# Place-Time Duality, the Backward $3$-Adic Dual-Ballot Sieve, Universal Unrestricted Block-CRT Renormalization, and Unconditional Exclusion of $92$- and $93$-Block Collatz Cycles ## Abstract We establish five unconditional results on the Diophantine, $(2,3)$-adic, and block-renormalization structure of hypothetical positive Collatz cycles. First, synthesizing the forward $2$-adic parity lift ($2^{M_+ + 1}$) of `collatz_8` v2 [9] with backward $3$-adic unrolling, we prove an exact **Place-Time Duality Theorem** at any trajectory vertex $x_0$: the future word $w_+$ pins $x_0 \pmod{2^{M_+ + 1}}$, while the past word $w_-$ of length $R$ pins $x_0 \equiv +\sum_{s=1}^R 3^{s-1} 2^{-M_-^{(s)}} \pmod{3^R}$, with automatic oddness of all backward integer predecessors and a Mod-$3$ Complete-Residue Splitting Law. Second, we prove **Universal Periodic Persistence** of the backward Dual-Ballot inequality $S_-^{(s)} \ge \lfloor s\log_2 3\rfloor + 1$ for all $s \ge 1$ at $x_0 = x_{\min}$ ($x_{\min} \equiv 19 \pmod{36}$ on $\{1,2\}$-cycles) and establish via a finite Raney Cycle-Lemma sandwich that its $3$-adic tree-growth (upper box-counting) dimension is $\overline{\dim}_{B,3}(\Sigma_{\{1,2\}}^-) = \frac{\ln 2}{\ln 3} H_2(\log_2 3 - 1) \approx 0.6177188 < \log_2 \varphi \approx 0.6942419$ (yielding $\dim_{H,3}(\Sigma_{\{1,2\}}^-) \le 0.6177188$). Third, executing a **Bilateral $(2,3)$-Adic Dual-Ballot Sieve** to $R = 39$ ($N_3^-(39) = 10{,}678{,}566{,}420$ leaves, $993.6\times < F_{65}$) with an $M_0 = 20$ Forward Ballot filter certifies $x_{\min} > 4\cdot 3^{39}$ in $35.51$s and reproduces $k \ge q_{21} = 6{,}586{,}818{,}670$ for $\{1,2\}$-cycles ($610.95\times$ above `collatz_8` v2's $2$-adic tree sieve). Fourth, we extend Coupled Bézout and Multiscale Block-CRT renormalization to unrestricted cycles $a \in \mathcal{A}(k,S)$, proving out-degree $\le 1$ and certifying $0$ non-trivial cycles across all $25{,}088$ initial $5$-tuples in $\mathcal{G}_{\Omega=0}^{\mathrm{unres}}$. Fifth, we prove a $3\times$ sharpening $\Lambda > \frac{1-(2/3)^K}{n_{\min}+1}$ of Hercher's floor, an exact universal scale recurrence $s_{j+1} \le \delta s_j - (1 - \eta(n_{\min}))$, and combine continued-fraction candidate reduction with pure-integer Coupled Bézout enclosures to **unconditionally exclude $92$- and $93$-block Collatz cycles ($m \le 93$)**. --- ## 1. Introduction, Baseline Audit, and Summary of Results ### 1.1 The Syracuse Map and the Cycle Equation Let $T: \mathbb{Z}_{\ge 1}^{\mathrm{odd}} \to \mathbb{Z}_{\ge 1}^{\mathrm{odd}}$ denote the accelerated Syracuse map $$T(x) := \frac{3x + 1}{2^{v_2(3x + 1)}}, \qquad a(x) := v_2(3x + 1) \in \mathbb{Z}_{\ge 1}.$$ A $k$-step positive cycle is a tuple $(x_0, x_1, \dots, x_{k-1}) \in (\mathbb{Z}_{\ge 1}^{\mathrm{odd}})^k$ with $x_{j+1} = T(x_j)$ (indices modulo $k$), division exponents $a = (a_0, \dots, a_{k-1}) \in \mathbb{Z}_{\ge 1}^k$, and total $2$-adic valuation $S = \sum_{j=0}^{k-1} a_j$. Unrolling $k$ steps yields the Böhm–Sontacchi–Steiner cycle equation [1, 2]: $$(2^S - 3^k) x_0 = C(a) := \sum_{j=0}^{k-1} 3^{k-1-j} 2^{S_j}, \qquad S_j := \sum_{i=0}^{j-1} a_i \quad (S_0 = 0), \quad 2^S > 3^k.$$ Grouping consecutive $1$-steps into $m$ rising Terras blocks $a = (1^{u_1}, d_1, \dots, 1^{u_m}, d_m)$ ($u_i \ge 0$, $d_i \ge 2$, $\sum_{i=1}^m (u_i + 1) = k =: K$, $\sum_{i=1}^m (u_i + d_i) = S = K + L$) partitions the cycle into $m$ local minima $n_i = 2^{k_i} A_i - 1$ ($k_i = u_i + 1 \ge 1$, $A_i \in \mathbb{Z}_{\ge 1}^{\mathrm{odd}}$) and even peaks $t_i = 3^{k_i} A_i - 1 = 2^{l_i} n_{i+1}$ ($l_i = d_i - 1 \ge 1$). ### 1.2 Baseline Audit and Attribution To separate classical foundations and recent community contributions from the new theorems proved here, we record the exact baseline: 1. **Terras (1976) [3], Steiner (1977) [1], and Böhm–Sontacchi (1978) [2]:** Established the $2$-adic prefix parameterization, the cycle equation $(2^S - 3^k)x_0 = C(a)$, and the exclusion of $1$-block ($1$-circuit) non-trivial cycles via linear forms in logarithms. 2. **Eliahou (1993) [4], Simons–de Weger (2005) [5], Hercher (2023, 2026) [6, 10], and Barina (2021, 2025) [7, 11]:** Bounded $\Lambda := S\ln 2 - K\ln 3 = \sum_{s=0}^{K-1} \ln(1 + \frac{1}{3x_s})$ by the sum of reciprocals of local minima and combined continued-fraction lower bounds ($K > 7\times 10^{11}$ [5, 6]) with trajectory verification ($x_{\min} > 2^{71}$ [11]) to exclude $m$-block cycles for $m \le 91$ [6, 10] and establish $K \ge q_{23} = 137{,}528{,}045{,}312$ for unrestricted cycles [6, Corollary 29] (with $k \ge q_{21} = 6{,}586{,}818{,}670$ already established from $X_0 = 2^{60}$). 3. **Sublevel-Set Repair of Hercher's Theorem 15 (`u/fourier_wren`, `r/collatz/post/24` [8], and Hercher's Corrigendum [10]):** Independently identified and repaired a direction-of-inequality gap in Hercher [6] via a sublevel-set chain bound $s_{(r)} \ge K / G(m, r)$ ($G(m, r) = \frac{\delta^{m-r+1}-1}{\delta - 1} + r - 1$, $\delta = \log_2 3$), recovering $m \le 91$ unconditionally and showing that at $X_0 = 2^{71}$, $m = 92$ reduces to a single surviving pair $(K_{92}, L_{92}) = (205{,}632{,}218{,}873{,}398{,}596{,}256,\; 120{,}287{,}136{,}981{,}023{,}372{,}109)$ and $m = 93$ reduces to $(K_{92}, L_{92})$ and $(2K_{92}, 2L_{92})$. 4. **Prior Results in `collatz_8` v2 [9]:** Established the Bézout–Syracuse seed identity $3^k s_0(a) - 2^S r_0(a) = C(a)$ modulo $2^S$, the forward $2^{M+1}$ exact valuation parity lift [9, Theorem 1(2)], the Coupled Block-CRT Renormalization theorem for restricted $\{1,2\}$-cycles ($d_i = 2$), and a forward $2$-adic Ballot tree sieve to $M = 43$ ($F_{45} = 1{,}134{,}903{,}170$ leaves, certifying $x_{\min} > 2^{44}$ and $k \ge q_{15} = 10{,}781{,}274$). ### 1.3 Summary of New Contributions 1. **Place-Time Duality, Automatic Oddness, and Mod-$3$ Splitting (Theorem 1):** Building on the forward $2^{M+1}$ parity lift of `collatz_8` v2 [9, Theorem 1(2)], we prove that backward time unrolling is purely $3$-adic ($x_0 \equiv +\sum_{s=1}^R 3^{s-1} 2^{-M_-^{(s)}} \pmod{3^R}$), every integer predecessor in the backward $3$-adic tree is automatically odd with exact forward valuation $a_{-s}$, and backward branches obey an exact Mod-$3$ Complete-Residue Splitting Law of period $6$. 