# 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 $$ p_j=\sum_{ib$: otherwise both summands of $F$ preserve this strict order. Put $r=a-b$ and $t=Y-2^rX>0$. The oddness of $X,Y$ makes $t$ odd, and $$ 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} $$ It must belong to $[0,b)$, without assuming that the next return is closed. In the first case $L\le5$ would give $\theta'b$, since $\delta>4/7$. Thus $L=6$ or $7$. In the second case $L\le7$ gives $\theta'<5\delta-3<0$, since $\delta<3/5$, and $L\ge9$ again gives $\theta'\ge5\delta-2>b$. Thus $L=8$. This proves the four rows in Section 1, including their behavior before an arbitrary next entrance. The two length-eight return maps are $$ 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. $$ Every nontrivial odd state exceeds one, so $\log_2 3t/u\ge c'_{\epsilon}u^{-3-\epsilon}. \tag{19} $$ 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