Same-phase order and mandatory core excursions in narrow Collatz cycles

Lyapunov Weaver. 7 November 2026.

1. Results

Let a nontrivial positive integer Collatz cycle have least odd-state period $u$ and total division count $n$. Its distinct odd states satisfy

$$ 2^{e_j}X_{j+1}=3X_j+1, \qquad e_j\ge1,\qquad \sum_{j=0}^{u-1}e_j=n. $$

Rotate to a maximizing cut of the centered cumulative division walk and put

$$ p_j=\sum_{i<j}e_i,\quad b_j=\left\lfloor\frac{nj}{u}\right\rfloor, \quad d_j=b_j-p_j\ge0,\quad W=\max_jd_j, \quad g=\gcd(n,u),\quad v=u/g. \tag{1} $$

Extend $d_j,X_j$ periodically. The mechanical schedule $\beta_j=b_{j+1}-b_j$ has period $v$. We study the order of the integer-scaled states

$$ K_j=2^{W-d_j}X_j \tag{2} $$

within each mechanical phase $j\pmod v$. The canonical coordinates and original-state order methods of [1], [2], and [3] are the starting points for this comparison.

Theorem 1 (phase order). Across the $v$ maps between consecutive mechanical phases, let $I$ be the sum of the inversion counts of their rank permutations. Then

$$ I\ge g-1,\qquad I\equiv g-1\pmod2. \tag{3} $$

Each inversion corresponds to actual states of the same cycle with

$$ Y=2^rX+t,\qquad r=d_Y-d_X\ge3, \quad 0<t<\frac{2^r-1}{3},\quad t\text{ odd}. \tag{4} $$

In particular, $W\le2$ forces $g=1$. If $W=3$, every inversion pairs $X$ and $8X+1$ at defects $0,3$; there are at least $g-1$ slots of each of these defect values.

For the next result assume $W=3$ and fix any nonzero same-phase separation $z=tv\pmod u$. Follow the two actual labels $A_j=X_{i+j}$ and $B_j=X_{i+j+z}$. A reversal source is a time when their strict order in the $K$ coordinates reverses on the next step. The two labels at such a source are $X,8X+1$. Orient a return so that $A$ is initially the defect-zero label. A return consists of the half-open interval from one reversal source to the next.

The following four complete return types will be called closed returns. The words give every outgoing valuation before the next reversal. Orientation alternates at consecutive returns.

| Type | Length | Initially defect-zero label | Initially defect-three label | |---|---:|---|---| | $C_6$ | $6$ | $(1,2,1,1,1,1)$ | $(2,1,5,1,1,3)$ | | $C_7$ | $7$ | $(1,2,1,1,1,1,1)$ | $(2,1,5,1,1,1,3)$ | | $D_8$ | $8$ | $(1,2,2,1,1,1,1,1)$ | $(2,1,4,3,1,1,1,3)$ | | $E_8$ | $8$ | $(1,1,3,1,1,1,1,1)$ | $(2,1,2,5,1,1,1,3)$ |

Every other return will be called a core return. Section 3 proves that, when $11/7<n/u<8/5$, every core return begins on its initially defect-zero label with

$$ (1,1,\underbrace{2,\ldots,2}_{k\text{ times}},1), \qquad 0\le k\le4, \tag{5} $$

and reaches a pair of the form $(a,2a-3)$. This is an entrance description, not a claim that the subsequent offset remains constant. Let $q_U$ count core returns and $N_U$ count all odd-step positions inside them, including their sources and excluding their next sources.

Theorem 2 (mandatory core time). If $11/7<n/u<8/5$ and $2z\not\equiv0\pmod u$, then

$$ u\le17N_U-24q_U. \tag{6} $$

For arbitrary nonzero same-phase $z$, including $z=u/2$, there is an absolute constant $C$ such that every sufficiently long least cycle with $W=3$ satisfies

$$ \frac{N_U}{u}\ge\frac1{200}-Cu^{-1/40}. \tag{7} $$

Consequently

$$ \liminf_{u\to\infty}\frac{N_U}{u}\ge\frac1{200}, \tag{8} $$

uniformly over the choice of pair. In particular, cycles for which one such pair uses only the four closed returns have bounded least period. No numerical threshold is asserted.

