# Bidirectional Residue-Class Translation: $2^k$ Cylinders and $3^u$ Progressions ## ——From the Collatz Local Affine Atlas to Exact Inverse Fibers, Odd Skeletons, and Double Helix Reconstruction **English Title:** *Bidirectional Residue-Class Translation: From $2^k$ Cylinders to $3^u$ Progressions in the Collatz Local Affine Atlas* **Author:** Neo.K **Institution:** EveMiss Technology Co., Ltd. (EveMissLab) **Series:** Collatz Operation Translation Series — Paper 04 **Version:** v0.1 **Date:** 2026-08-10 --- ## Abstract Paper 03 has proven that, for the modified Collatz map $$ T(n)= \begin{cases} n/2,&n\equiv0\pmod2,\\[2mm] (3n+1)/2,&n\equiv1\pmod2, \end{cases} $$ each admissible parity word $w$ of length $k$ corresponds to a unique source residue cylinder $$ \Omega_w = (r_w+2^k\mathbb Z)\cap\mathbb Z_{>0}, $$ and if $u=u(w)$, $m_w=T^k(r_w)$, then $$ \boxed{ T^k(r_w+2^ka) = m_w+3^ua. } $$ This paper elevates this unidirectional formula into a complete **bidirectional exact residue-class translation**. On the all-integer affine extension, define $$ \mathcal C_w=r_w+2^k\mathbb Z, $$ $$ \mathcal P_w=m_w+3^u\mathbb Z. $$ This paper proves: $$ \boxed{ F_w:\mathcal C_w\overset{\sim}{\longrightarrow}\mathcal P_w } $$ is a bijection, and its inverse is $$ \boxed{ F_w^{-1}(y) = r_w + 2^k\frac{y-m_w}{3^u}, \qquad y\equiv m_w\pmod{3^u}. } $$ Therefore, fixed-word Collatz transport can be understood as: $$ \boxed{ r_w+2^k\mathbb Z \;\longleftrightarrow\; m_w+3^u\mathbb Z, } $$ where both sides share the same exact quotient coordinate $$ a. $$ The source side uses $2^k$ as the lattice spacing, and the target side uses $3^u$ as the lattice spacing; in the chart coordinate, both forward and inverse are simply $$ a\leftrightarrow a. $$ This paper refers to this structure as **Bidirectional Residue Transport**. In the positive-integer Collatz domain, if we take the canonical representative $$ 0\le r_w<2^k, $$ then the lower bound for the valid quotient coordinate is $$ a_{\min}(w) = \begin{cases} 1,&r_w=0,\\ 0,&r_w>0. \end{cases} $$ So the true positive source/image are $$ \Omega_w = \{ r_w+2^ka:a\ge a_{\min}(w) \}, $$ $$ \Gamma_w = \{ m_w+3^ua:a\ge a_{\min}(w) \}. $$ The inverse in a fixed chart thus remains exact, single-valued, and lossless. This paper then reorganizes the author's early "double helix" research into two complementary levels: 1. **modified-map finite-word inverse transport**: composed of $2^k$ source cylinders and $3^u$ target progressions; 2. **accelerated odd-map inverse fibers**: for $$ S(n)=\frac{3n+1}{2^{v_2(3n+1)}} $$ define $$ R_\kappa(t)=\frac{2^\kappa t-1}{3}. $$ If $t$ is a positive odd integer, then $R_\kappa(t)$ is a valid odd predecessor of $t$ if and only if $$ \boxed{ 2^\kappa t\equiv1\pmod3. } $$ Therefore: - If $t\equiv1\pmod3$, the valid $\kappa$ must be even; - If $t\equiv2\pmod3$, the valid $\kappa$ must be odd; - If $3\mid t$, no accelerated odd predecessor exists. Specifically, for the terminal state $t=1$: $$ \kappa=2j $$ gives $$ \boxed{ R_{2j}(1) = \frac{4^j-1}{3}. } $$ Therefore, the $M_j=\frac{4^j-1}{3}$ from the author's previous research no longer needs to be