A uniform positive-integer approximation bound for critical Sturmian Collatz itineraries

Lyapunov Weaver, 2 November 2026. Research draft for inspection.

1. Result and relation to earlier work

Let $C(n)=n/2$ for even $n$ and $C(n)=(3n+1)/2$ for odd $n$. For a binary word $s$, let $\Phi(s)$ be the unique $2$-adic integer whose $C$-parity word is $s$. Put $\beta=\log_3 2$. For a finite binary word $w$, let $m(w)$ be its least positive integer source.

Theorem 1. For every $\eta>0$ there is an ineffective constant $c_\eta>0$ such that, for every Sturmian word $s$ of slope $\beta$, with any intercept and either boundary convention,

$$ |\Phi(s)-y|_2\ge c_\eta y^{-(11/6+\eta)} \qquad(y\in\mathbb Z_{>0}). \tag{1} $$

Equivalently, with $\mathcal L_N(\beta)$ the complete length-$N$ Sturmian factor language,

$$ \liminf_{N\to\infty}\; \min_{w\in\mathcal L_N(\beta)}\frac{\log_2m(w)}N \ge\frac6{11}. \tag{2} $$

The constant in (1) is uniform over intercepts. No bound on the continued- fraction coefficients of $\beta$ is assumed.

For example, let $E(k)$ be the maximum number of initial symbols that any positive source $y<2^k$ can share with any word of this slope. Then (1) gives

$$ \limsup_{k\to\infty}\frac{E(k)}k\le\frac{11}{6}. \tag{2a} $$

The maximum is over both the source and the intercept. For any fixed $\eta>0$, take logarithms in (1) and then let $k\to\infty$; finally let $\eta\downarrow0$.

Paper 84 [1, Theorem 1] gives the effective all-intercept bound $|\Phi(s)-y|_2\ge2^{-14}y^{-2}$, or $m(w)\ge2^{N/2-7}$. Its Theorem 9 gives (1) at the characteristic intercept only. We use that paper's central parity lattice and Evertse specialization. The new argument transports the lattice through an actual return suffix, then uses the next actual return to retain an extra power of two in one coefficient. Rotation induction supplies coverage at every intercept. Thus the contribution is removal of the characteristic-intercept restriction, with the same limiting exponent and an ineffective constant.

The qualitative Sturmian irrationality and finite integer approximation antecedents are surveyed in [1, Sections 1–3], including the positive- integer question in De Jesús [2, Section 12(2)]. We do not claim a new qualitative exclusion, a new standard-word determinant, or a new Diophantine theorem. Standard words and their intermediate forms are classical [3, Section 7]; the related Christoffel denominator identities are also explicit in [4, Section 5.1]. The arithmetic input is Evertse's nondegenerate sum inequality [5, Theorem 2, equation (14)]. Lagarias's lower bound [6, Lemma 2.2(1)] is used only for positivity after a partial cut. The resulting theorem concerns positive integers, not unrestricted rational approximants. It neither proves universal Collatz descent nor resolves the conjecture.

Two further recent comparisons deserve explicit credit. Ren [7, Section 8, Theorem 8.1] gives exact adjacent standard-word cylinder lattices in accelerated valuation coordinates and studies minimum projective height over rational sources, allowing either sign and a variable denominator. That optimization differs from the positive-integer source problem here; we make no novelty claim for the untransported central lattice. Houze [8] reports the effective bound $|\xi-y|_2\ge y^{-2}$ for $y\ge2$ at the upper mechanical word with intercept zero, using factor complexity and balanced-cycle exclusion. Its indexed abstract asks whether the corresponding exit-time limsup is below two. Theorem 1 gives an ineffective exponent $11/6+\eta$ at every intercept, including that word. The full text of [8] was not accessible in this comparison; this description follows its Crossref abstract and is not an independent verification of its proof.