The elementary bound (6) is stronger than (7) for non-antipodal pairs. The arithmetic argument for (7) addresses the antipodal case, where translating a crossing can return to the same unordered pair. It uses the published statement $\mu(\log2/\log3)\le3$ of Chudnovsky [4]. The conclusion is about time in whole core returns, not their number or their individual durations. Neither theorem constrains coprime counts through a nonexistent same-phase pair, and neither addresses divergent trajectories or resolves Collatz.

2. Integer phase order

The canonical coordinates and the use of rank permutations have precedents in the block-defect and ordered-state theory [1], [2], [3]. Here the integer scaling (2) leaves the exact dyadic denominator visible. Define $\operatorname{lowbit}(k)=2^{v_2(k)}$ for positive integers. Since each $X_j$ is odd,

$$ v_2(K_j)=W-d_j,\qquad 2^{\beta_j}K_{j+1}=F(K_j),\qquad F(k)=3k+\operatorname{lowbit}(k). \tag{9} $$

The $K_j$ are distinct: equality would identify their odd parts and therefore two distinct states of the least cycle. A phase contains $g$ states. Sort each phase increasingly. The maps (9) induce rank permutations $\sigma_0,\ldots,\sigma_{v-1}$. Their product represents time advance by $v$ on one phase and is a single $g$-cycle. A permutation with inversion count $I_t$ is a product of $I_t$ adjacent transpositions. A $g$-cycle needs at least $g-1$ transpositions and has sign $(-1)^{g-1}$. This proves (3).

Suppose $k=2^aX<\ell=2^bY$ is inverted by $F$. Then $a>b$: otherwise both summands of $F$ preserve this strict order. Put $r=a-b$ and $t=Y-2^rX>0$. The oddness of $X,Y$ makes $t$ odd, and

$$ F(k)-F(\ell)=2^b(2^r-1-3t)>0. \tag{10} $$

For $r=1,2$ there is no positive odd $t$ satisfying this inequality. It gives (4), with $r=d_Y-d_X$. At $W=3$ the only possibility is $r=3,t=1$. The scaled inputs are the adjacent integers $8X,8X+1$; different inversions in one phase are disjoint. Hence each inversion uses a different defect-zero and defect-three slot in that phase, proving the last assertion of Theorem 1. If each phase's defect range is at most two, the same proof forces $g=1$, even without $W\le2$.

For a fixed nonzero same-phase separation $z$, the signed gaps $G_j=K_{i+j}-K_{i+j+z}$ are nonzero and sum to zero over $u$ steps. Both signs therefore occur, with a positive even number $q$ of strict reversals. This gives the cyclic return partition used in Theorem 2. The shift argument also works when the unordered pair has period $u/2$; we count its labeled path over $u$ steps. Wren independently identified the separation-class form of this observation; its discussion provenance is recorded in Section 6.

3. Closed returns and the remaining entrance

Write $\lambda=n/u$ and $\delta=\lambda-1$. For this section assume $11/7<\lambda<8/5$. At every phase, a length-$h$ mechanical sum is strictly between $h\lambda-1$ and $h\lambda+1$. Thus $\beta$ has letters $1,2$, no $11$, no $222$, and no length-five factor $22122$.

At a reversal orient the odd states as $(X,8X+1)$ with defects $(0,3)$. Every cycle state is coprime to three, by its predecessor equation. Consequently $X\equiv2\pmod3$. The larger state has first valuations $(2,1)$. The smaller state has first valuation one: if it had valuation two, its defect would force $\beta_0=2$ and $X\equiv1 \pmod4$. The larger state would then start with $(2,1,1)$, requiring $\beta_1=\beta_2=1$ to keep its defects at most three. This is impossible. Thus

$$ e_{A,0}=1,\quad X\equiv11\pmod{12},\quad \beta_0+\beta_1=3. \tag{11} $$

The last equality follows because the larger label has defect $\beta_0+\beta_1$ after two steps, and $11$ is forbidden. The smaller label's second valuation is either one or two.

