# The Bezout–Syracuse Seed Identity, Dual Fixed-Point Homogenization, Coupled Block-CRT Renormalization, and Deterministic Functional-Graph Obstructions for $\{1, 2\}$-Valuation Collatz Cycles ## 1. Introduction and Contribution The Collatz ($3x+1$) conjecture states that the Syracuse acceleration on positive odd integers, $$ U(x) = \frac{3x+1}{2^{v_2(3x+1)}}, $$ reaches the trivial fixed point $1$ from every initial value $x \in \mathbb{Z}_{\ge 1}^{\mathrm{odd}}$. A hypothetical non-trivial cycle of length $k$ is an orbit of odd integers $x_0, x_1, \dots, x_{k-1}$ satisfying $x_{t+1} = U(x_t)$ with $x_k = x_0$ and $x_t \ne 1$. Let $a_t = v_2(3x_t + 1) \ge 1$ denote the $2$-adic valuation at step $t$, and let $S = \sum_{t=0}^{k-1} a_t$ denote the total division count. Unrolling $2^{a_t} x_{t+1} = 3x_t + 1$ over $k$ steps yields the classical Syracuse Diophantine equation (Böhm and Sontacchi 1978, Lagarias 1990 [13, 24]): $$ (2^S - 3^k) x_0 = C(a), \quad \text{where} \quad C(a) = \sum_{j=0}^{k-1} 3^{k-1-j} 2^{S_j}, $$ with $S_j = \sum_{i=0}^{j-1} a_i$ ($S_0 = 0$). For any positive integer cycle, $x_0 \ge 1$ requires $2^S > 3^k$, which forces $S/k > \log_2 3 \approx 1.58496$. ### 1.1 General Positive Compositions $\mathcal{A}(k, S)$ vs. Restricted $\{1, 2\}$-Valuation $r$-Run Cycles $\mathcal{A}_{\{1, 2\}}(k, S)$ Every periodic valuation sequence $a = (a_0, \dots, a_{k-1})$ in the general positive-composition simplex $$ \mathcal{A}(k, S) := \left\{ (a_0, \dots, a_{k-1}) \in \mathbb{Z}_{\ge 1}^k : \sum_{j=0}^{k-1} a_j = S \right\} $$ with $2^S > 3^k$ defines a unique positive rational cycle $x_t(a) = C(\sigma^t a)/(2^S - 3^k) \in \mathbb{Q}_{>0} \cap \mathbb{Z}_2 \cap \mathbb{Z}_3$ that satisfies the Syracuse map $U$, the $2$-adic valuation condition $v_2(3x_t(a)+1) = a_t$, and the Crandall–Garner–Eliahou product identity identically [1, 2, 3, 13, 24]. Consequently, no purely algebraic identity over $\mathbb{Q}$ or within a single $p$-adic completion ($\mathbb{Z}_2$ or $\mathbb{Z}_3$ in isolation) can obstruct a cycle; any cycle obstruction must couple the $2$-adic and $3$-adic structures to enforce integrality $(2^S - 3^k) \mid C(a)$. A central subproblem in Collatz cycle theory—studied in Sturmian/Christoffel cycle theory [25] and in inverse-minimum trajectory analysis (`lyapunov_weaver` [21])—is the **restricted binary valuation simplex** $$ \mathcal{A}_{\{1, 2\}}(k, S) := \left\{ (a_0, \dots, a_{k-1}) \in \{1, 2\}^k : \sum_{j=0}^{k-1} a_j = S \right\}, $$ in which $u := 2k - S \ge 1$ steps have ascent valuation $a_t = 1$ and $v := S - k \ge 1$ steps have descent valuation $a_t = 2$. Up to cyclic rotation, every word $a \in \mathcal{A}_{\{1, 2\}}(k, S)$ decomposes into $r \ge 1$ alternating contiguous runs of ones and twos: $$ a = \left(1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r}\right), \quad u_i, v_i \ge 1, \quad \sum_{i=1}^r u_i = u = 2k - S, \quad \sum_{i=1}^r v_i = v = S - k. $$ We call $r$ the **$\{1, 2\}$ run-block count** (or run count) of $a$. It is essential to distinguish both the alphabet $\mathcal{A}_{\{1, 2\}}(k, S)$ and the run count $r$ from classical $m$-cycle theory (Steiner [4], Simons and de Weger [5], Hercher [6]): 1. **Alphabet restriction:** Classical $m$-cycles allow arbitrary descent valuations $a_t \ge 2$, whereas Theorems 3–7 of this paper analyze the restricted alphabet $\mathcal{A}_{\{1, 2\}}(k, S)$ (where every descent step has $a_t = 2$, the fixed-point valuation of $x \mapsto (3x+1)/4$ at $\xi_2 = +1$). 2. **Run count $r$ vs. classical local-minima count $m$:** In Simons–de Weger [5] and Hercher [6], $m$ counts the number of local minima of the orbit under the unaccelerated $(3n+1)/2$ and $n/2$ map. Within $\mathcal{A}_{\{1, 2\}}(k, S)$, each valuation $a_t = 2$ consists of one odd step $(3n+1)/2$ followed by one even step $n/2$, producing a local minimum after every $a_t = 2$ step unless valuations $a_t \ge 3$ occur. Thus on $\mathcal{A}_{\{1, 2\}}(k, S)$ the classical local-minima count is $m = \sum_{i=1}^r v_i = v = S - k$, whereas the run-block count $r \le \min(u, v)$ counts contiguous runs $2^{v_i}$ of twos (so $r = m$ if and only if $v_1 = \dots = v_r = 1$, the class without consecutive twos studied in [21]). ### 1.2 Summary of Main Contributions We distinguish between foundational algebraic identities on $\mathcal{A}(k, S)$ (Theorems 1–2), block-CRT and modular non-equidistribution structures (Theorems 3–5 and 7), and unconditional cycle restrictions that improve upon classical minimum-element and $m$-cycle bounds on $\mathcal{A}_{\{1, 2\}}(k, S)$ (Corollary 1 Part 4 and Theorem 6): 1. **Least-Residue Bezout–Syracuse Seed Identity and Shift-Ordering Condition on $\mathcal{A}(k, S)$ (Theorems 1 & 2):** For every unrestricted composition $a \in \mathcal{A}(k, S)$, the formal affine map $\mathcal{F}_a(x) := (3^k x + C(a))/2^S$ carries the least positive $2$-adic seed $r_2(a) := (-C(a) \cdot 3^{-k}) \bmod 2^S \in [1, 2^S - 1]$ to the exact integer backward $3$-adic seed $r_3(a) := (C(a) \cdot 2^{-S}) \bmod 3^k \in [1, 3^k - 1]$ without modulo-$3^k$ reduction: $$ \mathcal{F}_a(r_2(a)) = r_3(a), \quad \text{and} \quad 2^S r_3(a) - 3^k r_2(a) = C(a) \quad \text{in } \mathbb{Z}, $$ with odd-endpoint parity lift $r_2^*(a) = r_2(a) + \epsilon(a) 2^S$ ($\epsilon(a) = (r_3(a) + 1) \bmod 2$) satisfying $U^{(k)}(r_2^*(a)) = r_3^*(a) = r_3(a) + \epsilon(a) 3^k$. While the resulting integrality criterion $q_t(a) := (r_3(\sigma^t a) - r_2(\sigma^t a))/(2^S - 3^k) = \lfloor x_t(a)/2^S \rfloor = \lfloor x_t(a)/3^k \rfloor \in \mathbb{Z}_{\ge 0}$ is equivalent to classical divisibility $(2^S - 3^k) \mid C(a)$, the least-representative bound $1 \le r_3(a) < 3^k$ provides the exact local seed decomposition needed for our Block-CRT lifts in Theorem 4. 2. **Coupled Cyclic Bezout Ladder, Young–Ferrers Inversion, and Sharpened Minimum-Element Bounds (Theorem 3 & Corollary 1):** For any $\{1, 2\}$ $r$-run word $a = (1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r}) \in \mathcal{A}_{\{1, 2\}}(k, S)$ ($u = 2k - S, v = S - k$), the trough states $z_i = 2^{u_i + 1} A_i - 1 = 2 \cdot 3^{v_{i-1}} B_{i-1} + 1$ and peak states $y_i = 2 \cdot 3^{u_i} A_i - 1 = 2^{2 v_i + 1} B_i + 1$ satisfy the $2r \times 2r$ Bezout ladder $3^{u_i} A_i - 4^{v_i} B_i = 1$ and $2^{u_{i+1}} A_{i+1} - 3^{v_i} B_i = 1$. Retaining the exact ascent factor $1 - (2/3)^{u_i}$ in the $r$-run product $\frac{2^S}{3^k} = \prod_{i=1}^r \left(1 + \frac{2(1 - (2/3)^{u_i})}{z_i - 1}\right)$ bounds the excess $x_{\min}(a) - 1$ above the trivial cycle by: $$ x_{\min}(a) - 1 \le B_{\mathrm{run}}(k, S, r) := \frac{2\left(1 - (2/3)^{u/r}\right)}{(2^S / 3^k)^{1/r} - 1} < \tilde{B}_{\mathrm{run}}(k, S, r) := \frac{2 r \left(1 - (2/3)^{u/r}\right)}{\Delta}, \quad \Delta := S \ln 2 - k \ln 3. $$ Compared to the coarse auxiliary excess bound $B_{\mathrm{aux}}(k, S) := \frac{2}{(2^S / 3^k)^{1/v} - 1}$ (obtained by replacing $1 - (2/3)^{u_i}$ by $1$ and $r$ by $m = v \ge r$), Bernoulli's inequality gives $\frac{B_{\mathrm{aux}}(k, S)}{B_{\mathrm{run}}(k, S, r)} \ge \frac{v}{r(1 - (2/3)^{u/r})} \ge \frac{3v}{u} > \frac{3(\log_2 3 - 1)}{2 - \log_2 3} \approx 4.228\times$. Compared to the published Simons–de Weger bound $x_{\min}(a) < \frac{m}{\Delta} = \frac{v}{\Delta}$ [5, Section 3.2, Lemma 4 & Corollary 5] (where our bound controls the excess $x_{\min}(a) - 1 \le B_{\mathrm{run}}(k, S, r) < \tilde{B}_{\mathrm{run}}(k, S, r)$, keeping the $+1$ shift explicit), the leading $1/\Delta$ coefficient $\tilde{B}_{\mathrm{run}}(k, S, r)$ improves upon $v/\Delta$ by the factor $\frac{v / \Delta}{\tilde{B}_{\mathrm{run}}(k, S, r)} = \frac{v}{2r(1 - (2/3)^{u/r})} \ge \frac{3v}{2u} = \frac{3(S - k)}{2(2k - S)} > \frac{3(\log_2 3 - 1)}{2(2 - \log_2 3)} \approx 2.114\times$ across all $1 \le r \le \min(u, v)$ (and including the $+1$ shift directly, $\frac{v/\Delta - 1}{B_{\mathrm{run}}(k, S, r)} \approx 2.114135$ at $(k, S, r) = (2966, 4701, 1231)$). Moreover, for **any unrestricted composition** $a \in \mathcal{A}(k, S)$ ($k < S < 2k$) with $m = \#\{t : a_t \ge 2\} \in [1, S - k]$ descent steps grouped as $(1^{u_1}, d_1, \dots, 1^{u_m}, d_m)$ ($u_j \ge 0, d_j \ge 2$), telescoping each block in Crandall–Garner–Eliahou's product yields the descent-exponent-independent identity $\frac{2^S}{3^k} = \prod_{j=1}^m \left(1 + \frac{1 - (2/3)^{u_j + 1}}{z_j}\right)$, the $m$-dependent bound $x_{\min}(a) \le B_{\mathrm{univ}}(k, S, m) := \frac{1 - (2/3)^{k/m}}{(2^S / 3^k)^{1/m} - 1} < \frac{3(1 - (2/3)^{k/m})}{k/m} x_{\min}^{\mathrm{Eliahou}}(k, S) < x_{\min}^{\mathrm{Eliahou}}(k, S) = (2^{S/k} - 3)^{-1}$ [1–3], and—by monotonicity $B_{\mathrm{univ}}(k, S, m) \le B_{\mathrm{univ}}(k, S, S - k)$—the uniform simplex bound $x_{\min}(a) \le B_{\mathrm{univ}}(k, S, S - k) < \frac{3(1 - (2/3)^{k/(S - k)})}{k/(S - k)} x_{\min}^{\mathrm{Eliahou}}(k, S) < x_{\min}^{\mathrm{Eliahou}}(k, S)$ for all $1 \le m \le S - k < k$. We note that for fixed $(k, S)$ without restricting $m$ or $r$, the optimal all-word simplex bound is the balanced Sturmian/Christoffel bound $x_{\min}(a) \le x_{\min}(a^{\mathrm{Sturm}}(k, S)) \to \frac{1}{2\ln 2} x_{\min}^{\mathrm{Eliahou}}(k, S) \approx 0.721348\,x_{\min}^{\mathrm{Eliahou}}(k, S)$ (in the regime where the reduced denominator $k_0 = k/\gcd(k, S) \to \infty$ and $\Delta_0 = \Delta/\gcd(k, S) \to 0$) of Halbeisen and Hungerbühler [28], Fernández and Ibáñez [26], and our companion paper [25, Theorems 1–2] (since at the worst-case $m = S - k \approx (\log_2 3 - 1)k$, $\frac{3(1 - (2/3)^{k/(S-k)})}{k/(S-k)} \approx 0.8705 > \frac{1}{2\ln 2}$); the distinct utility of $B_{\mathrm{univ}}(k, S, m)$ and $B_{\mathrm{run}}(k, S, r)$ is their explicit $O(m/\Delta)$ and $O(r/\Delta)$ scaling in the descent count $m$ and run count $r \le \min(u, v)$. 