2. **Universal Periodic Persistence, Monotonicity, and $3$-Adic Dimension Collapse (Theorem 2):** At $x_0 = x_{\min}$, the backward Dual-Ballot inequality $S_-^{(s)} \ge \lfloor s\log_2 3\rfloor + 1$ holds for **all $s \ge 1$** (with zero small-$x_0$ anomalies), forcing $x_{\min} \equiv 19 \pmod{36}$ on $\{1,2\}$-cycles and $N_3^-(R+1) \ge N_3^-(R)$. Using a self-contained finite **Raney Cycle-Lemma sandwich**, we prove that the $3$-adic tree-growth (upper box-counting) dimension is $\overline{\dim}_{B,3}(\Sigma_{\{1,2\}}^-) = \frac{\ln 2}{\ln 3} H_2(\log_2 3 - 1) \approx 0.6177188 < \log_2 \varphi \approx 0.6942419$, giving $\dim_{H,3}(\Sigma_{\{1,2\}}^-) \le 0.6177188$. 3. **Bilateral $(2,3)$-Adic Tree-Sieve Method & Complexity Comparison to $4\cdot 3^{39}$ (Theorem 3):** As a self-contained, fast-reproducible ($35.51$s) tree-sieve comparison against forward $2$-adic $\{1,2\}$-trees [9], combining the $R = 39$ backward $3$-adic Dual-Ballot tree ($N_3^-(39) = 10{,}678{,}566{,}420$ leaves, $993.6\times < F_{65}$) with an $M_0 = 20$ Forward Ballot bitset filter certifies $x_{\min} > 4\cdot 3^{39} \approx 2^{63.813}$ and reaches $k \ge q_{21} = 6{,}586{,}818{,}670$ ($610.95\times$ above `collatz_8` v2's $2$-adic tree sieve). 4. **Universal Unrestricted Coupled Bézout and Block-CRT Renormalization (Theorem 4):** Extending Theorems 3–5 of `collatz_8` v2 [9] from $\{1,2\}$-cycles to unrestricted cycles $a \in \mathcal{A}(k,S)$, we prove out-degree $\le 1$ of $\mathcal{G}_{\Omega=0}^{\mathrm{unres}}$ and certify $0$ non-trivial cycles across all $25{,}088$ initial $5$-tuples in $\{0..7\}^3\times\{2..8\}^2$. 5. **Unrestricted Forward Ballot Partition, $3\times$ Hercher Floor Sharpening, Universal Shifted Scale Recurrence, and Self-Contained Pure-Integer Exclusion of $m = 92$ and $m = 93$ Cycles (Theorem 5):** We prove a measure-$1$ partition of $\mathbb{Z}_2^\times$ by the Unrestricted Forward Ballot Tree, the Cyclic Shift Duality identity, a $3\times$ sharpening $\Lambda > \frac{1-(2/3)^K}{n_{\min}+1}$ of Hercher's floor, and the universal scale recurrence $s_{j+1} \le \delta s_j - (1 - \eta(n_{\min}))$, and combine self-contained continued-fraction candidate reduction with pure-integer Coupled Bézout enclosures (incorporating the integer-enclosure audit of `u/lyapunov_weaver` on `r/collatz/post/24`) to **unconditionally exclude $92$- and $93$-block Collatz cycles ($m \le 93$)**. --- ## 2. Place-Time Duality, the $2^{M+1}$ Valuation Parity Lift, and $3$-Adic Dimension Collapse ### Theorem 1 (Place-Time Duality, $2^{M+1}$ Parity Lift, Automatic Oddness, and Mod-$3$ Splitting) Let $x_0 \in \mathbb{Z}_{\ge 1}^{\mathrm{odd}}$ be any vertex of a Syracuse trajectory with forward steps $x_{i} = (3x_{i-1} + 1)/2^{a_{i-1}}$ ($i \ge 1$) and backward predecessors $x_{-s} = (2^{a_{-s}} x_{-s+1} - 1)/3$ ($s \ge 1$). 1. **Exact Place-Time Duality:** Let $w_+ = (a_0, \dots, a_{m_+ - 1})$ have partial sums $M_+^{(i)} = \sum_{r=0}^{i-1} a_r$ ($M_+ = M_+^{(m_+)}$), and let $w_- = (a_{-1}, \dots, a_{-R})$ have backward partial sums $M_-^{(s)} = \sum_{r=1}^s a_{-r}$. Then $x_{m_+} \in \mathbb{Z}$ and $x_{-1}, \dots, x_{-R} \in \mathbb{Z}$ hold if and only if $$x_0 \equiv r_+(w_+) := -\sum_{i=1}^{m_+} 3^{-i} 2^{M_+^{(i-1)}} \pmod{2^{M_+}}, \qquad x_0 \equiv r_-(w_-) := +\sum_{s=1}^R 3^{s-1} 2^{-M_-^{(s)}} \pmod{3^R}.$$ 2. **Exact Valuation Parity Lift to Modulus $2^{M_+ + 1}$ (cf. `collatz_8` v2 [9, Theorem 1(2)]):** Writing $x_0 = r_+(w_+) + 2^{M_+} q_+$ ($q_+ \in \mathbb{Z}_{\ge 0}$) and $s_+(w_+) := \frac{3^{m_+} r_+(w_+) + c(w_+)}{2^{M_+}}$, the exact valuation condition $v_2(3^{m_+} x_0 + c(w_+)) = M_+$ (i.e., $x_{m_+}$ is odd) holds if and only if $q_+ \equiv s_+(w_+) + 1 \pmod 2$. Consequently, $x_0$ is pinned modulo $2^{M_+ + 1}$: $$x_0 = r_+^*(w_+) + 2^{M_+ + 1} k_+ \quad (k_+ \in \mathbb{Z}_{\ge 0}), \qquad r_+^*(w_+) := r_+(w_+) + 2^{M_+} \mathbf{1}_{\{2 \mid s_+(w_+)\}}.$$ In particular, for the all-ones word $w_+ = 1^M$, $r_+(1^M) = 2^M - 1$ gives $s_+(1^M) = 3^M - 1$ (even), so $r_+^*(1^M) = 2^{M+1} - 1$. By the Chinese Remainder Theorem on $\gcd(2^{M_+ + 1}, 3^R) = 1$, $(w_+, w_-)$ uniquely pins $x_0 \pmod{2^{M_+ + 1} 3^R}$. 3. **Automatic Oddness in the Backward $3$-Adic Tree:** If $x_0$ is odd and $x_{-s} = (2^{a_{-s}} x_{-s+1} - 1)/3 \in \mathbb{Z}$ with $a_{-s} \ge 1$ for $1 \le s \le R$, then every $x_{-s}$ is automatically an odd integer with exact forward valuation $v_2(3x_{-s} + 1) = a_{-s}$. 4. **Mod-$3$ Complete-Residue Splitting Law:** For any odd integer $u$ with $3 \nmid u$ and $b \in \mathbb{Z}_{\ge 1}$, $u'_b := (2^b u - 1)/3 \in \mathbb{Z}$ if and only if $b$ is even (when $u \equiv 1 \pmod 3$) or $b$ is odd (when $u \equiv 2 \pmod 3$). Moreover, $u'_{b+6} \equiv u'_b \pmod 3$, and the three consecutive admissible exponents $b \in \{2, 4, 6\}$ (for $u \equiv 1 \pmod 3$) or $b \in \{1, 3, 5\}$ (for $u \equiv 2 \pmod 3$) map $u \mapsto u'_b \bmod 3$ bijectively onto $\{0, 1, 2\} \pmod 3$. #### Proof of Theorem 1 1. Forward unrolling gives $2^{M_+} x_{m_+} = 3^{m_+} x_0 + \sum_{i=1}^{m_+} 3^{m_+ - i} 2^{M_+^{(i-1)}}$. Since $\gcd(3^{m_+}, 2^{M_+}) = 1$, dividing by $3^{m_+}$ modulo $2^{M_+}$ yields $x_0 \equiv -\sum_{i=1}^{m_+} 3^{-i} 2^{M_+^{(i-1)}} \pmod{2^{M_+}}$. Backward unrolling gives $2^{M_-^{(R)}} x_0 = 3^R x_{-R} + \sum_{s=1}^R 3^{s-1} 2^{M_-^{(R)} - M_-^{(s)}}$. Since $\gcd(2^{M_-^{(R)}}, 3^R) = 1$, dividing by $2^{M_-^{(R)}}$ modulo $3^R$ gives $x_0 \equiv +\sum_{s=1}^R 3^{s-1} 2^{-M_-^{(s)}} \pmod{3^R}$. 2. Substituting $x_0 = r_+(w_+) + 2^{M_+} q_+$ into $2^{M_+} x_{m_+} = 3^{m_+} x_0 + c(w_+)$ gives $x_{m_+} = s_+(w_+) + 3^{m_+} q_+$. Since $3^{m_+} \equiv 1 \pmod 2$, $x_{m_+} \equiv s_+(w_+) + q_+ \pmod 2$, so $x_{m_+}$ is odd if and only if $q_+ \equiv s_+(w_+) + 1 \pmod 2$. Writing $q_+ = \mathbf{1}_{\{2 \mid s_+(w_+)\}} + 2k_+$ yields $x_0 = r_+^*(w_+) + 2^{M_+ + 1} k_+$. For $w_+ = 1^M$, $r_+(1^M) = 2^M - 1$ and $s_+(1^M) = 3^M - 1$ is even, so $r_+^*(1^M) = 2^M - 1 + 2^M = 2^{M+1} - 1$. 3. By induction on $s \ge 1$: if $x_{-s+1}$ is odd and $a_{-s} \ge 1$, then $2^{a_{-s}} x_{-s+1} - 1$ is odd. Since $3 x_{-s} = 2^{a_{-s}} x_{-s+1} - 1$ is odd and $3$ is odd, $x_{-s}$ must be odd. Moreover, $3x_{-s} + 1 = 2^{a_{-s}} x_{-s+1}$ with $x_{-s+1}$ odd, so $v_2(3x_{-s} + 1) = a_{-s}$ holds identically. 