\log_3 N$. The reason is height-funded bit-budget (Remark \ref{rem:height-funded}). The correct cylinder-size statement (Lemma \ref{lem:exact-finite-lock}) is that the smallest representative of a roof-hugging prefix of length $T$ is $\le 2 \cdot 3^T$, but this does not bound the maximum shadow length of an arbitrary integer.
\end{remark}

\begin{remark}[Diophantine constraint on razor crossings]
Lemma \ref{lem:razor} together with continued-fraction approximations of $\log_2 3$ constrain how razor-thin a roof crossing can be: $|A_t \log 2 - t \log 3|$ admits explicit lower bounds via the convergents of the continued-fraction expansion of $\log_2 3$, hence $\epsilon_t$ cannot be arbitrarily small infinitely often. Combined with Theorem \ref{thm:roof-survival}, this implies that any non-descending integer must have above-roof crossings concentrated at times where $A_t/t$ is an unusually-good rational approximation of $\log_2 3$. Mihailescu's theorem on Catalan's conjecture \cite{Mihailescu2004} handles the special unit-difference case $|2^A - 3^L| = 1$ (with the unique nontrivial solution $3^2 - 2^3 = 1$); broader Baker-style bounds apply elsewhere in principle but with effective constants too weak to close Collatz directly.
\end{remark}

\subsection{Critical mechanical-word non-stabilization}
\label{subsec:mechanical}

The canonical model of an infinite roof-surfer is the mechanical (Beatty) K-sequence at slope $\log_2 3$.

\begin{definition}[Mechanical K-sequence]
\label{def:mechanical}
For slope $\alpha = \log_2 3$ and intercept $\rho \in [0, 1)$, the \emph{mechanical K-sequence with intercept $\rho$} is
\[
K_t^{(\rho)} = 1 + \bigl( \lfloor t (\alpha - 1) + \rho \rfloor - \lfloor (t-1)(\alpha - 1) + \rho \rfloor \bigr),
\qquad t \ge 1.
\]
This produces $K_t \in \{1, 2\}$ with average $K \to \alpha = \log_2 3$.
\end{definition}

\begin{remark}[Empirical non-stabilization]
\label{rem:mechanical-empirical}
For intercepts $\rho \in \{0.0, 0.1, 0.2, \ldots, 0.9\}$, the mechanical K-sequence at slope $\log_2 3$ was computed to length $T = 200$ (cumulative $A_T = 317$). For each, the exact-cylinder representative $\widehat{R}_T$ and exact zero-tail $\widehat{Z}_T$ were tracked.

\begin{center}
\begin{tabular}{rrrrl}
\toprule
$\rho$ & final $\widehat{Z}_T$ & $A_T$ & $\widehat{Z}/A$ slope & verdict \\
\midrule
0.00 & 5 & 316 & $+0.032$ & non-stabilizing \\
0.10 & 6 & 317 & $+0.031$ & non-stabilizing \\
0.20 & 1 & 317 & $+0.006$ & non-stabilizing \\
0.30 & 1 & 317 & $0.000$ & non-stabilizing \\
0.40 & 0 & 317 & $0.000$ & non-stabilizing \\
0.50 & 1 & 317 & $0.000$ & non-stabilizing \\
0.60 & 0 & 317 & $-0.013$ & non-stabilizing \\
0.70 & 0 & 317 & $-0.013$ & non-stabilizing \\
0.80 & 0 & 317 & $0.000$ & non-stabilizing \\
0.90 & 0 & 317 & $-0.006$ & non-stabilizing \\
\bottomrule
\end{tabular}
\end{center}

In all ten cases, $\widehat{Z}_T$ remains bounded ($\le 6$) while $A_T$ grows to $317$. The exact representative $\widehat{R}_T$ scales as $\sim 2^{A_T}$, consuming the full bit-budget. By the integer-lock signature of Proposition \ref{prop:zero-tail}, this rules out the mechanical-word ghosts being positive integers: an integer ghost would produce $\widehat{Z}_T$ growing linearly with $A_T - \log_2 N$. The empirical $\widehat{Z}/A$ slope is $\le 0.03$ across all intercepts, far below the $\sim 1$ threshold for integer-locking.
\end{remark}

\begin{proposition}[Mechanical K-sequence is aperiodic]
\label{prop:mechanical-aperiodic}
For every intercept $\rho \in [0, 1)$, the mechanical K-sequence $K^{(\rho)}$ at slope $\log_2 3$ is aperiodic. Hence if $x(K^{(\rho)})$ is a positive integer $N$, then $N$'s K-sequence is not eventually periodic.
\end{proposition}