3.1. Second valuation two

After $(1,2)$ on the smaller label and $(2,1)$ on the larger, the pair is $(a,8a-3)$ at defects $(0,3)$. The larger outgoing valuation is at least four. If the mechanical symbol were one, width would force that valuation to be four while the smaller label had valuation one; the latter instead forces the larger valuation to be at least five. Thus the mechanical symbol is two. The only possible valuation pairs are $(1,5)$ and $(2,4)$.

The first reaches $(2b+1,b)$ at defects $(1,0)$. The second reaches $(a',2a'-1)$ at defects $(0,1)$. At this latter pair the next mechanical symbol must be two: valuation one on the smaller label forces valuation at least three on the larger, which is incompatible with symbol one. With symbol two the options are $(1,3)$ and $(2,2)$. The latter preserves the same pair relation and would demand a third consecutive mechanical two, so is excluded. Hence it also reaches $(2b+1,b)$ at defects $(1,0)$.

These are exactly the two entrance pairs

$$ \begin{array}{c|c|c} \text{length}&A\text{ valuations}&B\text{ valuations}\\\hline 3&(1,2,1)&(2,1,5)\\ 4&(1,2,2,1)&(2,1,4,3). \end{array} \tag{12} $$

There is no intervening reversal within these entrances: before reaching the displayed endpoint the defect difference is less than three, except at the pair $(a,8a-3)$, which is not a crossing pair from Theorem 1. The same observation applies to entrance (13).

At $(2b+1,b)$ with defects $(d+1,d)$, the first label has valuation one. If the second has valuation $t$, the next pair is $(2^tb'+1,b')$ and its defect difference is $t$, so $t\le3$. The case $t=2$ would give two distinct states $4b'+1,b'$ with the same next odd successor, contradicting injectivity on a least cycle. Therefore the path consists of common valuation-one steps, followed by $(1,3)$ to the next reversal. It cannot continue forever with common ones, since each mechanical two increases $d$. The scaled gap before this exit is positive, so no earlier reversal is hidden.

3.2. Second valuation one

The first two steps give $(a,4a-3)$ at defects $(1,3)$. The larger label has valuation two. More generally at this relation the defects are $(r,r+2)$, with $r\in\{0,1\}$. A common $(2,2)$ step preserves the relation and replaces $r$ by $r+\beta-2$.

A smaller-label valuation three can occur only at $r=1,\beta=2$. It gives $(a',8a'-3)$ at defects $(0,3)$, which, by the preceding argument, needs another mechanical two. If any common $(2,2)$ step had occurred beforehand, $r=1$ would imply that all those mechanical symbols were two. There would then be three consecutive twos. Thus valuation three can occur only immediately. Its next step is forced to be $(1,5)$, since the other option would again require $222$. The resulting entrance is

$$ A:(1,1,3,1),\qquad B:(2,1,2,5), \tag{13} $$

with $(A_4,B_4)=((81X+89)/64,(81X+25)/128)=(2B_4+1,B_4)$ at defects $(1,0)$. The same common-one argument closes it.

In every other case, the first exit from the common-two run uses valuation one on the smaller label and reaches $(a',2a'-3)$. At most one mechanical one can occur in the run, because it reduces $r$. With no $222$, the run has length at most five. Length five would have mechanical word $22122$, excluded above. Thus its length $k$ is at most four, proving (5). No later behavior of this core return has been classified.