interpreted as an independent "mysterious highway series"; it is precisely the valuation-labelled inverse fiber of the terminal state $1$ in the accelerated odd map. The number $5$ is simply $$ R_4(1)=5, $$ and $5\cdot2^q$ is its even $2$-ray in the ordinary Collatz inverse structure. This paper also reorganizes the odd skeleton. Any positive integer can be uniquely represented as $$ n=2^{v_2(n)}\operatorname{oddcore}(n), $$ so the ordinary Collatz inverse coverage can be decomposed into: $$ \boxed{ \text{odd inverse skeleton} + \text{even }2\text{-rays}. } $$ On the accelerated odd map, all nodes are odd states; the inverse fibers $R_\kappa(t)$ directly describe the edge labels of the odd skeleton. This paper particularly emphasizes that three different "inverses" must not be confused: - fixed-word local inverse; - inverse-tree predecessor relation; - global inverse coverage. The first two can be exact; the third remains equivalent to the global Collatz problem. In other words: $$ \boxed{ \text{local invertibility} \not\Rightarrow \text{global inverse-tree coverage}. } $$ This paper thus redefines the early "double helix" from a visual/graph-theoretic approach into a rigorous arithmetic framework: $$ \boxed{ \text{forward }2^k\text{-cylinder refinement} \quad\leftrightarrow\quad \text{backward }3^u\text{-progression / valuation fiber}. } $$ **Keywords:** Collatz conjecture, inverse iteration, residue class, $2$-adic cylinder, $3$-progression, accelerated Collatz, valuation fiber, odd skeleton, bidirectional transport, exact recovery --- # 1. Problem: Can the forward transport of Paper 03 be truly reversed? Paper 03 has obtained: $$ \boxed{ T^k(r_w+2^ka) = m_w+3^ua. } $$ This equation already implies: - the source quotient label is $a$; - the target quotient label is also $a$. So the most natural question is: > If only the target $y$ and the fixed chart $w$ are known, can the source $n$ be exactly recovered? The answer is affirmative. --- # 2. All-Integer Affine Extension To avoid interference from positive-integer boundaries on the algebraic structure, first consider on: $$ \mathbb Z $$ the fixed-word affine map: $$ F_w(x) = \frac{3^ux+b_w}{2^k}. $$ Define the source cylinder: $$ \boxed{ \mathcal C_w = r_w+2^k\mathbb Z. } $$ From Paper 03: $$ r_w \equiv -b_w3^{-u} \pmod{2^k}, $$ so for: $$ x\in\mathcal C_w, $$ $F_w(x)$ must be an integer. --- # 3. Target Progression Let: $$ m_w=F_w(r_w). $$ For: $$ x=r_w+2^ka, $$ we have: $$ F_w(x) = m_w+3^ua. $$ Therefore, the image is exactly: $$ \boxed{ \mathcal P_w = m_w+3^u\mathbb Z. } $$ So: $$ \boxed{ F_w(\mathcal C_w)=\mathcal P_w. } $$ --- # 4. Bidirectional Residue Transport Theorem ## Theorem 4.1 For any finite parity word $w$: $$ F_w: \mathcal C_w \to \mathcal P_w $$ is a bijection. Its inverse is: $$ \boxed{ F_w^{-1}(y) = r_w + 2^k \frac{y-m_w}{3^u}. } $$ ### Proof If: $$ y\in\mathcal P_w, $$ then there uniquely exists: $$ a\in\mathbb Z $$ such that: $$ y=m_w+3^ua. $$ Define: $$ x=r_w+2^ka. $$ Then: $$ F_w(x) = m_w+3^ua = y. $$ Uniqueness comes from $3^u\neq0$ and the uniqueness of the source coordinate $a$. This completes the proof. --- # 5. Inverse