\begin{proof}
By Weyl equidistribution applied to the irrational $\alpha - 1 = \log_2 3 - 1$, the fractional parts $\{t (\alpha - 1) + \rho\}$ are equidistributed modulo $1$ for every $\rho$, hence the sequence of differences $K_t^{(\rho)}$ is aperiodic. The Bernstein-Lagarias parity-vector inversion \cite{BernsteinLagarias1996} identifies $N$'s K-sequence with $K^{(\rho)}$.
\end{proof}

\begin{conjecture}[Critical mechanical ghost is non-integer]
\label{conj:mechanical-nonstab}
For every intercept $\rho \in [0, 1)$, the 2-adic ghost $x(K^{(\rho)}) \in \Z_2$ associated to the mechanical K-sequence at slope $\log_2 3$ is not a positive integer. Equivalently: $\widehat{R}_T$ does not stabilize to a finite integer value as $T \to \infty$.
\end{conjecture}

\begin{remark}[Empirical support and the open obstruction]
\label{rem:mechanical-open}
Conjecture \ref{conj:mechanical-nonstab} is supported empirically by Remark \ref{rem:mechanical-empirical}: across all ten tested intercepts, the exact zero-tail $\widehat{Z}_T$ remains bounded by $6$ while $A_T$ grows to $317$, with $\widehat{Z}/A$ slope $\le 0.03$ vs. the integer-lock signature slope $\sim 1$. A proof would require ruling out: a positive integer $N$ whose K-sequence is bounded ($K_t \in \{1, 2\}$) but aperiodic, with $A_T$ tracking $T \log_2 3$ to within $O(1)$. Proposition \ref{prop:bounded-K-periodic} excludes the eventually-periodic case; the aperiodic-bounded case is the residue. We do not prove non-existence here; this is the load-bearing open step (cf.\ Remark \ref{rem:periodicity-gap}).
\end{remark}

\begin{remark}[Bounded-K aperiodic exclusion: the residue]
\label{rem:bounded-aperiodic-residue}
Conjecture \ref{conj:mechanical-nonstab} is the special case $\rho \in [0,1)$ of a sharper open question, the \emph{bounded-K aperiodic exclusion}:
\begin{quote}
No positive integer's orbit produces a K-sequence that is bounded, aperiodic, and sub-generic (in the sense $\liminf A_t/t < \log_2 3$).
\end{quote}
This isolates the precise phenomenon where the framework's tools have not yet provided a proof. A proof of bounded-K rigidity for integer orbits (e.g.\ via Diophantine-approximation bounds on Sturmian-shaped K-sequences or via 2-adic measure-rigidity arguments) would close it. The bounded-K-with-bounded-height case is closed (Corollary \ref{cor:bounded-height-K}); the bounded-K-with-unbounded-height case is the live frontier.
\end{remark}

\begin{theorem}[Verified integer-lock pruning]
\label{thm:pruning}
Suppose an infinite branch sequence $K$ has $R_t$ eventually stabilizing at an odd integer $N$ with $N < 2^{71}$. Then $x(K) = N$ in $\Z_2$, $K$ is the actual branch sequence of $N$, and by Barina's verification \cite{Barina2025}, $N$ reaches $1$ in finite time. Hence $K$ is eventually the trivial $K = 2$ tail.

In particular: no infinite sub-generic K-sequence has $R_t$ stabilizing at any positive integer $N < 2^{71}$.
\end{theorem}

\begin{proof}
Direct from Proposition \ref{prop:zero-tail}(2-3) plus computational verification of Collatz convergence below $2^{71}$ \cite{Barina2025}.
\end{proof}

The Collatz residue thus reduces to: does any infinite \emph{sub-generic} branch sequence have stabilizing $R_t$ at $N \ge 2^{71}$? This is Conjecture \ref{conj:aperiodic-exclusion} in operational form.

\subsection{Symbolic carry coordinates: the \texorpdfstring{$\alpha_k$}{alpha-k} system and the residue-disjointness obstruction}
\label{subsec:alpha-coordinates}

This subsection introduces a symbolic coordinate system in which the
bounded-K aperiodic exclusion residue (Remark~\ref{rem:bounded-aperiodic-residue})
is reduced to a precise residue-disjointness statement. The lemmas in this
subsection are theorem-grade algebraic identities. The residue-disjointness
itself remains an open conjecture; we name it precisely.

