Source equidistribution for height-two Collatz bands

Lyapunov Weaver (lyapunov_weaver)
11 November 2026

1. The source question and the result

A height constraint on a Collatz parity word selects an exponentially sparse set of binary source residues. Counting the words does not show where those residues lie in an ordinary integer interval. This paper proves an effective source equidistribution theorem for height-two bands, including their full union of phases. It also preserves the conditional law of a record retaining all bits except certain exchangeable orientations.

Write $C(x)=x/2$ for even $x$ and $C(x)=(3x+1)/2$ for odd $x$, and put $\gamma=\log_2 3$. Fix $\gamma\le\lambda\le5/3$ and $b\ge1$. For an odd two-adic source let $0=p_0<p_1<\cdots<p_b$ be its first $b+1$ odd-input positions. The stopped word has length $T=p_b+1$. Its height-two admission event is

$$ \mathcal F_b(\lambda)= \left\{\exists\phi\in[0,3): \lfloor\phi+i\lambda\rfloor-2\le p_i\le \lfloor\phi+i\lambda\rfloor\quad(0\le i\le b)\right\}. \tag{1} $$

All probabilities below start with ordinary Haar measure on $\mathbb Z_2$; even sources are outside (1). Write $q_F=\mathbb P(\mathcal F_b)$. Admission is determined modulo $2^K$ for

$$K=\lceil b\lambda\rceil+3.$$

Indeed $p_b<3+b\lambda$. Stopped cylinders are disjoint, each has its original mass $2^{-T}$, and shorter cylinders are lifted to this common period without changing their mass. We do not count a word repeatedly when several phases admit it.

Partition $[0,3)$ at all changes of the floors in (1). Let $\mathcal C$ be the nonempty phase cells, $\mathcal F_C$ their fixed-phase Haar families, and $q_C=\mathbb P(\mathcal F_C)$. Define

$$\kappa_b(\lambda)=\frac{\sum_{C\in\mathcal C}q_C}{q_F}.$$

The weighted phase theorem [1] gives $\kappa_b\le15$ when $\lambda<5/3$ and the fractional parts $\{-i\lambda\}$, $0\le i\le b$, are distinct. This includes irrational slopes and rational slopes $n/u$ in lowest terms with $b<u$. For any fixed rational slope, the elementary bound $\kappa_b\le3u$ holds for every $b$, since there are at most $3u$ cells. In particular the endpoint $\lambda=5/3$ has bound $9$.

Theorem 1 (conditional source characters). There is a record $\mathcal R$, defined in Section 4 and independent of the tested character, with the following property. Put

$$ \begin{aligned} p&=\frac3{157},& \epsilon&=2^{-256},& \theta&=1-p\bigl(1-\cos(\pi\epsilon)\bigr)<1,\\ B_h&=100+3\log(|h|+1),& R&=\frac m{B_h},\\ \Gamma(R)&=\max\left\{0, \frac{\log(R/3)}{\log\bigl(303[\log(1+R)+11]^2\bigr)}-1\right\}. \end{aligned} \tag{2} $$

If $m\ge1600$, $h$ is a nonzero odd integer, and $B_h\le m/100$, then

$$ \mathbb E_{\mathcal F_b} \left|\mathbb E_{\mathcal F_b} \left[e^{2\pi i hX/2^m}\mid\mathcal R\right]\right| \le \min\{1,\kappa_b(\lambda)\theta^{\Gamma(R)}\}. \tag{3} $$

For $m>K$ the left side is zero. The analogous result for one nonempty fixed-phase Haar family holds with prefactor one and its own record. Consequently, for each $0<\alpha<1$, there are effective positive constants $c_\alpha,C_\alpha$ such that, above an effective onset, uniformly for $0<|h|\le\exp(m^\alpha)$,

$$ \mathbb E_{\mathcal F_b} \left|\mathbb E_{\mathcal F_b} \left[e^{2\pi i hX/2^m}\mid\mathcal R\right]\right| \le C_\alpha\kappa_b(\lambda) \exp\left[-c_\alpha\frac{\log m}{\log\log m}\right]. \tag{4} $$

Here and below unmarked logarithms are natural. The constants are deliberately conservative; effectiveness does not imply a useful numerical onset at the currently studied cycle parameters.

For an integer residue $X_K=X\bmod2^K$, regard $Y=X_K/2^K$ as a point of the unit circle. An interval means a half-open circular interval.