Legality Congruence From: $$ y=m_w+3^ua, $$ we obtain: $$ \boxed{ y\equiv m_w\pmod{3^u}. } $$ Conversely, if: $$ y\equiv m_w\pmod{3^u}, $$ then: $$ a=\frac{y-m_w}{3^u}\in\mathbb Z $$ and gives the unique source: $$ x=r_w+2^ka. $$ Therefore, the fixed-word inverse legality is exactly: $$ \boxed{ y\equiv m_w\pmod{3^u}. } $$ --- # 6. Dual Residue-Class Structure of Source and Target source: $$ \boxed{ x\equiv r_w\pmod{2^k}. } $$ target: $$ \boxed{ y\equiv m_w\pmod{3^u}. } $$ So a single chart simultaneously possesses: $$ \boxed{ (2^k,r_w) } $$ and: $$ \boxed{ (3^u,m_w) } $$ two sets of residual metadata. Which can be denoted as: $$ \boxed{ \mathcal R(w) = (2^k,r_w;3^u,m_w). } $$ --- # 7. Quotient Coordinate Conservation source coordinate: $$ a = \frac{x-r_w}{2^k}. $$ target coordinate: $$ a = \frac{y-m_w}{3^u}. $$ Therefore: $$ \boxed{ \frac{x-r_w}{2^k} = \frac{y-m_w}{3^u}. } $$ This is the core conservation equation of fixed-word transport. It is not that "the value $x$ is invariant". What is truly invariant is: $$ \boxed{ \text{chart quotient label }a. } $$ --- # 8. Cross-Multiplied Exact Relation The previous equation is equivalent to: $$ \boxed{ 3^u(x-r_w) = 2^k(y-m_w). } $$ This equation requires no division at all. Therefore, in an exact-integer backend, it can directly serve as a: $$ \boxed{ \text{transport certificate}. } $$ It also provides a bidirectional consistency check that does not rely on floating-point numbers. --- # 9. Positive-Integer Domain The true Collatz domain of Paper 03 is: $$ \Omega_w = \mathcal C_w\cap\mathbb Z_{>0}. $$ Take the canonical: $$ 0\le r_w<2^k. $$ If: $$ r_w>0, $$ then: $$ a\ge0 $$ guarantees: $$ r_w+2^ka>0. $$ If: $$ r_w=0, $$ then it must be that: $$ a\ge1. $$ Therefore, define: $$ \boxed{ a_{\min}(w) = \begin{cases} 1,&r_w=0,\\ 0,&r_w>0. \end{cases} } $$ --- # 10. Positive Source / Image Theorem So: $$ \boxed{ \Omega_w = \{ r_w+2^ka: a\ge a_{\min}(w) \}. } $$ Its positive-integer image is: $$ \boxed{ \Gamma_w = \{ m_w+3^ua: a\ge a_{\min}(w) \}. } $$ And: $$ \boxed{ T^k: \Omega_w \overset{\sim}{\longrightarrow} \Gamma_w } $$ remains a bijection. Therefore, on a fixed chart: $$ \boxed{ \text{positive-domain exact recovery} } $$ holds. --- # 11. Example: $w=U$ From Paper 03: $$ r_U=1, $$ $$ m_U=2, $$ $$ k=1, \qquad u=1. $$ Therefore: $$ \boxed{ 1+2a \longleftrightarrow 2+3a. } $$ inverse: $$ \boxed{ n = 1+2\frac{y-2}{3}. } $$ Valid target: $$ \boxed{ y\equiv2\pmod3. } $$ --- # 12. Single-step inverse branches of the modified Collatz map From: $$ T(n)=y. $$ There are two possibilities. ## even predecessor If $n$ is even: $$ n/2=y $$ so: $$ \boxed{ n=2y. } $$ This predecessor exists for all: $$ y>0 $$ --- ## odd predecessor If $n$ is odd: $$ \frac{3n+1}{2}=y. $$ So: $$ \boxed{ n=\frac{2y-1}{3}. } $$ To be an integer: $$ 2y-1\equiv0\pmod3. $$ That is: $$ \boxed{ y\equiv2\pmod3. } $$ In this case: $$ y=3q+2 $$ gives: $$ n=2q+1, $$ which is automatically odd. Therefore, the modified-map inverse relation is: $$ \boxed{ T^{-1}(y) = \{2y\} \cup \left\{ \frac{2y-1}{3} : y\equiv2\pmod3 \right\}. } $$ --- # 13. Relationship with the original Collatz inverse branch