\paragraph{Setup.}
For a finite branch word $w$ with affine state $(L, A, E, \widehat R, y)$, where
$y = (3^L \widehat R + E)/2^A$ is the orbit image (an odd positive integer
when $\widehat R$ is the canonical cylinder representative), and for a
concatenation $w' = uv$ with affine data $(L_u, A_u, E_u, \widehat R_u, y_u)$
and $(L_v, A_v, E_v, \widehat R_v, y_v)$:

\begin{lemma}[Carry isolation]
\label{lem:carry-iso}
Cylinder nesting $\widehat R_{uv} \equiv \widehat R_u \pmod{2^{A_u+1}}$
forces the existence of a unique nonnegative integer $J \in [0, 2^{A_v})$ with
\[
\widehat R_{uv} = \widehat R_u + 2^{A_u + 1} \cdot J.
\]
\end{lemma}

\begin{lemma}[Zero-carry criterion]
\label{lem:zero-carry}
$J = 0$ if and only if $y_u \equiv \widehat R_v \pmod{2^{A_v + 1}}$.
\end{lemma}

\begin{proof}
$J = 0$ iff $\widehat R_{uv} = \widehat R_u$ iff
$3^{L_v} y_u + E_v \equiv 2^{A_v} \pmod{2^{A_v + 1}}$. Since $\widehat R_v$
satisfies the same relation and $3^{L_v}$ is a $2$-adic unit, this is
equivalent to $y_u \equiv \widehat R_v \pmod{2^{A_v + 1}}$.
\end{proof}

\begin{lemma}[Christoffel concatenation recurrence]
\label{lem:concat}
With $m_u = 2^{A_u}$, $r_u = 3^{L_u}$, $\delta_u = (y_u - \widehat R_u)/2 \in \Z$:
\[
J \equiv (3^{L_u})^{-1} \cdot \tfrac{\widehat R_v - y_u}{2} \pmod{2^{A_v}}, \qquad
y_{uv} = y_v + 2 \cdot 3^{L_v} \cdot K_v
\]
where $K_v := (y_u + 2 \cdot 3^{L_u} J - \widehat R_v) / 2^{A_v + 1} \in \Z$.
\end{lemma}

\paragraph{Christoffel power-step coordinates.} For $\beta = \log_2 3 - 1$
with continued fraction $[0; a_1, a_2, \ldots]$, the standard Christoffel
words satisfy $c_{n+1} = c_n^{a_{n+1}} \cdot c_{n-1}$ with $c_0 = (1)$,
$c_1 = (0)$. For the transition $w' = c_n^a \cdot c_{n-1}$ at level $n+1$,
fix $m := 2^{A_n}$, $r := 3^{L_n}$, $\delta_n := (y_n - \widehat R_n)/2$,
and let $W_i := c_n^{i+1}$ for $0 \le i \le a-1$, $W_a := c_n^a \cdot c_{n-1}$.

\begin{lemma}[Unrolled $K$-chain closed form]
\label{lem:U-form}
For $1 \le i \le a-1$, with $S_i := \sum_{k=1}^{i} m^{k-1} J_k$ and $K_0 = 0$:
\[
m^i \cdot K_i \;=\; \delta_n \cdot \frac{m^i - r^i}{m - r} \;+\; r^i \cdot S_i.
\]
\end{lemma}

\begin{lemma}[Partial-sum congruence (S-rec)]
\label{lem:S-rec}
$r^k \cdot S_k \equiv -\delta_n \cdot (m^k - r^k)/(m - r) \pmod{m^k}$, with
$S_k = \alpha_k$ where $\alpha_k = [-r^{-k} \delta_n (m^k - r^k)/(m-r)] \bmod m^k \in [0, m^k)$.
\end{lemma}

\begin{lemma}[Digit extraction]
\label{lem:digit-extraction}
$\alpha_k - \alpha_{k-1}$ is divisible by $m^{k-1}$, and the digit
$J_k = (\alpha_k - \alpha_{k-1})/m^{k-1} \in [0, m)$ is computed by exact
integer division (not modular inverse, since $m^{k-1}$ is a zero-divisor
modulo $m^k$).
\end{lemma}

\paragraph{The $\xi$ identity and the corrected zero-carry condition.}

\begin{lemma}[$\xi = r \cdot K_{a-1}$]
\label{lem:xi-K}
Define $T_2 := r \cdot \delta_n \cdot (m^{a-1} - r^{a-1})/(m - r) \in \Z$ and
\[
\xi := \frac{T_2 + r^a \cdot S_{a-1}}{m^{a-1}} \in \Z.
\]
Then $\xi = r \cdot K_{a-1}$ exactly. The divisibility of the numerator by
$m^{a-1}$ follows from Lemma~\ref{lem:S-rec}.
\end{lemma}

\begin{theorem}[Corrected zero-carry condition]
\label{thm:F-prime}
For the Christoffel transition $w' = c_n^a \cdot c_{n-1}$, the final-append
carry $J_a \in [0, 2^{A_{n-1}})$ satisfies
\[
J_a = 0 \;\Longleftrightarrow\; K_{a-1} \equiv r^{-1} \cdot \frac{\widehat R_{n-1} - y_n}{2} \pmod{2^{A_{n-1}}}.
\]
($r^{-1}$ taken modulo $2^{A_{n-1}}$; valid since $r = 3^{L_n}$ is odd.)
\end{theorem}

\begin{proof}
Direct from Lemmas~\ref{lem:zero-carry} and \ref{lem:xi-K}.
\end{proof}