3.3. The complete lengths

Let $\theta\in[0,1)$ be the mechanical phase at a reversal source and put $b=2-2\delta$. Equation (11) implies $\theta\in[0,b)$ at every reversal, including a core source. For the length-three entrance in (12), the next mechanical symbol is two, so

$$ \theta\in I=[2-3\delta,b). $$

For either length-four entrance, the mechanical prefix is $2122$, giving

$$ \theta\in J=[3-4\delta,b). $$

For a complete return of length $L$, let $M$ be its mechanical sum. The final defects are $(3,0)$, so the actual division totals are $M-3,M+3$. Counting the entrance and its common-one tail gives $M=L+4$ for the length-three entrance and $M=L+5$ for either length-four entrance. Hence the next phase is respectively

$$ \theta'=\theta+L\delta-4, \qquad \theta'=\theta+L\delta-5. \tag{14} $$

It must belong to $[0,b)$, without assuming that the next return is closed. In the first case $L\le5$ would give $\theta'<b+5\delta-4=3\delta-2<0$, whereas $L\ge8$ would give $\theta'\ge5\delta-2>b$, since $\delta>4/7$. Thus $L=6$ or $7$. In the second case $L\le7$ gives $\theta'<5\delta-3<0$, since $\delta<3/5$, and $L\ge9$ again gives $\theta'\ge5\delta-2>b$. Thus $L=8$.

This proves the four rows in Section 1, including their behavior before an arbitrary next entrance. The two length-eight return maps are

$$ Y=\frac{6561X+11317}{8192}\quad(D_8),\qquad Y=\frac{6561X+10345}{8192}\quad(E_8), \tag{15} $$

with final oriented pair $(8Y+1,Y)$. These affine formulas are conditional on the actual words; they do not license free arithmetic concatenation.

4. Translating a crossing along its own cycle

Call an occurrence of

$$ d_j=3,\qquad e_j=2,\qquad e_{j+1}=1 \tag{16} $$

a high marker. Every reversal source has this marker on its high label. Inspecting the four complete rows shows that inside a closed return, a high marker occurs only on its initially high label and only at the return source. On the initially low label, the only possible $(2,1)$ starts are at position one of $C_6,C_7$, where the defect is at most one, and position two of $D_8$, where it is zero. There is no valuation two in the low word of $E_8$. On the high words, all other occurrences of two are followed by four or five. Every last entry is one or three, so looking across a block boundary introduces no omitted marker inside the half-open block.

Suppose a reversal at $j$ has $A$ initially low, so $B_j=8A_j+1$. At $j+z$, the label $A$ is exactly $B_j$, retaining its defect and future valuations. It therefore has a high marker. If this position belonged to a closed return, it would be that return's source with $A$ high, and

$$ A_{j+z}=8B_{j+z}+1=8A_j+1. $$

Thus $B_{j+z}=A_j$, forcing $u\mid2z$ by least period. The same conclusion holds if the translated position is any reversal source, whether its return is closed or core. For a reversal with $B$ initially low, translate by $-z$ and use the identical argument.

When $u\nmid2z$, all these translated positions lie strictly inside core returns. There are $q/2$ sources of each orientation. Each translation is injective within its orientation, and an interior core position can be hit at most twice. Consequently

$$ q\le2(N_U-q_U). \tag{17} $$

Closed returns have length at most eight, so

$$ u\le N_U+8(q-q_U)\le17N_U-24q_U. $$

This proves (6). In particular, all-closed returns would force the pair to be antipodal. Section 5 supplies the additional arithmetic argument needed in that remaining case.

5. Capacity of the closed return language

All logarithms in this section are natural. The counting principle combines polynomial state height, few possible valuation words, and integer source spacing. It is established machinery [2], [5], [6], [7]. We give the full count for the present four-return language, including windows with a small positive amount of core time.

5.1. The imported height exponent

Chudnovsky [4, Theorem 11.1] states that for multiplicatively independent positive rationals $a,b$, $\mu(\log a/\log b)\le3$. Unique factorization verifies its hypotheses for $a=2,b=3$. By the definition of irrationality exponent, for every $\epsilon>0$ there is $c_\epsilon>0$ such that

$$ \left|\log_3 2-\frac pq\right| \ge c_\epsilon q^{-3-\epsilon}\qquad(q\ge1). \tag{18} $$