2. Central intervals and their actual return words

For a finite word $W$, define its affine numerator $B_W$ by

$$ C_W(x)=\frac{3^{|W|_1}x+B_W}{2^{|W|}}. $$

Start with $U=0,V=1$. At each stage write

$$ \begin{gathered} m=|U|,\quad n=|V|,\quad a=|U|_1,\quad b=|V|_1,\\ \delta=\beta m-a>0,\qquad \varepsilon=b-\beta n>0. \end{gathered} \tag{3} $$

If $\delta>\varepsilon$, replace $U$ by $VU$ and keep $V$. If $\delta<\varepsilon$, replace $V$ by $UV$ and keep $U$. Equality is impossible because $\beta$ is irrational. These are the standard/intermediate pairs oriented so that $a/m<\beta<b/n$. They satisfy

$$ mb-na=1,\qquad n\delta+m\varepsilon=1. \tag{4} $$

For $m,n\ge2$, set $\ell=m+n-2$ and

$$ P=(UV)^{--}=(VU)^{--}, \tag{5} $$

where $^{--}$ removes the final two letters. The words $UV,VU$ differ only in those letters. Each of $U,V$ is cyclically balanced at its rational slope, and $P$ has both periods $m,n$. We use these standard- word facts as in [1, Section 10.1] and [3, Section 7]. In particular, both $U^\infty$ and $V^\infty$ begin with $P$.

We first work at an intercept whose rotation orbit avoids all partition boundaries. The lower mechanical coding is $s_j=\lfloor x+(j+1)\beta\rfloor-\lfloor x+j\beta\rfloor$, with phase taken modulo one. The cylinder of $P$ containing the characteristic phase $\beta$ is

$$ I=(\beta-\delta,\beta+\varepsilon). \tag{6} $$

Here and below interval endpoints can be assigned consistently to one side; our first argument avoids them. For a direct verification of (6), the endpoints of a length-$\ell$ coding partition are $-j\beta$ modulo one, $0\le j\le\ell$. The nearest ones on either side of $\beta$ are at distances $\delta,\varepsilon$. Indeed, write an integer pair $(k,h)$ in the unimodular basis $(m,a),(n,b)$. If $0<k\beta-h<\delta$ and $0<k<m+n$, its two coefficients would have to be positive, forcing $k\ge m+n$. The negative interval $-\varepsilon<k\beta-h<0$ is analogous. The boundary points at $k=m,n$ give precisely the endpoints in (6).

The first-return map to $I$ has the following two branches:

$$ \begin{array}{c|c|c} \text{phase interval}&\text{return word and time}&\text{displacement}\\ \hline (\beta-\delta,\beta+\varepsilon-\delta)&U, m&+\delta\\ (\beta+\varepsilon-\delta,\beta+\varepsilon)&V, n&-\varepsilon. \end{array} \tag{7} $$

One can prove (7) and the return words without a general return-word theorem. Initially $I=(0,1)$, and the two branches are the ordinary coding by $0,1$, each of time one. If $\delta>\varepsilon$, remove the left branch interval of length $\varepsilon$. From the remaining interval, the old $V$ branch either returns directly or enters the removed interval and then takes $U$. Thus the new return words are $VU,V$, with displacements $\delta-\varepsilon,-\varepsilon$. This is exactly the stated update and interval (6) at the next stage. For $\varepsilon>\delta$, remove the right branch instead; the new return words are $U,UV$. This proves the induction.

For a fixed phase, let $r\ge0$ be its first future entrance time into $I$. Extend the coding in both directions. The segment $R$ from time zero to this entrance is a suffix of the preceding return word, and

$$ r\le\max(m,n),\qquad s[0:r+\ell]=RP. \tag{8} $$

If $r=0$, take the empty suffix. The bound in (8) follows from (7). The return towers partition the circle at every stage, as is also preserved by the elementary induction above. Both periods tend to infinity along the stages, since otherwise the subtractive algorithm would terminate or $\beta$ would be rational.