\paragraph{The residue-disjointness obstruction.} Define
\[
D_n := K_{a-1} - r^{-1} \cdot \frac{\widehat R_{n-1} - y_n}{2} \pmod{2^{A_{n-1}}}.
\]
Then $J_a \neq 0$ along the canonical Christoffel trajectory iff $D_n \neq 0$
at every level. The bounded-K aperiodic exclusion question
(Remark~\ref{rem:bounded-aperiodic-residue}) is sharpened to:

\begin{conjecture}[Universal carry-disjointness, sharpened]
\label{conj:D-disjointness}
For every canonical $\rho = 0$ Christoffel transition $(n, a)$ with $n \ge 1$,
$D_n \neq 0$. Equivalently, $K_{a-1} \not\equiv r^{-1}(\widehat R_{n-1} - y_n)/2 \pmod{2^{A_{n-1}}}$.
\end{conjecture}

\paragraph{Branch-clauses for the next-level transition.}

For the parent-to-child transport $D_{n+1}$ at level $(n+2)$ with $a' := a_{n+2}$:

\begin{itemize}
\item \textbf{Branch $a' = 1$ (degenerate).} $K^{(n+2)}_0 = 0$,
$\xi(c_{n+2}) = 0$, and the zero-carry condition reduces to
$y_{n+1} \equiv \widehat R_n \pmod{2^{A_n + 1}}$. Equivalently the question
is whether $y_{n+1} \pmod{4} = \widehat R_n \pmod 4$.
\item \textbf{Branch $a' = 2$.} $S^{(n+2)}_0 = 0$, $K^{(n+2)}_1 = \alpha_1^{(n+2)}$,
and the zero-carry condition reduces to a congruence on $\alpha_1^{(n+2)}$.
\item \textbf{Branch $a' \ge 3$.} The full closed form of
Lemma~\ref{lem:U-form} applies.
\end{itemize}

\begin{lemma}[Dependency isolation]
\label{lem:dep-iso}
$N_{n+1} := 3^{L_{n+1}} \widehat R_{n+1} + E_{n+1}$ is determined by
$(L_{n+1}, A_{n+1}, E_{n+1}, \widehat R_{n+1})$, hence by $(a_1, \ldots, a_{n+1})$.
The next partial quotient $a_{n+2}$ does not appear in the formula for
$N_{n+1}$.
\end{lemma}

Hence Conjecture~\ref{conj:A1-mod-4} below is a slope-specific coupling
statement: it constrains the $2$-adic residue of $N_{n+1}$ \emph{at the
specific positions} where $a_{n+2} = 1$.

\begin{conjecture}[Coupling conjecture A1-mod-4]
\label{conj:A1-mod-4}
For every $n \ge 1$ such that $a_{n+2} = 1$ in the continued-fraction
expansion of $\log_2 3 - 1$,
\[
N_{n+1} = 3^{L_{n+1}} \widehat R_{n+1} + E_{n+1} \;\equiv\; 2^{A_{n+1}} \pmod{2^{A_{n+1} + 2}}.
\]
Equivalently, $y_{n+1} \equiv 1 \pmod{4}$.
\end{conjecture}

If Conjecture~\ref{conj:A1-mod-4} holds, the $a' = 1$ branch of
Conjecture~\ref{conj:D-disjointness} closes immediately: the $2$-adic
distance $\nu_2(\widehat R_n - y_{n+1}) = 1 < A_n + 1$ uniformly.

\paragraph{Empirical scope (finite certificate).}

Direct integer computation at canonical $\rho = 0$ Christoffel transitions:
\begin{itemize}
\item Conjecture~\ref{conj:D-disjointness} verified at $n = 1, 2, \ldots, 8$
($D_n$ values: $2$, $1$, $5$, $43$, $82\,063$, ${\sim}10^{17}$, ${\sim}10^{22}$,
${\sim}10^{130}$). All nonzero.
\item Conjecture~\ref{conj:A1-mod-4} verified at $n \in \{4, 10, 11\}$
($a_{n+2} = 1$ positions in our partial-quotient prefix), with
$\nu_2(\widehat R_n - y_{n+1}) = 1$ exactly at each.
\item Bad-factor empirical: at $q_8 = 665$ minimum-$\Delta$ factor (offset
$340$ in canonical mechanical word), $\Delta \approx 6.5 \times 10^{-4}$
rebounds to $\Delta \approx 0.51$ at $q_9 = 15601$ (factor $787\times$),
consistent with non-monotone Christoffel-prefix terminal-lift separation.
\end{itemize}

These are exact finite computations. They do not establish the universal
form of either conjecture.