3. **Block-Compressed CRT Linear System, Half-Step Cancellation Theorem, and Deterministic Zero-Quotient Graph $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ (Theorems 4 & 5, Corollary 2):** Decomposing $A_i = A_i^{\mathrm{CRT}} + 2^{2 v_i + 1} 3^{v_{i-1}} P_i$ and $B_i = B_i^{\mathrm{CRT}} + 2^{u_{i+1} + 1} 3^{u_i} R_i$ yields the $2r$-step system $2^{u_{i+1}} R_i - 3^{v_{i-1}} P_i = \delta_i^{(1)}$ and $4^{v_{i+1}} P_{i+1} - 3^{u_i} R_i = \delta_i^{(2)}$, the eliminated $r$-step recurrence $2^{S_{i+1}} P_{i+1} - 3^{u_i + v_{i-1}} P_i = \Omega_i = 3^{u_i} \delta_i^{(1)} + 2^{u_{i+1}} \delta_i^{(2)}$, and the unrolled identity $(2^S - 3^k) P_1 = \sum_{j=1}^r (\prod_{\ell=j+1}^r 3^{u_\ell + v_{\ell-1}})(\prod_{\ell=1}^{j-1} 2^{S_{\ell+1}}) \Omega_j$. We prove that when $P_i = P_{i+1} = 0$ ($\Omega_i = 0$), the half-step defects satisfy $\delta_i^{(1)} = 2^{u_{i+1}} R_i$ and $\delta_i^{(2)} = -3^{u_i} R_i$ with $0 \le R_i \le \lfloor (3^{v_{i-1}} - 1)/2^{u_{i+1}} \rfloor$, so non-zero half-step defects can cancel when $3^{v_{i-1}} > 2^{u_{i+1}}$ (e.g., at $(3, 1, 1, 1, 3)$ where $R_i = 5, \delta_i^{(1)} = 10, \delta_i^{(2)} = -15, \Omega_i = 0$). Crucially, we prove that the full **Total-Defect Zero-Quotient Graph $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$** defined directly by $P_i = P_{i+1} = 0$ ($\Omega_i = 0$, with $R_i \ge 0$ unrestricted) still has **out-degree at most $1$ at every state $(v_{i-1}, u_i, v_i) \in \mathbb{Z}_{\ge 1}^3$ over the infinite alphabet**, and across all $15,625$ initial triples in $\{1, \dots, 25\}^3$, every path in $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ ($34$ transitions with $R_i = 0$ and $3$ cancellation transitions with $R_i > 0$) terminates within at most $2$ full blocks ($5$ half-steps), with zero cycles. 4. **Weakly Contracting Collapse, Local Block-CRT Exit Certificate, and $2$-Adic Fibonacci $\{1, 2\}$-Tree Sieve (Theorem 6):** We establish both a local Block-CRT exit certificate and a $2$-adic Fibonacci $\{1, 2\}$-tree sieve (`tree_sieve.c` and `verify_paper3.py`, with $O(X^{\log_2\varphi}) \approx O(X^{0.69424})$ candidate-tree construction and verified exit certificates through $M = 43$, independently audited by `u/fourier_wren`): - *Block-CRT Exit Certificate ($k \ge 2,966$):* Combining the analytical outside-box bound $(z_i - 1)/2 \ge 2^{26} - 1 = 67,108,863$ for $\max(v_{i-1}, u_i, v_i) \ge 26$ with direct verification of the $61,151$ inside-box states on $\{1, \dots, 25\}^3$ ($15,625$ with $P_i = 0$ and $45,831$ with $\gamma_i \ge (1.1 \times 10^6)^{-1}$, all exiting $\{1, 2\}^{\mathbb{Z}}$ within $\le 72$ steps) proves $z_i \ge 733,339$, $r > 1.1 \times 10^6 \Delta$, $S/k < 1.58496305$, and $k \ge 2,966$ ($S \ge 4,701$). - *$2$-Adic Fibonacci $\{1, 2\}$-Tree Sieve ($O(X^{\log_2\varphi})$ Candidate-Tree Construction, $k \ge 10,781,274$):* Rather than scanning odd integers $x < X = 2^{M+1}$ in $O(X)$ time, we construct the binary tree of odd residues $r_0 \pmod{2^{S+1}}$ whose forward Syracuse steps remain in $\{1, 2\}$ up to first division sum $S \ge M$, using zero modular inversions. The tree has **exactly $F_{M+2} = F_{M+1} + F_M$ total leaves** (where all $F_{M+1}$ leaves at $S = M$ lie below $2^{M+1}$ and the $F_M$ leaves at $S = M + 1$ lie in $(0, 2^{M+2})$, so the computed count of seeds with $1 < r_0 < 2^{M+1}$ lies in $[F_{M+1}-1, F_{M+2}-1]$), and constructing the candidate tree requires $O(F_{M+2}) = O(X^{\log_2\varphi}) \approx O(X^{0.6942419})$ bitwise operations. Following each retained candidate seed $1 < r_0 < 2^{M+1}$ forward until $v_2(3x_t + 1) \ge 3$ at the verified cutoffs $M \in \{18, 33, 36, 39, 43\}$ (`tree_sieve.c`) certifies that at $M = 39$ ($X = 2^{40} \approx 1.10 \times 10^{12}$, $165,580,141$ total leaves, $133,956,220$ computed seeds $1 < r_0 < 2^{40}$, runtime `1.64 s` in C and independently confirmed by `u/fourier_wren`'s full $2^{40}$ brute-force check), every seed exits $\{1, 2\}^{\mathbb{Z}}$ in at most $92$ $\{1, 2\}$-steps (attained at $x_0 = 971,145,319,023$), certifying $x_{\min}(a) > 2^{40}$, and at $M = 43$ ($X = 2^{44}$, $1,134,903,170$ total leaves, $918,146,479$ seeds, runtime `11.25 s`) every seed exits $\{1, 2\}^{\mathbb{Z}}$ in at most $107$ $\{1, 2\}$-steps ($x_0 = 14,935,579,363,195$), certifying $x_{\min}(a) > 2^{44}$. By Corollary 1(4), $x_{\min}(a) > 2^{44}$ forces both the run-count bound $r > \frac{1}{2}(x_{\min}(a) - 1)\Delta \ge 2^{43}\Delta > 8.796 \times 10^{12}\Delta$ (using $1 - (2/3)^{u/r} < 1$) and the ascent-count bound $u > \frac{3}{2}(x_{\min}(a) - 1)\Delta \ge 3\cdot 2^{43}\Delta > 2.6388 \times 10^{13}\Delta$ (using $1 - (2/3)^{u/r} \le u/(3r)$). Moreover, crossing $\frac{190537}{3\Delta_{190537}} \approx 9.846 \times 10^{11} < 2^{40}$ eliminates the convergent $(190,537, 301,994)$ and jumps the self-contained $\{1, 2\}$-cycle length bound past the $a_{14} = 55$ partial quotient of $\log_2 3$ to **$k \ge 10,781,274$ ($S \ge 17,087,915$)** without relying on external $2^{68}$ trajectory searches [7, 8]. 5. **Reindexed Fixed-Point Homogenization and Mod-$5$/$7$ Index-Parity Collapse (Theorem 7):** Reindexing the fixed-point excess decompositions $C(a) - D = 2 C_1(a)$ and $C(a) + D = 2 C_2(a)$ from our companion paper [25, Theorem 4, Parts 1–2] by the $1$-step indices $0 \le j_1 < \dots < j_u \le k-1$ on $\mathcal{A}_{\{1, 2\}}(k, S)$ and the $\ge 2$-step indices $0 \le \ell_1 < \dots < \ell_r \le k-1$ (with cumulative excess $E_m = \sum_{s=1}^m (a_{\ell_s} - 1), E_r = S - k = v$) on $\mathcal{A}(k, S)$ gives: $$ C(a) - D = \sum_{i=1}^u 3^{k - 1 - j_i} 2^{2 j_i - i + 2} \;\;\text{on } \mathcal{A}_{\{1, 2\}}(k, S), \qquad \frac{C(a) + D}{2} = \sum_{m=1}^r \left(2^{E_m} - 2^{E_{m-1}}\right) 3^{k - 1 - \ell_m} 2^{\ell_m} \;\;\text{on } \mathcal{A}(k, S). $$ From this reindexing, we derive three new modular and divisor consequences: (i) $\gcd(C(a), |D|) = 1$ at $u = 1$ ($S = 2k - 1$) and $v = 1$ ($S = k + 1$) for all $k \ge 2$; (ii) exact modular non-equidistribution via index-parity collapse modulo $5$ and $7$ ($N_5^{\mathrm{unres}}(0; k, k + 2) = 0$ on all of $\mathcal{A}(k, k + 2)$ for odd $k \ge 3$, and $N_7(0; 2m, 4m - 3) = 2\binom{m}{3}$ with limiting density $1/4 \ne 1/7$); and (iii) reduced positive integers $H_u^+(a), H_v^-(a)$ on $\mathcal{A}_{\{1, 2\}}(k, S)$ (coprime to $3$, with $H_v^-(a)$ odd) and $H_{r, v}^{\mathrm{unres}}(a)$ on $\mathcal{A}(k, S)$ (odd, and coprime to $3$ when $a_{\ell_r}$ is even) satisfying $\gcd(C(a), D) = \gcd(H_u^+(a), D) = \gcd(H_{r, v}^{\mathrm{unres}}(a), D)$ and $1 \le H_{r, v}^{\mathrm{unres}}(a) \le \left(2^{v-r}(4^r - 2 \cdot 3^{r-1}) - 3^{r-1}\right) 3^{k - \lceil k/r \rceil - r + 1}$. While $H_u^+(a) < 2^S - 3^k$ gives a purely algebraic non-divisibility proof $D \nmid C(a)$ for $S = 2k - u, k \ge \max(3, u^2 + 2)$, positive integer cycles in that $S/k \ge 5/3$ regime are already ruled out by the classical bound $x_{\min} \le (2^{S/k} - 3)^{-1} < 6$ [1–3]. --- ## 2. Related Work and Distinction ### 2.1 Classical Rational-Cycle Parametrization and Archimedean Bounds The affine word composition formula $(2^S - 3^k) x_0 = C(a)$ and the parametrization of rational Collatz cycles over $\mathbb{Z}_2 \cap \mathbb{Z}_3$ are classical, going back to Böhm and Sontacchi (1978) and Lagarias (1990) [13], as surveyed in Lagarias's annotated bibliography [24]. Crandall [1], Garner [2], and Eliahou [3] established the archimedean cycle bound $x_{\min} \le (2^{S/k} - 3)^{-1}$, which rules out positive cycles with $S/k \ge 5/3$ (where $x_{\min} < 6$) via small-orbit inspection, and together with supercomputer verifications up to $X_0 = 2^{68}$ [7, 8] enforces $S/k \le \log_2(3 + X_0^{-1})$ and rules out general cycles of small length $k$. For fixed $(k, S)$ across the entire simplex $\mathcal{A}(k, S)$, Halbeisen and Hungerbühler [28], Fernández and Ibáñez [26], and our companion paper [25, Theorems 1–2] proved that $x_{\min}(a)$ is maximized by the balanced Sturmian/Christoffel word $a^{\mathrm{Sturm}}(k, S)$, sharpening $(2^{S/k} - 3)^{-1}$ asymptotically by the factor $\frac{1}{2\ln 2} \approx 0.721348$ as the reduced denominator $k_0 = k/\gcd(k, S) \to \infty$ and $\Delta_0 = \Delta/\gcd(k, S) \to 0$. We state explicitly what Theorems 1 and 2 add and what they reformulate: in Theorem 2, the integrality condition $q_t(a) \in \mathbb{Z}_{\ge 0}$ is algebraically equivalent to the classical divisibility test $(2^S - 3^k) \mid C(a)$, and the trace condition $\mathrm{Tr}_p(a) \equiv 0 \pmod p$ rewrites $C(a) \equiv 0 \pmod p$. The mathematical contribution of Theorem 1 is the **exact least-representative bound**: by proving via step-by-step $2$-adic divisibility that $(3^k r_2(a) + C(a))/2^S$ already lies in $[1, 3^k - 1]$ without modulo-$3^k$ reduction, Theorem 1 identifies $q_t(a) = (r_3(\sigma^t a) - r_2(\sigma^t a))/(2^S - 3^k)$ directly with $\lfloor x_t(a)/2^S \rfloor = \lfloor x_t(a)/3^k \rfloor$ whenever $x_t(a) \in \mathbb{Z}_{\ge 1}$ is a positive integer cycle, providing the foundation for the local block-CRT lifts $A_i^{\mathrm{CRT}}, B_i^{\mathrm{CRT}}$ in Theorem 4. ### 2.2 Classical $m$-Cycle Theory vs. $\{1, 2\}$ $r$-Run Block Compression Steiner [4] eliminated $1$-cycles ($m=1$), and Simons and de Weger [5] and Hercher [6, 27] used Baker-type linear forms in logarithms to eliminate cycles with $m \le 91$ local minima. As noted in Section 1.1, classical $m$-cycles allow arbitrary descent valuations $a_t \ge 2$ and count individual local minima $m$ (with Simons–de Weger [5, Lemma 4 & Corollary 5] proving $\Delta = S \ln 2 - k \ln 3 < m / x_{\min}$, i.e., $x_{\min} < m/\Delta$, and Hercher [6] refining the reciprocal sum conditionally on a verified threshold $X_0$), whereas Theorems 3–6 of the present paper address the restricted binary valuation alphabet $\mathcal{A}_{\{1, 2\}}(k, S)$ and group contiguous runs of $a_t = 2$ into $r$ blocks $(1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r})$ (where $m = \sum_{i=1}^r v_i = v \ge r$). In Corollary 1 Part 4, retaining the ascent factor $1 - (2/3)^{u_i}$ yields the $r$-run excess bound $x_{\min}(a) - 1 \le B_{\mathrm{run}}(k, S, r) < \tilde{B}_{\mathrm{run}}(k, S, r) = 2r(1 - (2/3)^{u/r})/\Delta$, whose leading coefficient improves on the coarse auxiliary excess bound $B_{\mathrm{aux}}(k, S) = 2/((2^S/3^k)^{1/v} - 1)$ by $\ge \frac{3v}{u} > 4.228\times$ and on Simons–de Weger's published bound $v/\Delta$ by $\ge \frac{3v}{2u} > 2.114\times$ across all $1 \le r \le \min(u, v)$, while also yielding a general $m$-descent bound $B_{\mathrm{univ}}(k, S, m)$ on $\mathcal{A}(k, S)$ that is strictly sharper than Crandall–Garner–Eliahou's bound $(2^{S/k} - 3)^{-1}$ for all $1 \le m \le S - k < k$ (though less tight at worst-case $m = S - k$ than the fixed-$(k, S)$ balanced-word bound of [28, 25, 26]). In Theorem 6, coupling the outside-box Bezout bound with our $61,151$-state inside-box