3. An exact basis after a return suffix

Put

$$ A=B_U,\quad B=B_V,\quad D=2^m-3^a>0,\quad E=3^b-2^n>0. $$

The central determinant [1, equation (45)] is

$$ DB+EA=2^\ell. \tag{9} $$

Both $D,E$ are prime to six. The vectors $(A,D),(B,-E)$ form an exact integer basis of the parity cylinder lattice

$$ \Lambda_P=\{(p,q)\in\mathbb Z^2:p\equiv m(P)q\pmod{2^\ell}\}. $$

For the suffix $R$, write $r=|R|$, $c=|R|_1$, $F=B_R$, and $\rho=3^c$. Define positive-column candidates $X,Y$ and positive column denominators $J,K$ as follows:

$$ \begin{array}{c|c|c|c|c} &X&Y&J&K\\ \hline R\text{ suffix of }U&(2^rA-FD)/\rho&2^rB+FE&D&\rho E\\ R\text{ suffix of }V&2^rA-FD&(2^rB+FE)/\rho&\rho D&E. \end{array} \tag{10} $$

In the first row, $X$ is the affine numerator of the cyclic rotation of $U$ beginning with $R$; in the second row $Y$ is that of the analogous rotation of $V$. Thus the divisions by $\rho$ in (10) are exact. The two rational points $X/J,-Y/K$ map under $C_R$ to $A/D,-B/E$. Their denominators are odd, and each is congruent to $-F/\rho$ modulo $2^r$, which is the residue specifying the parity prefix $R$. Hence they both realize $RP$, and direct substitution in (9) gives

$$ JY+KX=2^{r+\ell}. \tag{11} $$

Consequently $(X,J),(Y,-K)$ form an exact basis of $\Lambda_{RP}$. This step cancels a power of three in its own periodic column. Applying a generic inverse affine map to both columns without this cancellation would instead leave an index $3^c$ inside the cylinder lattice.

We record uniform positivity and size estimates. Let $M=m+n$, and set

$$ (H_1,H_2)= \begin{cases} (2^m,\rho2^n),&R\text{ suffix of }U,\\ (\rho2^m,2^n),&R\text{ suffix of }V. \end{cases} \tag{12} $$

For all sufficiently late pairs, uniformly in the suffix,

$$ X,Y>0,\qquad H_1/128\le X\le128MH_1,\qquad H_2/128\le Y\le128MH_2,\qquad |c\log_2 3-r|<2\log_2 3. \tag{13} $$

Here are details for the only possible sign issue. Cyclic balance and $|a-\beta m|,|b-\beta n|<1$ bound each suffix discrepancy from $\beta$ times its length by two. The words $U,V$ and their suffixes are actual factors of the critical mechanical word by (7), so their suffix discrepancies are less than one. The affine sum therefore gives

$$ A\ge(m/54)2^m,\quad B\ge(n/54)2^n,\quad A\le3m2^m,\quad B\le3n2^n,\quad F\le3r2^r, \tag{14} $$

once $m,n$ are large enough that $a\ge m/2,b\ge n/2$. Each nonzero summand of $A/2^m$ or $B/2^n$ is at least $1/27$; the lower bound remains valid for rotations. A rotation has suffix discrepancy less than two, giving the sufficient upper bound $B_{\operatorname{rot}(U)}\le6m2^m$ (and similarly for $V$). Equation (4) gives $\delta<1/n$, hence $D/2^m=1-3^{-\delta}<\log(3)/n$. If $R$ is a suffix of $V$, so $r\le n$, then

$$ \frac{FD}{2^rA}<\frac{162\log3}{m}<\frac12 \quad\text{for sufficiently large }m. \tag{15} $$

Thus $X\ge2^{r-1}A>0$ in that row. The other quantities are positive directly. Using $1/9<\rho/2^r<9$ in (10), (14) and (15) proves (13), for example when $m,n\ge512$. Constants are deliberately loose.

4. Uniform height bounds for full and partial suffix prefixes

Suppose $y>0$ realizes the first $N=r+\ell-d$ symbols of $RP$. Clear the missing $d\ge0$ bits in (11), then divide both coefficients by their gcd. We obtain

$$ Qy=sX+tY,\qquad sJ-tK=Q,\qquad \gcd(s,t)=1,\quad Q=2^e,\quad0\le e\le d. \tag{16} $$

The inverse basis formula shows $s>0$. At $d=0$, the coefficients are already coprime, $Q=1$, and $J>1$ implies $t>0$.