original map odd step: $$ n\mapsto3n+1. $$ The odd predecessor for target $y$ is: $$ n=\frac{y-1}{3}. $$ Its validity condition is: $$ \boxed{ y\equiv4\pmod6. } $$ The modified map merges the inevitable division by 2 after the odd step, so the target is changed to: $$ y_{\mathrm{mod}} = \frac{y_{\mathrm{orig}}}{2}. $$ Therefore: $$ y_{\mathrm{orig}}\equiv4\pmod6 $$ is exactly equivalent to: $$ y_{\mathrm{mod}}\equiv2\pmod3. $$ So the two inverse conditions are simply the same arithmetic restriction sampled at different times. --- # 14. Repositioning of early "branch points" Previous research called the original target: $$ y\equiv4\pmod6 $$ a branch point capable of generating an odd predecessor. In the modified map, it is more naturally written as: $$ \boxed{ y\equiv2\pmod3. } $$ Therefore, branch sparsity can be viewed as: $$ \boxed{ \text{target-domain inverse legality}. } $$ This is more precise than "decimal filtering" and completely independent of base-10. --- # 15. Difference between fixed-word inverse and inverse tree fixed-word inverse: $$ F_w^{-1}: \mathcal P_w \to \mathcal C_w $$ is single-valued. But the global Collatz inverse: $$ T^{-1}(y) $$ is generally: - one predecessor; - or two predecessors. So: $$ \boxed{ \text{fixed itinerary removes inverse branching}. } $$ This is an important local simplification. --- # 16. Why is the inverse single-valued after fixing the word? Because global inverse branching comes from: > Not knowing the branch history of the predecessor. Once $w$ is fixed, the branch history is known, so all branching decisions are eliminated. Therefore: $$ \boxed{ \text{inverse ambiguity} = \text{itinerary uncertainty}. } $$ In a fixed-word chart, itinerary uncertainty is zero, thus the inverse is exact and single-valued. --- # 17. Odd Core Any: $$ n\in\mathbb Z_{>0} $$ is uniquely represented as: $$ \boxed{ n=2^{v_2(n)}m, \qquad m\text{ odd}. } $$ Define: $$ \boxed{ \operatorname{oddcore}(n) = \frac{n}{2^{v_2(n)}}. } $$ Therefore, all positive integers are partitioned into disjoint $2$-rays: $$ \boxed{ \mathbb Z_{>0} = \bigsqcup_{m\text{ odd}} \{2^qm:q\ge0\}. } $$ --- # 18. Odd Skeleton of the Inverse Tree If a certain odd state: $$ m $$ is already in the inverse convergence tree, then: $$ m,2m,4m,8m,\ldots $$ are all automatically in the tree. Therefore, global inverse coverage is equivalent to odd coverage: $$ \boxed{ \text{all positive integers covered} \iff \text{all positive odd integers covered}. } $$ So the inverse tree can be divided into: $$ \boxed{ \text{odd skeleton} + \text{even }2\text{-rays}. } $$ This observation from previous research is preserved in this paper. --- # 19. Accelerated Odd Map For positive odd $n$, define: $$ \boxed{ S(n) = \frac{3n+1}{2^{\kappa(n)}}, } $$ where: $$ \boxed{ \kappa(n) = v_2(3n+1). } $$ Because: $$ 3n+1 $$ is even, so: $$ \kappa(n)\ge1. $$ And: $$ S(n) $$ is again odd. Therefore: $$ S: \mathbb Z_{>0}^{\mathrm{odd}} \to \mathbb Z_{>0}^{\mathrm{odd}}. $$ --- # 20. Accelerated Inverse Fiber Given an odd target: $$ t, $$ if: $$ S(n)=t, $$ then there exists: $$ \kappa\ge1 $$ such that: $$ 3n+1 = 2^\kappa t. $$ So: $$ \boxed{ n = R_\kappa(t) = \frac{2^\kappa t-1}{3}. } $$ This is the valuation-labelled inverse candidate. --- # 21. Inverse Fiber Legality Theorem ## Theorem 21.1 For positive odd $t$ and $\kappa\ge1$, $$ R_\kappa(t) = \frac{2^\kappa t-1}{3} $$ is a valid positive odd predecessor if and only if: $$ \boxed{ 2^\kappa t\equiv1\pmod3. } $$ ### Proof If the congruence holds, the numerator is divisible by 3. Since: $$ 2^\kappa t $$ is even, the numerator: $$ 2^\kappa t-1 $$ is odd. Dividing by the odd number 3 still yields an odd number. Positivity is obvious. The converse holds immediately. This completes the proof. --- # 22. Complete Classification Modulo 3 Since $t$ is odd, consider: $$ t\bmod3. $$ ### Case A If: $$ t\equiv1\pmod3, $$ it requires: $$ 2^\kappa\equiv1\pmod3. $$ And: $$ 2^\kappa \equiv (-1)^\kappa \pmod3. $$ So: $$ \boxed{ \kappa\text{ must be even}. } $$ ### Case B If: $$ t\equiv2\pmod3, $$ it requires: $$ 2^\kappa(-1)\equiv1\pmod3, $$ so: $$ \boxed{ \kappa\text{ must be odd}. } $$ ### Case C If: $$ t\equiv0\pmod3, $$ then: $$ 2^\kappa t\equiv0\pmod3 $$ cannot equal 1. So: $$ \boxed{ 3\mid t \Rightarrow S^{-1}(t)=\varnothing. } $$ --- # 23. Accelerated Map Image Avoids Multiples of 3 From the previous section: $$ \boxed{ S(n)\not\equiv0\pmod3 } $$ for all odd $n$. This can also be seen directly: $$ 3n+1\equiv1\pmod3, $$ and dividing by: $$ 2^\kappa $$ only multiplies by a unit mod 3. So the target states of the accelerated odd skeleton always fall in: $$ \boxed{ 1,2\pmod3. } $$ --- # 24. Terminal Fiber at $t=1$ Take: $$ t=1. $$ Since: $$ 1\equiv1\pmod3, $$ valid $\kappa$ must be even: $$ \kappa=2j, \qquad j\ge1. $$ Therefore: $$ R_{2j}(1) = \frac{2^{2j}-1}{3}. $$ That is: $$ \boxed{ R_{2j}(1) = \frac{4^j-1}{3}. } $$ This is exactly the: $$ \boxed{ M_j. } $$ from previous research. --- # 25. Reinterpretation of the old "highway" family Previous research: $$ M_j = \frac{4^j-1}{3} = 1,5,21,85,341,\ldots. $$ In the past, they were viewed as a special family for rapidly entering the powers-of-two spine. Now, their essence can be precisely written as: $$ \boxed{ S(M_j)=1. } $$ And: $$ \boxed{ v_2(3M_j+1)=2j. } $$ So: $$ M_j $$ is the complete valid even-valuation inverse fiber of the terminal odd state 1. This is more structured and more general than a "highway". --- # 26. Repositioning of the number 5 $$ 5 = \frac{4^2-1}{3} = R_4(1). $$ So: $$ \boxed{ 5 \text{ is simply an inverse-fiber member of }t=1,\kappa=4. } $$ Its ordinary Collatz trajectory is: $$ 5\to16\to8\to4\to2\to1. $$ And: $$ 5\cdot2^q $$ is just the even $2$-ray above the odd node 5. Therefore, there is no need to additionally assume that "5 is a super attractor". --- # 27. Inverse Highway Family for Arbitrary Targets For any positive odd: $$ t\not\equiv0\pmod3, $$ there are infinitely many parity-compatible $\kappa$: If: $$ t\equiv1\pmod3, $$ then: $$ \kappa=2,4,6,\ldots. $$ If: $$ t\equiv2\pmod3, $$ then: $$ \kappa=1,3,5,\ldots. $$ Therefore: $$ \boxed{ \mathcal R(t) = \left\{ \frac{2^\kappa t-1}{3}: 2^\kappa t\equiv1\pmod3 \right\} } $$ forms the accelerated inverse fiber of target $t$. So the early "highways" do not only