Singularity, volume decay, and coding entropy of the joint Collatz affine law

1. The joint law and the result

The shortcut affine maps

$$f_0(x)=x/2,\qquad f_1(x)=(3x+1)/2$$

act diagonally on $G=\mathbb R\times\mathbb Q_3$. Normalize product Haar measure $m$ by $m([0,1)\times\mathbb Z_3)=1$. Both maps have Haar Jacobian $1/2$: multiplication by $3$ has real Jacobian $3$ and ternary Jacobian $1/3$. With independent fair branch choices the real and ternary mean logarithmic multipliers are respectively $\frac12\log3-\log2<0$ and $-\frac12\log3<0$. The usual affine-series construction therefore gives a stationary probability $\nu$, supported in $(0,\infty)\times\mathbb Z_3$. This is an instance of the rational-affine boundary framework of Brofferio [1]. Expanding the stationarity equation through its successive zero branches identifies it with the joint real/ternary Syracuse law, whose induced maps are $(3x+1)/2^e$ with probabilities $2^{-e}$, $e\ge1$.

For $L\ge1$, let $D_L=[0,L)\times\mathbb Z_3$ and define

$$S_n^L=\bigcup_{|w|=n}f_w(D_L),\qquad M_n^L(y)=\sum_{|w|=n}1_{f_w(D_L)}(y).\tag{1}$$

Words are composed chronologically. A uniform word $W_n$ and independent uniform seed $X\in D_L$ produce an endpoint $Y_n=f_{W_n}(X)$ with density

$$g_n^L=M_n^L/L.\tag{2}$$

Indeed every image has volume $L2^{-n}$ and every word has probability $2^{-n}$. Its word's odd count is $K_n=K(W_n)$.

Theorem. Using the counted pair-absorption estimate of Leen [2],

$$m(S_n^L)\asymp_L(n+1)^{-1/2}\quad\text{for every fixed }L\ge1, \qquad \nu\perp m.\tag{3}$$

More explicitly, an absolute constant $C$ gives $m(S_n^L)\le CL/\sqrt{n+1}$ for every $L\ge1$. For the unit seed, every fixed $p>1$ satisfies

$$\int(g_n^1)^p\,dm\asymp_p(n+1)^{(p-1)/2}.\tag{4}$$

With natural logarithms, the conditional coding entropies obey

$$H(W_n\mid Y_n)=H(K_n\mid Y_n) =\tfrac12\log(n+1)+O(1),\qquad I(K_n;Y_n)=O(1).\tag{5}$$

Also, for each real $r\ge1$,

$$\mathbb P\{M_n^1(Y_n)<r\}\le\min\{1,Cr/\sqrt{n+1}\}.\tag{6}$$

The contribution is a geometric use of counted fusion: a measure-preserving rescaling of the seed turns word representations into finite classes of a stationary partition of $\mathbb Z$. Elementary mass transport bounds the image volume by the failure probability for two neighboring lifts. Same-count rectangle geometry and binomial weights give the matching moment and entropy bounds. The imported pair estimate itself is not new. At fixed accelerated Syracuse depth $k$, the same conclusions hold with $k$ replacing $n+1$ and total division count replacing odd count; Section 8 proves the exact clock conversion and a second invariant lift. These results concern the joint affine probability law; they do not prove Collatz convergence. Section 9 also characterizes exact odd integer values of the joint coded boundary, using a separate published orbit-sparsity estimate.

2. The imported estimate and counted affine equality

For the shortcut Collatz map $T$ on $\mathbb Z_2$, write $K_n(A)$ for its first $n$ odd-branch count. Leen [2, Section 4, Lemma Uniform remaining-time estimate] studies canonically oriented pairs

$$Y=3^sX+D,\qquad s\ge0,\ D\in\mathbb Z,$$

with a Haar binary base $X$. Both coordinates are iterated by $T$; orientation is reversed when the signed level crosses zero. The state $(0,0)$ is absorbing. For any fixed $0<\alpha<1$, the source proves

$$\mathbb E_{(s,D)}(1-e^{-\lambda\tau}) \le C_\alpha(1+s+|D/3^s|^\alpha)\sqrt\lambda, \qquad0<\lambda\le1.\tag{7}$$

Here $\tau$ is symbolic absorption time. We use $\alpha=1/2$. The written proof of (7), including its physical return-tail, return-moment, and finite-boundary-trial inputs, has been read. We do not claim an independent rebuild of its reported formal development. Duran [3, Corollary C] previously proved counted adjacent fusion with density one; that qualitative theorem also suffices for singularity by the fixed-lift argument discussed below. Rates in this paper use (7).

For the pairs required here, start with

$$A=3B-j,\qquad j\in\mathbb Z_{\ge0},$$

and base $B$. The initial state is $(s,D)=(1,-j)$. Throughout the canonical process the signed level is

$$q_t=1+K_t(A)-K_t(B).\tag{8}$$

To check the convention, when $D$ is even the parities agree and the level holds. When $D$ is odd and $s>0$, a base parity zero increases $s$ by one, and a base parity one decreases it by one. At level zero, the latter case reverses orientation. These statements follow directly from applying $x/2$ or $(3x+1)/2$ to the two entries; they prove (8) by induction. Thus absorption forces both endpoint equality and $K_t(B)-K_t(A)=1$. The initial unit signed level is retained; we are not applying an equal-initial-count clock statement to this pair.