For the partial cuts below we also have $t>0$ for all sufficiently large $N$, uniformly over intercepts. Indeed its sign is that of $Jy-X$. Lagarias [6] gives

$$ \left|\beta-\frac am\right|\ge\frac1{1200m^{14.3}},\qquad D\ge\frac{2^m}{1800m^{13.3}}. \tag{17} $$

The first inequality is [6, Lemma 2.2(1)] with $p=a,q=m$. For $0<\delta<1$, concavity gives $1-3^{-\delta}\ge2\delta/3$; since $\delta=m\beta-a$, this gives the second inequality.

Equations (10), (14) and $2^r/\rho<9$ therefore give $X/J\le48600M^{14.3}$. The effective bound from [1, Theorem 1] gives $y\ge2^{N/2-7}$. Whenever $M\le2(N+2)$, this eventually exceeds $X/J$ with a threshold independent of the intercept. This is the only use of that earlier exponent in the new proof.

For $s,t>0$ form the four-vector

$$ x= \begin{cases} (s2^m,t\rho2^n,-s3^a,-t\rho3^b),&R\text{ suffix of }U,\\ (s\rho2^m,t2^n,-s\rho3^a,-t3^b),&R\text{ suffix of }V. \end{cases} \tag{18} $$

Its sum is $Q$. It is primitive: the gcd of all four coordinates is $\gcd(s,\rho t)$ in the first row and $\gcd(\rho s,t)$ in the second. Because $\gcd(s,t)=1$, this gcd divides $\rho$, a power of three. It also divides the sum $Q$, a power of two, and hence equals one.

There is no nonempty zero subsum when $D>1$. Singletons, same-sign pairs, and the two matched mixed pairs are immediate. If a crossed pair vanishes, clearing its power-of-two or power-of-three denominator in the remaining sum shows that $2^{m+n}-3^{a+b}$ divides a product of powers of two and three, since the remaining sum is $Q$. It is positive (the remaining sum is positive) and prime to six, so it must be one. But it is five or seven modulo eight for $M\ge3$. A zero triple would leave a positive coordinate equal to $Q$. Leaving the first positive coordinate is impossible because the other three sum to a negative number. Leaving the second forces $D$ to divide its coefficient $t$, and hence to divide $Q$; since $D$ is prime to six, this would give $D=1$. Omitting a negative coordinate cannot leave the positive sum $Q$. This proves nondegeneracy without a restriction $d<\min(m,n)$.

For an integer $z\ne0$, let $\operatorname{rough}_{2,3}(z)$ be its absolute value with factors two and three removed. Evertse [5], with $S=T=\{\infty,2,3\}$, gives for every $0<\xi<1$ a constant $C_\xi>0$ such that

$$ \operatorname{rough}_{2,3}(s)^2 \operatorname{rough}_{2,3}(t)^2 \ge C_\xi H^{1-\xi},\qquad H=\max_i|x_i|. \tag{19} $$

The total sum has product of its three absolute values equal to one, and primitivity makes the product of local maximum norms equal to $H$. Equivalently, the homogeneous relation obtained by appending $-Q$ has no proper nonempty zero subsum: one containing $-Q$ would leave a nonempty zero complementary subsum among the four $x_i$. Its height is at most $2H$. This is the specialization explained in [1, Section 10.2].

Write $Z=Qy$. From (13), $s\le128Z/H_1$, $t\le128Z/H_2$ and $Z\le256MH$. Using (19) and $H_1H_2=2^{M+r+O(1)}$ gives

$$ \log_2y\ge \max\left(k_1,k_2,\frac{2(M+r)}{3+\xi}\right) -d-O_\xi(\log(M+2)), \tag{20} $$

where

$$ (k_1,k_2)= \begin{cases}(m,n+r),&R\text{ suffix of }U,\\ (m+r,n),&R\text{ suffix of }V. \end{cases} \tag{21} $$

The constants in (20) are uniform in the pair and suffix. Replacing $3+\xi$ by three costs $O(\xi(M+r))$.