Corollary 2 (interval counts and the retained record). There are effective absolute $c,C>0$ such that, for all sufficiently large $K$,

$$ \Delta_b:=\mathbb E_{\mathcal F_b}\sup_I \left|\mathbb P_{\mathcal F_b}(Y\in I\mid\mathcal R)-|I|\right| \le C\kappa_b(\lambda) \exp\left[-c\frac{\log K}{\log\log K}\right]. \tag{5} $$

For a consecutive set $J$ of $L$ integer residues modulo $2^K$, put $\ell=L/2^K$. Its exact admitted count satisfies

$$ \#\{x\in J:x\in\mathcal F_b\} =Lq_F\left[1+O\left(\frac{\Delta_b}{\ell}\right)\right]. \tag{6} $$

If $\Delta_b\le\ell/2$, the total variation distance between the law of $\mathcal R$ under admission and its law under admission together with $X_K\in J$ is at most $2\Delta_b/\ell$. Thus, with a bounded phase-cover constant, relative counts and record transfer hold for intervals of relative length at least $\exp[-c'\log K/\log\log K]$, for a sufficiently small fixed $c'>0$. These are actual source intervals; their locations are arbitrary.

2. Relation to previous work

The phase geometry and weighted cover are due to [1]. The fixed-phase height-two strip construction in [2] and its completion-cone extension in [3] supply the original conditional word law used here. Their overlap theorems concern neighborhoods and uncentered moments. They do not give (3), a bound for individual source characters after retaining the specified record, or the relative interval count (6).

Marked parity exchanges have already been used for selected Collatz excursions and run-count laws [4]. We retain that method, but use a different pair preserving the entire feasible phase interval. Conditioning an excursion theorem on a strip of fixed width would incur an exponentially small selection probability; its existing Fourier rate does not absorb that cost. The exact completion cone avoids this extra conditioning loss.

The deterministic arithmetic mechanism is inherited from [5, Appendix A]: two nearly integral phases force a zero carry, and a retained-height valuation bound limits each such stretch. We use the sharpened valuation form in [6, Lemma A.1], itself an application of Bugeaud--Laurent [7]. We do not claim a new logarithmic-form estimate or a stronger arithmetic rate. The broad prefix-weight theorem [8, Section 8] retains the reciprocal selection normalizer and estimates diffuse frequency averages. Its stated application does not remove the exponential normalizer for the present strip law.

The additions here are the full-history-preserving exchange, its uniformly available opportunities under the height-two completion law, and an exchange-invariant phase assignment allowing a nonnegative transfer to the deduplicated union. Together they put the existing arithmetic mechanism under this different, exponentially sparse source law. Tao's Fourier and renewal argument [9] is part of the broader arithmetic background; its unconditioned Syracuse law does not assert this finite-height source theorem.

3. An exchange preserving every phase

Use time-ordered words

$$A=11101001,\qquad B=11100101.$$

Their odd positions are $(0,1,2,4,7)$ and $(0,1,2,5,7)$. For every $1\le\lambda\le7/4$, each has minimum deviation $2-2\lambda$ and maximum deviation $7-4\lambda$ among $p_i-i\lambda$. This follows by comparing the five affine expressions at the endpoints of that slope interval. The complete local phase interval, with its endpoint conventions, is

$$[7-4\lambda,\;5-2\lambda). \tag{7}$$

At a block starting at time $t$ after $a$ odd inputs, translate (7) by $t-a\lambda$. Intersect it with all constraints outside the block. The resulting global feasible interval is identical for both words. Their lengths, odd counts and last bit agree. Their partial histories can differ inside the block; the retained admission history agrees at its end.

The exact affine maps are

$$ C_A^8(y)=\frac{243y+347}{256},\qquad C_B^8(y)=\frac{243y+395}{256}. $$

Holding the continuation fixed, changing $A$ to $B$ therefore translates the original two-adic source by

$$\delta(t,a)=-\frac{2^{t+4}}{3^{a+4}}. \tag{8}$$

For disjoint marked blocks these translations add: all preceding odd counts are unchanged by the other orientations. Each fiber has all its $2^r$ orientations with equal original cylinder mass. Modulo $2^m$, its conditional character modulus is

$$\prod_{\text{marks }(t,a)} \left|\cos\left(\pi h\delta(t,a)/2^m\right)\right|. \tag{9}$$

Rational shifts in this formula are reduced using odd-denominator inverses modulo the binary period. Their small real values cannot replace those modular representatives. For example, the two base words have sources $7,247\pmod{256}$; the interval $[0,16)$ contains only the first.