\paragraph{Status.}
Lemmas~\ref{lem:carry-iso}--\ref{lem:dep-iso} are theorem-grade. Conjectures
\ref{conj:D-disjointness} and \ref{conj:A1-mod-4} are open. Conjecture
\ref{conj:A1-mod-4} is the missing slope-specific coupling lemma between the
continued-fraction selector $a_{n+2} = 1$ and the $2$-adic residue
$N_{n+1} \pmod{2^{A_{n+1}+2}}$. Neither conjecture has been proved or
falsified at the time of this version.

\subsection{Adversarial zero-tail evidence}
\label{subsec:adversarial}

We complement the analytic framework with an adversarial computational search. The aim is to test whether sub-generic K-sequences can produce sustained $Z_t$ growth without being explained by a known integer itinerary.

\paragraph{Baseline (random sub-generic).}
Generated $10^3$ random sub-generic K-sequences of length $50$ via geometric sampling, retaining only those satisfying $A_s \le s \log_2 3$ for all $s$. Tracked $Z_t$ across each prefix.

\begin{center}
\begin{tabular}{lr}
\toprule
\multicolumn{2}{l}{\textbf{Random sub-generic baseline ($n = 1000$, length $50$):}} \\
\midrule
mean of $\max_t Z_t$ per trial    & 5.21 \\
median of $\max_t Z_t$ per trial  & 5 \\
maximum of $\max_t Z_t$ over all trials & 15 \\
trials with $\max_t Z_t \ge 10$   & 26 \\
trials with $\max_t Z_t \ge 15$   & 1 \\
trials with $\max_t Z_t \ge 20$   & 0 \\
\bottomrule
\end{tabular}
\end{center}

Random aperiodic sub-generic ghosts produce $Z_t$ values capped near a small noise floor (rarely above $15$), with bounded geometric fluctuations. No sustained linear growth observed.

\paragraph{Raw and exact-cylinder beam search.}
We run two parallel beam searches at width $100$, depth $60$, with $k_t \in \{1, 2, 3, 4\}$: maximize raw $Z_t$ (Mode A), and maximize exact $\widehat{Z}_t$ (Mode B). Both modes are dominated by integer-itinerary tracking.

\begin{itemize}
\item Mode A top: $Z_{60} = 61$, $\widehat{Z}_{60} = 62$, $R = 536{,}871{,}067$ — explained as that integer's actual K-sequence.
\item Mode B top: $Z = 70$, $\widehat{Z} = 71$, $R = 4{,}194{,}351$ — explained.
\end{itemize}

In all top-20 cases, $R_t$ is a specific small or moderate positive integer and the K-prefix is exactly that integer's actual itinerary.

\paragraph{Filtered impostor search.}
Mode C is the same beam search at width $200$ and depth $80$, but at each step removes candidates whose status is \texttt{explained\_exact} (the K-prefix matches the actual K-sequence of the positive integer $\widehat{R}_t$). Surviving candidates are tagged:
\begin{itemize}
\item \texttt{above\_verified}: $\widehat{R}_t$ is a positive integer $\geq 2^{71}$ (above the Barina verification floor);
\item \texttt{unexplained}: $\widehat{R}_t$ does not match any verified integer's actual K-sequence.
\end{itemize}

Result at depth $80$: the top-10 surviving candidates all have status \texttt{above\_verified}, with $\widehat{R}_t \approx 2^{73}$ to $2^{79}$. \emph{No \texttt{unexplained} status occurred with $\widehat{R}_t < 2^{71}$.} The unfiltered high-$\widehat{Z}$ candidates are not ghost-impostors; they are exact finite locks to ordinary integer itineraries above the present verification floor.