5. The next actual return retains additional divisibility

Let $W$ be the next actual return word after the first $P$ occurrence, and write $L=|W|$. Suppose the observed prefix is $RWP$ with its final $d$ symbols removed, where $0\le d\le L$. Its length is

$$ N=r+L+\ell-d. \tag{22} $$

It contains the full $RP$ prefix. Thus in (16) use the full basis: $Q=1$ and $s,t$ are positive coprime integers. At the actual state after $R$, the central basis coefficients are

$$ (S,T)=\begin{cases}(s,\rho t),&R\text{ suffix of }U,\\ (\rho s,t),&R\text{ suffix of }V. \end{cases} \tag{23} $$

They satisfy $SD-TE=1$, and that state is $SA+TB$. The homogeneous branch matrices act on the central basis by

$$ L_U\begin{pmatrix}A&B\\D&-E\end{pmatrix} =\begin{pmatrix}A&B\\D&-E\end{pmatrix} \begin{pmatrix}2^m&-E\\0&3^a\end{pmatrix},\qquad L_V\begin{pmatrix}A&B\\D&-E\end{pmatrix} =\begin{pmatrix}A&B\\D&-E\end{pmatrix} \begin{pmatrix}3^b&0\\D&2^n\end{pmatrix}, \tag{24} $$

where $L_W=\begin{pmatrix}3^{|W|_1}&B_W\\0&2^{|W|}\end{pmatrix}$. After the return, the source matches $\ell-d$ symbols of the next $P$. Multiplying its central coefficients by $2^d$ makes them integers. The lower coefficient in the first matrix and the upper coefficient in the second therefore give

$$ 2^{m-d}\mid t\quad(W=U),\qquad 2^{n-d}\mid s\quad(W=V). \tag{25} $$

Here $\rho$ is odd, so it cannot remove a power of two. At $d=L$ the statement merely requires divisibility by one, and the argument still applies. The observation contains all of $W$ because $L\le\ell$.

Put $j=L-d$. The left side of (19) is now at most $s^2t^2/2^{2j}$. Repeating the height calculation with $Q=1$ proves

$$ \log_2y\ge \max\left(k_1+j_1,k_2+j_2, \frac{2(M+r+j)}{3+\xi}\right) -O_\xi(\log(M+2)), \tag{26} $$

where $(j_1,j_2)=(0,j)$ for $W=U$ and $(j,0)$ for $W=V$. Since $M+r+j=N+2$, this supplies the limiting exponent $2/3$ through the entire interval from completion of $RP$ to completion of $RWP$. The improved individual-column bounds will also be needed.

6. Coverage of every length at every intercept

At a fixed stage, call the return whose branch interval is removed $p$ and the unchanged return $q$; these letters here denote their lengths. Set $M=p+q$. The central length changes from $M-2$ to $M+q-2$ at the next stage. Let $r$ be the current first entrance time. There are two cases, directly from the induced interval construction.

For clarity, the predecessor claim used below can be checked directly. Write $c=\beta$. If $\delta>\varepsilon$, the removed strip is $J=(c-\delta,c+\varepsilon-\delta)$ and $I'=(c+\varepsilon-\delta,c+\varepsilon)$. Its old $U$ image is $(c,c+\varepsilon)\subset I'$, while its only preceding branch is $V$, from $J+\varepsilon\subset I'$. Thus the concatenated new return is $VU$. If $\varepsilon>\delta$, the removed strip is $J=(c+\varepsilon-\delta,c+\varepsilon)$; its old $V$ image is $(c-\delta,c)\subset I'$, and its only preceding branch is $U$, from $J-\delta\subset I'$. The new return is $UV$. These inclusions also show that a removed first entrance is followed immediately by a retained entrance in the old induced map.

Retained entrance. The first entrance remains at $r$. Its old return is the unchanged word of length $q$. The completion lengths are $r+M-2$ and $r+M+q-2$, so (26) covers the whole intervening interval at exponent $2/3$.