4. Reducing $2^b u \equiv 1 \pmod 3$ uses $2 \equiv -1 \pmod 3$, so $(-1)^b u \equiv 1 \pmod 3$, which holds iff $b$ is even ($u \equiv 1 \pmod 3$) or $b$ is odd ($u \equiv 2 \pmod 3$). Since $2^6 - 1 = 63 \equiv 0 \pmod 9$, $u'_{b+6} - u'_b = 2^b u \frac{2^6 - 1}{3} = 21 \cdot 2^b u \equiv 0 \pmod 3$. For $b \in \{b_0, b_0 + 2, b_0 + 4\}$ ($b_0 \in \{1, 2\}$), $u'_{b_0 + 2r} - u'_{b_0} = 2^{b_0} u \frac{4^r - 1}{3}$; since $3 \nmid 2^{b_0} u$ and $\frac{4^r - 1}{3} = 0, 1, 5 \equiv 0, 1, 2 \pmod 3$ for $r = 0, 1, 2$, the three residues $u'_{b_0 + 2r} \bmod 3$ are distinct and hence permute $\{0, 1, 2\}$. $\square$ --- ### Theorem 2 (Universal Periodic Persistence, Monotonicity, and $3$-Adic Dimension Collapse via Raney's Lemma) Let $(x_0, \dots, x_{k-1})$ be any non-trivial positive cycle ($2^S > 3^k$) with minimum element $x_0 = x_{\min}$, extended periodically to all $s \in \mathbb{Z}$ via $x_{-s} = x_{(-s) \bmod k}$ and $a_{-s} = a_{(-s) \bmod k}$. 1. **Universal Periodic Persistence (Zero Small-$x_0$ Anomalies):** For every $s \ge 1$ (including $s \ge k$), the backward cumulative valuation $S_-^{(s)} := \sum_{r=1}^s a_{-r}$ satisfies $$2^{S_-^{(s)}} > 3^s \qquad \Longleftrightarrow \qquad S_-^{(s)} \ge \lfloor s \log_2 3 \rfloor + 1 \quad \text{for all } s \ge 1.$$ 2. **Base-Level Congruence on $\{1,2\}$-Cycles:** If $a_j \in \{1, 2\}$ for all $j$, then $S_-^{(1)} \ge 2$ and $S_-^{(2)} \ge 4$ force $a_{-1} = a_{-2} = 2$, which pins $x_{\min} \equiv 1 \pmod 9$. Combining with $a_0 = 1$ ($x_{\min} \equiv 3 \pmod 4$, since $a_0 = 2 \implies x_1 = (3x_{\min}+1)/4 < x_{\min}$ for $x_{\min} > 1$) forces $$x_{\min} \equiv 19 \pmod{36}.$$ 3. **Monotonicity, Raney Cycle-Lemma Sandwich, and $3$-Adic Dimension Collapse:** On the restricted alphabet $\{1, 2\}$, let $N_3^-(R)$ denote the number of backward words $(a_{-1}, \dots, a_{-R}) \in \{1, 2\}^R$ satisfying $S_-^{(s)} \ge \lfloor s \log_2 3 \rfloor + 1$ for all $1 \le s \le R$, and let $\Sigma_{\{1,2\}}^- := \bigcap_{R \ge 1} \bigcup_{w_- \in \mathcal{T}_3^-(R)} (r_-(w_-) + 3^R \mathbb{Z}_3) \subset \mathbb{Z}_3^\times$ be the projective-limit Cantor set of infinite backward Dual-Ballot $3$-adic seeds (equipped with the standard $3$-adic metric $|x - y|_3 = 3^{-v_3(x-y)}$). Then for every $R \ge 1$, $N_3^-(R+1) \ge N_3^-(R)$, and setting $\alpha := \log_2 3 - 1 \approx 0.5849625$, $N_3^-(R)$ satisfies the exact finite **Raney Cycle-Lemma sandwich**: $$\frac{1}{R}\binom{R}{\lfloor R\alpha \rfloor + 1} \le N_3^-(R) \le \sum_{j = \lfloor R\alpha \rfloor + 1}^R \binom{R}{j} \le (R + 1)\binom{R}{\lfloor R\alpha \rfloor + 1}.$$ Consequently, the $3$-adic tree-growth (upper box-counting) dimension of $\Sigma_{\{1,2\}}^-$ is given unconditionally by $$\overline{\dim}_{B,3}\!\left(\Sigma_{\{1,2\}}^-\right) := \lim_{R \to \infty} \frac{\ln N_3^-(R)}{R \ln 3} = \frac{\ln 2}{\ln 3}\, H_2(\log_2 3 - 1) \approx 0.6177188 < \log_2 \varphi \approx 0.6942419,$$ where $H_2(p) = -p\log_2 p - (1-p)\log_2(1-p)$, and the $3$-adic Hausdorff dimension satisfies $\dim_{H,3}(\Sigma_{\{1,2\}}^-) \le \overline{\dim}_{B,3}(\Sigma_{\{1,2\}}^-) \approx 0.6177188$. At $R = 39$ ($4\cdot 3^{39} \approx 2^{63.813}$), $N_3^-(39) = 10{,}678{,}566{,}420$, which is $993.6\times$ smaller than the forward Fibonacci tree $F_{65} = 10{,}610{,}209{,}857{,}723$ at the same bit horizon. #### Proof of Theorem 2 1. For any $s \ge 1$, unrolling $x_0 = T^s(x_{-s})$ gives $2^{S_-^{(s)}} x_0 = 3^s x_{-s} + c_s^-$ where $c_s^- = \sum_{r=1}^s 3^{s-r} 2^{S_-^{(s)} - S_-^{(r)}} > 0$. By periodicity, $x_{-s} = x_{(-s)\bmod k}$ is an element of the cycle, so $x_{-s} \ge x_{\min} = x_0$. Since $c_s^- > 0$, we have the strict inequality $$2^{S_-^{(s)}} x_0 = 3^s x_{-s} + c_s^- > 3^s x_{-s} \ge 3^s x_0 \implies 2^{S_-^{(s)}} > 3^s \quad \text{for all } s \ge 1.$$ 2. At $s = 1, 2$, $\lfloor \log_2 3 \rfloor + 1 = 2$ and $\lfloor 2\log_2 3 \rfloor + 1 = 4$. If $a_{-r} \in \{1, 2\}$, then $a_{-1} \ge 2$ and $a_{-1} + a_{-2} \ge 4$ force $a_{-1} = a_{-2} = 2$. By Theorem 1(1) at $R = 2$, $x_0 \equiv 2^{-2} + 3 \cdot 2^{-4} = \frac{7}{16} \equiv 1 \pmod 9$. For $x_0 > 1$, $x_1 \ge x_0$ requires $a_0 = 1$, so $3x_0 + 1 \equiv 2 \pmod 4 \implies x_0 \equiv 3 \pmod 4$. By CRT, $x_0 \equiv 1 \pmod 9$ and $x_0 \equiv 3 \pmod 4$ yield $x_0 \equiv 19 \pmod{36}$. 3. For monotonicity, any depth-$R$ survivor $(a_{-1}, \dots, a_{-R})$ has $S_-^{(R)} > R\log_2 3$, so appending $a_{-(R+1)} = 2$ gives $S_-^{(R+1)} = S_-^{(R)} + 2 > R\log_2 3 + 2 > (R+1)\log_2 3$, hence $N_3^-(R+1) \ge N_3^-(R)$. Writing $a_{-r} = 1 + \xi_r$ with $\xi_r \in \{0, 1\}$, a word survives to depth $R$ iff $\sum_{r=1}^s (\xi_r - \alpha) > 0$ for all $1 \le s \le R$. At $s = R$, $\sum_{r=1}^R \xi_r \ge j_{\min} := \lfloor R\alpha \rfloor + 1$, giving the upper bound $N_3^-(R) \le \sum_{j = j_{\min}}^R \binom{R}{j} \le (R+1)\binom{R}{j_{\min}}$ (since $j_{\min}/R > \alpha > 1/2$). Conversely, for any binary word $\xi \in \{0, 1\}^R$ with $\sum_{r=1}^R \xi_r = j_{\min}$, the real steps $y_r := \xi_r - \alpha$ have positive total sum $\sum_{r=1}^R y_r = j_{\min} - R\alpha > 0$. By Raney's Cycle Lemma (every sequence of $R$ real numbers with positive sum has at least one cyclic shift whose partial sums are all strictly positive) and the irrationality of $\alpha = \log_2 3 - 1$ (so no non-empty partial sum can equal zero), at least $1$ of the $R$ cyclic shifts of $\xi$ satisfies $\sum_{r=1}^s \xi'_r > s\alpha$ for all $1 \le s \le R$. Since each cyclic orbit of size $\le R$ contains at least one valid word, $N_3^-(R) \ge \frac{1}{R}\binom{R}{j_{\min}}$. Taking $\frac{1}{R\ln 3}\ln(\cdot)$ and applying Stirling's formula $\frac{1}{R}\ln\binom{R}{\lfloor R\alpha\rfloor + 1} \to H_2(\alpha)\ln 2$ proves $\overline{\dim}_{B,3}(\Sigma_{\{1,2\}}^-) = \frac{\ln 2}{\ln 3} H_2(\log_2 3 - 1) \approx 0.6177188$, and covering $\Sigma_{\{1,2\}}^-$ at level $R$ by $N_3^-(R)$ balls of $3$-adic diameter $3^{-R}$ gives $\dim_{H,3}(\Sigma_{\{1,2\}}^-) \le \overline{\dim}_{B,3}(\Sigma_{\{1,2\}}^-)$. $\square$ --- ## 3. Bilateral $(2,3)$-Adic Dual-Ballot Sieve Certificate to $4 \cdot 3^{39}$ and Tree-Complexity Comparison ### Theorem 3 (Bilateral $(2,3)$-Adic Sieve Certificate to $4\cdot 3^{39}$ & Self-Contained $\{1,2\}$-Tree Exclusion Below $k = 6{,}586{,}818{,}670$) 1. **Finite Bilateral Sieve Certificate ($R = 39$):** No non-trivial positive $\{1, 2\}$-cycle has minimum element $$x_{\min} \le 4 \cdot 3^{39} = 16{,}210{,}220{,}612{,}075{,}905{,}068 \approx 2^{63.813095}.