exist at 1. Every valid odd target has its own valuation-labelled inverse family. --- # 28. Odd Skeleton as a Valuation-Labeled Graph Therefore, the accelerated inverse graph can be represented as: Nodes: $$ t\in\mathbb Z_{>0}^{\mathrm{odd}}, \qquad 3\nmid t $$ and their admissible predecessors. Edge label: $$ \boxed{ \kappa=v_2(3n+1). } $$ Edge relation: $$ \boxed{ n \xrightarrow{\;\kappa\;} t \iff n=\frac{2^\kappa t-1}{3}. } $$ So the odd skeleton is a valuation-labelled directed graph. --- # 29. Relationship with the early double helix The old "double helix" mainly viewed: - the forward orbit; - the backward convergence tree; as two paths moving toward each other. After this paper, it can be rewritten more precisely. ## Forward local strand $$ \boxed{ r_w+2^k a \to m_w+3^u a. } $$ ## Backward local strand $$ \boxed{ m_w+3^u a \to r_w+2^k a. } $$ ## Accelerated odd inverse strand $$ \boxed{ t \leftarrow \frac{2^\kappa t-1}{3}. } $$ Therefore, the "double helix" is no longer just a visual diagram, but: $$ \boxed{ \text{two compatible exact coordinate directions}. } $$ --- # 30. Why is it called $2^k\leftrightarrow3^u$? In a fixed finite word: source spacing: $$ 2^k. $$ target spacing: $$ 3^u. $$ So: $$ \boxed{ 2^k } $$ controls the admissible source residue resolution, while: $$ \boxed{ 3^u } $$ controls the target progression resolution. The quotient label for these two scales: $$ a $$ is exactly the same. --- # 31. This does not mean "$2$-adic = $3$-adic" Over-interpretation must be avoided. This paper only proves the fixed-word arithmetic progression transport: $$ r_w+2^k\mathbb Z \leftrightarrow m_w+3^u\mathbb Z. $$ It does not automatically establish: $$ \mathbb Z_2\cong\mathbb Z_3. $$ In fact: $$ \mathbb Z_2 $$ and: $$ \mathbb Z_3 $$ have different local-field / topological structures. So: $$ \boxed{ \text{$2^k$/$3^u$ bidirectional residue transport} } $$ is not: $$ \boxed{ \text{global $2$-adic/$3$-adic isomorphism}. } $$ --- # 32. Local Inverse ≠ Global Coverage For a fixed $w$: $$ F_w^{-1} $$ exists and is exact. But the Collatz conjecture requires: $$ \forall n>0, $$ its forward orbit eventually enters the terminal cycle. The inverse formulation equivalently requires: $$ \boxed{ \text{inverse tree rooted at 1 covers all positive integers}. } $$ The local inverse theorem only answers: > If the itinerary / target congruence is known, how can we exactly solve for the inverse? It does not answer: > Does every integer appear in the terminal inverse tree? So: $$ \boxed{ \text{local exact inversion} \not\Rightarrow \text{global inverse coverage}. } $$ --- # 33. Merge and Inversion are also different If different charts: $$ w\neq v $$ satisfy: $$ \Gamma_w\cap\Gamma_v\neq\varnothing, $$ there may exist different sources that merge to the same target after a fixed number of steps. This does not violate fixed-chart injectivity, because injectivity is internal to: $$ F_w|_{\mathcal C_w} $$ Across charts: $$ F_w(x)=F_v(z) $$ is entirely possible. Therefore: $$ \boxed{ \text{local bijection} \neq \text{global one-to-one dynamics}. } $$ --- # 34. Fixed-Word Fiber Intersection If: $$ y\in\Gamma_w\cap\Gamma_v, $$ then there exist: $$ a,b $$ such that: $$ y=m_w+3^{u_w}a $$ and: $$ y=m_v+3^{u_v}b. $$ This transforms into a linear Diophantine congruence: $$ m_w-m_v = 3^{u_v}b-3^{u_w}a. $$ Therefore, the target merge problem itself can also be reduced to an arithmetic progression intersection. This will reappear in the finite certificate frontier of Paper 09. --- # 35. Relationship with the 3x+1 Semigroup Existing research on the 3x+1 semigroup has already used rational multiplicative generators to encode backward iteration. This paper does not claim that "backward algebraic encoding" is a new discovery. The specific work of this paper is to write the finite parity chart of Paper 03 as: $$ \boxed{ \text{source residue class} \overset{F_w}{\longleftrightarrow} \text{target progression} } $$ and provide exact forward/inverse recovery using the same quotient coordinate. This is the atlas formulation of the Operation Translation Series. --- # 36. Relationship with the $2$-adic inverse parity transform Existing $2$-adic Collatz research has studied: $$ \text{$2$-adic integer} \leftrightarrow \text{infinite parity sequence}. $$ This paper only deals with finite words: $$ w\in\{D,U\}^k $$ and their: $$ r_w\bmod2^k. $$ Therefore, the finite residue atlas is compatible with $2$-adic parity coding, but this paper further appends: $$ \boxed{ m_w\bmod3^u } $$ as target-side metadata, for exact inverse recovery. --- # 37. Bidirectional Chart Object This paper denotes the complete bidirectional data of a fixed word as: $$ \boxed{ \mathcal B_w = ( w, k, u, b_w, r_w, m_w, \mathcal C_w, \mathcal P_w, \phi_w, \psi_w ). } $$ where: $$ \phi_w(n)=\frac{n-r_w}{2^k}, $$ $$ \psi_w(y)=\frac{y-m_w}{3^u}. $$ and: $$ \boxed{ \psi_wF_w\phi_w^{-1} = \operatorname{id}. } $$ --- # 38. Bidirectional Certificate Given: $$ (x,y,w), $$ it can be verified using three exact conditions: ### Source legality $$ \boxed{ x\equiv r_w\pmod{2^k}. } $$ ### Target legality $$ \boxed{ y\equiv m_w\pmod{3^u}. } $$ ### Transport consistency $$ \boxed{ 3^u(x-r_w) = 2^k(y-m_w). } $$ If all three hold and satisfy the positive-domain quotient bound, then the fixed-word transport can be machine-exactly verified. --- # 39. Significance for finite verification Traditional finite verification often stores: $$ n\to T(n)\to T^2(n)\to\cdots. $$ This paper shows that for a known word: $$ w $$ one only needs to store: $$ \boxed{ (r_w,m_w,k,u,b_w). } $$ All starting states of the entire cylinder can be batch-described by the quotient label $a$. Therefore: $$ \boxed{ \text{trajectory storage} \to \text{chart certificate storage}. } $$ This will be formalized in Paper 09. --- # 40. Summary of Main Theorems in this Paper ## Theorem A — Bidirectional Residue Transport $$ \boxed{ r_w+2^k\mathbb Z \overset{\sim}{\longleftrightarrow} m_w+3^u\mathbb Z. } $$ ## Theorem B — Exact Inverse $$ \boxed{ x = r_w+ 2^k\frac{y-m_w}{3^u}. } $$ ## Theorem C — Target Legality $$ \boxed{ y\equiv m_w\pmod{3^u}. } $$ ## Theorem D — Quotient Conservation $$ \boxed{ \frac{x-r_w}{2^k} = \frac{y-m_w}{3^u}. } $$ ## Theorem E — Accelerated Inverse Fiber $$ \boxed{ R_\kappa(t) = \frac{2^\kappa