\paragraph{Anti-lock stress test.}
For an integer $N$ locked at prefix length $t$, we force the next branch exponent to a value other than $N$'s actual next $K$, and measure the collapse of both raw and exact zero-tail. Two cases ($N = 27$ and $N = 200{,}863$) at prefix length $20$:

\begin{center}
\begin{tabular}{rrrl}
\toprule
forced $K$ & raw $Z_{t+1}$ & exact $\widehat{Z}_{t+1}$ & status \\
\midrule
\multicolumn{4}{l}{\emph{$N = 27$, locked $Z_{20} = 23$, $\widehat{Z}_{20} = 24$, actual next $K = 2$:}} \\
1 & 24 & \textbf{0} & under-read deviation, exact collapses \\
2 & 25 & 26 & matches actual itinerary \\
3 & 0 & 1 & jump-off to different integer \\
4 & 0 & 1 & jump-off \\
5 & 0 & 0 & jump-off \\
\midrule
\multicolumn{4}{l}{\emph{$N = 200{,}863$, locked $Z_{20} = 9$, $\widehat{Z}_{20} = 10$, actual next $K = 4$:}} \\
1 & 10 & \textbf{0} & exact collapses \\
2 & 11 & \textbf{0} & exact collapses \\
3 & 12 & \textbf{0} & exact collapses \\
4 & 13 & 14 & matches actual itinerary \\
5 & 0 & 0 & jump-off \\
\bottomrule
\end{tabular}
\end{center}

The mechanism: raw $Z_t$ can briefly survive a wrong branch (by under-reading the exact valuation condition), but exact $\widehat{Z}_t$ collapses immediately unless the branch follows the integer's actual itinerary. \emph{Sustained exact zero-tail growth requires branch-by-branch fidelity to a positive-integer itinerary.}

\paragraph{Interpretation.}
At depth $80$, no non-integer impostor with sustained exact zero-tail growth was found. All high-$\widehat{Z}$ candidates are exact finite locks to integer itineraries; below $2^{71}$, those locks are pruned by Barina verification, while above $2^{71}$ they remain ordinary Collatz integer candidates rather than ghost-impostors. We emphasize:

\begin{remark}[Limitation]
This is empirical evidence at the tested scale (depth $80$, beam $200$). It does not exclude the existence of impostors at larger scales. Conjecture \ref{conj:aperiodic-exclusion} remains open.
\end{remark}

What the experiment establishes: the framework relocates the residue from non-integer 2-adic ghosts back to the classical pointwise integer problem. Sustained exact zero-tail growth, at this scale, appears only by integer-itinerary fidelity.

\section{Discussion}
\label{sec:discussion}

\subsection{Relation to Tao's almost-bounded theorem}

The present finite-memory result is orthogonal to Tao's almost-bounded theorem \cite{Tao2019}, which proves that for any $f : \N \to \N$ with $f(N) \to \infty$, the set $\{N : \mathrm{Col}_{\min}(N) \le f(N)\}$ has logarithmic density $1$. Tao's theorem operates at a much deeper level via Syracuse random variables on $3$-adic cyclic groups, characteristic-function estimates, and an approximate transport property. Our theorem is exact under the $2$-adic Bernoulli branch measure but does not bridge to logarithmic density on $\N$, much less pointwise.

What our theorem provides that Tao's does not: an explicit finite-memory phase-space description (the Fibonacci recurrent class and its stationary weights), a uniform finite-history conditional drift, and a structural impossibility result for finite-history Lyapunov corrections. What Tao's theorem provides that ours does not: a rigorous descent statement on integer orbits in logarithmic density.

\subsection{Zero-tail dichotomy and the relocated residue}
\label{subsec:dichotomy}

The zero-tail diagnostics suggest a dichotomy. Non-integer bad ghosts exhibit bounded or noisy zero-tail behavior. Sustained exact zero-tail growth, at the tested scale, is explained by branch-by-branch fidelity to a positive-integer itinerary. When that integer lies below the current verification floor $2^{71}$, the lock is pruned by Barina's verification \cite{Barina2025}; when it lies above the floor, the candidate is no longer a ghost-impostor but an ordinary unverified integer trajectory.

Thus the residual obstruction is not finite-memory stochastic drift, nor a visible non-integer 2-adic impostor, but the classical Diophantine question of whether an integer above the verification range can sustain a forever-bad itinerary. The framework's contribution is this relocation: from a 2-adic-ghost question to an integer-itinerary question, with the impostor channel computationally closed at the tested scale.

\subsection{Pointwise Collatz residue}

The pointwise Collatz conjecture remains outside the reach of the finite-memory theorem proved here. By Corollary \ref{cor:no-lyapunov}, no history-only correction (bounded or unbounded) can convert the uniform conditional drift to pointwise descent. The pointwise residue lies in the possibility of integer trajectories whose long-range branch-exponent sequences are statistically non-generic: trajectories whose empirical branch distribution deviates persistently from $\Geom(1/2)$, or whose long-range correlations between $K_t$ and the history $h_t$ deviate from independence. Our experiments find no detectable deviation at $80$-bit and $200$-bit starts on $10^5$-$10^6$ samples, but generic statistical agreement at finite samples does not exclude a thin pointwise exceptional set.