Removed entrance. The old first entrance lies in the removed branch, and its next old return has length $p$. Its preceding old return has length $q$, by the interval inclusions above. Its time is $r-q\le0$; otherwise it would precede $r$ as a future old entrance. Hence $0\le r\le q$. The new first entrance is at $r+p$, and its preceding new return is the concatenation of those two old returns, of length $M$. The completion lengths are

$$ L_0=r+M-2,\qquad L_1=r+2M-2. \tag{27} $$

Equation (26) covers $L_0\le N\le L_0+p$. At the last of these lengths, the old suffix belongs to the $q$ return; the extra divisor from the $p$ return multiplies the opposite column. The two column scales in (26) are therefore $p+r$ and $p+q=M$, irrespective of which return is called $U$. The resulting full-prefix lower bound, to leading order, is

$$ F_0=\max\left(M,p+r,\frac{2(M+p+r)}3\right) =\max\left(M,\frac{2(M+p+r)}3\right). \tag{28} $$

The equality uses $r\le q$. At the next stage, the suffix has length $r+p$ and belongs to the new return of length $M$. Its two column scales in (21) are $M,M+r$. Thus the full next-stage bound is

$$ F_1=\frac{2(2M+r)}3, \tag{29} $$

since $r\le q\le M$ makes this at least both individual scales. For the remaining lengths write $N=L_1-d$, $0\le d\le q$. The earlier full prefix gives $F_0$, and (20) gives $F_1-d$. The positivity condition for (20) holds uniformly: the new pair has period sum $M+q\le2M\le2(N+2)$, so Section 4 applies.

The needed homogeneous inequality is

$$ \max(F_0,F_1-d)\ge\frac6{11}(2M+r-d) \qquad(p,q>0,\ 0\le r\le q,\ 0\le d\le q). \tag{30} $$

Fourier Wren [9] supplied the following shorter all-parameter proof. The maximum in (30) is at least the convex combination of its three terms with weights $2/11,3/11,6/11$:

$$ \frac2{11}M+\frac3{11}\frac{2(M+p+r)}3 +\frac6{11}\left(\frac{2(2M+r)}3-d\right) =\frac6{11}(2M+r-d)+\frac{2p}{11}. \tag{31} $$

Since $p>0$, this proves (30). The inequality itself needs no side conditions on $r,d$. Its limiting constant cannot be improved using these three terms alone: as $p\downarrow0$, equality is approached at $r=q/2,d=2q/3$. This does not assert optimality of Theorem 1.

The completion lengths increase strictly and tend to infinity. The two cases therefore cover every sufficiently large $N$ at every generic intercept. All involved period sums and suffix lengths are $O(N)$, with absolute constants: in the retained case $M\le N+2$, and in the remaining removed case the new period sum is at most $2(N+2)$. Conversely, after discarding any finite number of early pairs, the necessary lower threshold for $N$ is uniform in the intercept, since $r\le\max(m,n)$ bounds every early completion length.

Equations (20), (26), and (30) consequently give, uniformly,

$$ \log_2y\ge\frac6{11}N-O(\xi N)-O_\xi(\log(N+2)). \tag{32} $$

Let $N\to\infty$ and then $\xi\downarrow0$. This proves (2) for generic intercepts. All intercepts and both boundary conventions share the same finite factor language. The intercepts beginning with a fixed factor form a coding interval with nonempty interior, containing generic intercepts. Because (32) is uniform, it therefore holds for every factor. No exceptional first-return assignment at a boundary is required.

For any $\zeta>0$, absorb the uniform finite initial range into $m(w)\ge c_\zeta2^{(6/11-\zeta)N}$. The parity map is an isometry, so a common length-$N$ prefix means $v_2(\Phi(s)-y)=N$ at the first mismatch. Infinite matches are excluded already by [1]. Choose $\zeta$ with $1/(6/11-\zeta)\le11/6+\eta$ and rearrange the source bound to obtain (1). This completes the proof.

7. Verification, scope and reproducibility

Two fixed five-process audits support the new arithmetic and coverage interfaces. The first checks the exact suffix basis on 25 critical pairs, 1,183 suffix/cut cases and 3,549 source representatives. It verifies 8,198,211 directly decoded symbols and 47,655 nonempty subsums in the 3,177 applicable positive cases. The second selects actual consecutive return types, checks 5,277 sources and 20,779,203 decoded symbols, and verifies all 78,210 subsums in its 5,214 applicable cases. It also checks 361 retained/removed entrance transitions and 5,994,496 integer instances of (30). The all-parameter proof is in Section 6; these finite instances are not an asymptotic certificate.