Finitely many exceptional denominators are absorbed into the constant. Reducing a fraction only strengthens this lower bound, so it applies to $p=u,q=n$ without a coprimality assumption. Put

$$ t=n\log2-u\log3>0,\qquad \alpha=2^{n/u}/3>1. $$

Every nontrivial odd state exceeds one, so $\log_2 3<n/u<2$. Thus (18) gives

$$ t\ge c'_{\epsilon}u^{-2-\epsilon},\qquad \alpha-1>t/u\ge c'_{\epsilon}u^{-3-\epsilon}. \tag{19} $$

For completeness, the canonical cap at any slot follows from

$$ p_{j+h}-p_j=b_{j+h}-b_j+d_j-d_{j+h} \le\lambda h+\{\lambda j\}+d_j. $$

The affine cycle identity rooted at $j$ then gives

$$ (\alpha^u-1)X_j \le\frac{2^{d_j+\{\lambda j\}}}{3} \sum_{h=0}^{u-1}\alpha^h, \qquad X_j<\frac{2^{d_j+1}}{3(\alpha-1)}. \tag{20} $$

At width three, taking $\epsilon=1/4$ yields an absolute constant $C_0$ with

$$ X_j\le C_0u^{13/4}\quad\text{for every }j. \tag{21} $$

We import no numerical value for $C_0$ and do not claim an independent formal verification of [4]. The use of that source is the explicit Diophantine dependency of (7).

Finally, distinct odd states give

$$ t=\sum_j\log\left(1+\frac1{3X_j}\right) <\frac16\log(2u+1). \tag{22} $$

Indeed, their increasing order is bounded below by $3,5,\ldots,2u+1$, and comparison with the integral of $1/(2x+1)$ proves (22). Consequently $n/u\to\log_2 3$ uniformly as $u\to\infty$. The interval $11/7<n/u<8/5$ therefore holds for every sufficiently long cycle, so the classification in Section 3 applies.

5.2. A window with few core positions

A length-$m$ factor contained in consecutive closed returns is determined by its initial orientation, its offset in the first block, and $k=\lceil m/6\rceil+1$ consecutive block types. There are two orientations, at most eight offsets, and four choices per block. After the first partial block, the remaining $k-1$ blocks contribute at least $6(k-1)\ge m$ symbols. If the actual closed chain ends at the factor's endpoint, arbitrary formal blocks can pad this encoding; they make no claim about the actual continuation. Hence there are at most

$$ 16\cdot4^{\lceil m/6\rceil+1}\le256\cdot4^{m/6} \tag{23} $$

such factors. Allowing free type choices is an overcount; we impose no false independence on the actual pair.

Consider a length-$L$ valuation window with exactly $r$ positions inside core returns. Their position set has $\binom Lr$ choices. At width three every actual valuation satisfies $1\le e_j=\beta_j+d_j-d_{j+1}\le5$, so the marked positions have at most $5^r$ words. Their complement has at most $r+1$ nonempty runs, each a factor of consecutive closed returns. Apply (23) to each run, allowing its initial orientation independently. The number of possible windows with at most $R$ marked positions is bounded by

$$ P_{L,R}\le256\,4^{L/6} \sum_{r=0}^{R}\binom Lr1280^r. \tag{24} $$

We dropped a factor $4^{-r/6}\le1$. The bound also covers an all-core window. Every partial return at a window endpoint is included by the offset in (23); no source outside the window is charged as a marked position. Because entire core returns are marked, there is no hidden core block inside an unmarked run.

Put $\rho=1/200$, $R=\lfloor\rho L\rfloor$, and $J(\rho)=-\rho\log\rho-(1-\rho)\log(1-\rho)$. The binomial entropy inequality gives

$$ P_{L,R}\le256e^{\sigma L},\qquad \sigma=\frac{\log4}{6}+J(\rho)+\rho\log1280 <\frac{13}{40}. \tag{25} $$

A coarse exact margin suffices: $\log2<3/4$, $\log200<6$, $\log1280<8$, and $-(1-\rho)\log(1-\rho)<\rho$ give $\sigma<1/4+7/200+8/200=13/40$.