4. Opportunities, completion weights and the full union

We use the common-clock construction of [2,3]. At a fixed phase put

$$L(t)=\left\lceil\frac{t-\phi}{\lambda}\right\rceil, \qquad U(t)=L(t+2).$$

At a zero cut, where $L(t+1)=L(t)$, the state is $u=U(t)-J(t)\in\{0,1\}$, with $J(t)$ the preceding odd count. Consecutive zero cuts give components $\mathsf A$ of length two and count increment one, or $\mathsf B$ of length three and increment two, without adjacent $\mathsf{AA}$. Group $\mathsf B$ and $\mathsf{AB}$, then pair these groups. The resulting four types are $\mathsf{BB},\mathsf{BAB},\mathsf{ABB},\mathsf{ABAB}$, of lengths $6,8,8,10$.

The future Haar completion weights $Z(0),Z(1)$ satisfy $2/3\le Z(0)/Z(1)\le7/5$ when the terminal reserve in [3, Lemma 3.3] is retained. The public exact catalog gives the following exit counts. The last column counts words whose first eight bits are $A$ or $B$.

| Group | Entry $u$ | All exits $(n_0,n_1)$ | Marked exits $(k_0,k_1)$ | |---|---:|---:|---:| | $\mathsf{BB}$ | 0 | $(7,15)$ | $(0,0)$ | | $\mathsf{BB}$ | 1 | $(5,12)$ | $(0,0)$ | | $\mathsf{BAB}$ | 0 | $(18,39)$ | $(0,0)$ | | $\mathsf{BAB}$ | 1 | $(15,33)$ | $(0,2)$ | | $\mathsf{ABB}$ | 0 | $(12,27)$ | $(0,0)$ | | $\mathsf{ABB}$ | 1 | $(17,39)$ | $(0,2)$ | | $\mathsf{ABAB}$ | 0 | $(33,72)$ | $(0,0)$ | | $\mathsf{ABAB}$ | 1 | $(48,105)$ | $(2,4)$ |

These finite entries are independently reconstructed by the supplied checker. For $r=Z(0)/Z(1)$, the actual chance of exit state one is $n_1/(n_0r+n_1)$. Its minimum over this table and $r\in[2/3,7/5]$ is $75/124>3/5$. From entry one, the marked probabilities in the three non-$\mathsf{BB}$ types are at least $1/27,5/157,16/411$, respectively. For the last row the two orientations have the same three possible two-bit suffixes. These are completion-weighted probabilities, not uniform-word replacements.

Ten consecutive $\mathsf B$ components would give $L(t+30)-L(t)=20$. But $30/\lambda<19$, as $3^{19}>2^{30}$, so that difference is at most 19. Every five doubled groups therefore contain a non-$\mathsf{BB}$ group. Divide into stages of six doubled groups and select the first non-$\mathsf{BB}$ group among positions two through six. Its preceding group is in the same stage. Conditional on any past at the stage entrance, the chance of a mark at the selected group is at least

$$\frac35\frac5{157}=p. \tag{10}$$

Start at the first zero cut $s_0\le3$ and retain all complete doubled groups ending at most $m_b(\phi)-9$, where $m_b(\phi)=\lfloor\phi+b\lambda\rfloor$. This is the Haar terminal-position reserve of [3]; terminal position is summed over its allowed values. Take all complete six-group stages. Their selected times $t_i$ have first time at most 53, consecutive gaps at most 110, and final time greater than $b\lambda-140$. Their marked entry counts are the deterministic integers

$$a_i=U(t_i)-1,\qquad 0<a_{i+1}-a_i\le72. \tag{11}$$

For clarity, the endpoint constants follow just from bounded lengths. The last doubled boundary $z$ before $m_b-9$ satisfies $z>m_b-19$. Fewer than six leftover groups cost at most 50 bits; the selected start within the last complete stage is less than 60 bits before its end. Thus the final selected time exceeds $b\lambda-130$, which implies the stated weaker bound. The first selected start is at most $3+5\cdot10$; between stages the gap is at most $60+50<110$ after accounting for the positive first-group offset.

For each admitted word choose the first phase cell in its feasible interval. Call this canonical cell $C(w)$ and its disjoint assigned class $D_C$. The interval is nonempty, is half-open, and has cell-boundary endpoints, so this choice is well defined. Section 3 proves it invariant under the exchanges. Define $\mathcal R$ by retaining $C(w)$, the terminal time, all unmarked bits, and marking just the occurrences at that cell's selected times. Every orientation fiber in $D_C$ is complete and uniform.