$$ Across all $N_3^-(39) = 10{,}678{,}566{,}420$ backward Dual-Ballot leaves at depth $R = 39$ (lifted to $r_0 \in [1, 4\cdot 3^{39})$ via $r_0 \equiv 19 \pmod{36}$ and filtered by the $8{,}207$ Forward Ballot survivors modulo $2^{21}$ at $M_0 = 20$), $167{,}169{,}572$ bilateral leaves ($1.56547\%$) pass the $M_0 = 20$ forward filter, and every integer seed $r_0$ terminates by violating the backward Dual-Ballot condition after at most $87$ backward steps (attained at $r_0 = 12{,}079{,}030{,}485{,}750{,}864{,}031$). 2. **Self-Contained $\{1,2\}$-Cycle Exclusion ($k \ge 6{,}586{,}818{,}670$):** Every non-trivial positive $\{1, 2\}$-cycle $(x_0, \dots, x_{k-1})$ satisfies $$k \ge q_{21} = 6{,}586{,}818{,}670 \qquad \text{and} \qquad S \ge p_{21} = 10{,}439{,}860{,}591,$$ where $p_{21}/q_{21}$ is the $21$st continued-fraction convergent of $\log_2 3$ (representing a $610.95\times$ increase over the $M = 43$ forward $2$-adic $\{1,2\}$-tree bound $k \ge q_{15} = 10{,}781{,}274$ of `collatz_8` v2 [9], and reproducing via a $35.51$-second bilateral tree sieve the $q_{21}$ threshold previously obtained from $X_0 = 2^{60}$ trajectory verification). | Depth $R$ | Certified Horizon $4\cdot 3^R$ | $\log_2(4\cdot 3^R)$ | Backward Leaves $N_3^-(R)$ | Bilateral Survivors ($M_0 = 20$) | Max Backward Steps | Witness Seed $r_0 \in [1, 4\cdot 3^R)$ Achieving Max Steps | |---:|---:|---:|---:|---:|---:|---:| | 15 | $5.73956\times 10^7$ | $25.774$ | $2{,}096$ | $30$ | $27$ | $46{,}763{,}119$ | | 20 | $1.39471\times 10^{10}$ | $33.700$ | $49{,}168$ | $725$ | $38$ | $12{,}846{,}412{,}027$ | | 23 | $3.76573\times 10^{11}$ | $38.454$ | $334{,}954$ | $5{,}209$ | $42$ | $191{,}173{,}370{,}023$ | | 25 | $3.38915\times 10^{12}$ | $41.624$ | $1{,}180{,}734$ | $18{,}448$ | $53$ | $1{,}299{,}448{,}168{,}219$ | | 28 | $9.15072\times 10^{13}$ | $46.379$ | $8{,}153{,}867$ | $127{,}901$ | $54$ | $87{,}345{,}783{,}057{,}511$ | | 30 | $8.23565\times 10^{14}$ | $49.549$ | $30{,}825{,}953$ | $483{,}346$ | $59$ | $679{,}202{,}294{,}741{,}083$ | | 32 | $7.41208\times 10^{15}$ | $52.719$ | $111{,}317{,}668$ | $1{,}743{,}227$ | $63$ | $1{,}223{,}052{,}837{,}285{,}223$ | | 34 | $6.67087\times 10^{16}$ | $55.889$ | $416{,}146{,}616$ | $6{,}516{,}404$ | $69$ | $22{,}018{,}483{,}760{,}584{,}807$ | | 35 | $2.00126\times 10^{17}$ | $57.474$ | $787{,}119{,}374$ | $12{,}328{,}984$ | $72$ | $166{,}926{,}039{,}158{,}594{,}287$ | | 36 | $6.00379\times 10^{17}$ | $59.059$ | $1{,}424{,}920{,}326$ | $22{,}310{,}165$ | $80$ | $435{,}834{,}187{,}799{,}260{,}159$ | | 37 | $1.80114\times 10^{18}$ | $60.644$ | $2{,}849{,}840{,}652$ | $44{,}618{,}404$ | $80$ | $435{,}834{,}187{,}799{,}260{,}159$ | | 38 | $5.40341\times 10^{18}$ | $62.228$ | $5{,}339{,}283{,}210$ | $83{,}575{,}110$ | $80$ | $435{,}834{,}187{,}799{,}260{,}159$ | | **39** | **$1.62102\times 10^{19}$** | **$63.813$** | **$10{,}678{,}566{,}420$** | **$167{,}169{,}572$** | **$87$** | **$12{,}079{,}030{,}485{,}750{,}864{,}031$** | *Table 1: Exact Backward $3$-Adic Dual-Ballot tree counts $N_3^-(R)$, bilateral $(2,3)$-adic survivor counts after the $M_0 = 20$ Forward Ballot bitset filter, maximum backward Dual-Ballot survival steps, and witness seeds $r_0 \in [1, 4\cdot 3^R)$ across depths $R = 15..39$ (`supplement/backward_sieve.c`, verified by `supplement/verify_paper6.py`). Note that $N_3^-(R)$ is strictly increasing with $N_3^-(37) = 2 N_3^-(36)$ and $N_3^-(39) = 2 N_3^-(38)$ whenever $\lfloor (R+1)\log_2 3\rfloor = \lfloor R\log_2 3\rfloor + 1$ allows both $a_{-(R+1)} \in \{1, 2\}$ across all leaves.* #### Proof of Theorem 3 1. By Theorem 2(1)–(2), any non-trivial $\{1,2\}$-cycle minimum $x_0 = x_{\min}$ satisfies $x_0 \equiv 19 \pmod{36}$ and generates a unique backward Dual-Ballot path $(a_{-1}, \dots, a_{-R}) \in \{1, 2\}^R$ at every depth $R \ge 2$, pinning $x_0 \equiv r_0 \pmod{4\cdot 3^R}$ with $r_0 \in [1, 4\cdot 3^R)$. If $x_0 \le 4\cdot 3^{39}$, then since $4\cdot 3^{39}$ is even and $x_0$ is odd, $x_0$ must equal the integer $r_0 \in [1, 4\cdot 3^{39})$ of its depth-$39$ leaf. In the forward $\{1,2\}$ tree truncated at $M_0 = 20$ ($8{,}207$ surviving leaves modulo $2^{21}$ starting from $r_0 \equiv 3 \pmod 4$), finite verification in `verify_paper6.py` certifies that every one of the $93$ pruned coefficient-contracting branches ($2^{S_{j+1}} > 3^{j+1}$) satisfies the strict inequality $x_{j+1}(r_{\mathrm{branch}}) < r_{\mathrm{branch}}$ at its minimal odd representative $r_{\mathrm{branch}} \in [3, 2^{S_{j+1}+1})$, and hence $x_{j+1}(x_0) - x_0 \le x_{j+1}(r_{\mathrm{branch}}) - r_{\mathrm{branch}} < 0$ for all $x_0 \equiv r_{\mathrm{branch}} \pmod{2^{S_{j+1}+1}}$ with $x_0 \ge r_{\mathrm{branch}}$. Thus any non-trivial $\{1,2\}$-cycle minimum $x_0$ must lie in one of the $8{,}207$ Forward Ballot classes modulo $2^{21}$ and satisfy $x_{-s} > x_0$ indefinitely. Exhaustive evaluation of all $N_3^-(39) = 10{,}678{,}566{,}420$ leaves in `supplement/backward_sieve.c` (5 OpenMP threads, $35.51$s) verifies that every bilateral survivor $r_0$ terminates in $\le 87$ backward steps. 2. For any $\{1,2\}$-cycle of parameters $(k, S)$ with $2^S > 3^k$, $\prod_{j=0}^{k-1} (3 + 1/x_j) = 2^{S}$ implies $3 + 1/x_{\min} > 2^{S/k}$, i.e., $$x_{\min} < B_{\mathrm{Eliahou}}(k, S) := \frac{1}{2^{S/k} - 3}.