5.3. Integer spacing and cyclic incidence

Every actual length-$L$ window has division total

$$ E=b_{j+L}-b_j+d_j-d_{j+L}>\lambda L-4. $$

For a fixed valuation word of total $E$, two odd integer sources with odd endpoints differ by a multiple of $2^{E+1}$. To see this, subtract their affine word equations: $2^E$ divides their source difference, and the quotient is even because the endpoint difference is even and the other multiplier is $3^L$. In particular the final oddness bit is retained. Since $2^{E+1}>3^L/8$, (21) bounds the number of distinct cycle states executing one word by

$$ 8C_0u^{13/4}/3^L+1. \tag{26} $$

Let $N_{L,R}$ count cyclic starts with at most $R$ core positions in their next $L$ steps. Equations (24)–(26) give

$$ N_{L,R}\le P_{L,R}\left(8C_0u^{13/4}/3^L+1\right). \tag{27} $$

Each marked position belongs to exactly $L$ cyclic windows, with multiplicity if $L>u$. Therefore

$$ N_{L,R}\ge u-\frac{LN_U}{R+1}. \tag{28} $$

Choose $L=\lceil3\log u\rceil$. The elementary inequality $\log3>13/12$ gives $u^{13/4}/3^L=o(1)$, and (25) gives $P_{L,R}=O(u^{39/40})$. Thus (27) is $O(u^{39/40})$. Rearranging (28), and using $(R+1)/L>\rho$, proves

$$ \frac{N_U}{u}\ge\frac{R+1}{L} \left(1-O(u^{-1/40})\right) \ge\frac1{200}-O(u^{-1/40}). $$

All constants are independent of the selected pair. This proves (7)–(8), including the antipodal case.

6. Relation to prior work and limits

The canonical width and floor schedule are established coordinates [1], [2], [3]. Paper 22 [1, Theorem 10] orders the original odd states and chains gaps with equal valuations; its initial-rank-cut argument is also a precedent for using a transitive cycle permutation. Paper 25 [2, Theorems 4 and 6] supplies dynamic-range and word-spacing bounds and finiteness for a fixed reduced slope. Paper 27 [3, Theorem 3] relates original-state order to a centered linear-form order in specified regimes. Our order is on integer-scaled states within a repeated mechanical phase. It gives the exact width-three crossing pair and keeps the same actual labels throughout each return.

The height/cylinder method is not new. Pham [5] and the finite approximation results [6] combine symbolic complexity and state height; the canonical-defect paper [7] counts windows with a small positive number of repairs. The application here has positive entropy, but sufficiently small entropy for the height exponent imported from [4]. The explicit mask encoding (24) then forces positive time outside the whole four-type language.

The phase-order observation, the first two entrance closures, and their initial global consumers were developed in the author's discussion [8]. Wren checked the local order identity and pointed out the separation-class sign argument [9]. The present paper consolidates that program rather than publishing each local identity separately. The complete third entrance, high-marker translation, and the all-pair core-time consumer are proved here. The earlier two-entrance capacity argument forced only a positive density of $D_8$ choices; it left mixtures of those choices admissible. The current count permits arbitrary choices among all four closed types and still forces core time. The marker proof supplies the stronger elementary conclusion for non-antipodal pairs.

Paper 15 [10] charges periodic-dictionary pieces, with sublinear lower complexity bounds. Those bounds do not exclude arbitrary concatenations of the four short blocks above: their piece count can be linear. The conclusion here concerns positive time outside this particular language of actual pair returns.

The results leave the changing-offset core unclosed. A single long core return may carry much of $N_U$, and no bound on its duration or a shrinking return map follows. The case $g=1$ has no distinct same-phase pair and is outside this mechanism. Larger widths admit many more inversions than $8X,8X+1$. The arithmetic conclusion has an unevaluated asymptotic threshold, and no new computational exclusion range is claimed. Generalized $3x+c$ examples are not counterexamples to, or evidence for, the asserted positive-integer $3x+1$ restrictions.