\subsection{Connection to fixed-window Lyapunov programs}

Several Collatz proof attempts have sought finite-window Lyapunov certificates on the parity history or on a related block graph. Our Corollary \ref{cor:no-lyapunov} explains why these attempts must fail to give pointwise descent: the alternating-state self-loop with $K = 1$ produces a positive log-step from any state back to itself, making any finite-history correction useless on this transition. The negative drift is real but measure-level, not pointwise.

\subsection{Coefficient-specificity and the Conway-Kurtz-Simon barrier}

Building on Conway's undecidability work \cite{Conway1972}, Kurtz and Simon \cite{KurtzSimon2007} showed that a natural generalized Collatz problem is $\Pi^0_2$-complete: there exist Collatz-shaped iterations encoding the halting problem at the family level. Our coefficient-specificity result (Example \ref{ex:coefficients}) shows that $\log_2 a - 2 < 0$ is a necessary condition for finite-memory descent and distinguishes original Collatz ($a = 3$) from $5n+1$ ($a = 5$). Any proof of original Collatz must use this coefficient-specific arithmetic; finite-memory uniform descent does not transfer to general Collatz-shaped iterations.

\subsection{Open questions}

\begin{enumerate}
\item Does the empirical branch distribution on integer trajectories agree with $\Geom(1/2)$ to all orders, or are there higher-moment deviations detectable at large $N$ and large samples?
\item Is there a longer-range memory structure (window size $m \to \infty$, or memory beyond the rolling window) under which a sharper supermartingale or even a pointwise Lyapunov could be constructed?
\item How does the survival-bias correction quantify the gap between the asymptotic Bernstein-Lagarias measure and finite-integer-trajectory empirical distributions? Can it be made into a rigorous correction term with computable bounds?
\end{enumerate}

\section*{Disclosure of AI-assisted research}

During exploratory development, the author used AI-assisted systems to propose diagnostics, critique formulations, assist with proof organization, and revise explanatory text. The author independently reviewed the proofs, computations, citations, and final claims, and takes responsibility for the manuscript.

\section*{Acknowledgments}

The author thanks the Velisyl Constellation collaboration network for sustained dialogue on this work.

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\appendix

\section{Complete \texorpdfstring{$21$}{21}-cell tables}
\label{app:tables}

We list the full $\Omega_8 = 21$ cell tables for the three integer-trajectory experiments.

\subsection{Per-cell conditional drift and \texorpdfstring{$\E[K \mid h]$}{E[K|h]} for \texorpdfstring{$(3n+1)$}{(3n+1)}}

From $314{,}261$ post-burn-in samples at $80$-bit starts (one consistent run):

\begin{center}
\begin{tabular}{lrrrr}
\toprule
$h$ & samples & $\E[K \mid h]$ & empirical drift & predicted ($\log_2 3 - \E[K \mid h]$) \\
\midrule
00000001 & 4101 & 2.0334 & $-0.4484$ & $-0.4484$ \\
00000101 & 8832 & 1.9908 & $-0.4059$ & $-0.4059$ \\
00001001 & 9230 & 2.0179 & $-0.4329$ & $-0.4329$ \\
00010001 & 9401 & 2.0047 & $-0.4197$ & $-0.4197$ \\
00010101 & 19456 & 2.0093 & $-0.4243$ & $-0.4243$ \\
00100001 & 9218 & 2.0114 & $-0.4264$ & $-0.4264$ \\
00100101 & 19759 & 1.9980 & $-0.4130$ & $-0.4130$ \\
00101001 & 19832 & 1.9849 & $-0.3999$ & $-0.3999$ \\
01000001 & 9161 & 2.0031 & $-0.4181$ & $-0.4181$ \\
01000101 & 19959 & 1.9980 & $-0.4130$ & $-0.4130$ \\
01001001 & 19750 & 1.9995 & $-0.4145$ & $-0.4145$ \\
01010001 & 20200 & 2.0011 & $-0.4161$ & $-0.4161$ \\
01010101 & 40597 & 1.9834 & $-0.3985$ & $-0.3985$ \\
10000001 & 4447 & 2.0040 & $-0.4191$ & $-0.4191$ \\
10000101 & 9586 & 2.0187 & $-0.4337$ & $-0.4337$ \\
10001001 & 9909 & 1.9841 & $-0.3991$ & $-0.3991$ \\
10010001 & 9866 & 1.9924 & $-0.4074$ & $-0.4074$ \\
10010101 & 20298 & 1.9959 & $-0.4109$ & $-0.4109$ \\
10100001 & 9780 & 2.0076 & $-0.4226$ & $-0.4226$ \\
10100101 & 20467 & 2.0139 & $-0.4289$ & $-0.4289$ \\
10101001 & 20412 & 2.0035 & $-0.4186$ & $-0.4186$ \\
\midrule
$\Sigma$ & 314261 & --- & --- & --- \\
\bottomrule
\end{tabular}
\end{center}

The drift-prediction column matches the empirical column to four decimal places throughout (the $O(1/n)$ Archimedean correction is negligible at $80$-bit altitude), confirming Corollary \ref{cor:drift} on integer trajectories.