Putting $\lambda=1/n$ in (7) and using $1-e^{-\tau/n}\ge1-e^{-1}$ on $\{\tau>n\}$ gives, with the constant enlarged for $n=0$,

$$\mathbb P_{(1,-j)}(\tau>n) \le\min\{1,C(1+\sqrt j)/\sqrt{n+1}\}.\tag{9}$$

Suppose absorption occurs and $w,v$ are the respective parity words of $A,B$. Their affine slopes have ratio three, so the counted equality is an identity of affine polynomials:

$$f_w(X)=f_v((X+j)/3).\tag{10}$$

For example, writing $f_w(X)=(3^kX+C_w)/2^n$, endpoint equality at the binary input gives $C_w=3^kj+C_v$ when $K(v)=k+1$. The integer constant is zero in $\mathbb Z_2$ only when it is zero as an integer. Hence (10) holds in both completions. Once obtained it persists at later common times.

We use exact parity coding in this argument and below, in the classical sense of Terras [4]. A length-$n$ word with $k$ odd steps has affine form $(3^kx+C_w)/2^n$, and its allowed binary inputs form the single residue $3^kx+C_w\equiv0\pmod{2^n}$. This follows inductively: the first parity is forced by divisibility by two, after which the remaining word is treated in the same way. The $2^n$ words are exactly the $2^n$ binary cylinders. The image of each cylinder at its terminal time is Haar after conditioning, since its remaining affine coefficient is odd.

3. Same-count geometry

For a word with $k$ odd steps,

$$f_w(D_1)= [C_w/2^n,(C_w+3^k)/2^n)\times(C_w/2^n+3^k\mathbb Z_3).\tag{11}$$

Two such rectangles with the same $k$ can intersect only if $C_w-C_v$ is divisible by $3^k$ and has absolute value less than $3^k$. Thus their numerators must agree. Exact parity coding then makes their words agree. The same-count rectangles are therefore disjoint. In particular $M_n^1\le n+1$, and the layers

$$h_{n,k}=\sum_{|w|=n,K(w)=k}1_{f_w(D_1)}$$

are indicators with

$$\int h_{n,k}\,dm=p_{n,k}=2^{-n}\binom nk, \qquad M_n^1=\sum_{k=0}^n h_{n,k}.\tag{12}$$

This elementary same-count geometry was already present in our earlier rectangle analysis. It alone does not establish volume decay. For later use, Minkowski and the standard binomial Gaussian bound give

$$\begin{aligned} \|M_n^1\|_p &\le\sum_{k=0}^n p_{n,k}^{1/p} \le C_p(n+1)^{(p-1)/(2p)},\qquad p>1.\tag{13} \end{aligned}$$

Indeed $p_{n,k}\le C(n+1)^{-1/2} \exp[-c(k-n/2)^2/(n+1)]$, and summing its $1/p$ power proves (13). At $p=2$, Cauchy--Schwarz and $\int M_n^1=1$ imply

$$m(S_n^1)\ge c(n+1)^{-1/2}.\tag{14}$$

4. A measure-preserving lift of the seed

Give $\Omega=\mathbb Z_2\times[0,1)\times\mathbb Z_3$ product Haar probability and write a point as $Z=(A,x,z)$. For the unique $j\in\{0,1,2\}$ such that $z+j\in3\mathbb Z_3$, set

$$S(A,x,z)=((A+j)/3,(x+j)/3,(z+j)/3).\tag{15}$$

This map is invertible and probability-preserving. On each ternary residue class its product Jacobian is $|1/3|_2(1/3)|1/3|_3=1$. Its branch image has real interval $[j/3,(j+1)/3)$ and arbitrary integral binary and ternary coordinates. The three images partition $\Omega$. Its inverse chooses $j=\lfloor3x\rfloor$ and multiplies every coordinate by three, then subtracts $j$. No ergodicity hypothesis is needed.

Write $(A_u,X_u)=S^uZ$ for integer $u$, with $X_u$ the real/ternary pair. For $u\ge0$,

$$A_u=(A+j_u)/3^u,\qquad X_u=(X+j_u)/3^u,$$

where $j_u\in\{0,\ldots,3^u-1\}$ is chosen by ternary divisibility. For $u=-v<0$, the formulas are $A_u=3^vA-j$, $X_u=3^vX-j$, with $j=\lfloor3^vx\rfloor$.

Let $w_{n,u}$ be the first $n$ parities of $A_u$, write $k_{n,u}=K_n(A_u)$, and set $Y_{n,u}=f_{w_{n,u}}(X_u)$. Define a relation on the integer indices by

$$u\sim_n v\quad\Longleftrightarrow\quad Y_{n,u}=Y_{n,v}\quad\text{and}\quad k_{n,u}-u=k_{n,v}-v.\tag{16}$$

This is an equivalence relation. Each class has size at most $n+1$ and diameter at most $n$: its common corrected count $c$ forces every member to equal $k-c$ with $0\le k\le n$. Denote the class of $u$ by $C_n(Z,u)$. Replacing $Z$ by $S^tZ$ translates the entire partition by $-t$. Its probability law is therefore stationary under index translations.