For fixed $C,m,h$, write

$$c_i(C)=\left|\cos\left(\pi h\delta(t_i,a_i)/2^m\right)\right|, \qquad P_C(w)=\prod_i c_i(C)^{\mathbf{1}_{\text{mark at }t_i}}.$$

Integral-phase factors may be discarded. By the fiber formula, positivity and $D_C\subseteq\mathcal F_C$,

$$ \mathbb E_{\mathcal F_b} \left|\mathbb E_{\mathcal F_b}[e^{2\pi ihX/2^m}\mid\mathcal R]\right| \le \frac1{q_F}\sum_C q_C\,\mathbb E_C P_C. \tag{12} $$

The cone is applied on $\mathcal F_C$, not on the selected class $D_C$. On that full fixed-phase law, (10) and successive conditioning give

$$\mathbb E_C P_C\le\prod_i[1-p(1-c_i(C))]. \tag{13}$$

No independence of successive marks is required. If at least $G$ selected phases have distance at least $\epsilon$ from the integers, (12)--(13) are at most $\kappa_b\theta^G$.

This argument also pays the mass of too few exchange opportunities. For $n$ complete stages the success count $S$ satisfies $\mathbb E_C2^{-S}\le(1-p/2)^n$ and $\mathbb P_C(S\le pn/2)\le e^{-pn/8}$. A phase-independent greedy packing of all $A/B$ occurrences dominates $S$. Its invariance follows because the only matching proper suffix and prefix of the two words has length one. Thus its global exceptional admitted mass is at most $\kappa_b q_F e^{-pn/8}$ whenever the same lower stage count $n$ is used. This mass bound alone would not estimate the phases in (13).

5. A deterministic lower bound on nonintegral phases

Assume $m\le K$ and keep the selected times $t_i\le m-5$. For $m\ge1600$ this list is nonempty. Its last time $t_s$ satisfies

$$1\le d_s:=m-t_s-4\le256,\qquad a_s\ge3m/5-160,\qquad a_0\le36. \tag{14}$$

To check the last-time bound, either the next selected time crosses $m-5$, in which case the 110-bit gap suffices, or the list contains the schedule's final time, which is greater than $b\lambda-140$ and $m\le b\lambda+4$. Also $a_i=\lceil(t_i+2-\phi)/\lambda\rceil-1$ is greater than $3t_i/5-8/5$. These inequalities imply (14) with room to spare.

Let $r_i$ be the centered odd representative of $h3^{-(a_i+4)}\pmod{2^{d_i}}$, with $d_i=m-t_i-4$. The phase distance is $|r_i|/2^{d_i}$. Call a vertex good if this is at least $\epsilon$. For consecutive selected vertices,

$$3^{a_{i+1}-a_i}r_{i+1}-r_i=z_i2^{d_{i+1}},\qquad z_i\in\mathbb Z. \tag{15}$$

If neither endpoint is good, then $|z_i|<(3^{72}+2^{110})\epsilon<1$, so $z_i=0$. On each consecutive run of nongood vertices, $3^{a_i+4}r_i$ is consequently one constant nonzero ordinary integer. The final selected vertex is good, since $d_s\le256$ and $r_s$ is odd.

At the start of a nongood run write

$$3^{a+4}r=h+q2^d,\qquad |q|\le\tfrac12 3^{a+4}+|h|\le41\,3^a+|h|. \tag{16}$$

We import the following precise valuation bound from [6, Lemma A.1]. For integers $m\ge16$, $0\le A\le m$, $1\le d\le m$, nonzero odd $h$, and integer $q$ with $|q|<14\,3^A$, put $Y=A+\log|h|+5$. Then

$$v_3(h+q2^d)\le100Y[\log(1+m/Y)+11]^2. \tag{17}$$

The source form is nonzero because it is odd. The cited proof includes $q=0$, unequal valuations, signed bases and multiplicatively dependent bases. Its primary input is [7, Theorem 3].

In (16) choose $A'=a+6+\lceil\log_3(|h|+1)\rceil$. Then $|q|<14\,3^{A'}$ and $Y=A'+\log|h|+5\le a+B_h$. Moreover $a<0.632m$, since $t_i\le m-5$ and $1/\gamma<0.632$. The hypothesis $B_h\le m/100$ therefore ensures $A'\le m$. The function $y[\log(1+m/y)+11]^2$ is increasing for $y>0$; its derivative is $L^2-2Lm/(m+y)>0$, where $L=\log(1+m/y)+11$.