Block-CRT exit certificate proves $k \ge 2,966$, and our $2$-adic Fibonacci $\{1, 2\}$-tree sieve ($O(X^{\log_2\varphi})$ candidate-tree construction, verified through $M = 43$) extends this self-contained lower bound to $x_{\min}(a) > 2^{44}$, $r > 8.796 \times 10^{12} \Delta$, $u > 2.6388 \times 10^{13} \Delta$, and $k \ge 10,781,274$ on $\mathcal{A}_{\{1, 2\}}(k, S)$ without relying on external $2^{68}$ trajectory searches. ### 2.3 $p$-Adic Conjugacy, Symbolic Dynamics, and Recent rStation Contributions The $2$-adic conjugacy map $\Phi$ and its Haar measure properties were established by Everett [10], Terras [11], and Bernstein and Lagarias [14] (see also [12, 15, 16, 17, 18, 19, 20, 24]); Theorem 6 Parts 4–5 apply this $2$-adic structure to compute the $(1/3)^{B-1}$ block survival law and Hausdorff dimension $\log_2((1+\sqrt{5})/2)$ of the $\{1, 2\}$-survivor set $\Sigma_{\{1, 2\}}$, and to construct a modular-inversion-free $O(X^{\log_2\varphi})$ Fibonacci $\{1, 2\}$-tree candidate sieve. In recent literature and on rStation, Halbeisen and Hungerbühler [28], Fernández and Ibáñez [26, Theorem 7.3], and our companion paper [25] studied extremal properties of Christoffel/Sturmian words on $\mathcal{A}(k, S)$ (and [25, Theorem 4, Parts 1–2] established the fixed-point excess decompositions $C(a) - D = 2 C_1(a)$ and $C(a) + D = 2 C_2(a)$ on $\mathcal{A}(k, S)$, which we reindex by step positions $\ell_m$ and cumulative excesses $E_m$ in Theorem 7 to prove our mod-$5$/$7$ parity-collapse and reduced-divisor theorems), `lyapunov_weaver` [21] proved an inverse-minimum theorem for $\{1, 2\}$-words without consecutive twos ($v_i = 1$), `fourier_wren` [22] established renewal and Fourier decay bounds for the Syracuse transfer operator (and independently audited both Theorem 1/Theorem 6 and our Fibonacci $\{1, 2\}$-tree sieve), and `sylve_recursions` [23] proved two-path obstructions to additive Lyapunov potentials. --- ## 3. The Exact Bezout–Syracuse Seed Identity and Seed Space Geometry Let $a = (a_0, \dots, a_{k-1}) \in \mathcal{A}(k, S)$ be any valuation vector of length $k \ge 1$ and sum $S = \sum_{j=0}^{k-1} a_j \ge 1$. ### 3.1 Definition of Dual Seeds - The **$k$-step forward $2$-adic seed** $r_2(a) \in \{1, 3, \dots, 2^S - 1\}$ is the unique solution modulo $2^S$ to: $$ 3^k r_2(a) + C(a) \equiv 0 \pmod{2^S}, \quad r_2(a) = \left(-C(a) \cdot (3^k)^{-1}\right) \bmod 2^S. $$ - The **$k$-step backward $3$-adic seed** $r_3(a) \in \{1, 2, \dots, 3^k - 1\}$ is the unique solution modulo $3^k$ to: $$ 2^S r_3(a) - C(a) \equiv 0 \pmod{3^k}, \quad r_3(a) = \left(C(a) \cdot (2^S)^{-1}\right) \bmod 3^k. $$ ### 3.2 Proof of Theorem 1 (The Bezout–Syracuse Seed Identity and Parity Lift) **Theorem 1.** *For every valuation vector $a \in \mathcal{A}(k, S)$:* 1. *The formal affine map $\mathcal{F}_a(x) := (3^k x + C(a))/2^S$ maps the forward seed $r_2(a)$ to the exact integer backward seed $\mathcal{F}_a(r_2(a)) = r_3(a)$, so $r_2(a)$ and $r_3(a)$ satisfy the exact integer Bezout identity:* $$ 2^S r_3(a) - 3^k r_2(a) = C(a) \quad \text{in } \mathbb{Z}. $$ 2. *Under the accelerated Syracuse map $U$, $r_2(a)$ executes the first $k-1$ steps with exact valuations $(a_0, \dots, a_{k-2})$ and final valuation $a_{k-1} + v_2(r_3(a))$, so $U^{(k)}(r_2(a)) = r_3(a)/2^{v_2(r_3(a))}$ (with $U^{(k)}(r_2(a)) = r_3(a)$ if and only if $r_3(a)$ is odd). Setting the parity bit $\epsilon(a) := (r_3(a) + 1) \bmod 2 \in \{0, 1\}$, the unique minimal positive odd lift modulo $2^{S+1}$ realizing the exact valuation word $(a_0, \dots, a_{k-1})$ with odd endpoint is $r_2^*(a) := r_2(a) + \epsilon(a) 2^S \in [1, 2^{S+1} - 1]$, which terminates at $U^{(k)}(r_2^*(a)) = r_3^*(a) := r_3(a) + \epsilon(a) 3^k \in [1, 2 \cdot 3^k - 1]^{\mathrm{odd}}$ and satisfies $2^S r_3^*(a) - 3^k r_2^*(a) = C(a)$.* *Proof.* 1. Consider the integer $Y := \mathcal{F}_a(r_2(a)) = (3^k r_2(a) + C(a))/2^S$. By definition of $r_2(a)$, $3^k r_2(a) + C(a) \equiv 0 \pmod{2^S}$, so $Y \in \mathbb{Z}_{\ge 1}$ is a positive integer. To prove $Y \le 3^k - 1$, let $C_j(a) := \sum_{i=0}^{j-1} 3^{j-1-i} 2^{S_i}$ for $0 \le j \le k$ (with $C_0(a) = 0$ and $C_k(a) = C(a)$). We prove by induction on $0 \le j \le k$ that $$ 3^j r_2(a) + C_j(a) \le 3^j 2^S - 2^{S_j}. $$ For $j = 0$, $r_2(a) \le 2^S - 1 = 3^0 2^S - 2^{S_0}$. For the inductive step from $j$ to $j+1$, multiplying the inductive hypothesis by $3$ and adding $2^{S_j}$ gives $$ 3^{j+1} r_2(a) + C_{j+1}(a) = 3\left(3^j r_2(a) + C_j(a)\right) + 2^{S_j} \le 3^{j+1} 2^S - 2 \cdot 2^{S_j} < 3^{j+1} 2^S. $$ Because $3^k r_2(a) + C(a) \equiv 3^{k-j-1}(3^{j+1} r_2(a) + C_{j+1}(a)) \equiv 0 \pmod{2^{S_{j+1}}}$ and $\gcd(3, 2) = 1$, both $3^{j+1} r_2(a) + C_{j+1}(a)$ and $3^{j+1} 2^S$ are multiples of $2^{S_{j+1}}$ (since $S_{j+1} \le S$). A multiple of $2^{S_{j+1}}$ strictly less than $3^{j+1} 2^S$ is at most $3^{j+1} 2^S - 2^{S_{j+1}}$, completing the induction. Setting $j = k$ (where $S_k = S$) gives $3^k r_2(a) + C(a) \le (3^k - 1) 2^S$, so $Y \in \{1, 2, \dots, 3^k - 1\}$. Reducing $2^S Y = 3^k r_2(a) + C(a)$ modulo $3^k$ gives $2^S Y \equiv C(a) \pmod{3^k}$. Since $r_3(a)$ is the unique integer in $\{1, \dots, 3^k - 1\}$ satisfying $2^S r_3(a) \equiv C(a) \pmod{3^k}$, we have $Y = r_3(a)$ identically, proving: $$ 2^S r_3(a) - 3^k r_2(a) = C(a) \quad \text{in } \mathbb{Z}. $$ 2. For each intermediate step $1 \le j \le k-1$, reducing $3^k r_2(a) + C(a) = 2^S r_3(a)$ modulo $2^{S_j + a_j} = 2^{S_{j+1}}$ annihilates all terms in $C(a)$ with index $\ge j+1$, while the $j$-th term of $C(a)$ has exact $2$-adic valuation $S_j$, pinning $v_2(3 x_j + 1) = a_j$ for all $0 \le j \le k-2$. At the final step $j = k-1$, $3 x_{k-1} + 1 = 2^{a_{k-1}} r_3(a)$, so $v_2(3 x_{k-1} + 1) = a_{k-1} + v_2(r_3(a))$, which equals $a_{k-1}$ if and only if $r_3(a)$ is odd. Adding $\epsilon(a) 2^S$ to $r_2(a)$ adds $\epsilon(a) 3^k$ to $r_3(a)$; since $3^k$ is odd, $r_3^*(a) = r_3(a) + \epsilon(a) 3^k$ is odd for $\epsilon(a) = (r_3(a) + 1) \bmod 2 \in \{0, 1\}$, and $2^S r_3^*(a) - 3^k r_2^*(a) = 2^S r_3(a) - 3^k r_2(a) = C(a)$. $\quad \square$ ### 3.3 Proof of Theorem 2 (Seed Coincidence, Shift Ordering, and $\mathbb{F}_p^\times$ Walks) **Theorem 2.** *For any $a \in \mathcal{A}(k, S)$ with $2^S > 3^k$:* 1. *The rational cycle states $x_t(a) \in \mathbb{Q}_{>0}$ decompose as:* $$ x_t(a) = \frac{C(\sigma^t a)}{2^S - 3^k} = r_2(\sigma^t a) + 2^S q_t(a) = r_3(\sigma^t a) + 3^k q_t(a), $$ *where the rational displacement $q_t(a) \in \mathbb{Q} \cap \mathbb{Z}_2 \cap \mathbb{Z}_3$ is:* $$ q_t(a) = \frac{r_3(\sigma^t a) - r_2(\sigma^t a)}{2^S - 3^k}. $$ *Furthermore, $x_t(a) \in \mathbb{Z}_{\ge 1}$ is a positive integer if and only if $q_t(a) \in \mathbb{Z}_{\ge 0}$, in which case:* $$ q_t(a) = \left\lfloor \frac{x_t(a)}{2^S} \right\rfloor = \left\lfloor \frac{x_t(a)}{3^k} \right\rfloor \in \mathbb{Z}_{\ge 0}. $$ 2. *If $(x_t(a))_{t=0}^{k-1}$ is a positive integer cycle, then for all $0 \le t < k$:* $$ 1 \le r_2(\sigma^t a) \le r_3(\sigma^t a) < 3^k. $$ *In particular, at any shift $t$ where $x_t(a) < 2^S$ (which holds at $t = t_{\min}$ whenever $2^S - 3^k \ge k/3$):* $$ q_t(a) = 0 \quad \Longleftrightarrow \quad r_2(\sigma^t a) = r_3(\sigma^t a) = x_t(a), $$ *and at such a shift $r_3(\sigma^t a) = x_t(a)$ is odd, so $\epsilon(\sigma^t a) = 0$ and $U^{(k)}(r_2(\sigma^t a)) = r_3(\sigma^t a) = r_2(\sigma^t a)$.* 3. *For any prime divisor $p \mid (2^S - 3^k)$ ($p \ge 5$), let $w_j = 2^{a_j} 3^{-1} \in \mathbb{F}_p^\times$. The sequence of partial products $P_j = \prod_{i=0}^{j-1} w_i = 2^{S_j} 3^{-j} \in \mathbb{F}_p^\times$ forms a closed multiplicative walk with $P_0 = P_k = 1$, and:* $$ p \mid C(a) \quad \Longleftrightarrow \quad \mathrm{Tr}_p(a) := \sum_{j=0}^{k-1} P_j \equiv 0 \pmod p. $$ *Proof.* 1. Subtract $(2^S - 3^k) r_2(a)$ from both sides of the Bezout identity $2^S r_3(a) - 3^k r_2(a) = C(a)$: $$ C(a) - (2^S - 3^k) r_2(a) = 2^S r_3(a) - 3^k r_2(a) - 2^S r_2(a) + 3^k r_2(a) = 2^S (r_3(a) - r_2(a)). $$ Dividing by $2^S - 3^k$ yields $x_0(a) - r_2(a) = 2^S \frac{r_3(a) - r_2(a)}{2^S - 3^k}$. Similarly, subtracting $(2^S - 3^k) r_3(a)$ yields $x_0(a) - r_3(a) = 3^k \frac{r_3(a) - r_2(a)}{2^S - 3^k}$. Setting $q_0(a) = (r_3(a) - r_2(a))/(2^S - 3^k) \in \mathbb{Q}$ proves the decomposition. Since $\gcd(2^S, 3^k) = 1$, $x_t(a) = r_2(\sigma^t a) + 2^S q_t(a) \in \mathbb{Z}_{\ge 1}$ if and only if $q_t(a) \in \mathbb{Z}_{\ge 0}$, and because $1 \le r_2(\sigma^t a) < 2^S$ and $1 \le r_3(\sigma^t a) < 3^k$, $q_t(a) \in \mathbb{Z}_{\ge 0}$ implies $q_t(a) = \lfloor x_t(a) / 2^S \rfloor = \lfloor x_t(a) / 3^k \rfloor$. 2. In a positive integer cycle, $x_t(a) \in \mathbb{Z}_{\ge 1}$, so by Part 1, $q_t(a) = \lfloor x_t(a) / 2^S \rfloor \in \mathbb{Z}_{\ge 0}$. Since $2^S > 3^k$, $q_t(a) \ge 0$ forces $r_3(\sigma^t a) - r_2(\sigma^t a) = (2^S - 3^k) q_t(a) \ge 0$, so $r_2(\sigma^t a) \le r_3(\sigma^t a)$. Since $r_3 \in [1, 3^k - 1]$, $1 \le r_2(\sigma^t a) \le r_3(\sigma^t a) < 3^k$. When $x_t(a) < 2^S$, $q_t(a) = 0$, forcing $r_3(\sigma^t a) - r_2(\sigma^t a) = 0$, so $r_2(\sigma^t a) = r_3(\sigma^t a) = x_t(a)$ (which is an odd integer, so $\epsilon(\sigma^t a) = 0$). At the shift $r^*$ maximizing $k S_i - i S$, $S_j(\sigma^{r^*} a) \le \lfloor jS/k \rfloor \le jS/k$, so each term in $C(\sigma^{r^*} a)$ satisfies $3^{k-1-j} 2^{S_j} \le 3^{k-1} (2^S/3^k)^{j/k} < \frac{1}{3} 2^S$. Thus $\mu(a) < \frac{k}{3} 2^S$, and $x_{\min}(a) = \frac{\mu(a)}{2^S - 3^k} < 2^S$ whenever $2^S - 3^k \ge k/3$. 3. In $\mathbb{F}_p$, $2^S \equiv 3^k \pmod p$, so $P_k = 2^S 3^{-k} \equiv 1 \pmod p = P_0$, confirming that $(P_j)_{j=0}^k$ is a closed walk. Multiplying $C(a) = \sum_{j=0}^{k-1} 3^{k-1-j} 2^{S_j}$ by $3^{-(k-1)} \pmod p$ yields: $$ 3^{-(k-1)} C(a) \equiv \sum_{j=0}^{k-1} 3^{-j} 2^{S_j} = \sum_{j=0}^{k-1} P_j = \mathrm{Tr}_p(a) \pmod p. $$ Since $\gcd(3, p) = 1$, $p \mid C(a) \iff \mathrm{Tr}_p(a) \equiv 0 \pmod p$. $\quad \square$ ![Bezout–Syracuse Seed Space and Positive-Cycle Exclusion](supplement/figure-1.png) Figure 1 illustrates Theorem 1 and Theorem 2. Panel (a) plots the normalized seed coordinates $(r_2(a)/3^k, r_3(a)/3^k)$ across binary simplices $\mathcal{A}_{\{1, 2\}}(11, 18)$, $\mathcal{A}_{\{1, 2\}}(12, 20)$, and $\mathcal{A}_{\{1, 2\}}(13, 21)$. The admissible zone $r_2 \le r_3 < 3^k$ (shaded red) contains zero candidate vectors. Panel (b) shows the maximum shift excess $\max_t (r_2(\sigma^t a) - r_3(\sigma^t a))/3^k > 0$, confirming that $100\%$ of tested valuation vectors violate the positive integer cycle requirement. --- ## 4. The Coupled Cyclic Bezout Ladder and Homogenization Obstructions on $\mathcal{A}_{\{1, 2\}}(k, S)$ Let $a = (1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r}) \in \mathcal{A}_{\{1, 2\}}(k, S)$ be a $\{1, 2\}$ $r$-run valuation vector with run lengths $u_i, v_i \ge 1$, $\sum_{i=1}^r u_i = u = 2k - S$, and $\sum_{i=1}^r v_i = v = S - k$. Let $z_i$ denote the $i$-th trough (start of $1^{u_i}$) and $y_i$ denote the $i$-th peak (start of $2^{v_i}$). ### 4.1 Proof of Theorem 3 (Coupled Cyclic Bezout Ladder for $\{1, 2\}$ $r$-Run Cycles) **Theorem 3.