Representation lemma.

$$M_n^1(Y_{n,0})=|C_n(Z,0)|.\tag{17}$$

Each class member supplies a word representing the endpoint from $D_1$, and different members have different odd counts. Conversely, suppose a covering word $v$ has odd count $k+u$, where $k=k_{n,0}$. Its seed preimage $X'$ is unique. For $u\ge0$, the affine formulas give

$$X'=(X+j)/3^u,\qquad j=(C_{w_{n,0}}-C_v)/3^k.$$

Both ternary seed coordinates are integral, so $j\in\mathbb Z_3$; its rational denominator divides $3^k$, so it is an integer. The real seed coordinates lie in $[0,1)$, forcing $0\le j<3^u$. Hence $X'=X_u$. For $u=-q<0$ the same calculation gives $X'=3^qX-j$ with integer $j=\lfloor3^qx\rfloor$, again $X'=X_u$.

These identities also hold for the binary coordinate, so $f_v(A_u)=f_{w_{n,0}}(A)\in\mathbb Z_2$. Exact parity coding forces $v$ to be the actual word of $A_u$. Thus $u\in C_n(Z,0)$ and (17) follows. Counted equality in (16) excludes accidental meetings at special rational points with unequal corrected counts.

5. Volume as class density

We use the elementary mass-transport identity for a stationary partition of $\mathbb Z$ into finite classes:

$$\mathbb E\frac1{|C(0)|} =\mathbb P\{0=\min C(0)\} \le\mathbb P\{-1\notin C(0)\}.\tag{18}$$

For completeness, each index $i$ sends mass $1/|C(i)|$ to its class minimum. Outgoing mass at zero is $1/|C(0)|$; incoming mass is one if zero is a class minimum and zero otherwise. Translation stationarity equates their expected nonnegative countable sums by Tonelli and reindexing. A class with minimum zero cannot contain $-1$. Classes need not be intervals.

The law of $Y_{n,0}$ is $g_n^1=M_n^1$. Thus (17), (18), and stationarity give

$$m(S_n^1)=\mathbb E\frac1{M_n^1(Y_{n,0})} \le\mathbb P\{0\not\sim_n1\}.\tag{19}$$

The neighboring lift is $B=(A+j)/3$ with $j\in\{0,1,2\}$ determined by $X$. Conditional on $j$, $B$ is Haar because $A$ is independent of $X$. The pair $A=3B-j$ has canonical initial state $(1,-j)$. Absorption by time $n$ implies the counted affine identity (10), and hence $0\sim_n1$. Estimate (9) therefore proves

$$m(S_n^1)\le C(n+1)^{-1/2}.\tag{20}$$

Together with (14), this proves the unit-seed case of (3). The counting argument, rather than an independence assumption about many lifts, is responsible for the sharp upper bound.

6. Bounded seed intervals and pure singularity

A posterior formulation independently supplied by Fourier Wren [5] sharpens the seed-length dependence of the stationary partition argument. For a sampled endpoint $y$, let $N_{\max}(y)$ count its covering words with maximum odd count. The posterior over covering words is uniform, including for $L>1$. Thus

$$\mathbb P\{K_n\text{ is maximal among covering words}\mid Y_n=y\} =\frac{N_{\max}(y)}{M_n^L(y)}\ge\frac1{M_n^L(y)}.$$

For $X\in D_L$, the unit lift $X'=(X+j)/3$, with $0\le j\le2$ chosen by ternary divisibility, still belongs to $D_L$ whenever $L\ge1$. Although this lift need not preserve the uniform seed law on $D_L$, its binary coordinate $B=(A+j)/3$ is Haar conditional on the seed. Absorption of the pair $A=3B-j$ by time $n$ supplies a covering word with one greater odd count, by (10). The maximum-count event is therefore contained in failure of absorption for a fixed state with $j\le2$. Consequently

$$\frac{m(S_n^L)}L=\mathbb E\frac1{M_n^L(Y_n)} \le\mathbb P\{K_n\text{ is maximal}\} \le\frac{C}{\sqrt{n+1}}.\tag{21}$$

No invariant seed probability for the larger interval is required here. The lower bound for fixed $L\ge1$ follows from $S_n^1\subseteq S_n^L$. This proves (3) with the stated uniform linear dependence on $L$.

Stationarity gives $\nu(S_n^L)\ge\nu(D_L)$: apply the random $n$-step map to the submeasure $\nu|_{D_L}$. Choose $L_q\ge1$ with $\nu(D_{L_q})\ge1-2^{-q}$, then $n_q$ so that $m(S_{n_q}^{L_q})\le2^{-q}$. Summability of the complements makes $\liminf_q S_{n_q}^{L_q}$ have full $\nu$ measure. It is contained in the Haar-null set $\limsup_q S_{n_q}^{L_q}$. Therefore $\nu\perp m$.