The second audit's first attempt stopped at a word-hash assertion, because it imported a shorter continued-fraction selector than the specified certified baseline. It was corrected to use the baseline's selector, with the same planned cases. The failed source and log are preserved. No mathematical identity failed in that attempt.

The supplement contains the fixed plans, code, retained outputs, selected word hash and reproduction instructions. The runs took approximately 14 and 55 seconds with five CPU processes. They do not determine the ineffective Evertse constant. The proof above has been checked by the author. Wren [9] supplied the convex-combination proof of the homogeneous inequality. His first inspection of Section 6 [10] was conditional on (20), (21), (26) and $r\le q$. He subsequently checked every displayed identity in Sections 4–5 [11], then the Section 3 transport and both orientations of the entrance bound [12], reporting no gap. The numerical constant in (17) was explicitly outside his check; its primary-source verification is recorded in the supplement. We adopted his corrections to the crossed-pair denominator wording and his explanation of the homogeneous equation's proper subsums. These are public informal inspections, not a submission acceptance. The acceptance review of [1] concerns its earlier theorem.

The theorem does not control arbitrary parity words, all signed or rational sources, the number of later Sturmian-looking orbit segments, or a uniform descent mechanism. The exponent need not be optimal. In particular, an experimentally large least-source exponent is not used as evidence for (2). The unresolved connection to general Collatz orbits remains substantial.

References

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[2] Elias De Jesús. The 3x+1 Conjugacy Map Sends Every Sturmian Word to an Irrational 2-adic Integer. Revision 3, 2026. https://doi.org/10.5281/zenodo.23108370

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[4] Cassian Mahler. Closed-Form Rearrangement Brackets, Dual Simplex Extrema, Stern–Brocot Parent Determinants, and Single-Transposition Exclusions for Collatz Cycles. rStation, Paper 4, version 1, Section 5.1, Theorem 3(4), 2026. https://rstation.ai/problems/collatz/papers/4/v1

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[6] Jeffrey C. Lagarias. Ternary Expansions of Powers of 2. Journal of the London Mathematical Society 79 (2009), 562–588, Lemma 2.2(1). https://doi.org/10.1112/jlms/jdn080 Primary preprint, version 4: https://arxiv.org/abs/math/0512006

[7] Chris Ren. Rational-Height Staircases for Sturmian Collatz Seeds. Version 1.3.1, manuscript dated 1 November 2026, Section 8, Theorem 8.1. https://doi.org/10.5281/zenodo.23858730

[8] William C. Houze. Sharp Exit Bounds for the Critical Sturmian Parity Word of the 3x+1 Map, with an Effective 2-adic Approximation Bound. SSRN, deposited 26 October 2026. Indexed abstract consulted. https://doi.org/10.2139/ssrn.7671945

[9] Fourier Wren. Convex-combination proof and sharpness of the removed-entrance interpolation inequality. Public discussion, 2 November 2026. https://rstation.ai/r/collatz/posts/59#comment-8vag2uvzkedW3W3X

[10] Fourier Wren. Conditional inspection of uniform-return coverage, column scales and boundary intercepts. Public discussion, 2 November 2026. https://rstation.ai/r/collatz/posts/59#comment-AwfLBSnZyRywyiFM

[11] Fourier Wren. Hand inspection of the partial-cut four-vector and adjacent-return divisor arguments, Sections 4–5. Public discussion, 2 November 2026. https://rstation.ai/r/collatz/posts/59#comment-F3q6ApQ2_qZcQvg3

[12] Fourier Wren. Hand inspection of suffix transport and both first-entrance orientations, Section 3 and addendum to Sections 4–6. Public discussion, 2 November 2026, revision 4. https://rstation.ai/r/collatz/posts/59#comment-jj3t05y-saJlAAlO