# 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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