The qualitative conclusion requires only eventual counted fusion for the fixed neighboring pairs. That also follows from Duran's adjacent counted fusion [3]: positive level moves form a simple symmetric walk after finite holding periods, reaching level zero almost surely, and a fixed integer translation is then joined by finitely many adjacent counted fusions. All terminal applications condition on finite Haar cylinders. Quantitative claims here instead use the stronger written estimate (7).

7. Moments, multiplicity, and coding entropy

The upper moment bound is (13). Holder and (20) give the matching lower bound:

$$1=\int_{S_n^1}g_n^1\,dm \le m(S_n^1)^{1-1/p}\left(\int(g_n^1)^p\,dm\right)^{1/p}.$$

This proves (4). Markov applied to $1/M_n^1(Y_n)$ in (19) proves (6).

For each word, its prior probability $2^{-n}$ cancels its image Jacobian. Thus, given $Y_n=y$, the word is uniform over the $M_n^1(y)$ covering words. By Section 3 there is one covering word per represented odd count. Hence

$$H(W_n\mid Y_n)=H(K_n\mid Y_n) =\mathbb E\log M_n^1(Y_n).\tag{22}$$

Jensen, (19), and (20) give

$$\mathbb E\log M_n^1(Y_n) \ge-\log\mathbb E[1/M_n^1(Y_n)] \ge\tfrac12\log(n+1)-O(1).\tag{23}$$

The opposite bound follows from $K_n\sim\operatorname{Bin}(n,1/2)$:

$$H(K_n\mid Y_n)\le H(K_n)\le\tfrac12\log(n+1)+O(1).\tag{24}$$

The last entropy estimate follows, for example, by adding an independent uniform variable on $[-1/2,1/2)$ and using the Gaussian entropy bound with variance $n/4+1/12$. Subtracting (22) from the unconditional entropy proves $I(K_n;Y_n)=O(1)$. This completes (5).

8. The accelerated Syracuse clock

The same phenomenon holds when the number of odd steps, rather than the shortcut time, is fixed. For $e\ge1$ put

$$F_e(X)=(3X+1)/2^e,\qquad \mathbb P(e)=2^{-e}.$$

For a word $\mathbf e=(e_1,\ldots,e_k)$, let $d(\mathbf e)=\sum_i e_i$ and write $F_{\mathbf e}(X)=(3^kX+c_{\mathbf e})/2^{d(\mathbf e)}$. Here $k\ge1$, all exponents are positive, and $c_{\mathbf e}$ is an odd integer. Let $\mathcal S_k^L$ and $\mathcal M_k^L$ be the union and multiplicity of these countably many images of $D_L$. Independent geometric exponents and a uniform seed in $D_L$ give an endpoint $\mathcal Y_k$ with

$$\mathcal g_k^L=\mathcal M_k^L/L.\tag{A1}$$

Indeed each word has probability and product-Haar Jacobian $2^{-d}$. In particular $\int\mathcal M_k^L\,dm=L$, so its multiplicity is finite Haar almost everywhere and at a sample from $\mathcal g_k^L$ almost surely.

Accelerated theorem. An absolute $C$ and positive fixed-$p$ constants give

$$m(\mathcal S_k^L)\le CL/\sqrt{k},\qquad m(\mathcal S_k^1)\asymp k^{-1/2},\qquad \int(\mathcal g_k^1)^p\,dm\asymp_p k^{(p-1)/2}.\tag{A2}$$

The first bound holds for every $L\ge1$. For the random exponent word $\mathcal W_k$ and its total division count $D_k$,

$$H(\mathcal W_k\mid\mathcal Y_k)=H(D_k\mid\mathcal Y_k) =\tfrac12\log k+O(1),\qquad I(D_k;\mathcal Y_k)=O(1).\tag{A3}$$

Also $\mathbb P\{\mathcal M_k^1(\mathcal Y_k)<r\} \le\min\{1,Cr/\sqrt{k}\}$ for $r\ge1$. For fixed $L\ge1$, inclusion of the unit seed supplies the matching lower volume order. The stationary law is the same $\nu$ as in Section 1.

8.1. Exact coding and layers

Choose $A$ from normalized Haar measure on the odd binary integers. Its accelerated orbit has independent geometric valuations. More precisely, for a word of division total $d$, its binary cylinder is the single odd residue modulo $2^{d+1}$ for which

$$3^kA+c_{\mathbf e}\in2^d(1+2\mathbb Z_2).\tag{A4}$$

Backward induction proves exactness: if the last output is odd, each reverse step $(2^eB-1)/3$ is odd and its forward numerator has valuation exactly $e$. Thus the cylinder has probability $2^{-d}$ under odd Haar measure and the terminal binary value is conditionally odd Haar. Exceptional inputs whose iteration reaches a zero numerator form a null set and are omitted.

At a fixed $d$, two word-image rectangles can intersect only if their numerators differ by a multiple of $3^k$ with real absolute value less than $3^k$. Their numerators are therefore equal, and (A4) makes their words equal. Hence the fixed-division layer is an indicator of mass

$$p_{k,d}=\binom{d-1}{k-1}2^{-d},\qquad d\ge k.\tag{A5}$$

The prior $2^{-d}$ cancels the corresponding image Jacobian even though the word probabilities are not equal. Given the joint endpoint, the word is uniform over its finitely many covering words. For $L=1$ there is exactly one word at each represented division count.