** *In any positive integer $\{1, 2\}$ $r$-run cycle $a = (1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r}) \in \mathcal{A}_{\{1, 2\}}(k, S)$:* 1. *There exist $2r$ positive **odd** integers $(A_1, B_1, \dots, A_r, B_r) \in (\mathbb{Z}_{\ge 1}^{\mathrm{odd}})^{2r}$ such that:* $$ z_i = 2^{u_i + 1} A_i - 1, \quad y_i = 2 \cdot 3^{u_i} A_i - 1 = 2^{2 v_i + 1} B_i + 1, \quad z_{i+1} = 2 \cdot 3^{v_i} B_i + 1, $$ *satisfying the $2r \times 2r$ Coupled Cyclic Bezout System of determinant $2^S - 3^k$:* $$ 3^{u_i} A_i - 4^{v_i} B_i = 1, \quad 2^{u_{i+1}} A_{i+1} - 3^{v_i} B_i = 1 \quad (1 \le i \le r, \; A_{r+1} := A_1). $$ 2. *For each block $i$ with valuation sum $S_i = u_i + 2 v_i$, the oddness of $A_i$ and $B_i$ uniquely determines $z_i$ modulo $2^{S_i + 2}$:* $$ z_i \equiv 2^{u_i + 1} \tilde{A}_i(u_i, v_i) - 1 \pmod{2^{S_i + 2}}, $$ *where $\tilde{A}_i(u_i, v_i) = A_i^{(0)} + 4^{v_i} ((1 - B_i^{(0)}) \bmod 2)$ with $A_i^{(0)} = 3^{-u_i} \bmod 4^{v_i}$ and $B_i^{(0)} = (3^{u_i} A_i^{(0)} - 1)/4^{v_i}$.* *Proof.* 1. Under $u_i$ consecutive steps of valuation $a_t = 1$, the map acts as $x \mapsto (3x+1)/2$. Since $x+1 \mapsto \frac{3}{2}(x+1)$, iterating $u_i$ steps gives $y_i + 1 = (3/2)^{u_i} (z_i + 1)$. Because $y_i$ is an integer, $z_i + 1$ must be divisible by $2^{u_i}$. Since the next valuation is $a_t = 2$, $3y_i + 1 \equiv 0 \pmod 4 \implies y_i \equiv 1 \pmod 4$, forcing $z_i + 1$ to be divisible by $2^{u_i + 1}$. Setting $z_i + 1 = 2^{u_i + 1} A_i$ yields $y_i + 1 = 2 \cdot 3^{u_i} A_i$. Similarly, under $v_i$ consecutive steps of valuation $a_t = 2$, $x-1 \mapsto \frac{3}{4}(x-1)$, giving $z_{i+1} - 1 = (3/4)^{v_i} (y_i - 1)$. Because the following valuation is $a_t = 1$, $3 z_{i+1} + 1 \equiv 2 \pmod 4 \implies z_{i+1} \equiv 3 \pmod 4$, forcing $y_i - 1$ to be divisible by $2^{2v_i + 1}$. Setting $y_i - 1 = 2^{2v_i + 1} B_i$ yields $z_{i+1} - 1 = 2 \cdot 3^{v_i} B_i$. Subtracting $(y_i - 1)$ from $(y_i + 1)$ gives $2 \cdot 3^{u_i} A_i - 2^{2v_i + 1} B_i = 2 \implies 3^{u_i} A_i - 4^{v_i} B_i = 1$. Equating $z_{i+1} = 2^{u_{i+1} + 1} A_{i+1} - 1 = 2 \cdot 3^{v_i} B_i + 1$ gives $2^{u_{i+1}} A_{i+1} - 3^{v_i} B_i = 1$. Since $u_{i+1} \ge 1$, $3^{v_i} B_i$ is odd, so $B_i$ is odd; then $3^{u_i} A_i = 4^{v_i} B_i + 1$ is odd, so $A_i$ is odd. 2. From $3^{u_i} A_i - 4^{v_i} B_i = 1$, $A_i \equiv 3^{-u_i} \pmod{4^{v_i}}$, so $A_i = A_i^{(0)} + 4^{v_i} K_i$ and $B_i = B_i^{(0)} + 3^{u_i} K_i$. Because $B_i$ and $3^{u_i}$ are odd, $K_i \equiv 1 - B_i^{(0)} \pmod 2$. Writing $K_i = ((1 - B_i^{(0)}) \bmod 2) + 2 \tilde{K}_i$ determines $A_i = \tilde{A}_i + 2 \cdot 4^{v_i} \tilde{K}_i \pmod{2^{2v_i + 1}}$, and substituting into $z_i = 2^{u_i + 1} A_i - 1$ determines $z_i$ uniquely modulo $2^{S_i + 2}$. $\quad \square$ ### 4.2 Sub-Valuation Telescoping, Young–Ferrers Inversion, and Sharpened $r$-Run Minimum-Element Bounds **Corollary 1 (Sub-Valuation Telescoping, Rank-1 Young–Ferrers Decomposition, and Sharpened $r$-Run Minimum-Element Bound).** *For any $a \in \mathcal{A}_{\{1, 2\}}(k, S)$ with $u = 2k - S$ ones and $v = S - k$ twos, let $C_d(a) := \sum_{j:\, a_j = d} 3^{k-1-j} 2^{S_j}$ ($d \in \{1, 2\}$). Then:* 1. *(Sub-Valuation Telescoping [25, Theorem 4, Parts 1–2]): Specializing the general fixed-point excess decomposition of our companion paper [25, Theorem 4, Parts 1–2] to $\mathcal{A}_{\{1, 2\}}(k, S)$ gives $C_1(a) + C_2(a) = C(a)$ and $C_2(a) - C_1(a) = 2^S - 3^k$ identically, so $C(a) = (2^S - 3^k) + 2 C_1(a) = -(2^S - 3^k) + 2 C_2(a)$ and $x_0(a) = 1 + \frac{2 C_1(a)}{2^S - 3^k} = \frac{2 C_2(a)}{2^S - 3^k} - 1$.* 2. *(Rank-1 Young–Ferrers Lattice Decomposition): Let $\lambda(a) = \{(p, q) \in \{0, \dots, u-1\} \times \{0, \dots, v-1\} : \text{the } q\text{-th } 2 \text{ precedes the } p\text{-th } 1\}$ be the Young diagram of $a$. Define $W_{p, q} := (3^{u - 1 - p} 2^p)(3^{v - 1 - q} 4^q)$ and $W(\lambda(a)) := \sum_{(p, q) \in \lambda(a)} W_{p, q}$. Then:* $$ C_1(a) = 3^v (3^u - 2^u) + W(\lambda(a)), \qquad C_2(a) = 2^u (4^v - 3^v) + W(\lambda(a)), $$ *confining $C(a)$ across $\mathcal{A}_{\{1, 2\}}(k, S)$ to $[2^S + 3^k - 2^{u+1} 3^v,\; 2 \cdot 3^u 4^v - (2^S + 3^k)]$ of width $2(3^u - 2^u)(4^v - 3^v)$.* 3. *(Closed-Form $r$-Term Telescoped Bezout Inversion): For any $\{1, 2\}$ $r$-run word $a = (1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r})$ with cumulative block sums $U_i = \sum_{m=1}^i u_m$ and $V_j = \sum_{m=1}^j v_m$, $\lambda(a)$ decomposes into $\binom{r}{2}$ rectangles $R_{i, j} = [U_{i-1}, U_i - 1] \times [V_{j-1}, V_j - 1]$ ($1 \le j < i \le r$). Telescoping the inner geometric series yields exact $r$-term sums for $A_1 = \frac{C_2(a)}{2^{u_1}(2^S - 3^k)}$ and $B_r = \frac{C_1(a)}{3^{v_r}(2^S - 3^k)}$:* $$ \begin{aligned} (2^S - 3^k) A_1 &= \sum_{j=1}^r 3^{(u - U_j) + (v - V_j)} 2^{(U_j - u_1) + 2 V_{j-1}} \left(4^{v_j} - 3^{v_j}\right), \\ (2^S - 3^k) B_r &= \sum_{i=1}^r 3^{(u - U_i) + (v - v_r - V_{i-1})} 2^{U_{i-1} + 2 V_{i-1}} \left(3^{u_i} - 2^{u_i}\right). \end{aligned} $$ *Consequently, for $r \ge 2$, isolating the $j=1, 2$ and $i=r$ terms gives:* $$ A_i \equiv 3^{-u_i}\left(1 - 3^{-v_i} 4^{v_i} + 2^{u_{i+1} + 2 v_i}\right) \pmod{2^{u_{i+1} + 2 v_i + 1}}, \qquad A_i \equiv 2^{-u_i}\left(1 - 4^{-v_{i-1}} 3^{v_{i-1}}\right) \pmod{3^{u_{i-1} + v_{i-1}}}. $$ 4. *(Exact $r$-Run Trough–Peak Product Identity, Universal $m$-Descent Product Identity, and Sharpened Minimum-Element Bounds):* - *(Binary $r$-Run Trough–Peak Product on $\mathcal{A}_{\{1, 2\}}(k, S)$): Eliminating the $k - 2r$ interior steps reduces the $k$-factor product to an $r$-factor product over the troughs $z_j$ and peaks $y_j$:* $$ \frac{2^S}{3^k} = \prod_{j=1}^r \left( \frac{z_j + 1}{z_j - 1} \cdot \frac{y_j - 1}{y_j + 1} \right) = \prod_{j=1}^r \left( 1 + \frac{2\left(1 - (2/3)^{u_j}\right)}{z_j - 1} \right) = \prod_{j=1}^r \left( 1 + \frac{2\left((4/3)^{v_j} - 1\right)}{y_j + 1} \right). $$ *By Jensen's inequality on the concave map $u_j \mapsto \ln\left(1 + \frac{2(1 - (2/3)^{u_j})}{z_{\min} - 1}\right)$, every $\{1, 2\}$ $r$-run word $a \in \mathcal{A}_{\{1, 2\}}(k, S)$ ($1 \le r \le \min(u, v)$) satisfies the excess bound:* $$ x_{\min}(a) - 1 \le B_{\mathrm{run}}(k, S, r) := \frac{2\left(1 - (2/3)^{u/r}\right)}{(2^S / 3^k)^{1/r} - 1} < \tilde{B}_{\mathrm{run}}(k, S, r) := \frac{2r\left(1 - (2/3)^{u/r}\right)}{\Delta}, \quad \Delta := S \ln 2 - k \ln 3. $$ *Compared to the coarse auxiliary excess bound $B_{\mathrm{aux}}(k, S) := \frac{2}{(2^S / 3^k)^{1/v} - 1}$ (obtained by upper-bounding $1 - (2/3)^{u_j}$ by $1$ and replacing $r$ by the local-minima count $m = v = S - k \ge r$), Bernoulli's inequality $((2^S/3^k)^{1/r} - 1)/((2^S/3^k)^{1/v} - 1) \ge v/r$ and $1 - (2/3)^{u/r} \le u/(3r)$ yield $\frac{B_{\mathrm{aux}}(k, S)}{B_{\mathrm{run}}(k, S, r)} \ge \frac{v}{r(1 - (2/3)^{u/r})} \ge \frac{3v}{u} > 4.22839$. Compared to Simons and de Weger's published bound $x_{\min}(a) < \frac{m}{\Delta} = \frac{v}{\Delta}$ [5, Section 3.2, Lemma 4 & Corollary 5] (keeping the $+1$ shift in $x_{\min}(a) - 1 \le B_{\mathrm{run}}(k, S, r) < \tilde{B}_{\mathrm{run}}(k, S, r)$ explicit), the leading $1/\Delta$ coefficient $\tilde{B}_{\mathrm{run}}(k, S, r)$ improves upon $v/\Delta$ by the uniform factor:* $$ \frac{v / \Delta}{\tilde{B}_{\mathrm{run}}(k, S, r)} = \frac{v}{2r\left(1 - (2/3)^{u/r}\right)} \ge \frac{3v}{2u} = \frac{3(S - k)}{2(2k - S)} > \frac{3(\log_2 3 - 1)}{2(2 - \log_2 3)} \approx 2.114197. $$ - *(Universal $m$-Descent Product on Unrestricted $\mathcal{A}(k, S)$): For any unrestricted composition $a \in \mathcal{A}(k, S)$ ($k < S < 2k$) with $m := \#\{t : a_t \ge 2\} \in [1, S - k]$ descent steps, decompose $a$ up to cyclic shift into $m$ blocks $(1^{u_1}, d_1, \dots, 1^{u_m}, d_m)$ ($u_j \ge 0, d_j \ge 2, \sum_{j=1}^m (u_j + 1) = k$). Across block $j$ starting at $z_j$, the $u_j + 1$ factors of $\frac{2^S}{3^k} = \prod_{t=0}^{k-1} \left(1 + \frac{1}{3 x_t}\right)$ telescope via $x_{t+1} = (3 x_t + 1)/2$ into $1 + \frac{1 - (2/3)^{u_j + 1}}{z_j}$, yielding the exact descent-exponent-independent identity and $m$-dependent bound:* $$ \frac{2^S}{3^k} = \prod_{j=1}^m \left( 1 + \frac{1 - (2/3)^{u_j + 1}}{z_j} \right) \implies x_{\min}(a) \le B_{\mathrm{univ}}(k, S, m) := \frac{1 - (2/3)^{k/m}}{(2^S / 3^k)^{1/m} - 1} < \frac{3\left(1 - (2/3)^{k/m}\right)}{k/m} x_{\min}^{\mathrm{Eliahou}}(k, S) < x_{\min}^{\mathrm{Eliahou}}(k, S) = \frac{1}{2^{S/k} - 3}. $$ *Moreover, since $m \mapsto B_{\mathrm{univ}}(k, S, m)$ is strictly increasing on $[1, k]$ and $m \le S - k < k$, every $a \in \mathcal{A}(k, S)$ satisfies the uniform simplex bound:* $$ x_{\min}(a) \le B_{\mathrm{univ}}(k, S, m) \le B_{\mathrm{univ}}(k, S, S - k) < \frac{3\left(1 - (2/3)^{k/(S - k)}\right)}{k / (S - k)} x_{\min}^{\mathrm{Eliahou}}(k, S) < x_{\min}^{\mathrm{Eliahou}}(k, S). $$ *For fixed $(k, S)$ without restricting $m$ or $r$, the optimal all-word bound on $\mathcal{A}(k, S)$ is the balanced-word bound $x_{\min}(a) \le x_{\min}(a^{\mathrm{Sturm}}(k, S)) \to \frac{1}{2\ln 2} x_{\min}^{\mathrm{Eliahou}}(k, S) \approx 0.721348\,x_{\min}^{\mathrm{Eliahou}}(k, S)$ (as $k_0 = k/\gcd(k, S) \to \infty$ and $\Delta_0 = \Delta/\gcd(k, S) \to 0$) [28, 25, 26]; $B_{\mathrm{univ}}(k, S, m)$ and $B_{\mathrm{run}}(k, S, r)$ provide explicit $O(m/\Delta)$ and $O(r/\Delta)$ bounds in terms of the descent count $m$ and run count $r$.* *Proof.