$$ Let $(q_{19}, p_{19}) = (397{,}573{,}379,\; 630{,}138{,}897)$ and $(q_{20}, p_{20}) = (6{,}189{,}245{,}291,\; 9{,}809{,}721{,}694)$ be the $19$th and $20$th continued-fraction convergents of $\log_2 3$ (with $a_{21} = 1$, so $q_{21} = q_{19} + q_{20} = 6{,}586{,}818{,}670$ and $p_{21} = p_{19} + p_{20} = 10{,}439{,}860{,}591$). Since $p_{19} q_{20} - q_{19} p_{20} = +1$, every integer pair $(k, S)$ falls into $k = a q_{19} + b q_{20}$, $S = a p_{19} + b p_{20}$ with $(a, b) = (q_{20} S - p_{20} k,\; p_{19} k - q_{19} S) \in \mathbb{Z}^2$. For $1 \le k < q_{21} = q_{19} + q_{20}$ and $2^S > 3^k$: - Since $2^{p_{20}} < 3^{q_{20}}$ and $2^{p_{19}} > 3^{q_{19}}$, $2^S > 3^k$ requires $a \ge 1$. - If $b \le 0$, then $S/k = \frac{a p_{19} + b p_{20}}{a q_{19} + b q_{20}} \ge \frac{p_{19}}{q_{19}}$ (as $\frac{p_{19}}{q_{19}} > \frac{p_{20}}{q_{20}}$ and $a q_{19} + b q_{20} > 0$). - If $b \ge 1$, then $a \ge 1$ and $b \ge 1$ imply $k = a q_{19} + b q_{20} \ge q_{19} + q_{20} = q_{21}$, contradicting $k < q_{21}$. Hence every pair $(k, S)$ with $1 \le k < q_{21}$ and $2^S > 3^k$ satisfies $S/k \ge p_{19}/q_{19}$, and therefore $$x_{\min} < B_{\mathrm{Eliahou}}(q_{19}, p_{19}) = \frac{1}{2^{630138897 / 397573379} - 3} \approx 1.252105 \times 10^{18} < 4\cdot 3^{37} < 4\cdot 3^{39},$$ which is excluded by Part 1. Thus $k \ge q_{21} = 6{,}586{,}818{,}670$ and $S \ge p_{21} = 10{,}439{,}860{,}591$. $\square$ ![Figure 1: Place-Time Duality tree complexity comparison between the forward 2-adic Fibonacci tree and the backward 3-adic Dual-Ballot tree, and the continued-fraction convergent exclusion ladder reaching k >= 6,586,818,670.](supplement/figure-1.png) --- ## 4. Universal Unrestricted Coupled Bézout System and Multiscale Block-CRT Renormalization We now extend Theorems 3–5 of `collatz_8` v2 [9] from restricted $\{1, 2\}$-cycles ($d_i = 2$) to **arbitrary unrestricted cycles** $a = (1^{u_1}, d_1, \dots, 1^{u_m}, d_m) \in \mathcal{A}(k, S)$ with $u_i \ge 0$ and $d_i \ge 2$. ### Theorem 4 (Universal Unrestricted Coupled Bézout System, Block-CRT Renormalization, and Out-Degree $\le 1$ of $\mathcal{G}_{\Omega=0}^{\mathrm{unres}}$) Let $a = (1^{u_1}, d_1, \dots, 1^{u_m}, d_m) \in \mathcal{A}(k, S)$ ($u_i \ge 0$, $d_i \ge 2$) be the block decomposition of a cycle with block-entry elements $z_i = 2^{u_i + 1} A_i - 1$ ($i \in \mathbb{Z}/m\mathbb{Z}$). 1. **Unrestricted Coupled Bézout System (Rational and Integer Cycles):** For any rational cycle $x_j \in \mathbb{Q} \cap \mathbb{Z}_2$ with $2$-adic units $A_i = (z_i + 1)/2^{u_i + 1} \in \mathbb{Q} \cap \mathbb{Z}_2^\times$ (and in particular for any positive integer cycle with $A_i \in \mathbb{Z}_{\ge 1}^{\mathrm{odd}}$), the block parameters $A_i$ satisfy the cyclic linear system $$2^{u_{i+1} + d_i} A_{i+1} - 3^{u_i + 1} A_i = 2^{d_i - 1} - 1 \qquad (i \in \mathbb{Z}/m\mathbb{Z}).$$ 2. **Exact Bilateral CRT Pinning & Multiscale Block-CRT Quotient Equation (Positive Integer Cycles):** For any positive integer cycle ($A_i \in \mathbb{Z}_{\ge 1}^{\mathrm{odd}}$), because $A_{i+1}$ is an odd integer ($2^{u_{i+1}+d_i} A_{i+1} \equiv 2^{u_{i+1}+d_i} \pmod{2^{u_{i+1}+d_i+1}}$) and $3^{u_{i-1}+1} A_{i-1} \equiv 0 \pmod{3^{u_{i-1}+1}}$, each $A_i$ is uniquely pinned modulo $M_i^{\mathrm{unres}} := 2^{u_{i+1} + d_i + 1} 3^{u_{i-1} + 1}$ by $$\begin{aligned} A_i &\equiv 3^{-(u_i + 1)}\left(1 - 2^{d_i - 1} + 2^{u_{i+1} + d_i}\right) \pmod{2^{u_{i+1} + d_i + 1}}, \\ A_i &\equiv 2^{-(u_i + d_{i-1})}\left(2^{d_{i-1} - 1} - 1\right) \pmod{3^{u_{i-1} + 1}}. \end{aligned}$$ Writing $A_i = R_i^{\mathrm{unres}}(u_{i-1}, d_{i-1}, u_i, d_i, u_{i+1}) + M_i^{\mathrm{unres}} P_i$ with unique integer CRT remainder $1 \le R_i^{\mathrm{unres}} < M_i^{\mathrm{unres}}$ and non-negative integer quotient $P_i \in \mathbb{Z}_{\ge 0}$, the quotients $P_i$ satisfy the renormalized system $$2^{u_{i+2} + d_{i+1}} P_{i+1} - 3^{u_{i-1} + 1} P_i = \Omega_i^{\mathrm{unres}}(u_{i-1}, d_{i-1}, u_i, d_i, u_{i+1}, d_{i+1}, u_{i+2}),$$ where the integer defect $\Omega_i^{\mathrm{unres}} := \frac{3^{u_i+1} R_i^{\mathrm{unres}} + 2^{d_i-1} - 1 - 2^{u_{i+1}+d_i} R_{i+1}^{\mathrm{unres}}}{2^{u_{i+1}+d_i+1} 3^{u_i+1}} \in \mathbb{Z}$ satisfies the strict two-sided bound $$-2^{u_{i+2} + d_{i+1}} < \Omega_i^{\mathrm{unres}} < 3^{u_{i-1} + 1}.$$ 3. **Out-Degree $\le 1$ of the Unrestricted Zero-Quotient Graph $\mathcal{G}_{\Omega=0}^{\mathrm{unres}}$:** On the zero-quotient locus $P_i = P_{i+1} = 0$ ($\Omega_i^{\mathrm{unres}} = 0$) for positive integer cycles, every $5$-tuple state $(u_{i-1}, d_{i-1}, u_i, d_i, u_{i+1}) \in \mathbb{Z}_{\ge 0}^3 \times \mathbb{Z}_{\ge 2}^2$ has **out-degree at most $1$** across the entire infinite alphabet $(d_{i+1}, u_{i+2}) \in \mathbb{Z}_{\ge 2} \times \mathbb{Z}_{\ge 0}$: setting $A_i = R_i^{\mathrm{unres}}$ uniquely determines $A_{i+1} = (3^{u_i+1} A_i + 2^{d_i-1} - 1)/2^{u_{i+1}+d_i}$ and $W_{i+1} := 3^{u_{i+1}+1} A_{i+1} - 1$, which uniquely forces $$d_{i+1} = v_2(W_{i+1}) + 1, \qquad u_{i+2} = v_2\!\left(\frac{W_{i+1}}{2^{d_{i+1} - 1}} + 1\right) - 1,$$ and a valid transition to $(u_i, d_i, u_{i+1}, d_{i+1}, u_{i+2})$ exists if and only if $A_{i+1} < 2^{u_{i+2}+d_{i+1}+1} 3^{u_i+1}$. Across all $8^3 \times 7^2 = 25{,}088$ initial $5$-tuples in $\{0..7\}^3 \times \{2..8\}^2$, $\mathcal{G}_{\Omega=0}^{\mathrm{unres}}$ contains **zero** non-trivial cycles (besides the fixed point $(0,2,0,2,0)$ corresponding to $x_0 = 1$) and every non-trivial path terminates in at most $18$ block steps. #### Proof of Theorem 4 1. In block $i$, $u_i$ consecutive steps with $a = 1$ map $z_i = 2^{u_i+1} A_i - 1$ to $2\cdot 3^{u_i} A_i - 1$, and the terminal step with $a = d_i \ge 2$ maps $2\cdot 3^{u_i} A_i - 1$ to $$z_{i+1} = \frac{3(2\cdot 3^{u_i} A_i - 1) + 1}{2^{d_i}} = \frac{2\cdot 3^{u_i+1} A_i - 2}{2^{d_i}} = \frac{3^{u_i+1} A_i - 1}{2^{d_i - 1}}.$$ Substituting $z_{i+1} = 2^{u_{i+1}+1} A_{i+1} - 1$ and multiplying by $2^{d_i - 1}$ yields $2^{u_{i+1}+d_i} A_{i+1} - 2^{d_i - 1} = 3^{u_i+1} A_i - 1$, i.e., $2^{u_{i+1}+d_i} A_{i+1} - 3^{u_i+1} A_i = 2^{d_i - 1} - 1$. 2. For positive integer cycles ($A_i \in \mathbb{Z}_{\ge 1}^{\mathrm{odd}}$), $A_{i+1} \equiv 1 \pmod 2$, so $3^{u_i+1} A_i = 2^{u_{i+1}+d_i} A_{i+1} - (2^{d_i-1} - 1) \equiv 2^{u_{i+1}+d_i} + 1 - 2^{d_i-1} \pmod{2^{u_{i+1}+d_i+1}}$. Shifting $i \mapsto i - 1$ gives $2^{u_i+d_{i-1}} A_i = 3^{u_{i-1}+1} A_{i-1} + 2^{d_{i-1}-1} - 1 \equiv 2^{d_{i-1}-1} - 1 \pmod{3^{u_{i-1}+1}}$. By CRT on $\gcd(2^{u_{i+1}+d_i+1}, 3^{u_{i-1}+1}) = 1$, $A_i \equiv R_i^{\mathrm{unres}} \pmod{M_i^{\mathrm{unres}}}$ with $1 \le R_i^{\mathrm{unres}} < M_i^{\mathrm{unres}}$ and $P_i = (A_i - R_i^{\mathrm{unres}})/M_i^{\mathrm{unres}} \in \mathbb{Z}_{\ge 0}$. Substituting $A_i = R_i^{\mathrm{unres}} + M_i^{\mathrm{unres}} P_i$ and $A_{i+1} = R_{i+1}^{\mathrm{unres}} + M_{i+1}^{\mathrm{unres}} P_{i+1}$ into $2^{u_{i+1}+d_i} A_{i+1} - 3^{u_i+1} A_i = 2^{d_i-1} - 1$ and dividing by $2^{u_{i+1}+d_i+1} 3^{u_i+1}$ gives $2^{u_{i+2}+d_{i+1}} P_{i+1} - 3^{u_{i-1}+1} P_i = \Omega_i^{\mathrm{unres}}$. Since $1 \le R_i^{\mathrm{unres}} < 2^{u_{i+1}+d_i+1} 3^{u_{i-1}+1}$ and $1 \le R_{i+1}^{\mathrm{unres}} < 2^{u_{i+2}+d_{i+1}+1} 3^{u_i+1}$, the numerator of $\Omega_i^{\mathrm{unres}}$ lies strictly between $-2^{u_{i+1}+d_i} \cdot 2^{u_{i+2}+d_{i+1}+1} 3^{u_i+1}$ and $3^{u_i+1} \cdot 2^{u_{i+1}+d_i+1} 3^{u_{i-1}+1}$, giving $-2^{u_{i+2}+d_{i+1}} < \Omega_i^{\mathrm{unres}} < 3^{u_{i-1}+1}$. 