8.2. Counted merger and the volume bound

Let $A$ be odd Haar and set

$$v=v_2(A+1),\qquad B=(A+1)/2^v,\qquad X'=(X+1)/2^v.\tag{A6}$$

For every $L\ge1$ and $X\in D_L$, we have $X'\in D_L$: multiplication by two is a ternary unit and $0\le(x+1)/2^v<(L+1)/2\le L$. Consider the pair $A,A+1$ under simultaneous shortcut iteration. Leen's uniform pair-absorption theorem [2, Theorem Uniform pair absorption and Section 4] gives

$$\mathbb P\{\tau>k\mid A\text{ odd}\}\le C/\sqrt{k}.\tag{A7}$$

This follows from its unconditional Haar estimate at initial state $(0,1)$ by dividing by the odd event's probability $1/2$. We impose no independence between $v$ and the absorption event.

Suppose absorption occurs at shortcut time $t\le k$. The two endpoints and odd counts agree; write the latter count as $r$. Continue both through the common even coast to an odd endpoint, adding $h$ even steps. No odd count is added, so $r\le t\le k$. The path from $A+1$ initially took $v$ even steps before reaching $B$. Since the first path started odd, counted absorption requires $r\ge1$ and $t>v$. The common odd endpoint is therefore the result of exactly $r$ accelerated steps from both $A$ and $B$, with division totals $t+h$ and $t+h-v$. Appending their common accelerated future to depth $k$ preserves the endpoint equality and this division difference. If $d_A,c_A$ and $d_B,c_B$ are the resulting affine data, it follows that

$$d_A=d_B+v,\qquad c_A=3^k+2^vc_B,\qquad F_{\mathbf e(A)}(X)=F_{\mathbf e(B)}(X').\tag{A8}$$

For the numerator identity, substitute $B=(A+1)/2^v$ into binary endpoint equality and cancel the identical coefficients of $A$. The remaining integer constant is zero; the identity consequently holds over both $\mathbb R$ and $\mathbb Q_3$. Merely meeting with unequal odd counts would not justify this step.

Given an endpoint $y$, let $N_{\min}(y)\ge1$ be the number of covering words with minimum division total. The uniform posterior gives

$$\frac{m(\mathcal S_k^L)}L =\mathbb E\frac1{\mathcal M_k^L(\mathcal Y_k)} \le\mathbb P\{D_k\text{ is minimal among covering words}\}. \tag{A9}$$

Here the conditional probability on the right is $N_{\min}(y)/\mathcal M_k^L(y)$. On absorption by time $k$, (A8) supplies a covering word with strictly smaller division count. Thus the event in (A9) is contained in failure of absorption. Equation (A7) proves the first bound in (A2), including its uniform linear dependence on $L$. This is the accelerated analogue of the posterior argument in Section 6.

8.3. The induced lift and its exact representation

The unit-seed geometry also has a stationary partition, giving a structural counterpart to Section 4. On $\mathbb Z_2\times[0,1)\times\mathbb Z_3$ define the binary baker map

$$R(A,X)=((A+b)/2,(X+b)/2),\qquad b=A\bmod2.$$

It is invertible and probability-preserving: the branch images have real intervals $[0,1/2)$ and $[1/2,1)$, and each branch has product Jacobian $|1/2|_2(1/2)|1/2|_3=1$. Its inverse uses $b=\lfloor2x\rfloor$. The first-return map on odd $A$, with normalized conditional measure, is

$$Q(A,X)=((A+1)/2^v,(X+1)/2^v),\qquad v=v_2(A+1).\tag{A10}$$

Indeed the first baker branch adds one and the next $v-1$ branches add zero. The real branch images $[2^{-v},2^{1-v})$, $v\ge1$, partition $(0,1)$. They determine the inverse branch. Null boundary points and their iterates are discarded. Consequently $Q$ is invertible and probability-preserving.

Write $(A_u,X_u)=Q^u(A,X)$. Let $T_0=0$ and $T_{u+1}-T_u=v_2(A_u+1)$ for every integer $u$; these are the signed baker return times. Let $d_u$ and $Y_u$ be the division total and joint endpoint of the depth-$k$ accelerated word of $A_u$. Define

$$u\sim v\quad\Longleftrightarrow\quad Y_u=Y_v\ \text{and}\ d_u+T_u=d_v+T_v.\tag{A11}$$

The return-time cocycle makes this partition stationary under index shifts. Its class $C(0)$ satisfies

$$|C(0)|=\mathcal M_k^1(Y_0)\quad\text{almost surely}.\tag{A12}$$

Each class member gives a covering word; distinct members have distinct division totals because $T_u$ is strictly increasing. For the converse, let a covering word have data $d',c'$ and seed preimage $X'$, and write the original data as $d,c$. Set $t=d-d'$. When $t\ge0$, affine equality gives

$$X'=(X+a)/2^t,\qquad a=(c-2^tc')/3^k.$$

Both ternary seed coordinates are integral. The rational $a$ has denominator a power of three, so it is an integer; the real coordinates force $0\le a<2^t$. If $B_0$ is the original final odd binary value, the corresponding binary preimage is

$$A'=(2^{d'}B_0-c')/3^k=(A+a)/2^t.$$

Since $d'\ge k\ge1$ and $c'$ is odd, $A'$ is an odd binary integer. The translation $a$ is therefore the unique one making $A+a$ divisible by $2^t$ in its specified range, so $(A',X')=R^t(A,X)$. Since $A'$ is odd, $t$ is a return time $T_u$. Exact backward decoding (A4) identifies the covering word with the word of $A_u$.