t-1}{3} } $$ Valid iff: $$ \boxed{ 2^\kappa t\equiv1\pmod3. } $$ ## Theorem F — Terminal Fiber $$ \boxed{ R_{2j}(1) = \frac{4^j-1}{3}. } $$ --- # 41. Limitations of this Paper First, the fixed-word inverse relies on a known chart $w$. Second, target progressions can overlap across charts. Third, this paper does not prove that the inverse tree rooted at 1 covers all positive integers. Fourth, the $2^k\leftrightarrow3^u$ in this paper is a finite arithmetic progression transport, not a global isomorphism between $\mathbb Z_2$ and $\mathbb Z_3$. Fifth, accelerated inverse fibers describe odd skeleton edge candidates, but global coverage remains an unsolved problem. --- # 42. Conclusion Paper 03 proved: $$ \text{parity word} \longleftrightarrow \text{unique }2^k\text{ source cylinder}. $$ This paper further proves: $$ \boxed{ \text{source }2^k\text{ cylinder} \longleftrightarrow \text{target }3^u\text{ progression}. } $$ Both sides are connected by the same exact quotient coordinate: $$ a $$ Therefore, fixed-word Collatz dynamics can not only be forward-compressed, but also allow for exact inverse recovery. This gives the author's early "double helix" research a more rigorous new form: $$ \boxed{ \text{forward residue refinement} + \text{backward valuation / progression fibers}. } $$ On the other hand, the: $$ \frac{4^j-1}{3} $$ from previous research is also repositioned as: $$ \boxed{ \text{the accelerated inverse fiber of terminal state }1. } $$ So: - powers of two; - odd skeleton; - $M_j$; - 5; - $5\cdot2^q$; no longer need to be treated as mutually isolated "special structures", but can be unified into: $$ \boxed{ \text{odd inverse skeleton} + \text{valuation-labelled fibers} + \text{even }2\text{-rays}. } $$ At this point, the early double helix framework has completed its reconstruction from a graphical method to exact residue transport. The next paper will turn to another question: > Among all finite parity charts, which charts inevitably descend at a sufficiently large scale? and establish the contraction bound for: $$ \boxed{ 3^u<2^k } $$ the word-order threshold, and the binomial Cylinder Law. --- # References 1. David Applegate, Jeffrey C. Lagarias, *The 3x+1 Semigroup*, Journal of Number Theory 117 (2006), arXiv:math/0411140. 2. Olivier Rozier, *Parity sequences of the 3x+1 map on the 2-adic integers and Euclidean embedding*, arXiv:1805.00133. 3. Tong Niu, *Parity vectors and paradoxical sequences in the accelerated Collatz map*, arXiv:2605.13886. 4. Terence Tao, *Almost all orbits of the Collatz map attain almost bounded values*, Forum of Mathematics, Pi 10 (2022), arXiv:1909.03562. 5. Collatz Operation Translation Series — Paper 01, *Reclassification and Correction of Existing Research on the Collatz Conjecture*. 6. Collatz Operation Translation Series — Paper 02, *Collatz Local Affine Atlas: Exact Affinization of Finite Parity Words*. 7. Collatz Operation Translation Series — Paper 03, *Parity Words, Residue Cylinders, and Local Identitization*. --- ## Next Paper **Paper 05 — *Finite-Word Contraction Bounds and the Binomial Cylinder Law*** Core tasks: 1. Prove $$ T_w(n)