3. If $P_i = P_{i+1} = 0$, then $A_i = R_i^{\mathrm{unres}}$ is fixed by $(u_{i-1}, d_{i-1}, u_i, d_i, u_{i+1})$, which fixes $A_{i+1} = (3^{u_i+1} A_i + 2^{d_i-1} - 1)/2^{u_{i+1}+d_i}$. In turn, the block-$(i+1)$ equation $2^{d_{i+1}-1}(2^{u_{i+2}+1} A_{i+2} - 1) = 3^{u_{i+1}+1} A_{i+1} - 1 =: W_{i+1}$ with $A_{i+2}$ odd uniquely fixes $d_{i+1} - 1 = v_2(W_{i+1})$ and $u_{i+2} + 1 = v_2(W_{i+1}/2^{d_{i+1}-1} + 1)$. Finally, $P_{i+1} = 0$ holds iff $A_{i+1} = R_{i+1}^{\mathrm{unres}} < M_{i+1}^{\mathrm{unres}} = 2^{u_{i+2}+d_{i+1}+1} 3^{u_i+1}$ (since $A_{i+1}$ automatically satisfies both defining congruences of $R_{i+1}^{\mathrm{unres}}$). Exhaustive tracking of all $25{,}088$ initial $5$-tuples in `supplement/verify_paper6.py` confirms $0$ non-trivial cycles and maximum path length $18$. $\square$ --- ## 5. Unrestricted Forward Ballot Partition, $3\times$ Sharpening of the Hercher Method Floor, Universal Shifted Scale Recurrence, and Self-Contained Exclusion of $92$- and $93$-Block Collatz Cycles ### Theorem 5 (Unrestricted Forward Ballot Partition, Cyclic Shift Duality, Universal Shifted Scale Recurrence, and Exclusion of $m \le 93$ Cycles) 1. **Measure-$1$ Partition of $\mathbb{Z}_2^\times$ by the Unrestricted Coefficient-Stopping-Time Tree & Finite Strict-Descent Certificate:** In the forward $2$-adic tree over the unrestricted alphabet $\mathbb{Z}_{\ge 1}$, at each node $(j, S_j, P_j = 3^j)$ with $2^{S_r} < 3^r$ ($1 \le r \le j$) and $w_{j+1} = 3^j x_0 + c_j$, let $d_0(j, S_j) := \lfloor (j+1)\log_2 3 \rfloor + 1 - S_j \ge 1$. For each $1 \le d < d_0(j, S_j)$, the non-contracting branch $a_j = d$ ($2^{S_j + d} < 3^{j+1}$) has $2$-adic Haar measure $2^{-(S_j + d)}$ in $\mathbb{Z}_2^\times$, while the combined coefficient-contracting terminal branch $a_j \ge d_0(j, S_j)$ ($2^{S_{j+1}} > 3^{j+1}$, corresponding to Terras's coefficient stopping time $\tau(x_0) = j + 1$ [3, 13]) has measure $2^{-(S_j + d_0 - 1)}$, partitioning $\mathbb{Z}_2^\times$ with total pruned + surviving measure identically equal to $1$ at every truncation depth $M \ge 1$. As distinguished by Rozier and Terracol [13, Definition 1.2], coefficient contraction $2^{S_{j+1}} > 3^{j+1}$ implies strict trajectory descent $x_{j+1}(x_0) < x_0$ if and only if $(2^{S_{j+1}} - 3^{j+1})x_0 > c_{j+1}$; if this strict inequality holds at the minimal representative $r_0 \in [1, 2^{S_{j+1}+1})$ of a pruned branch, then $x_{j+1}(x_0) - x_0 \le x_{j+1}(r_0) - r_0 < 0$ holds for all $x_0 \equiv r_0 \pmod{2^{S_{j+1}+1}}$ with $x_0 \ge r_0$. At the truncation depth $M = 18$ ($14{,}990$ surviving leaves), finite verification in `verify_paper6.py` certifies that all $1{,}750$ non-trivial pruned branches ($r_0 > 1$, i.e., excluding the trivial cycle $r_0 = 1$ at $j = 0, a_0 = 2$) satisfy the strict descent inequality $x_{j+1}(r_0) < r_0$ (`cur_w < cur_r`), and likewise all $93$ pruned branches of the $\{1,2\}$ Forward Ballot filter at $M_0 = 20$ ($8{,}207$ surviving leaves) satisfy $x_{j+1}(r_0) < r_0$. 2. **Cyclic Shift Duality Identity:** For any $m$-block cycle with local minima $n_i = 2^{k_i} A_i - 1$ ($k_i = u_i + 1 \ge 1$), peak divisions $l_i = d_i - 1 \ge 1$, and logarithmic excess $\Lambda = S\ln 2 - K\ln 3 = \sum_{i=1}^m T(n_i)$ ($K = \sum_{i=1}^m k_i$, $L = \sum_{i=1}^m l_i$, $S = K + L$), the block contributions satisfy $$\frac{1 - (2/3)^{k_i}}{n_i + 1} < T(n_i) < \frac{1 - (2/3)^{k_i}}{n_i}, \qquad \frac{(2/3)^{k_i}}{n_i + 1} = \frac{2^{-l_i}}{n_{i+1} + 2^{-l_i}},$$ and summing around the cycle yields the exact dual representation $$\sum_{i=1}^m \frac{1 - (2/3)^{k_i}}{n_i + 1} = \sum_{i=1}^m \frac{1 - 2^{-l_{i-1}}}{n_i + 1} - \sum_{i=1}^m \frac{2^{-l_i}(1 - 2^{-l_i})}{(n_{i+1} + 1)(n_{i+1} + 2^{-l_i})}.$$ 3. **$3\times$ Sharpening of the Hercher Method Floor, Universal Shifted Scale Recurrence, and Non-Circular Sublevel Bound:** - **(a) $K$-Step Geometric Floor ($3\times$ / $+\log_2 3$ bits):** Along any positive Syracuse trajectory starting at $x_0 = n_{\min}$, every step $s \ge 0$ satisfies $x_{s+1} + 1 = \frac{3x_s + 1}{2^{a_s}} + 1 \le \frac{3}{2}(x_s + 1)$ and thus $x_s + 1 \le (3/2)^s (n_{\min} + 1)$. Consequently, $$\Lambda = \sum_{s=0}^{K-1} \ln\!\left(1 + \frac{1}{3x_s}\right) > \sum_{s=0}^{K-1} \frac{1}{3x_s + 3} \ge \frac{1 - (2/3)^K}{n_{\min} + 1} \implies n_{\min} > \frac{1 - (2/3)^K}{\Lambda} - 1,$$ improving the $1$-step method floor $n_{\min} > \frac{1}{3\Lambda} - 1$ by a factor of $3$ ($+\log_2 3 \approx +1.58496$ bits). - **(b) Exact Universal Peak-Division Scale Recurrence & Non-Circular Sublevel Bound:** For any rising Syracuse block from $n_j = 2^{k_j} A_j - 1$ to $n_{j+1} = (3^{k_j} A_j - 1)/2^{l_j}$ ($l_j \ge 1$, $A_j \ge 1$ odd), since $3^{k_j} A_j = 2^{l_j} n_{j+1} + 1 \ge 2 n_{j+1} + 1$ and $3^{k_j} A_j \le (n_j + 1)^\delta = 2^{\delta s_j}$ ($s_j := \log_2(n_j + 1)$, $\delta = \log_2 3$), we have $$n_{j+1} + 1 = \frac{3^{k_j} A_j + 2^{l_j} - 1}{2^{l_j}} \le \frac{3^{k_j} A_j + 1}{2} = \frac{1}{2} 3^{k_j} A_j \left(1 + \frac{1}{3^{k_j} A_j}\right) \le 2^{\delta s_j - 1}\left(1 + \frac{1}{2 n_{j+1} + 1}\right),$$ with exact equality iff $A_j = 1$ and $l_j = 1$ (e.g., at $n_j = 7 \to n_{j+1} = 13$). Thus along any cycle with minimum $n_{\min}$, every block $j$ satisfies $$s_{j+1} \le \delta s_j - \bigl(1 - \eta(n_{\min})\bigr) \iff s_{j+1} - \sigma(n_{\min}) \le \delta\bigl(s_j - \sigma(n_{\min})\bigr), \qquad \sigma(n_{\min}) := \frac{1 - \eta(n_{\min})}{\delta - 1},$$ where $\eta(n) := \log_2(1 + \frac{1}{2n+1})$. Let $s_{(1)} \le s_{(2)} \le \dots \le s_{(m)}$ be the sorted block scales. For any $r \in \{1, \dots, m\}$, the $r-1$ smallest blocks satisfy $s_{(q)} - \sigma(n_{\min}) \le s_{(r)} - \sigma(n_{\min})$ ($1 \le q \le r-1$), while the $m - r + 1$ blocks with $s_i \ge s_{(r)}$ decompose along the cycle into $d \ge 1$ chains of lengths $c_1 + \dots + c_d = m - r + 1$ entering from elements $\le s_{(r)}$, so their shifted scales are bounded by $\sum_{u=1}^d \sum_{t=0}^{c_u - 1} \delta^t (s_{(r)} - \sigma(n_{\min})) \le \sum_{t=0}^{m-r} \delta^t (s_{(r)} - \sigma(n_{\min}))$ (since $\delta^{a+b} \ge \delta^a + \delta^b - 1$ for $\delta > 1$). Summing all $m$ blocks yields the partition-independent, non-circular lower bound $$s_{(r)} \ge \sigma(n_{\min}) + \frac{K - m\sigma(n_{\min})}{G(m, r)}, \qquad G(m, r) := \frac{\delta^{m-r+1} - 1}{\delta - 1} + r - 1 \qquad (1 \le r \le m).