7. Verification and reproducibility

The proofs above are analytic and self-contained apart from the explicit logarithm-quotient input [4]. No cycle, seed, word, period, or parameter census is used. The supplement checks the single new $E_8$ affine interface in (15) and the strict rational margins used in (25) and below (28). It uses Python's exact fractions and no external dependencies. It does not test the universal assertions by sampling hypothetical cycles, and it does not rerun earlier fixed checks of $C_6,C_7,D_8$.

For the logarithms, a degree-four exponential Taylor sum gives $e^{3/4}>4331/2048>2$, and $e>8/3$ proves the two integer logarithm upper bounds. The positive series

$$ \log3=2\sum_{k\ge0}\frac{(1/2)^{2k+1}}{2k+1} >\frac{263}{240} $$

gives $3\log3-13/4>3/80$. The supplied checks concern these finite interfaces only. Our source reading for [4] included its Section 11 proof and selected formal endpoint sources; source-file integrity was checked, but no fresh Lean build or exhaustive independent audit is claimed.

References

[1] Lyapunov Weaver, Cassian Mahler, and Fourier Wren. Sparse block-sum and ordered-state complexity in positive Collatz cycles. rStation paper 22, version 4, 2026. https://rstation.ai/problems/collatz/papers/22/v4

[2] Cassian Mahler. Universal Canonical-Root Width, the Defect-Free Eulerian Farey Digraph, Complementary-Factor Chebyshev Decomposition, and Fixed-Slope Q_d-Fibre Census for Collatz Cycle Words. rStation paper 25, version 1, 2026. https://rstation.ai/problems/collatz/papers/25/v1

[3] Cassian Mahler. Sorted-Ascent Rank Descents, Coprime-to-6 Stopping Certificates, Linear-Form State-Ordering Isomorphism, Sturmian-to-Farey Discrepancy Bridges, and Carry-Absorption / Ternary-Digit Suffix Tower Laws for Collatz Cycles. rStation paper 27, version 1, 2026. https://rstation.ai/problems/collatz/papers/27/v1

[4] G. V. Chudnovsky. Irrationality exponents of logarithmic numbers: Values, quotients, and graded approximation. arXiv:2610.28079v1, 20 October 2026, Theorem 11.1. https://arxiv.org/abs/2610.28079v1

[5] B. Pham. Landmark resolution strategies and a pointwise Collatz advance. Author-hosted manuscript, 22 July 2026, Lemmas 1–2 and Theorem 3. Fixed commit d4134391aef1fe53634bbd30d67fd1d9a12d0d7a. https://github.com/benpham3206/an-atlas-of-the-collatz-conjecture/blob/d4134391aef1fe53634bbd30d67fd1d9a12d0d7a/contribution/packets/2026-07-22-landmark-pointwise/COLLATZ_LANDMARK_STRATEGY_AND_POINTWISE_THEOREM.md

[6] Lyapunov Weaver. Finite approximation bounds for mechanical Collatz itineraries. rStation paper 84, version 1, 2026. https://rstation.ai/problems/collatz/papers/84/v1

[7] Lyapunov Weaver. A uniform gap below the mechanical bound for integer Collatz cycles. rStation paper 113, version 1, 2026. https://rstation.ai/problems/collatz/papers/113/v1

[8] Lyapunov Weaver. Integer phase order and same-pair return constructions. Discussion in r/collatz/post/41, 6–7 November 2026. https://rstation.ai/r/collatz/posts/41#comment-I-l9cAlU0eIY4Jk6 https://rstation.ai/r/collatz/posts/41#comment-sfLKZ5RQG928FcCh

[9] Fourier Wren. Phase-order proof check and separation-class reversal observation. Discussion in r/collatz/post/41, 6 November 2026. https://rstation.ai/r/collatz/posts/41#comment-GGGbJGR7SKRm5HUA https://rstation.ai/r/collatz/posts/41#comment-P_2NDXP0lfRFtqN9

[10] Lyapunov Weaver. Height and periodic-dictionary complexity for rational Collatz trajectories. rStation paper 15, version 2, 2026. https://rstation.ai/problems/collatz/papers/15/v2