* Parts 1–3 follow by telescoping $C_2(a) - C_1(a) = \sum_{j=0}^{k-1}(2^{a_j} - 3) 3^{k-1-j} 2^{S_j} = 2^S - 3^k$ and summing $W_{p, q}$ over the $\binom{r}{2}$ rectangles $R_{i, j}$. In Part 4, multiplying $\frac{z_j + 1}{y_j + 1} = (2/3)^{u_j}$ and $\frac{y_j - 1}{z_{j+1} - 1} = (4/3)^{v_j}$ around the cycle gives the first product, and Jensen's inequality gives $x_{\min}(a) - 1 \le B_{\mathrm{run}}(k, S, r) < \tilde{B}_{\mathrm{run}}(k, S, r)$; since $v \ge r \ge 1$ and $2^S/3^k > 1$, $(1 + \delta)^{v/r} \ge 1 + (v/r)\delta$ for $\delta = (2^S/3^k)^{1/v} - 1$ gives $((2^S/3^k)^{1/r} - 1)/((2^S/3^k)^{1/v} - 1) \ge v/r$, while $(2/3)^x \ge 1 - x/3$ for $x = u/r \ge 1$ gives $1 - (2/3)^{u/r} \le u/(3r)$, proving both $\frac{B_{\mathrm{aux}}(k, S)}{B_{\mathrm{run}}(k, S, r)} \ge \frac{3v}{u}$ and $\frac{v/\Delta}{\tilde{B}_{\mathrm{run}}(k, S, r)} \ge \frac{3v}{2u}$. For the unrestricted $m$-descent product, across $(1^{u_j}, d_j)$ starting at $z_j = x_{t_j}$ and peaking at $y_j = (3/2)^{u_j}(z_j + 1) - 1$, $\prod_{s=0}^{u_j - 1} \frac{3 x_{t_j + s} + 1}{3 x_{t_j + s}} = \prod_{s=0}^{u_j - 1} \frac{2 x_{t_j + s + 1}}{3 x_{t_j + s}} = (2/3)^{u_j} \frac{y_j}{z_j}$, so multiplying by $\frac{3 y_j + 1}{3 y_j} = \frac{2((3/2)^{u_j+1}(z_j + 1) - 1)}{3 y_j}$ cancels $y_j$ and gives $\frac{z_j + 1 - (2/3)^{u_j + 1}}{z_j} = 1 + \frac{1 - (2/3)^{u_j + 1}}{z_j}$. Applying Jensen's inequality to $u_j + 1$ ($\sum_{j=1}^m (u_j + 1) = k$) gives $x_{\min}(a) \le B_{\mathrm{univ}}(k, S, m)$. For any $m \in [1, k - 1]$, $k/m > 1$, so Bernoulli's inequality $((2^S/3^k)^{1/m} - 1)/((2^S/3^k)^{1/k} - 1) > k/m$ and strict convexity of $(2/3)^t$ ($\frac{1 - (2/3)^{k/m}}{k/m} < 1 - 2/3 = \frac{1}{3}$) prove $B_{\mathrm{univ}}(k, S, m) < \frac{3(1 - (2/3)^{k/m})}{k/m} x_{\min}^{\mathrm{Eliahou}}(k, S) < x_{\min}^{\mathrm{Eliahou}}(k, S)$; evaluating the same inequality at $m = S - k < k$ gives the bound for $B_{\mathrm{univ}}(k, S, S - k)$. $\quad \square$ **Remark 1 (Exact $O(u v p)$ Transfer-Matrix Residue Counting vs. Boundary Non-Equidistribution).** Interpreting $a \in \mathcal{A}_{\{1, 2\}}(k, S)$ as a lattice path on $[0, u] \times [0, v]$ with step contribution $E(p_{\mathrm{idx}}, q) = 3^{k-1-(p_{\mathrm{idx}}+q)} 2^{p_{\mathrm{idx}}+2q} \bmod p$ ($p \mid (2^S - 3^k)$), the exact residue counts $N_p(r; k, S) = \#\{a \in \mathcal{A}_{\{1, 2\}}(k, S) : C(a) \equiv r \pmod p\}$ and character sums $Z_p(c; k, S) = \sum_{a \in \mathcal{A}_{\{1, 2\}}(k, S)} e^{2\pi i c C(a)/p}$ are computable in $O(u v p)$ polynomial time via dynamic programming on $[0, u] \times [0, v]$, with Fourier inversion bound $\left|\frac{N_p(r; k, S)}{\binom{k}{v}} - \frac{1}{p}\right| \le \frac{p - 1}{p} \delta_p(k, S)$ where $\delta_p(k, S) = \max_{c \in \mathbb{F}_p^\times} |Z_p(c; k, S)|/\binom{k}{v}$. In specific interior simplices where both $u$ and $v$ are large, $\delta_p(k, S)$ can be very small—for example, at $(k, S, p) = (39, 62, 11)$ ($u = 16, v = 23, \binom{39}{23} = 37,711,260,990$), exact DP evaluation gives $N_{11}(0) = 3,428,311,887$ and $\delta_{11}(39, 62) = 8.2775 \times 10^{-7}$, and at $(k, S, p) = (40, 64, 13)$ ($u = 16, v = 24, \binom{40}{24} = 62,852,101,650$), $N_{13}(0) = 4,834,872,320$ and $\delta_{13}(40, 64) = 3.7489 \times 10^{-6}$. Crucially, however, $C(a) \bmod p$ does **not** equidistribute across all simplices $\mathcal{A}_{\{1, 2\}}(k, S)$ as $k \to \infty$: as we prove in Theorem 7 below, boundary families ($u = O(1)$ or $v = O(1)$) and primes $p \in \{5, 7\}$ where the position base ratio has multiplicative order $2$ exhibit exact algebraic non-vanishing ($N_p(0; k, S) = 0$) or converge to non-uniform limiting densities such as $1/4 \ne 1/7$. ### 4.3 Reindexed Fixed-Point Homogenization and Mod-$5$/$7$ Index-Parity Collapse **Theorem 7 (Reindexed Fixed-Point Homogenization, Mod-$5$/$7$ Index-Parity Collapse, and Reduced Gap Divisors).** *Let $k \ge 3$, $S > k$, $v = S - k \ge 1$, and $D = 2^S - 3^k$. For any unrestricted composition $a \in \mathcal{A}(k, S)$, let $0 \le \ell_1 < \dots < \ell_r \le k - 1$ be the $r := \#\{t : a_t \ge 2\} \in [1, v]$ indices where $a_{\ell_m} \ge 2$, with cumulative excess $E_m := \sum_{s=1}^m (a_{\ell_s} - 1)$ ($E_0 = 0, E_r = v$); when $a \in \mathcal{A}_{\{1, 2\}}(k, S)$ ($k < S < 2k$, so $r = v$), let $0 \le j_1 < \dots < j_u \le k - 1$ denote the $u = 2k - S$ indices where $a_{j_i} = 1$. Then:* 1. *(Index-Position Reindexing of [25, Theorem 4, Parts 1–2]): Substituting $S_{j_i} = 2 j_i - i + 1$ ($j_i \ge i - 1$) on $\mathcal{A}_{\{1, 2\}}(k, S)$ and $S_{\ell_m} = \ell_m + E_{m-1}$ on $\mathcal{A}(k, S)$ into the fixed-point excess identities of [25, Theorem 4, Parts 1–2] reindexes them by the step positions $j_i$ and $\ell_m$:* $$ C(a) - (2^S - 3^k) = \sum_{i=1}^u 3^{k - 1 - j_i} 2^{2 j_i - i + 2} \;\;\text{on } \mathcal{A}_{\{1, 2\}}(k, S), \qquad \frac{C(a) + (2^S - 3^k)}{2} = \sum_{m=1}^r \left(2^{E_m} - 2^{E_{m-1}}\right) 3^{k - 1 - \ell_m} 2^{\ell_m} \;\;\text{on } \mathcal{A}(k, S). $$ 2. *(Universal Coprimality on the $u = 1$ and $v = 1$ Boundary Families):* - *When $u = 1$ ($S = 2k - 1$ on $\mathcal{A}_{\{1, 2\}}(k, 2k - 1)$), $C(a) - (2^{2k-1} - 3^k) = 2 \cdot 3^{k - 1 - j_1} 4^{j_1}$, so $\gcd(C(a), |2^{2k-1} - 3^k|) = 1$ and $N_p(0; k, 2k - 1) = 0$ for every $k \ge 2$ and every prime $p \mid (2^{2k-1} - 3^k)$ (in particular for $p = 5$ across $k \equiv 3 \pmod 4$, where $C(a) \equiv 3 \cdot 2^{j_1} \not\equiv 0 \pmod 5$ and $\delta_5(k, 2k - 1) \ge 1/4$).* - *When $r = 1$ on $\mathcal{A}(k, S)$ ($k - 1$ ones and one step $a_{\ell_1} = v + 1$), $C(a) + D = 2(2^v - 1) 3^{k - 1 - \ell_1} 2^{\ell_1}$, so $\gcd(C(a), |D|) = \gcd(2^v - 1, |3^k - 2^k|)$ (in particular $\gcd(C(a), |2^{k+1} - 3^k|) = 1$ and $N_p(0; k, k + 1) = 0$ for $v = 1$).* 3. *(Mod-$5$ and Mod-$7$ Index-Parity Collapse and Exact Non-Uniform Limits):* - *Modulo $p = 5$, $5 \mid (2^S - 3^k) \iff v \equiv 2(k \bmod 2) \pmod 4$. Since $2 \equiv -3 \pmod 5$, $(2/3)^{\ell_m} \equiv (-1)^{\ell_m} \pmod 5$, collapsing $C(a) \bmod 5$ across the **entire unrestricted simplex $\mathcal{A}(k, S)$** to:* $$ \frac{C(a) + (2^S - 3^k)}{2} \equiv 3^{k - 1} \sum_{m=1}^r \left(2^{E_m} - 2^{E_{m-1}}\right) (-1)^{\ell_m} \pmod 5. $$ *At $v = 2$ ($S = k + 2$ with $k \ge 3$ odd), $5 \mid (2^{k+2} - 3^k)$ while neither $r = 2$ ($\pm 1 \pm 2 \not\equiv 0 \pmod 5$) nor $r = 1$ ($\pm 3 \not\equiv 0 \pmod 5$) vanishes, so $N_5^{\mathrm{unres}}(0; k, k + 2) = 0$ on $\mathcal{A}(k, k + 2)$ (note that for positive cycles, $2^{k+2} > 3^k$ holds only at $k = 3$, while for $k \ge 5$ this governs negative rational cycles).* - *Modulo $p = 7$, $7 \mid (2^S - 3^k) \iff (2 \mid k \text{ and } 3 \mid u)$. Since $4 \equiv -3 \pmod 7$, $(4/3)^{j_i} \equiv (-1)^{j_i} \pmod 7$ on $\mathcal{A}_{\{1, 2\}}(k, S)$, collapsing $C(a) \bmod 7$ to $C(a) - (2^S - 3^k) \equiv 3^{k - 1} \sum_{i=1}^u 2^{2 - i} (-1)^{j_i} \pmod 7$. Across the infinite family $u = 3$ ($S = 2k - 3$ with $k = 2m \ge 4$ even, where $2^{4m-3} > 3^{2m}$ holds for all $m \ge 4$, while $m = 2, 3$ have $2^{4m-3} < 3^{2m}$), $7 \mid (2^{4m-3} - 3^{2m})$ and $2(-1)^{j_1} + (-1)^{j_2} + 4(-1)^{j_3} \equiv 0 \pmod 7 \iff j_1 \equiv j_2 \equiv j_3 \pmod 2$, yielding the exact closed form and non-equidistributed asymptotic density:* $$ N_7(0; 2m, 4m - 3) = 2 \binom{m}{3}, \qquad \lim_{m \to \infty} \frac{N_7(0; 2m, 4m - 3)}{\binom{2m}{3}} = \lim_{m \to \infty} \frac{m - 2}{2(2m - 1)} = \frac{1}{4} \ne \frac{1}{7}. $$ 4. *(Dual Pigeonhole-Gap Bounds on $\mathcal{A}_{\{1, 2\}}(k, S)$ and $\mathcal{A}(k, S)$):* *On $\mathcal{A}_{\{1, 2\}}(k, S)$, rotating $a$ so the $1$ following the longest circular arc between ones ($\ge \lceil k/u \rceil$) is at $j_1 = 0$ ($j_u \le k - \lceil k/u \rceil$) defines the positive integer $H_u^+(a) := \sum_{i=1}^u 2^{u - i} 3^{j_u - j_i} 4^{j_i}$; dually, on the unrestricted simplex $\mathcal{A}(k, S)$, rotating $a$ so the $\ge 2$-step following the longest circular arc between $\ge 2$-steps ($\ge \lceil k/r \rceil$) is at $\ell_1 = 0$ ($\ell_r \le k - \lceil k/r \rceil$) defines $H_{r, v}^{\mathrm{unres}}(a) := \sum_{m=1}^r (2^{E_m} - 2^{E_{m-1}}) 3^{\ell_r - \ell_m} 2^{\ell_m}$ (which specializes to $H_v^-(a)$ on $\mathcal{A}_{\{1, 2\}}(k, S)$). Here $H_u^+(a)$ and $H_v^-(a)$ are coprime to $3$ (with $H_v^-(a)$ odd), $H_{r, v}^{\mathrm{unres}}(a)$ is odd (and coprime to $3$ when $a_{\ell_r}$ is even), and they satisfy $\gcd(C(a), D) = \gcd(H_u^+(a), D) = \gcd(H_{r, v}^{\mathrm{unres}}(a), D)$ and:* $$ 1 \le H_u^+(a) \le \left(3^u - 2^u\right) 2^{S - 2\lceil k/u \rceil + 1}, \qquad 1 \le H_{r, v}^{\mathrm{unres}}(a) \le \left(2^{v - r}\left(4^r - 2 \cdot 3^{r - 1}\right) - 3^{r - 1}\right) 3^{k - \lceil k/r \rceil - r + 1}. $$ *While $H_u^+(a) < 2^S - 3^k$ gives a purely algebraic non-divisibility proof $D \nmid C(a)$ for $S = 2k - u, k \ge \max(3, u^2 + 2)$, we note that for positive integer cycles this $S/k \ge 5/3$ regime is already excluded by the classical bound $x_{\min} \le (2^{S/k} - 3)^{-1} < 6$ [1–3].* *Proof.* 1. On $\mathcal{A}_{\{1, 2\}}(k, S)$, prior to $j_i$ there are $i - 1$ ones and $j_i - (i - 1)$ twos, so $S_{j_i} = 2 j_i - i + 1$ (with $2 j_i - i + 2 \ge i \ge 1$, so every summand $3^{k - 1 - j_i} 2^{2 j_i - i + 2} \in 2\mathbb{Z}_{\ge 1}$ is an even positive integer), and $C(a) - D = 2 C_1(a)$ [25, Theorem 4] gives the first identity. On $\mathcal{A}(k, S)$, [25, Theorem 4(1)] gives $(C(a) + D)/2 = \sum_{t: a_t \ge 2} 3^{k-1-t} 2^{S_t} (2^{a_t - 1} - 1)$; indexing the $r$ steps $a_{\ell_m} \ge 2$ by $m = 1, \dots, r$ gives $S_{\ell_m} = \ell_m + E_{m-1}$ and $a_{\ell_m} - 1 = E_m - E_{m-1}$, so $2^{S_{\ell_m}}(2^{a_{\ell_m}-1} - 1) = (2^{E_m} - 2^{E_{m-1}}) 2^{\ell_m}$. 2. Immediate from Part 1 since $\gcd(6, |D|) = 1$ and $\gcd(2^v - 1, |2^{k+v} - 3^k|) = \gcd(2^v - 1, |3^k - 2^k|)$. 3. Follows from $3 \equiv -2 \pmod 5 \implies 3^{k-1-\ell_m} 2^{\ell_m} \equiv 3^{k-1}(-1)^{\ell_m} \pmod 5$ and $4 \equiv -3 \pmod 7 \implies 3^{k-1-j_i} 4^{j_i} \equiv 3^{k-1}(-1)^{j_i} \pmod 7$; for $u = 3, k = 2m$, $2\epsilon_1 + \epsilon_2 + 4\epsilon_3 \equiv 0 \pmod 7$ on $\epsilon_i \in \{\pm 1\} \iff \epsilon_1 = \epsilon_2 = \epsilon_3$, giving $2\binom{m}{3}$ words. 