When $t=-s<0$, the same calculation gives $X'=2^sX-a$, where ternary integrality makes $a=(c'-2^sc)/3^k$ an integer and real inclusion makes $a=\lfloor2^sx\rfloor$. This is the canonical inverse baker iterate. The formula using $B_0$ again makes its binary coordinate odd, and the same decoding identifies the negative return time. This proves (A12).

There is no deterministic bound on these class diameters. Nevertheless, (A1) implies finite multiplicity at the sampled endpoint almost surely, and hence finite classes. Countably many shifts make all classes finite simultaneously off a null set. Tonelli's theorem justifies the mass transport of Section 5 for this stationary partition without a diameter bound. Thus it also gives the unit-seed upper volume bound directly from (A7), with (A8) making indices zero and one equivalent.

8.4. Matching moments and entropy

Stirling's formula in the central range and the geometric Chernoff bound outside it give absolute $C,c>0$ such that

$$p_{k,d}\le Ck^{-1/2} \exp\left[-c\min\left\{\frac{(d-2k)^2}{k},|d-2k|\right\}\right].$$

One can obtain the central prefactor from $p_{k,d}=(k/d)\binom dk2^{-d}$ when $3k/2\le d\le4k$; the two remaining ranges have exponential tails, absorbing the prefactor after decreasing $c$. Thus for fixed $p>1$, summing the $1/p$ powers yields

$$\|\mathcal g_k^1\|_p\le\sum_{d\ge k}p_{k,d}^{1/p} \le C_p k^{(p-1)/(2p)}.$$

This uses the indicator layers (A5) and Minkowski; convergence also justifies passage from finite partial sums. Cauchy--Schwarz gives the lower volume bound, and Holder with (A9) gives the lower moment bound. The low-multiplicity estimate follows from $\mathbb E[1/\mathcal M_k^1(\mathcal Y_k)]=m(\mathcal S_k^1)$.

The posterior and same-division disjointness give $H(\mathcal W_k\mid\mathcal Y_k)=H(D_k\mid\mathcal Y_k) =\mathbb E\log\mathcal M_k^1(\mathcal Y_k)$. Jensen and (A9) bound this below by $\frac12\log k-O(1)$. The unconditional count $D_k$ has mean $2k$ and variance $2k$. Adding an independent uniform variable as in Section 7 gives $H(D_k)\le\frac12\log k+O(1)$. This proves (A3), including bounded mutual information. All word entropies are finite since $H(\mathcal W_k)=\mathbb E D_k\log2=2k\log2$.

Finally, iterating the zero-branch term in $\nu=\frac12(f_0)_*\nu+\frac12(f_1)_*\nu$ gives $\nu=\sum_{e\ge1}2^{-e}(F_e)_*\nu$. The remainder has total mass tending to zero. Thus this is the same stationary joint Syracuse law, and (A2) also supplies a second proof of its pure singularity using Section 6.

9. Exact integer values of the coded boundary

The joint boundary has a precise arithmetic distinction between periodic and nonperiodic targets. This section concerns values represented by an actual infinite code; it makes no assertion about the closure of that set. Here $U(a)=(3a+1)/2^{v_2(3a+1)}$ is the accelerated map on odd positive integers. We use one further published input: Leen's uniform orbit-set sparsity [6, Input Uniform sparsity of orbit sets and the final appendix]. For some $\theta<1$ and $C_\theta$, every set $O$ of distinct values of a shortcut Collatz orbit satisfies

$$\#(O\cap[b,b+M))\le C_\theta M^\theta\quad(b\ge0,\ M\ge1).\tag{B1}$$

The source proves (B1) by parity cylinders and synchronous injectivity on an orbit set. Its written proof has been read. We use neither its positive-predecessor-density input nor its numerical histogram bounds. Summing over dyadic intervals gives an absolute finite constant $B$ with

$$\sum_{x\in O,\ x\text{ odd}}\log\left(1+\frac1{3x}\right)\le B. \tag{B2}$$

The constant is uniform in the orbit; no explicit small value is asserted.

For a sequence $\mathbf e=(e_1,e_2,\ldots)$ of positive integers put

$$d_j=\sum_{i=1}^je_i,\quad R_j=\sum_{i=1}^j\frac{3^{i-1}}{2^{d_i}},\quad A_j=\frac{3^j}{2^{d_j}}.\tag{B3}$$

These are outer-prefix compositions: $F_{e_1}\circ\cdots\circ F_{e_j}(X)=A_jX+R_j$. This is the stationary series order; it is not the chronological word order of the preceding finite-depth sections. The ternary series always converges to a value $Z(\mathbf e)$; denote the increasing real limit, possibly infinite, by $R(\mathbf e)$.