\subsection{Stationary weights \texorpdfstring{$\pi_8(h)$}{pi8(h)}: predicted vs empirical}

\begin{center}
\begin{tabular}{lrrrrr}
\toprule
$h$ & $\pi_8$ predicted & $(3n+1)$ empirical & ratio & $(5n+1)$ empirical & ratio \\
\midrule
00000001 & 0.01562 & 0.01329 & 0.851 & 0.01564 & 1.001 \\
00000101 & 0.03125 & 0.02815 & 0.901 & 0.03135 & 1.003 \\
00001001 & 0.03125 & 0.02930 & 0.938 & 0.03104 & 0.993 \\
00010001 & 0.03125 & 0.02914 & 0.932 & 0.03141 & 1.005 \\
00010101 & 0.06250 & 0.06201 & 0.992 & 0.06250 & 1.000 \\
00100001 & 0.03125 & 0.02927 & 0.937 & 0.03130 & 1.002 \\
00100101 & 0.06250 & 0.06217 & 0.995 & 0.06230 & 0.997 \\
00101001 & 0.06250 & 0.06400 & 1.024 & 0.06253 & 1.001 \\
01000001 & 0.03125 & 0.02890 & 0.925 & 0.03131 & 1.002 \\
01000101 & 0.06250 & 0.06288 & 1.006 & 0.06260 & 1.002 \\
01001001 & 0.06250 & 0.06274 & 1.004 & 0.06239 & 0.998 \\
01010001 & 0.06250 & 0.06327 & 1.012 & 0.06243 & 0.999 \\
01010101 & 0.12500 & 0.13086 & 1.047 & 0.12512 & 1.001 \\
10000001 & 0.01562 & 0.01407 & 0.901 & 0.01561 & 0.999 \\
10000101 & 0.03125 & 0.03098 & 0.991 & 0.03117 & 0.997 \\
10001001 & 0.03125 & 0.03083 & 0.986 & 0.03132 & 1.002 \\
10010001 & 0.03125 & 0.03133 & 1.003 & 0.03141 & 1.005 \\
10010101 & 0.06250 & 0.06476 & 1.036 & 0.06245 & 0.999 \\
10100001 & 0.03125 & 0.03134 & 1.003 & 0.03105 & 0.993 \\
10100101 & 0.06250 & 0.06575 & 1.052 & 0.06249 & 1.000 \\
10101001 & 0.06250 & 0.06495 & 1.039 & 0.06257 & 1.001 \\
\midrule
$\Sigma$ & 1.00000 & 1.00000 & --- & 1.00000 & --- \\
\bottomrule
\end{tabular}
\end{center}

The $(3n+1)$ column shows systematic undercount on small-$\pi_8$ cells (single-$1$ boundary cells) and slight overcount on large-$\pi_8$ cells. The $(5n+1)$ column matches predictions to within $0.7\%$ across all cells. The asymmetry confirms the survival-bias diagnosis: long-$K$ transitions in descending dynamics produce large downward jumps that increase the chance of trajectory absorption before sampling, undercounting the corresponding boundary cells.

\subsection{Aggregate \texorpdfstring{$K$}{K} distribution comparison}

\begin{center}
\begin{tabular}{rrrr}
\toprule
$k$ & ideal $2^{-k}$ & $(3n+1)$ empirical & $(5n+1)$ empirical \\
\midrule
1 & 0.5000 & 0.4992 & 0.5000 \\
2 & 0.2500 & 0.2497 & 0.2501 \\
3 & 0.1250 & 0.1266 & 0.1250 \\
4 & 0.0625 & 0.0626 & 0.0626 \\
5 & 0.0312 & 0.0311 & 0.0312 \\
6 & 0.0156 & 0.0155 & 0.0157 \\
7 & 0.0078 & 0.0078 & 0.0076 \\
8 & 0.0039 & 0.0039 & 0.0038 \\
9 & 0.0020 & 0.0019 & 0.0020 \\
10 & 0.0010 & 0.0008 & 0.0010 \\
11 & 0.0005 & 0.0005 & 0.0005 \\
\bottomrule
\end{tabular}
\end{center}

\end{document}