$$ - **(c) Self-Contained Continued-Fraction Candidate Reduction ($m \le 93$):** Partition $(7\times 10^{11}, K_{\mathrm{up}}(m)]$ ($K_{\mathrm{up}}(m) := \lceil 1.4784\,m\,\delta^m \rceil$ [6, Lemma 2]) into consecutive convergent intervals $[q_n, q_{n+1}]$ of $\delta = \log_2 3$ ($23 \le n \le 44$). Because $|p_n q_{n+1} - q_n p_{n+1}| = 1$, $(q_n, p_n)$ and $(q_{n+1}, p_{n+1})$ form a unimodular $\mathbb{Z}$-basis of $\mathbb{Z}^2$, so every $(K, P) \in \mathbb{Z}^2$ is uniquely $(K, P) = c_n(q_n, p_n) + c_{n+1}(q_{n+1}, p_{n+1})$ with $\Lambda = c_n \lambda_n + c_{n+1} \lambda_{n+1}$ and $\lambda_n \lambda_{n+1} < 0$. On $K \in [q_n, \min(q_{n+1}, K_{\mathrm{up}})]$, if $c_n, c_{n+1} \ge 0$, then $q_n, q_{n+1} > 0$ forces $0 \le c_n \le \min(a_{n+1}, \lfloor K_{\mathrm{up}}/q_n\rfloor)$ and $c_{n+1} \in \{0, 1\}$; if $c_n c_{n+1} < 0$, then $c_n \lambda_n$ and $c_{n+1} \lambda_{n+1}$ have the same sign, so $|\Lambda| = |c_n|\,|\lambda_n| + |c_{n+1}|\,|\lambda_{n+1}| < B_{\max}(m) := \frac{97m}{162 X_0}$ ($X_0 = 2^{71}$ [11]) forces $1 \le |c_n| \le \lfloor B_{\max}(m)/|\lambda_n| \rfloor$ and $1 \le |c_{n+1}| \le \lfloor B_{\max}(m)/|\lambda_{n+1}| \rfloor$. Exhaustive evaluation over these complete coefficient ranges under $0 < \Lambda < S_{\mathrm{Hercher}}(K, m)/3$ (`verify_paper6.py`) certifies **$0$ surviving $(K, L)$ pairs for $m \le 91$**, the **unique surviving pair** $(K_{92}, L_{92}) = (q_{43}, p_{43} - q_{43}) = (205{,}632{,}218{,}873{,}398{,}596{,}256,\; 120{,}287{,}136{,}981{,}023{,}372{,}109)$ for $m = 92$, and **exactly two surviving pairs** $(K_{92}, L_{92})$ and $(2K_{92}, 2L_{92})$ for $m = 93$. - **(d) Directed Outward-Rounded Enclosures & Pure-Integer Step Enclosures (`u/lyapunov_weaver` [12]):** Using $120$-term rational Taylor series with directed rounding (`ROUND_FLOOR` / `ROUND_CEILING` in `verify_paper6.py`) to enclose $\ln 2, \ln 3, \delta, \Lambda$, and $2^{s_{(r)}} \ge 2^{\lfloor s_{(r)} \rfloor} \sum_{j=0}^{44} \frac{(\{s_{(r)}\}\ln 2)^j}{j!}$, we convert all window endpoints to certified integers before modular enumeration. For any block step $n \mapsto n' = (3^k A - 1)/2^l$ from $n = 2^k A - 1 \le N$, let $A_{\max}(N, k)$ be the largest odd integer $\le \lfloor (N+1)/2^k \rfloor$; then $n' \le \mathrm{endpoint\_hi}(N, k, l) := \lfloor (3^k A_{\max}(N, k) - 1)/2^l \rfloor$, and reaching $n' \ge L$ from $n \le N$ requires $3^k(N+1) \ge 2^k(2L + 1)$ (`least_possible_rise`). Because the Syracuse map is deterministic and no non-trivial cycle has $d \le 91$ blocks, all $m$ local minima of any $92$- or $93$-block cycle are pairwise distinct: $n_{(1)} < n_{(2)} < \dots < n_{(m)}$. 4. **Unconditional Pure-Integer Exclusion of $92$-Block Collatz Cycles ($m = 92$):** For $(K_{92}, L_{92})$ ($\Lambda_{92} \in [8.900755224565\times 10^{-23},\; 8.900755224566\times 10^{-23}]$): - By Part 3(a), $n_1 = n_{(1)} \ge \mathrm{LO}_{92} := 11{,}235{,}001{,}691{,}093{,}021{,}257{,}300$ ($\sigma(n_1) \ge \sigma_{92,\mathrm{lo}} > 1.70951129135145477697608$). - By Part 3(b) at $r = 2, 3$, outward interval evaluation certifies $s_{(2)} \ge 77.304143685$ ($n_{(2)} \ge L_{2,92} := 186{,}580{,}408{,}057{,}834{,}151{,}310{,}096$) and $s_{(3)} \ge 121.524168859$ ($n_{(3)} \ge L_{3,92} := 3{,}823{,}138{,}529{,}586{,}028{,}606{,}781{,}685{,}560{,}173{,}673{,}738$). Bounding all $92$ minima explicitly via $\Lambda_{92} < \frac{1}{n_1} + \frac{1}{L_{2,92}} + \frac{90}{L_{3,92}}$ certifies $n_1 \le \mathrm{HI}_{92} := 11{,}954{,}868{,}071{,}484{,}238{,}211{,}756 \approx 2^{73.340016}$ (collapsing the window width by $21.2\times$). - For every $1 \le k_1 \le 74$, $\mathrm{endpoint\_hi}(\mathrm{HI}_{92}, k_1, 1) < L_{3,92} \le n_{(3)}$, so $n_2 < n_{(3)}$, forcing $n_2 = n_{(2)} \ge L_{2,92}$ and $n_3 \ge n_{(3)} \ge L_{3,92}$. - Exhaustive pure-integer enumeration of all $(k_1, l_1, k_2)$ with $\mathrm{endpoint\_hi}(\mathrm{HI}_{92}, k_1, l_1) \ge L_{2,92}$ and $3^{k_2}(n_{2,\mathrm{hi}} + 1) \ge 2^{k_2}(2 L_{3,92} + 1)$ yields **$43{,}971$ triples**, and the one-step Coupled Bézout congruence $A_1 \equiv 3^{-k_1}(1 - 2^{l_1} + 2^{l_1+k_2}) \pmod{2^{l_1+k_2+1}}$ gives $$\min_{(k_1, l_1, k_2)} \left(2^{k_1} A_{1,\min}(k_1, l_1, k_2) - 1\right) = 10{,}171{,}858{,}727{,}578{,}545{,}185{,}357{,}823 \approx 2^{83.072786} > \mathrm{HI}_{92},$$ attained at $(k_1, l_1, k_2) = (35, 1, 52)$ (**$0$ survivors**). 5. **Two-Step Coupled Bézout Identity and Pure-Integer Exclusion of $93$-Block Collatz Cycles ($m \le 93$):** - **(a) Pair $(2K_{92}, 2L_{92})$ at $m = 93$:** Here $\Lambda = 2\Lambda_{92}$, giving its own dedicated lower bound $n_1 \ge \mathrm{LO}_{93}^{(2)} := 5{,}617{,}500{,}845{,}546{,}510{,}628{,}649$ and shift $\sigma_{93,\mathrm{lo}}^{(2)} > 1.70951129135145477697597$. Outward evaluation of Part 3(b) with $\sigma_{93,\mathrm{lo}}^{(2)}$ certifies $s_{(2)} \ge 97.0993168$ ($n_{(2)} \ge L_{2,93}^{(2)} := 169{,}748{,}887{,}054{,}766{,}119{,}338{,}550{,}345{,}647$) and $s_{(3)} \ge 152.8987757$ ($n_{(3)} \ge L_{3,93}^{(2)} := 1{,}064{,}431{,}946{,}643{,}019{,}121{,}037{,}518{,}157{,}608{,}632{,}805{,}616{,}345{,}5202$). Explicit tail bounding $\Lambda < \frac{1}{n_1} + \frac{1}{L_{2,93}^{(2)}} + \frac{91}{L_{3,93}^{(2)}}$ certifies $n_1 \le \mathrm{HI}_{93}^{(2)} := 5{,}617{,}501{,}031{,}446{,}502{,}914{,}481$. Since $\mathrm{endpoint\_hi}(\mathrm{HI}_{93}^{(2)}, k_1, 1) < L_{3,93}^{(2)}$ for all $1 \le k_1 \le 73$, $n_2 = n_{(2)} \ge L_{2,93}^{(2)}$ and $n_3 \ge n_{(3)} \ge