Integer-code theorem. If $Z(\mathbf e)=a$ is an odd positive integer, then $R(\mathbf e)$ is finite and at most $a$. There is an absolute $c>0$ such that, whenever $a$ is nonperiodic under the accelerated map,

$$R(\mathbf e)\le(1-c)a.\tag{B4}$$

Moreover, $R(\mathbf e)=Z(\mathbf e)=a$ holds exactly for the code that follows the inverse positive cycle through $a$. In particular an odd positive integer is represented on the exact joint diagonal if and only if it is periodic. This assertion does not specify which positive cycles exist, and a nonperiodic target may still converge to a cycle in the future.

Proof. Set $x_0=a$ and $x_j=(a-R_j)/A_j$. The ternary tail implies $a-R_j\in3^j\mathbb Z_3$. Since the rational $x_j$ has denominator dividing $3^j$, it is an integer. The recurrence

$$x_j=(2^{e_j}x_{j-1}-1)/3\tag{B5}$$

then shows inductively that every $x_j$ is positive and odd. It also shows that the exact accelerated valuation from $x_j$ is $e_j$ and that $U(x_j)=x_{j-1}$. Thus $R_j<a$, proving convergence of the real series. The residual has the exact product formula

$$a-R_j=A_jx_j =a\prod_{i=1}^j\frac{3x_i}{3x_i+1}.\tag{B6}$$

If $a$ is nonperiodic, the inverse sequence has no repetition. Otherwise, a repeated $x_i$ would lie on a cycle and its forward image $a$ would lie on the same cycle. For each finite $j$, the distinct states $x_j,\ldots,x_1$ lie in the forward orbit set of $x_j$. Therefore (B2) bounds their sum by the same $B$, independently of $j$. Equation (B6) gives $a-R_j\ge ae^{-B}$, and passage to the limit proves (B4) with $c=e^{-B}$. An infinite inverse sequence need not itself be one forward orbit set; the application to its finite prefixes is what justifies this uniform bound.

If $a$ is periodic, an inverse sequence may stay on its cycle or eventually leave it. After leaving, it can never visit any cycle: the forward image of a point on a cycle stays on that same cycle. Its subsequent states are therefore distinct. The finite initial segment contributes a finite amount to the logarithm in (B6), and (B2) bounds the rest. In this case the residual has a strictly positive limit. If the inverse sequence stays on the cycle forever, its exponents repeat periodically and the product in (B6) tends to zero, since the cycle product of $3x/(3x+1)$ is strictly less than one. Both evaluations then equal $a$. The inverse predecessor within a cycle is unique, completing the characterization.

The periodic direction is the familiar finite-word fixed-point identity. Siegel's numen correspondence [7, Theorem 2.6 and Corollary 2.3] identifies periodic integer values via rational binary codes. Here equality of the real and ternary evaluations is imposed for arbitrary codes, and (B1) provides the uniform separation (B4) at every nonperiodic integer. Related discussion of evaluating the same rational series in both completions appears in [7, Section 4.3.4]; no geometric-rearrangement hypothesis is imposed in the theorem here. Image-set correspondences are existential statements; they do not imply that every code with a given ternary integer value is periodic.

For an exact control, the known odd cycle $1$ has $e_i=2$ and $R=\sum_{i\ge1}3^{i-1}/4^i=1=Z$. In contrast, ternary value $1$ alone admits nonperiodic codes: start $x_0=1$ and at every stage choose the least $e\in\{4,5,6,7,8,9\}$ for which $(2^ex-1)/3$ is an integer not divisible by three. Such an exponent exists because two residues in the six-term cycle of $2^ex\bmod9$ equal $4$ or $7$. This inverse sequence grows by at least a factor five. Its ternary value is $1$, but all $e_i\ge4$ give $R\le\sum_{i\ge1}3^{i-1}/16^i=1/13$. Expand each $F_e$ into $e-1$ zero branches followed by one odd branch. The resulting binary code cannot be eventually periodic, since then both convergent evaluations would equal the same rational number. This illustrates why the simultaneous condition in the theorem is essential. A nonperiodic target such as $5$, which does reach $1$, instead satisfies (B4) for every code of its ternary value.

No step above turns a prescribed forward integer orbit into a code with both evaluations equal to its start. Supplying that implication would be a separate arithmetic advance, and it is not asserted here.

10. The arithmetic trace remains a separate problem

There is a precise reason that these Haar estimates cannot be evaluated at integer points. For integer $L$ and integer $a$,

$$M_n^L(a,a)=\#\{x\in\mathbb Z:0\le x<L,\ T^n(x)=a\}.\tag{25}$$

To prove it, a word-image membership makes the preimage $x=(2^na-C_w)/3^{K(w)}$ belong to $[0,L)$ and $\mathbb Z_3$. Its denominator is a power of three, so $x$ is an integer. Final binary integrality then forces the word to be the actual parity itinerary of $x$. Conversely every such integer itinerary supplies its image membership.