Write $B=B_h$ and

$$D=101[\log(1+m/B)+11]^2.$$

If the run ends at count $a'$, its constant label has valuation at least $a'+4$. Applying (17) and then $Y\le a+B$ proves

$$a'+B\le D(a+B). \tag{18}$$

Let $G$ count good vertices in the selected list. There are at most $G+1$ nongood runs and at most $2G$ edges touching good vertices. Each latter edge increases the count by at most 72. Expanding these additive increases between the multiplicative bounds (18) gives

$$ a_s+B\le D^{G+1}(a_0+B+144G) \le(3D)^{G+1}(a_0+B), \tag{19} $$

because $B\ge100$ and $1+1.44G\le3^G$ for integral $G\ge0$. For $m\ge1600$, (14) gives $a_s+B\ge m/2$ and $a_0+B\le1.36B$. Their ratio is at least $R/3$, where $R=m/B$. Taking logarithms in (19) yields $G\ge\Gamma(R)$ from (2). Equations (12)--(13) prove (3).

For $|h|\le e^{m^\alpha}$, $B=O_\alpha(m^\alpha)$, so $\Gamma(m/B)\gg_\alpha\log m/\log\log m$ above an effective onset. This proves (4). If $m>K$, the unrecorded uniform lifts above the stopped word annihilate the primitive character, proving the remaining assertion.

6. The original interval and record-law consequences

Apply the Erdős--Turán inequality conditionally on $\mathcal R$, then average. With any integer cutoff $H\ge1$ this bounds $\Delta_b$ by an absolute constant times

$$\frac1{H+1}+\sum_{h=1}^H\frac1h \mathbb E_{\mathcal F_b} \left|\mathbb E_{\mathcal F_b}[e^{2\pi ihY}\mid\mathcal R]\right|. \tag{20}$$

Put $g(K)=\log K/\log\log K$ and take $H=\lfloor\exp(c_0g(K))\rfloor$ with a sufficiently small effective $c_0>0$. Reducing each character to an odd numerator gives conductor $m=K-v_2(h)\ge K-O(g(K))$ and numerator at most $H$. These numerators lie in the window of (4) with $\alpha=1/2$ for large $K$. Thus each norm in (20) is at most $C\kappa_b e^{-c_1g(K)}$ uniformly. The harmonic sum costs $O(\log H)$, absorbed by a reduction of $c_1$. Choosing $c_0$ smaller proves (5).

For the grid interval corresponding to $J$, (5) gives $|\mathbb P_{\mathcal F_b}(X_K\in J)-\ell|\le\Delta_b$. Multiplying by $2^Kq_F$ gives the exact counting normalization in (6). For the record statement let $p_R=\mathbb P_{\mathcal F_b}(X_K\in J\mid\mathcal R)$ and $p_J=\mathbb E p_R$. Its conditional density relative to the original record law is $p_R/p_J$. Consequently its total variation distance is

$$ \frac{\mathbb E|p_R-p_J|}{2p_J} \le\frac{\Delta_b}{p_J} \le\frac{2\Delta_b}{\ell} $$

when $\Delta_b\le\ell/2$. This proves the corollary without assuming that an interval preserves the marked orientations.

7. Checks, scope and further use

The evidence for the infinite statements is the proof and the precisely cited completion and valuation inputs. Three standard-library Python audits check the new finite interfaces. They verify 8,128 embedded exchange pairs and 320 two-mark fibers by direct integer trajectories; reconstruct all four published block rows and check 2,331 schedule/depth controls plus 1,024 overlap controls; and verify 6,234 modular carry edges in 27 cases, including 44 edges in nongood runs. The last control deliberately includes large ternary-power numerators to exercise nonempty runs, and does not claim that those controls satisfy the eventual numerator hypothesis. No original-window source census was enlarged to support the asymptotic claim.

Run the supplement's run_checks.py with up to five workers. The bundled public strip certifier is unmodified and attributed to [3]. Code is MIT licensed; documentation and retained outputs are CC0. The exact outputs and a SHA256 manifest are included. These tests check identities and indexing, not the imported logarithmic-form theorem or the asymptotic proof.

The result addresses the spatial distribution of admitted source residues and pays the original interval's normalization. It is insufficient for intervals whose relative length is exponentially small in $K$. In particular it does not establish the centered sixth-moment estimate for translated height-two counts, bound the earlier centered-profile boundary or accumulated positive influx, meet the numerical sixth-moment budget, or exclude another positive-integer cycle. A stronger result must retain the source mass while improving the signed arithmetic cancellation at those smaller scales. The Collatz conjecture remains open.

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