L_{3,93}^{(2)}$. Pure-integer Coupled Bézout checking across all **$3{,}752$ triples** $(k_1, l_1, k_2)$ yields **$0$ survivors** ($\min n_1 = 519{,}299{,}313{,}705{,}111{,}194{,}890{,}542{,}344{,}385{,}436{,}225{,}372{,}159 > \mathrm{HI}_{93}^{(2)}$ at $(64, 1, 79)$). - **(b) Pair $(K_{92}, L_{92})$ at $m = 93$:** Outward evaluation of Part 3(b) at $r = 3, 4, 5$ with $\sigma_{92,\mathrm{lo}}$ certifies $n_{(3)} \ge L_{3,93} := 186{,}580{,}408{,}057{,}834{,}147{,}634{,}785$, $n_{(4)} \ge L_{4,93} := 3{,}823{,}138{,}529{,}586{,}028{,}419{,}141{,}028{,}692{,}252{,}800{,}567$, and $n_{(5)} \ge L_{5,93} > 4.94827\times 10^{57}$. Using $n_{(2)} > n_1$, explicit tail bounding $\Lambda_{92} < \frac{2}{n_1} + \frac{1}{L_{3,93}} + \frac{1}{L_{4,93}} + \frac{89}{L_{5,93}}$ certifies $n_1 \le \mathrm{HI}_{93} := 23{,}909{,}736{,}142{,}961{,}822{,}350{,}321 \approx 2^{74.340016}$. For every $1 \le k_1 \le 75$, $\mathrm{endpoint\_hi}(\mathrm{HI}_{93}, k_1, 1) < L_{4,93} \le n_{(4)}$, so $n_2 \in \{n_{(2)}, n_{(3)}\}$. - **Case A ($n_3 \ge n_{(4)} \ge L_{4,93}$):** Pure-integer Coupled Bézout checking across all **$49{,}609$ triples** $(k_1, l_1, k_2)$ with $n_{2,\mathrm{hi}} \ge \mathrm{LO}_{93}$ and $3^{k_2}(n_{2,\mathrm{hi}}+1) \ge 2^{k_2}(2 L_{4,93}+1)$ yields **$0$ survivors** ($\min n_1 = 2{,}410{,}065{,}382{,}792{,}568{,}026{,}169{,}343 \approx 2^{80.995} > \mathrm{HI}_{93}$ at $(23, 1, 62)$). - **Case B ($n_3 < n_{(4)}$, so $\{n_1, n_2, n_3\} = \{n_{(1)}, n_{(2)}, n_{(3)}\}$ and $n_4 \ge n_{(4)} \ge L_{4,93}$):** Since $\mathrm{endpoint\_hi}(L_{3,93}, k_3, 1) < L_{4,93}$ for all $1 \le k_3 \le 78$, reaching $n_4 \ge L_{4,93}$ requires $n_3 \ge L_{3,93}$ (so $n_2 = n_{(2)}$ and $n_3 = n_{(3)} \ge L_{3,93}$). Eliminating $A_2$ between $3^{k_1} A_1 = 2^{l_1+k_2} A_2 - (2^{l_1} - 1)$ and $3^{k_2} A_2 = 2^{l_2+k_3} A_3 - (2^{l_2} - 1)$ yields the **Two-Step Coupled Bézout Identity**: $$3^{k_1 + k_2} A_1 = 2^{l_1 + k_2 + l_2 + k_3} A_3 - 2^{l_1 + k_2}(2^{l_2} - 1) - 3^{k_2}(2^{l_1} - 1).$$ Three-stage pure-integer $2$-adic pruning [12] modulo $2^{M_1+1}$ ($M_1 = l_1+k_2$: **$25{,}736$ surviving triples**), modulo $2^{M_2+1}$ ($M_2 = M_1+l_2$: **$234{,}068$ surviving quadruples**), and modulo $2^{M_2+k_3+1}$ across all **$7{,}713{,}429$ quintuples** $(k_1, l_1, k_2, l_2, k_3)$ yields **$0$ survivors**: both the minimum Stage-3 bitmask lower bound $\min 2^{k_1}(X \bmod 2^{M_2+k_{3,\min}}) - 1$ (attained at $(k_1, l_1, k_2, l_2, k_{3,\min}) = (8, 1, 13, 1, 65)$, where the full congruence at $k_3 = 65$ gives $n_1 = 310{,}206{,}356{,}439{,}444{,}899{,}438{,}632{,}703$) and the exact full-congruence minimum over all $7{,}713{,}429$ quintuples (attained at $(k_1, l_1, k_2, l_2, k_3) = (8, 1, 13, 1, 67)$, where bit $M_2 + 67$ of $X$ is $1$ and bits $M_2 + 65, M_2 + 66$ are $0$) equal $$\min_{(k_1, l_1, k_2, l_2, k_3)} \left(2^{k_1} A_{1,\min} - 1\right) = 721{,}346{,}618{,}099{,}830{,}713{,}851{,}647 \approx 2^{79.255039} > \mathrm{HI}_{93}.$$ Thus **no non-trivial Collatz cycle has $m \le 93$ local minima**. #### Proof of Theorem 5 1. At each surviving node $(j, S_j)$, $w_{j+1} = 3^j x_0 + c_j \equiv 0 \pmod{2^{S_j}}$ has odd quotient $y_j = w_{j+1}/2^{S_j}$ by Theorem 1(2). Writing $x_0 = r_0 + 2^{S_j + 1} t$, the next numerator $3^{j+1} x_0 + c_{j+1}$ divides by $2^{S_j + d}$ ($1 \le d < d_0$) on exactly one parity bit at each level, with measure $2^{-(S_j + d)}$. Since $\sum_{d=1}^{d_0-1} 2^{-d} + 2^{-(d_0-1)} = 1$, measure is preserved inductively. On any pruned branch $a_j \ge d_0$, $2^{S_{j+1}} > 3^{j+1}$ implies that $x_{j+1}(x_0) - x_0 = \frac{c_{j+1} - (2^{S_{j+1}} - 3^{j+1})x_0}{2^{S_{j+1}}}$ is strictly decreasing in $x_0$, so verifying $x_{j+1}(r_0) < r_0$ (`cur_w < cur_r`) at the minimal representative $r_0 > 1$ of each of the $1{,}750$ pruned branches at $M = 18$ (and $93$ pruned branches at $M_0 = 20$) certifies strict descent for all $x_0 \ge r_0$ in those residue classes. 2. Inside block $i$, the $k_i$ odd steps are $x_{i,r} = (3/2)^r (n_i + 1) - 1$ for $0 \le r \le k_i - 1$. Using $\frac{1}{3x + 3} < \ln(1 + \frac{1}{3x}) < \frac{1}{3x}$ and $x_{i,r} \ge (3/2)^r n_i$ gives $\frac{1-(2/3)^{k_i}}{n_i+1} < T(n_i) < \frac{1-(2/3)^{k_i}}{n_i}$. Since $2^{l_i} n_{i+1} + 1 = 3^{k_i} A_i = (3/2)^{k_i}(n_i + 1)$, dividing by $2^{l_i}$ gives $n_{i+1} + 2^{-l_i} = 2^{-l_i}(3/2)^{k_i}(n_i + 1)$, so $\frac{(2/3)^{k_i}}{n_i + 1} = \frac{2^{-l_i}}{n_{i+1} + 2^{-l_i}}$. Decomposing this fraction as $$\frac{2^{-l_i}}{n_{i+1} + 2^{-l_i}} = \frac{2^{-l_i}}{n_{i+1} + 1} + \frac{2^{-l_i}(1 - 2^{-l_i})}{(n_{i+1} + 1)(n_{i+1} + 2^{-l_i})},$$ subtracting both terms from $\frac{1}{n_i + 1}$, and re-indexing $\sum_{i=1}^m \frac{2^{-l_i}}{n_{i+1}+1} = \sum_{i=1}^m \frac{2^{-l_{i-1}}}{n_i+1}$ around the cycle yields the exact identity with a minus sign before the second sum. 3. Parts 3(a)–(d), 4, and 5 are proved in the theorem body and verified end-to-end in `supplement/verify_paper6.py` in $3.32$ seconds. $\square$ ![Figure 2: Extinction curve of the unrestricted zero-quotient graph G_{Omega=0}^unres on 25,088 initial 5-tuples, and the 21.2x collapse and Coupled Bezout exclusion of the lone m = 92 candidate window.](supplement/figure-2.png) --- ## 6. Limitations, Scope, and Open Problems 1. **Scope of the Backward $3$-Adic Sieve (Theorems 2–3):** The box-counting dimension collapse $\overline{\dim}_{B,3}(\Sigma_{\{1,2\}}^-) \approx 0.6177188$ and the $35.51$-second bilateral tree-sieve certificate ($x_{\min} > 4\cdot 3^{39}$) apply to cycles over the restricted $\{1, 2\}$ valuation alphabet (where unrestricted cycles already satisfy $K \ge q_{23} = 137{,}528{,}045{,}312$ [6, 11]). On the unrestricted alphabet $\mathbb{Z}_{\ge 1}$, the Mod-$3$ Complete-Residue Splitting Law (Theorem 1(4)) still restricts admissible backward exponents to a single parity of $b$, but large backward jumps $b \ge 3$ expand the tree branching unless paired with an upper bound on $\max a_j$. 2. **Scope of the Unrestricted Zero-Quotient Graph (Theorem 4):** Theorem 4 proves out-degree $\le 1$ and certifies $0$ non-trivial cycles on the zero-quotient locus $P_i = 0$ ($\Omega_i^{\mathrm{unres}} = 0$) for $\{0..7\}^3 \times \{2..8\}^2$. 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