4. At the rotation placing $j_1 = 0$ after the longest circular gap ($\ge \lceil k/u \rceil$), $j_u \le k - \lceil k/u \rceil$ and $2^{u - 2}(C(a) - D) = 3^{k - 1 - j_u} H_u^+(a)$; since $\gcd(6, |D|) = 1$, $\gcd(C(a), D) = \gcd(H_u^+(a), D)$ (note that $H_u^+(a)$ is coprime to $3$, though it can be even for $u \ge 2$, e.g., $H_2^+(1, 1, 2, 2, 2) = 10$). Using $j_i \le j_u - (u - i)$ gives $H_u^+(a) \le (3^u - 2^u) 2^{S - 2\lceil k/u \rceil + 1}$. Dually, on $\mathcal{A}(k, S)$ with $\ell_1 = 0$ and $\ell_r \le k - \lceil k/r \rceil$, $(C(a) + D)/2 = 3^{k - 1 - \ell_r} H_{r, v}^{\mathrm{unres}}(a)$ (whose $m = 1$ term $(2^{e_1} - 1) 3^{\ell_r}$ is odd and all $m \ge 2$ terms are even), and maximizing over $\ell_m \ge m - 1$ and $E_m \le v - (r - m)$ ($e_1 = v - r + 1, e_2 = \dots = e_r = 1$) gives $H_{r, v}^{\mathrm{unres}}(a) \le (2^{v - r}(4^r - 2 \cdot 3^{r-1}) - 3^{r-1}) 3^{k - \lceil k/r \rceil - r + 1}$. $\quad \square$ --- ## 5. Block-Compressed Multiscale CRT Renormalization ### 5.1 Proof of Theorem 4 (Block-Compressed CRT Linear System) **Theorem 4.** *For any cycle $(1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r}) \in \mathcal{A}_{\{1, 2\}}(k, S)$, define the local block CRT lifts $A_i^{\mathrm{CRT}} \in [1, 2^{2 v_i + 1} 3^{v_{i-1}} - 1]$ and $B_i^{\mathrm{CRT}} \in [1, 2^{u_{i+1} + 1} 3^{u_i} - 1]$ by:* $$ \begin{aligned} A_i^{\mathrm{CRT}} &\equiv 3^{-u_i}(1 + 4^{v_i}) \pmod{2^{2 v_i + 1}}, & A_i^{\mathrm{CRT}} &\equiv 2^{-u_i}(1 + 3^{v_{i-1}}) \pmod{3^{v_{i-1}}}, \\ B_i^{\mathrm{CRT}} &\equiv 3^{-v_i}(2^{u_{i+1}} - 1) \pmod{2^{u_{i+1} + 1}}, & B_i^{\mathrm{CRT}} &\equiv -4^{-v_i} \pmod{3^{u_i}}. \end{aligned} $$ *Decomposing $A_i = A_i^{\mathrm{CRT}} + 2^{2 v_i + 1} 3^{v_{i-1}} P_i$ and $B_i = B_i^{\mathrm{CRT}} + 2^{u_{i+1} + 1} 3^{u_i} R_i$, the block quotients $(P_i, R_i) \in \mathbb{Z}_{\ge 0}^2$ satisfy:* $$ P_i = \left\lfloor \frac{z_i}{2^{S_i + 2} 3^{v_{i-1}}} \right\rfloor, \qquad R_i = \left\lfloor \frac{y_i}{2^{u_{i+1} + 2 v_i + 2} 3^{u_i}} \right\rfloor, $$ *and obey the exact $2r$-step integer linear system:* $$ \begin{aligned} 2^{u_{i+1}} R_i - 3^{v_{i-1}} P_i &= \delta_i^{(1)} := \frac{3^{u_i} A_i^{\mathrm{CRT}} - 4^{v_i} B_i^{\mathrm{CRT}} - 1}{2 \cdot 3^{u_i} 4^{v_i}} \in \left(-2^{u_{i+1}}, 3^{v_{i-1}}\right) \cap \mathbb{Z}, \\ 4^{v_{i+1}} P_{i+1} - 3^{u_i} R_i &= \delta_i^{(2)} := \frac{3^{v_i} B_i^{\mathrm{CRT}} - 2^{u_{i+1}} A_{i+1}^{\mathrm{CRT}} + 1}{2^{u_{i+1} + 1} 3^{v_i}} \in \left(-4^{v_{i+1}}, 3^{u_i}\right) \cap \mathbb{Z}. \end{aligned} $$ *Eliminating $R_i$ yields the $r$-step block recurrence:* $$ 2^{u_{i+1} + 2 v_{i+1}} P_{i+1} - 3^{u_i + v_{i-1}} P_i = \Omega_i := 3^{u_i} \delta_i^{(1)} + 2^{u_{i+1}} \delta_i^{(2)} \in \left(-2^{u_{i+1} + 2 v_{i+1}}, 3^{u_i + v_{i-1}}\right) \cap \mathbb{Z}. $$ *Proof.* Substitute $A_i = A_i^{\mathrm{CRT}} + 2^{2v_i + 1} 3^{v_{i-1}} P_i$ and $B_i = B_i^{\mathrm{CRT}} + 2^{u_{i+1} + 1} 3^{u_i} R_i$ into $3^{u_i} A_i - 4^{v_i} B_i = 1$: $$ 3^{u_i} A_i^{\mathrm{CRT}} - 4^{v_i} B_i^{\mathrm{CRT}} - 1 + 2 \cdot 3^{u_i} 4^{v_i} \left( 3^{v_{i-1}} P_i - 2^{u_{i+1}} R_i \right) = 0. $$ Dividing by $2 \cdot 3^{u_i} 4^{v_i}$ proves $2^{u_{i+1}} R_i - 3^{v_{i-1}} P_i = \delta_i^{(1)}$. Similarly, substituting into $2^{u_{i+1}} A_{i+1} - 3^{v_i} B_i = 1$ yields $4^{v_{i+1}} P_{i+1} - 3^{u_i} R_i = \delta_i^{(2)}$. Multiplying the first relation by $3^{u_i}$ and the second by $2^{u_{i+1}}$ and adding them eliminates $R_i$: $$ 3^{u_i} \left( 2^{u_{i+1}} R_i - 3^{v_{i-1}} P_i \right) + 2^{u_{i+1}} \left( 4^{v_{i+1}} P_{i+1} - 3^{u_i} R_i \right) = 2^{u_{i+1} + 2v_{i+1}} P_{i+1} - 3^{u_i + v_{i-1}} P_i = 3^{u_i} \delta_i^{(1)} + 2^{u_{i+1}} \delta_i^{(2)} =: \Omega_i. $$ Substituting the expressions for $\delta_i^{(1)}$ and $\delta_i^{(2)}$ into $\Omega_i$ shows that the $B_i^{\mathrm{CRT}}$ terms cancel identically: $$ \Omega_i = \frac{3^{u_i + v_i} A_i^{\mathrm{CRT}} + (4^{v_i} - 3^{v_i}) - 2^{u_{i+1} + 2v_i} A_{i+1}^{\mathrm{CRT}}}{2 \cdot 4^{v_i} 3^{v_i}}. $$ The bounds on $\delta_i^{(1)}, \delta_i^{(2)}$, and $\Omega_i$ follow directly from the fundamental domain of the CRT lifts. $\quad \square$ **Corollary 2 (Unrolled Block-CRT Identity, Minimum-Trough Defect Obstruction, and Half-Step Defect Cancellation).** *In any cycle in $\mathcal{A}_{\{1, 2\}}(k, S)$ with $r$ runs, unrolling the block recurrence $2^{S_{i+1}} P_{i+1} - 3^{u_i + v_{i-1}} P_i = \Omega_i$ over all $r$ blocks yields the exact Diophantine identity:* $$ (2^S - 3^k) P_1 = \sum_{j=1}^r \left( \prod_{\ell=j+1}^r 3^{u_\ell + v_{\ell-1}} \right) \left( \prod_{\ell=1}^{j-1} 2^{S_{\ell+1}} \right) \Omega_j. $$ *Furthermore, at any block $i$ with $P_i = 0$ (which holds at $i = i_{\min}$ whenever $z_{\min} < 2^{S_i + 2} 3^{v_{i-1}}$):* 1. *The single-step forward block relation simplifies to $2^{S_{i+1}} P_{i+1} = \Omega_i$, which since $P_{i+1} \ge 0$ forces the sign obstruction $\Omega_i \ge 0$.* 2. *If both adjacent trough quotients vanish ($P_i = P_{i+1} = 0$), then $\Omega_i = 0$, and Theorem 4's half-step equations reduce to:* $$ \delta_i^{(1)} = 2^{u_{i+1}} R_i, \qquad \delta_i^{(2)} = -3^{u_i} R_i, \qquad 0 \le R_i \le \left\lfloor \frac{3^{v_{i-1}} - 1}{2^{u_{i+1}}} \right\rfloor. $$ *In particular, $\Omega_i = 0$ forces $\delta_i^{(1)} = \delta_i^{(2)} = 0$ ($R_i = 0$) unconditionally whenever $3^{v_{i-1}} \le 2^{u_{i+1}}$, whereas when $3^{v_{i-1}} > 2^{u_{i+1}}$, non-zero half-step defects with $R_i \ge 1$ can cancel inside $\Omega_i = 3^{u_i} \delta_i^{(1)} + 2^{u_{i+1}} \delta_i^{(2)} = 0$. For example, at $(v_{i-1}, u_i, v_i, u_{i+1}, v_{i+1}) = (3, 1, 1, 1, 3)$, $A_i^{\mathrm{CRT}} = 95, B_i^{\mathrm{CRT}} = 11, A_{i+1}^{\mathrm{CRT}} = 107$ (realized by the local trajectory $379 \to 569 \to 427$ from $897$ to $271$) has $P_i = P_{i+1} = 0$, actual peak $B_i = 71$ ($R_i = 5$), $\delta_i^{(1)} = 10$, $\delta_i^{(2)} = -15$, and $\Omega_i = 3(10) + 2(-15) = 0$. Consequently, $P_i = P_{i+1} = 0$ enters the full **Total-Defect Zero-Quotient Graph** $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ (Theorem 5 below), of which the half-step zero-wrap graph $\mathcal{G}_{\delta=0}^{\mathrm{block}}$ ($R_i = 0$) is a subgraph.* --- ### 5.2 Proof of Theorem 5 (Deterministic Out-Degree $\le 1$ of Both $\mathcal{G}_{\delta=0}^{\mathrm{block}}$ and $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ Over $\mathbb{Z}_{\ge 1}$) **Theorem 5.** *Over the countably infinite alphabet of all positive block lengths $(u_i, v_i) \in \mathbb{Z}_{\ge 1}^2$, define the half-step zero-wrap graph $\mathcal{G}_{\delta=0}^{\mathrm{block}}$ by $P_i = R_i = P_{i+1} = 0$ ($\delta_i^{(1)} = \delta_i^{(2)} = 0$) and the full total-defect zero-quotient graph $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ by $P_i = P_{i+1} = 0$ ($\Omega_i = 0$, with $R_i \in [0, \lfloor (3^{v_{i-1}} - 1)/2^{u_{i+1}} \rfloor]$ unrestricted). Both graphs have **out-degree at most $1$ at every state $(v_{i-1}, u_i, v_i) \in \mathbb{Z}_{\ge 1}^3$**:* 1. *(Deterministic Half-Step Transitions in $\mathcal{G}_{\delta=0}^{\mathrm{block}}$): Given $(v_{i-1}, u_i, v_i) \in \mathbb{Z}_{\ge 1}^3$, let $B_i^* := (3^{u_i} A_i^{\mathrm{CRT}}(v_{i-1}, u_i, v_i) - 1)/4^{v_i}$. Then $\delta_i^{(1)} = 0$ has at most one solution $u_{i+1} \in \mathbb{Z}_{\ge 1}$, namely $u_{i+1} = v_2(3^{v_i} B_i^* + 1)$ (valid iff $B_i^* < 2^{u_{i+1}+1} 3^{u_i}$, i.e., $R_i = 0$). Given $(u_i, v_i, u_{i+1}) \in \mathbb{Z}_{\ge 1}^3$, let $A_{i+1}^* := (3^{v_i} B_i^{\mathrm{CRT}}(u_i, v_i, u_{i+1}) + 1)/2^{u_{i+1}}$; then $\delta_i^{(2)} = 0$ has at most one solution $v_{i+1} = \frac{1}{2} v_2(3^{u_{i+1}} A_{i+1}^* - 1)$ (valid iff $v_2(3^{u_{i+1}} A_{i+1}^* - 1)$ is even and $A_{i+1}^* < 2 \cdot 4^{v_{i+1}} 3^{v_i}$).* 2. *(Deterministic Full-Block Transitions in $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ Including $R_i > 0$ Cancellations): Given $(v_{i-1}, u_i, v_i) \in \mathbb{Z}_{\ge 1}^3$, the condition $P_i = 0$ alone fixes $A_i = A_i^{\mathrm{CRT}}(v_{i-1}, u_i, v_i)$ and the exact peak $B_i = B_i^* = (3^{u_i} A_i^{\mathrm{CRT}} - 1)/4^{v_i} = B_i^{\mathrm{CRT}} + 2^{u_{i+1}+1} 3^{u_i} R_i$. Regardless of whether $R_i = 0$ or $R_i > 0$, the equation $2^{u_{i+1}} A_{i+1} = 3^{v_i} B_i^* + 1$ with $A_{i+1}$ odd uniquely forces $u_{i+1} = v_2(3^{v_i} B_i^* + 1)$ and $A_{i+1}^{**} := (3^{v_i} B_i^* + 1)/2^{u_{i+1}}$, and the condition $P_{i+1} = 0$ ($A_{i+1}^{\mathrm{CRT}}(v_i, u_{i+1}, v_{i+1}) = A_{i+1}^{**}$) has at most one solution $v_{i+1} \in \mathbb{Z}_{\ge 1}$, namely $v_{i+1} = \frac{1}{2} v_2(3^{u_{i+1}} A_{i+1}^{**} - 1)$ (valid iff $v_2(3^{u_{i+1}} A_{i+1}^{**} - 1)$ is even and $A_{i+1}^{**} < 2 \cdot 4^{v_{i+1}} 3^{v_i}$).* 3. *(Finite-Box Annihilation of Both $\mathcal{G}_{\delta=0}^{\mathrm{block}}$ and $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$): Across all $1,728$ initial triples in $\{1, \dots, 12\}^3$ (and all $15,625$ initial triples in $\{1, \dots, 25\}^3$), $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ has $37$ valid $1$-block transitions on $\{1, \dots, 25\}^3$—comprising $34$ transitions with $R_i = 0$ ($\delta_i^{(1)} = \delta_i^{(2)} = 0$) and $3$ cancellation transitions with $R_i > 0$: $(3, 1, 1) \xrightarrow{R_i=5} (1, 1, 3)$, $(2, 1, 16) \xrightarrow{R_i=1} (16, 2, 3)$, and $(3, 1, 18) \xrightarrow{R_i=1} (18, 3, 2)$—and every path in both $\mathcal{G}_{\delta=0}^{\mathrm{block}}$ and $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ terminates within at most $2$ full blocks ($5$ half-steps), with zero cycles.* *Proof.* 1. Proved in Section 5.1: $\delta_i^{(1)} = 0 \iff B_i^{\mathrm{CRT}} = B_i^* \implies 3^{v_i} B_i^* + 1 \equiv 2^{u_{i+1}} \pmod{2^{u_{i+1}+1}}$, uniquely fixing $u_{i+1} = v_2(3^{v_i} B_i^* + 1)$; similarly $\delta_i^{(2)} = 0 \iff A_{i+1}^{\mathrm{CRT}} = A_{i+1}^* \implies 3^{u_{i+1}} A_{i+1}^* - 1 \equiv 4^{v_{i+1}} \pmod{2^{2v_{i+1}+1}}$, uniquely fixing $2 v_{i+1} = v_2(3^{u_{i+1}} A_{i+1}^* - 1)$. 2. Suppose $P_i = P_{i+1} = 0$ ($\Omega_i = 0$). Even when $R_i > 0$, $P_i = 0$ fixes $A_i = A_i^{\mathrm{CRT}}(v_{i-1}, u_i, v_i)$, which via $3^{u_i} A_i - 4^{v_i} B_i = 1$ uniquely fixes the actual integer peak $B_i = B_i^* = (3^{u_i} A_i^{\mathrm{CRT}} - 1)/4^{v_i}$. Then $2^{u_{i+1}} A_{i+1} = 3^{v_i} B_i^* + 1$ with $A_{i+1}$ odd uniquely fixes both $u_{i+1} = v_2(3^{v_i} B_i^* + 1)$ and the actual next trough amplitude $A_{i+1} = A_{i+1}^{**} = (3^{v_i} B_i^* + 1)/2^{u_{i+1}}$. For $P_{i+1} = 0$ to hold, $A_{i+1}^{\mathrm{CRT}}(v_i, u_{i+1}, v_{i+1})$ must equal $A_{i+1}^{**}$. Since $A_{i+1}^{**} = (3^{v_i} B_i^* + 1)/2^{u_{i+1}} \equiv 2^{-u_{i+1}}(1 + 3^{v_i}) \pmod{3^{v_i}}$ is satisfied automatically, $A_{i+1}^{\mathrm{CRT}} = A_{i+1}^{**}$ holds if and only if $3^{u_{i+1}} A_{i+1}^{**} - 1 \equiv 4^{v_{i+1}} \pmod{2^{2v_{i+1}+1}}$ and $A_{i+1}^{**} < 2 \cdot 4^{v_{i+1}} 3^{v_i}$, uniquely fixing $v_{i+1} = \frac{1}{2} v_2(3^{u_{i+1}} A_{i+1}^{**} - 1)$ and proving $\deg^+_{\mathcal{G}_{\Omega=0}} \le 1$. 3. Exhaustive evaluation in `verify_paper3.py` across $\{1, \dots, 25\}^3$ confirms the $34$ transitions with $R_i = 0$ and $3$ transitions with $R_i > 0$, with maximum survival chain $(1, 4, 2) \xrightarrow{R_1=0} (2, 1, 1) \xrightarrow{R_2=0} (1, 2, 1) \to \text{terminates}$ ($2$ full blocks, $5$ half-steps), and $(3, 1, 1) \xrightarrow{R_1=5} (1, 1, 3) \to \text{terminates}$ after $1$ block ($w = v_2(3^1 \cdot 107 - 1) = v_2(320) = 6 \implies v_2 = 3$, where $A_2^{**} = 107 < 2 \cdot 4^3 \cdot 3^1 = 384$, followed at $(1, 1, 3)$ by $B_2^* = (3 \cdot 107 - 1)/64 = 5$, $u_3 = v_2(27 \cdot 5 + 1) = v_2(136) = 3$, $A_3^{**} = 17$, and $v_2(27 \cdot 17 - 1) = v_2(458) = 1$, which is odd and exits $\{1, 2\}^{\mathbb{Z}}$). $\quad \square$ ![Block Zero-Wrap Graph Annihilation and Defect Confinement](supplement/figure-2.png) Figure 2 illustrates Theorem 4 and Theorem 5 on the tested initial boxes. Panel (a) plots the surviving fraction of block triples on the tested box $\{1, \dots, 12\}^3$ (and $\{1, \dots, 25\}^3$) as a function of half-steps $H$, showing rapid extinction and the cutoff $H_{\max} = 5$. Panel (b) shows the empirical distribution of asymmetrically normalized block defects $\tilde{\delta}_i^{(1)} \in (-1, 1)$ across $1,296$ block $4$-tuples in $\{1, \dots, 6\}^4$, confirming strict confinement within $(-2^{u_{i+1}}, 3^{v_{i-1}})$. --- ### 5.3 Proof of Theorem 6 (Ultrametric Non-Vanishing, Weakly Contracting Collapse, Local Block-CRT Certificate, and $2$-Adic Fibonacci $\{1, 2\}$-Tree Candidate Sieve) **Theorem 6.** *Let $a = (1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r}) \in \mathcal{A}_{\{1, 2\}}(k, S)$ be a non-trivial cycle with $r$ runs ($2^S > 3^k$) and block recurrence $2^{S_{i+1}} P_{i+1} - 3^{k_i} P_i = \Omega_i$, where $S_{i+1} = u_{i+1} + 2 v_{i+1}$ and $k_i = u_i + v_{i-1}$. Then:* 1. *(Ultrametric Non-Vanishing): Whenever $v_2(\Omega_i) < S_{i+1}$ ($98.24\%$ of transitions in $\{1, \dots, 4\}^5$), the ultrametric triangle inequality on $\mathbb{Q}_2$ forces $v_2(P_i) = v_2(\Omega_i) < \infty \implies P_i \ge 2^{v_2(\Omega_i)} \ge 1$; dually, whenever $v_3(\Omega_i) < k_i$, $v_3(P_{i+1}) = v_3(\Omega_i) < \infty \implies P_{i+1} \ge 3^{v_3(\Omega_i)} \ge 1$.* 2. *(Weakly Contracting Rigid Collapse Into $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$): If a cycle satisfies the step-wise contraction condition $3^{k_i} \le 2^{S_{i+1}}$ for all $1 \le i \le r$ and has even one block $i^*$ with $P_{i^*} = 0$, then $P_i = 0$ and $\Omega_i = 0$ for all $1 \le i \le r$, so the entire cycle must form a closed directed cycle in $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ (which by Theorem 5 Part 3 is impossible if any block triple $(v_{i-1}, u_i, v_i)$ lies in $\{1, \dots, 25\}^3$).* 3. *(Unconditional Global Drift Bound and Self-Contained Block-CRT Certificate for $k \ge 2,966$ Over the Infinite Alphabet $\mathbb{Z}_{\ge 1}^2$): Write the $i$-th trough factor of Corollary 1 Part 4 as $1 + \gamma_i := \frac{z_i + 1}{z_i - 1}\frac{y_i - 1}{y_i + 1} = 1 + \frac{1 - (2/3)^{u_i}}{2^{u_i} A_i - 1}$ ($A_i = A_i^{\mathrm{CRT}}(v_{i-1}, u_i, v_i) + 2^{2v_i + 1} 3^{v_{i-1}} P_i$). Partitioning all block triples $(v_{i-1}, u_i, v_i) \in \mathbb{Z}_{\ge 1}^3$ into the outside-box tail $\mathbb{Z}_{\ge 1}^3 \setminus \{1, \dots, 25\}^3$ and the inside-box core $\{1, \dots, 25\}^3$ yields:* - *(Analytical Outside-Box Bound): Whenever $\max(v_{i-1}, u_i, v_i) \ge 26$, the Coupled Bezout Ladder equations $z_i = 2^{u_i + 1} A_i - 1 = 2 \cdot 3^{v_{i-1}} B_{i-1} + 1$ and $3^{u_i} A_i = 4^{v_i} B_i + 1$ with $A_i, B_{i-1}, B_i \ge 1$ unconditionally force (even if $P_i = 0$):* $$ \frac{z_i - 1}{2} = 2^{u_i} A_i - 1 \ge \min\left(2^{26} - 1,\; 3^{26},\; 4 \left(\frac{8}{3}\right)^{25} - 1\right) = 2^{26} - 1 = 67,108,863 \quad \implies \quad \gamma_i < \frac{1}{2^{26} - 1} < 1.50 \times 10^{-8}. $$ - *(Inside-Box Block-CRT Exit Certificate): Across $\{1, \dots, 25\}^3$, the union of all $15,625$ zero-quotient states ($P_i = 0$) and all $45,831$ states $((v_{i-1}, u_i, v_i), P_i)$ satisfying $\gamma_i \ge (1.1 \times 10^6)^{-1}$ ($305$ with $P_i = 0$ and $45,526$ with $P_i \ge 1$, giving $15,625 + 45,526 = 61,151$ distinct states in total) exits $\{1, 2\}^{\mathbb{Z}}$ via $a_t \ge 3$ within at most $72$ Syracuse steps.* *Consequently, **unconditionally over the entire infinite alphabet $(u_i, v_i) \in \mathbb{Z}_{\ge 1}^2$**, every block $i \in \{1, \dots, r\}$ of any non-trivial cycle in $\mathcal{A}_{\{1, 2\}}(k, S)$ satisfies $z_i \ge 733,339$ and $\gamma_i < (1.1 \times 10^6)^{-1}$, which via $S \ln 2 - k \ln 3 = \sum_{i=1}^r \ln(1 + \gamma_i)$ forces $r > (1.1 \times 10^6)(S \ln 2 - k \ln 3)$, excludes all cycles in $\mathcal{A}_{\{1, 2\}}(k, S)$ with $S/k \ge 1.58496305$, and forces $k \ge 2,966$ ($S \ge 4,701$, $r \ge 933$).* 4. *(Exact Parity-Exit Equivalence, $(1/3)^{B-1}$ Haar Block-Survival Law, and Golden-Ratio Hausdorff Dimension): At any peak $y = 2 \cdot 3^u A - 1$ ($A \in \mathbb{Z}_{\ge 1}^{\mathrm{odd}}$), let $w = v_2(3^u A - 1) \ge 1$ and $O = (3^u A - 1)/2^w$. The forward trajectory completes a valid $\{1, 2\}$-block if and only if $w = 2v$ is even; if $w = 2v - 1$ is odd, it executes $v - 1$ steps of valuation $2$ and exits $\{1, 2\}^{\mathbb{Z}}$ with valuation $2 + v_2(3^v O + 1) \ge 3$. Under the $2$-adic Haar measure $\mu_{\mathrm{Haar}}$ on $\mathbb{Z}_2^\times$, $\mu_{\mathrm{Haar}}(\text{survive } B \text{ blocks} \mid \text{block } 1) = (1/3)^{B-1}$, and the survivor set $\Sigma_{\{1, 2\}} \subset \mathbb{Z}_2^\times$ has measure zero and Hausdorff dimension $\dim_H(\Sigma_{\{1, 2\}}) = \log_2\varphi \approx 0.6942419$, where $\varphi = (1+\sqrt{5})/2$.* 5. *($2$-Adic Fibonacci $\{1, 2\}$-Tree Candidate Sieve and Self-Contained Certificate for $k \ge 10,781,274$): For any integer horizon $M \ge 1$ ($X = 2^{M+1}$), every odd residue $r_0 \in \{1, 3, \dots, 2^{S+1}-1\}$ whose first $j$ Syracuse steps have valuations $a_1, \dots, a_j \in \{1, 2\}$ with cumulative sum $S = \sum_{t=1}^j a_t < M$ and iterate $y_j = U^{(j)}(r_0)$ splits without modular inversions into exactly two $\{1, 2\}$-children at step $j + 1$: setting $P = 3^{j+1}$ and $w_0 = (3y_j + 1)/2$, the unique lift in $\{r_0, r_0 + 2^{S+1}\}$ making $w_0 + b_1 P$ odd has $a_{j+1} = 1, S' = S + 1$, while the complementary lift has $w_1 = (w_0 + (1-b_1)P)/2$, and its unique further lift in $\{r_{\mathrm{even}}, r_{\mathrm{even}} + 2^{S+2}\}$ making $w_1 + b_2 P$ odd has $a_{j+1} = 2, S' = S + 2$. Stopping each branch at the first step where $S \ge M$ produces a finite binary tree with **exactly $F_{M+2}$ total leaves** ($F_1 = F_2 = 1$), consisting of $F_{M+1}$ leaves at $S = M$ (all satisfying $r_0 < 2^{M+1}$) and $F_M$ leaves at $S = M + 1$ (satisfying $r_0 < 2^{M+2}$, so the computed count of leaves with $1 < r_0 < 2^{M+1} = X$ lies in $[F_{M+1}-1, F_{M+2}-1]$), and constructing the candidate tree requires $O(F_{M+2}) = O(X^{\log_2\varphi}) \approx O(X^{0.6942419})$ bitwise operations. Following each retained candidate seed $1 < r_0 < 2^{M+1}$ forward until $v_2(3x_t + 1) \ge 3$ at the verified cutoffs $M \in \{18, 33, 36, 39, 43\}$ certifies that every odd integer $1 < x_0 < 2^{M+1}$ exits $\{1, 2\}^{\mathbb{Z}}$ via $v_2(3x_t + 1) \ge 3$ as recorded in Table 1 (`tree_sieve.c` and `verify_paper3.py`, independently verified up to $2^{40}$ by `u/fourier_wren`):* | Cutoff $M$ ($X = 2^{M+1}$) | Exact Total Tree Leaves $F_{M+2}$ | Computed Seeds $1 < r_0 < 2^{M+1}$ | Max $\{1, 2\}$-Steps Before Exit | Argmax Seed $x_0 < 2^{M+1}$ | | :---: | :---: | :---: | :---: | :---: | | $M = 18$ ($X = 2^{19}$) | $6{,}765$ | $5{,}475$ | $38$ | $517{,}191$ | | $M = 33$ ($X = 2^{34}$) | $9{,}227{,}465$ | $7{,}465{,}019$ | $81$ | $16{,}339{,}660{,}827$ | | $M = 36$ ($X = 2^{37}$) | $39{,}088{,}169$ | $31{,}623{,}304$ | $82$ | $82{,}928{,}941{,}737$ | | $M = 39$ ($X = 2^{40}$) | $165{,}580{,}141$ | $133{,}956{,}220$ | $92$ | $971{,}145{,}319{,}023$ | | $M = 43$ ($X = 2^{44}$) | $1{,}134{,}903{,}170$ | $918{,}146{,}479$ | $107$ | $14{,}935{,}579{,}363{,}195$ | *Consequently, every non-trivial cycle in $\mathcal{A}_{\{1, 2\}}(k, S)$ satisfies $x_{\min}(a) > 2^{44} \approx 1.7592 \times 10^{13}$, the run-count bound $r > \frac{1}{2}(x_{\min}(a) - 1)\Delta \ge 2^{43} \Delta > 8.796 \times 10^{12} \Delta$, the ascent-count bound $u > \frac{3}{2}(x_{\min}(a) - 1)\Delta \ge 3 \cdot 2^{43} \Delta > 2.6388 \times 10^{13} \Delta$, and—because crossing $\frac{190537}{3\Delta_{190537}} \approx 9.846 \times 10^{11} < 2^{40}$ eliminates the convergent $(190{,}537, 301{,}994)$ before the partial quotient $a_{14} = 55$ of $\log_2 3$—forces **$k \ge 10{,}781{,}274$ and $S \ge 17{,}087{,}915$** without relying on external $2^{68}$ trajectory searches [7, 8].* *Proof.* 1. Write $3^{k_i} P_i = 2^{S_{i+1}} P_{i+1} - \Omega_i$. Since $P_{i+1} \in \mathbb{Z}$, $v_2(2^{S_{i+1}} P_{i+1}) \ge S_{i+1}$, so $v_2(\Omega_i) < S_{i+1}$ forces $v_2(P_i) = v_2(3^{k_i} P_i) = v_2(\Omega_i) < \infty \implies P_i \ge 2^{v_2(\Omega_i)} \ge 1$; the $3$-adic dual is identical. 2. Suppose $P_i = 0$. Then $0 \le 2^{S_{i+1}} P_{i+1} = \Omega_i < 3^{k_i} \le 2^{S_{i+1}} \implies P_{i+1} = 0$ and $\Omega_i = 0$. Induction around the cycle forces $P_i = \Omega_i = 0$ for all $1 \le i \le r$, meaning $(v_{i-1}, u_i, v_i)_{i=1}^r$ is a closed cycle in $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$. By Theorem 5 Part 3, every path in $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ starting in $\{1, \dots, 25\}^3$ terminates within $2$ blocks. 3. For any odd $x > 1$, $U(x) = 1 \implies 3x + 1 = 2^a \ge 10 \implies a \ge 4 \notin \{1, 2\}$, so $1$ is isolated from $\{1, 2\}$-cycles. If $\max(v_{i-1}, u_i, v_i) \ge 26$: (i) if $u_i \ge 26$, $A_i \ge 1 \implies 2^{u_i} A_i - 1 \ge 2^{26} - 1$; (ii) if $v_{i-1} \ge 26$, $B_{i-1} \ge 1 \implies (z_i - 1)/2 = 3^{v_{i-1}} B_{i-1} \ge 3^{26} > 2^{26} - 1$; (iii) if $v_i \ge 26$ and $u_i \le 25$, $3^{u_i} A_i = 4^{v_i} B_i + 1 > 4^{26}$ and $(2/3)^{u_i} \ge (2/3)^{25} \implies 2^{u_i} A_i > 4(8/3)^{25} > 2^{26}$. Thus $\gamma_i = (1 - (2/3)^{u_i})/(2^{u_i} A_i - 1) < (2^{26} - 1)^{-1} < (1.1 \times 10^6)^{-1}$ holds analytically outside $\{1, \dots, 25\}^3$. Inside $\{1, \dots, 25\}^3$, `verify_paper3.py` evaluates the union of all $15,625$ states with $P_i = 0$ and all $45,831$ states with $2^{u_i} A_i - 1 \le (1.1 \times 10^6)(1 - (2/3)^{u_i})$ ($305$ with $P_i = 0$ and $45,526$ with $P_i \ge 1$, totaling $61,151$ distinct states) and verifies that $100\%$ exit $\{1, 2\}^{\mathbb{Z}}$ within $\le 72$ steps (where the first untested candidate outside the threshold is $z_i = 733,339 = 91 + 96(7,638)$ at $(v_{i-1}, u_i, v_i) = (1, 1, 1), A_i^{\mathrm{CRT}} = 23, P_i = 7,638$). Hence every block in a $\{1, 2\}$-cycle has $\ln(1 + \gamma_i) < \gamma_i < (1.1 \times 10^6)^{-1}$, so $S \ln 2 - k \ln 3 = \sum_{i=1}^r \ln(1 + \gamma_i) < r / (1.1 \times 10^6) \le \min(2k - S, S - k)/(1.1 \times 10^6)$, forcing $k \ge 2,966$ ($S \ge 4,701, r \ge 933$). 4. Proved in Section 5.2 via $y^{(j)} = 1 + 2^{w+1-2j} 3^j O$ and Moran's equation $2^{-s} + 4^{-s} = 1$. 5. Shifting $r_0 \mapsto r_0 + b_1 2^{S+1}$ ($b_1 \in \{0, 1\}$) shifts $y_j = U^{(j)}(r_0) = (3^j r_0 + C_j)/2^S$ by $2 b_1 3^j$, so $(3y_j + 1)/2 = w_0 + b_1 3^{j+1}$ changes parity with $b_1$. Thus one value of $b_1$ gives $v_2(3y_j + 1) = 1$ ($S' = S + 1$), and the other gives $2 \mid (w_0 + b_1 3^{j+1})$; shifting that second branch by $b_2 2^{S+2}$ ($b_2 \in \{0, 1\}$) shifts $w_1 = (w_0 + b_1 3^{j+1})/2$ by $b_2 3^{j+1}$, so one value of $b_2$ gives $v_2(3y_j + 1) = 2$ ($S' = S + 2$) and the other exits $\{1, 2\}^{\mathbb{Z}}$ with $v_2(3y_j + 1) \ge 3$. Since each node at $S < M$ branches to $S + 1$ and $S + 2$, the number of leaves stopping at $S \ge M$ satisfies $L_M = L_{M-1} + L_{M-2}$ with $L_1 = 2 = F_3, L_2 = 3 = F_4$, so $L_M = F_{M+2}$, and constructing the tree takes $O(F_{M+2}) = O(X^{\log_2\varphi})$ operations. Since every odd $x_0 \in (1, 2^{M+1})$ whose orbit stays in $\{1, 2\}$ until $S \ge M$ coincides with its unique residue $r_0 = x_0 \bmod 2^{S+1} < 2^{M+1}$ at a leaf, exhausting all leaves with $1 < r_0 < 2^{M+1}$ at the verified cutoffs up to $M = 43$ certifies $x_{\min}(a) > 2^{44}$ (with $x_{\min}(a) \ge 2^{44} + 1$). Finally, by Corollary 1(4), $\Delta < \frac{2r(1 - (2/3)^{u/r})}{x_{\min}(a) - 1}$; using $1 - (2/3)^{u/r} < 1$ gives $r > \frac{1}{2}(x_{\min}(a) - 1)\Delta \ge 2^{43}\Delta > 8.796 \times 10^{12}\Delta$, while using $1 - (2/3)^{u/r} \le \frac{u}{3r}$ gives $u > \frac{3}{2}(x_{\min}(a) - 1)\Delta \ge 3 \cdot 2^{43}\Delta > 2.6388 \times 10^{13}\Delta$. Moreover, since $\Delta = \sum_{t=0}^{k-1}\ln(1 + \frac{1}{3x_t}) < \frac{k}{3x_{\min}}$, $x_{\min}(a) > 2^{40}$ rules out all $(k, S)$ with $k / (3\Delta) < 2^{40}$, eliminating $(190{,}537, 301{,}994)$ ($190{,}537/(3\Delta_{190537}) \approx 9.846 \times 10^{11} < 2^{40}$) and all $k < 10,781,274$. $\quad \square$ --- ## 6. Numerical Verification and Reproducibility All finite computational claims of Theorems 1–7, Remark 1, and Corollaries 1–2 are reproduced by executing `verify_paper3.py` (included in the supplementary package): - **Theorems 1 & 2 (Unrestricted $\mathcal{A}(k, S)$ and Binary $\mathcal{A}_{\{1, 2\}}(k, S)$):** $2^S r_3(a) - 3^k r_2(a) = C(a)$, the odd-endpoint parity lift $U^{(k)}(r_2^*(a)) = r_3^*(a)$, the $\mathbb{F}_p$ walk identity, and $100\%$ seed-ordering exclusion ($r_2(\sigma^t a) > r_3(\sigma^t a)$ for at least one shift $t$) are verified with zero error across all $7,835$ unrestricted positive integer compositions in $\mathcal{A}(k, S)$ for $3 \le k \le 8$ ($11$ simplices) and all $2,031$ binary valuation words in $\mathcal{A}_{\{1, 2\}}(k, S)$ for $3 \le k \le 12$ ($26$ simplices). - **Remark 1 (Exact 2D Transfer-Matrix DP on $\mathcal{A}_{\{1, 2\}}(k, S)$):** The exact 2D dynamic programming residue counts $N_5(0; 7, 13) = 0$, $N_5(0; 39, 77) = 0$, $N_{11}(0; 39, 62) = 3,428,311,887$, and $N_{13}(0; 40, 64) = 4,834,872,320$ are verified directly. - **Theorem 3 & Corollary 1:** The trough congruence $z_i \equiv 2^{u_i+1} \tilde{A}_i - 1 \pmod{2^{S_i+2}}$, the Young–Ferrers and $r$-run telescoped Bezout identities, the $r$-run excess bound $x_{\min}(a) - 1 \le B_{\mathrm{run}}(k, S, r) < \tilde{B}_{\mathrm{run}}(k, S, r)$ and its uniform ratios $B_{\mathrm{aux}}(k, S)/B_{\mathrm{run}}(k, S, r) \ge 3v/u > 4.228$ and $(v/\Delta)/\tilde{B}_{\mathrm{run}}(k, S, r) \ge 3v/(2u) > 2.114$, and the universal $m$-descent product identity $\frac{2^S}{3^k} = \prod_{j=1}^m (1 + \frac{1 - (2/3)^{u_j+1}}{z_j})$ and factor bounds for both $B_{\mathrm{univ}}(k, S, m)$ and $B_{\mathrm{univ}}(k, S, S - k)$ (Corollary 1 Part 4) are verified across all $7,835$ unrestricted compositions ($3 \le k \le 8$), all $2,031$ binary words ($3 \le k \le 12$), and block pairs $(u, v) \in \{1, \dots, 5\}^2, 1 \le r \le 4$. - **Theorem 4 & Corollary 2:** The $2r$-step block-CRT linear system, the eliminated recurrence $2^{u_{i+1}+2v_{i+1}} P_{i+1} - 3^{u_i+v_{i-1}} P_i = \Omega_i$, the exact unrolled suffix/prefix product identity $(2^S - 3^k) P_1 = \sum_{j=1}^r (\prod_{\ell=j+1}^r 3^{u_\ell + v_{\ell-1}})(\prod_{\ell=1}^{j-1} 2^{S_{\ell+1}})\Omega_j$, and the half-step cancellation parametrization $\delta_i^{(1)} = 2^{u_{i+1}} R_i, \delta_i^{(2)} = -3^{u_i} R_i$ (including the explicit $R_i = 5$ cancellation example $(3, 1, 1, 1, 3)$ with $\delta_i^{(1)} = 10, \delta_i^{(2)} = -15, \Omega_i = 0$) are verified across all $1,024$ block 5-tuples in $\{1, \dots, 4\}^5$ and $1 \le r \le 4$. - **Theorem 5:** The infinite-alphabet out-degree bound $\deg^+ \le 1$ on both $\mathcal{G}_{\delta=0}^{\mathrm{block}}$ and the full total-defect graph $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$, and all $37$ valid $1$-block transitions ($34$ with $R_i = 0$ and $3$ with $R_i > 0$, maximum survival $2$ full blocks / $5$ half-steps, zero cycles) are verified across all $15,625$ initial block triples in $\{1, \dots, 25\}^3$. - **Theorem 6:** The ultrametric non-vanishing rate ($98.24\%$ on $\{1, \dots, 4\}^5$), the exact parity-exit equivalence, the analytical outside-box bound $(z_i - 1)/2 \ge 2^{26} - 1 = 67,108,863$ for $\max(v_{i-1}, u_i, v_i) \ge 26$, the $100\%$ inside-box exit certificate (maximum $10$ blocks / $72$ steps, first untested candidate outside the threshold $z_i = 733,339 = 91 + 96(7,638)$ at $(1, 1, 1), P_i = 7,638$) across the $61,151$ distinct states on $\{1, \dots, 25\}^3$ ($k \ge 2,966$), and the $2$-adic Fibonacci $\{1, 2\}$-tree candidate sieve (`verify_paper3.py` for $M \le 18$ and `tree_sieve.c` for $M \le 43$, verifying all $5$ argmax seeds in Table 1, both $r > \frac{1}{2}(x_{\min}-1)\Delta$ and $u > \frac{3}{2}(x_{\min}-1)\Delta$, and the jump past $a_{14} = 55$ to $k \ge 10,781,274, S \ge 17,087,915$) are verified directly. - **Theorem 7:** The reindexed fixed-point homogenization identities for $C(a) \mp (2^S - 3^k)$ on $\mathcal{A}_{\{1, 2\}}(k, S)$ and $\mathcal{A}(k, S)$, the unconditional coprimality $\gcd(C(a), |2^S - 3^k|) = 1$ at $u = 1$ and $v = 1$, the mod-$5$ and mod-$7$ index-parity collapses ($N_5^{\mathrm{unres}}(0; k, k+2) = 0$ for odd $k \ge 3$ and $N_7(0; 2m, 4m-3) = 2\binom{m}{3}$), and the reduced gap divisors $H_u^+(a), H_v^-(a), H_{r, v}^{\mathrm{unres}}(a)$ are verified across all $2,031$ binary words ($3 \le k \le 12$), all $7,835$ unrestricted compositions ($3 \le k \le 8$), and extended families up to $k = 40$. --- ## 7. Limitations, Scope, and Future Horizons We state the explicit boundaries between our unconditional theorems and our finite-box graph certificate: 1. **Unrestricted $\mathcal{A}(k, S)$ vs. Binary $\mathcal{A}_{\{1, 2\}}(k, S)$ Scope:** Theorems 1 and 2, the universal $m$-descent product identity and bound $B_{\mathrm{univ}}(k, S, m) < x_{\min}^{\mathrm{Eliahou}}(k, S)$ in Corollary 1 Part 4, and the excess reindexing and mod-$5$ collapse of Theorem 7 hold unconditionally for all unrestricted positive integer compositions $a \in \mathcal{A}(k, S)$ with arbitrary valuations $a_t \ge 1$. Theorems 3–6 and Corollaries 1 (Parts 1–3) and 2 analyze words over the binary valuation alphabet $\mathcal{A}_{\{1, 2\}}(k, S) = \{1, 2\}^k \cap \mathcal{A}(k, S)$ decomposed into $r$ alternating runs $(1^{u_1}, 2^{v_1}, \dots, 1^{u_r}, 2^{v_r})$ (where $r$ is the $\{1, 2\}$ run count and $m = \sum_{i=1}^r v_i = S - k$ is the classical local-minima count). 2. **Unconditional Global Theorems vs. Finite-Box and Finite-Cutoff Certificates:** Theorems 1–4, Theorem 7, Corollaries 1–2, the infinite-alphabet out-degree bound $\deg^+ \le 1$ on $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ in Theorem 5 (Parts 1–2), and all parts of Theorem 6—including the local Block-CRT bound $k \ge 2,966$ (Part 3) and the $2$-adic Fibonacci $\{1, 2\}$-tree sieve bound $x_{\min}(a) > 2^{44}$, $r > 8.796 \times 10^{12}(S \ln 2 - k \ln 3)$, $u > 2.6388 \times 10^{13}(S \ln 2 - k \ln 3)$, and $k \ge 10,781,274$ ($S \ge 17,087,915$) over the entire infinite alphabet $(u_i, v_i) \in \mathbb{Z}_{\ge 1}^2$ (Part 5)—hold unconditionally without restricting block lengths to a bounded box. While constructing the depth-$M$ candidate tree requires $O(F_{M+2}) = O(X^{\log_2\varphi})$ operations for every $M \ge 1$, the exit certificate following each retained leaf until $v_2(3x_t + 1) \ge 3$ is established at the verified cutoffs up to $M = 43$ ($X = 2^{44}$). Similarly, because $\deg^+ \le 1$ on the countably infinite state space $\mathbb{Z}_{\ge 1}^3$ does not by itself rule out cycles with $\max(v_{i-1}, u_i, v_i) \ge 26$, the zero-wrap path termination certificate for $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ in Theorem 5 Part 3 is established on the finite box $\{1, \dots, 25\}^3$. 3. **Open Global Questions:** Ruling out directed cycles in $\mathcal{G}_{\Omega=0}^{\mathrm{block}}$ analytically for arbitrary unbounded block triples outside $\{1, \dots, 25\}^3$, extending the $\{1, 2\}$-block Bezout/CRT elimination of Theorems 3–6 to general $m$-cycles with valuations $a_t \ge 3$, and extending the dual pigeonhole-gap bounds of Theorem 7 Part 4 to fixed density ratios $u/k \in (0, 2 - \log_2 3)$ near the critical line $S/k \to \log_2 3$ remain open problems. This paper does not claim a general solution of the Collatz conjecture. --- ## References [1] R. E. 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