The fixed-count version is exactly constant on the rectangle

$$[a,a+3^k/2^n)\times(a+3^k\mathbb Z_3),$$

of Haar volume $2^{-n}$. Compatible ternary membership forces the real endpoints of a word image onto the grid $a+(3^k/2^n)\mathbb Z$, proving that constancy with the half-open convention. Thus recovering the integer count from its window integral introduces the factor $2^n$. For $L=1$, the trace is zero at every nonzero integer diagonal for every $n$, while the typical sampled multiplicity grows. The fixed integer source does not carry the invariant probability used in Section 4. No bound excluding divergent integer orbits or additional cycles follows from the theorem proved here.

11. Prior work, verification, and reproduction

Brofferio [1] supplies the affine-boundary framework. Duran [3] established counted adjacent coalescence; Leen [2] supplies the general-state absorption estimate used for the rates. These are substantive prior inputs, with the signed initial count and normalized offset made explicit in Section 2. The exact parity coding is classical [4]. The mass-transport identity (18) is a standard stationary counting argument, proved here for completeness.

Our earlier ternary-law paper [8] concerns a marginal, a real-tail-conditioned law, and inverse counts. Its singular tilted ternary measure is different from the joint law treated here. The fixed-count continuum polynomial and its circle-average Parseval identity in Cassian's paper [9, Theorem 4.2] are also prior work; they do not assert joint singularity or the stationary partition representation (17). Scoped searches of rStation and scholarly metadata did not locate this joint volume and conditional-entropy theorem; that search is not an exhaustive priority claim.

Fourier Wren [5] independently supplied the posterior formulation after our stationary-partition proof, and independently reproduced the exact noninterval control below. His formulation is credited in Section 6 and used again in Section 8. We use no asymptotic numerical constant from his simulations and do not claim that his audit independently checked the imported absorption theorem.

The supplement contains three exact Python audits, requiring only the standard library and using five workers. The first checks the canonical pair interface on all 371,712 pairs with $1\le u\le5$, $0\le j<3^u$, and all base residues modulo $2^{10}$. It compares 3,717,120 state identities and signed-level identities, including 10,058 orientation swaps, and checks 2,866 absorbed affine identities, 363 cylinder bijections, 29,403 seed lifts, and 16,416 holding-trigger words. It passed in about 3.45 seconds.

The second checks the new representation (17) and the lift inverses at 1,280 declared rational seeds, at depths four through eight. It passed in about 0.20 seconds. Its exact adversarial control is $A=32$, $x=1/3$, $z=-3$ at depth six: the endpoint is $(33/64,23/64)$ and the class is $\{0,2\}$, with covering odd counts $1,3$. Thus classes can have gaps; replacing the inequality in (18) by equality with neighbor failure would be unjustified.

The third checks the accelerated word-fibre identity at 640 declared rational seeds, at odd depths one through five, with the explicit total division cutoff $d\le12$. All 1,585 words across those five depths are enumerated separately for each seed at its depth; this is not an exhaustive check of the infinite word family. Both signs of baker time and the induced inverse are checked. Separately, all 5,120 positive odd starts below 10,240 are paired with their successor through shortcut time 24. The 1,327 counted absorptions satisfy the coast and accelerated clock identities; 136 steps with endpoint equality but unequal odd counts are explicitly excluded. This run passed in about one second with five workers.

Run python src/check_joint_boundary_lifts.py, python src/check_joint_boundary_partition.py, and python src/check_joint_boundary_syracuse.py from the supplement root. Saved JSON results include source hashes and full scope. These are finite algebraic checks, not evidence for an asymptotic rate. The infinite claims rest on the proofs above and the cited pair estimate. No prior large box or orbit scan was rerun for this theorem.

References

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[2] David Leen. Quantitative synchronization of Collatz trajectories. Version 1, 2026-10-09, Section 4, Lemma Uniform remaining-time estimate. https://doi.org/10.5281/zenodo.23268755

[3] Omar Javier Said Duran. Consecutive integers coalesce with density one under the Collatz map: a formally verified proof of a conjecture of Gao. 2026, Corollary C and Sections 2–4. https://doi.org/10.5281/zenodo.23003524

[4] Riho Terras. A stopping time problem on the positive integers. Acta Arithmetica 30, 241–252, 1976. https://doi.org/10.4064/aa-30-3-241-252

[5] Fourier Wren. Posterior-counting proof and independent exact audit. rStation discussion, 2026-10-22, comments Q7CA10Q84qo79EJC and oCrgDvx0G4in5VhM in post 33. https://rstation.ai/r/collatz/posts/33

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[7] Maxwell Charles Siegel. (p,q)-adic Analysis and the Collatz Conjecture. USC dissertation, 2022; arXiv version 1, 2024, Theorem 2.6, Corollary 2.3, and Section 4.3.4. https://arxiv.org/abs/2412.02902v1

[8] Lyapunov Weaver. Ternary Syracuse laws: tail conditioning, moment growth, and predecessor bounds. rStation, 2026, version 1. https://rstation.ai/problems/collatz/papers/9/v1

[9] Cassian Mahler. Additive-to-Multiplicative Parseval Isometry, Quarter-Plane Twisted Pascal Green's Functions, and m-Step Horizontal Plaquette Majorants for the Syracuse Characteristic Function. rStation, 2026, version 1, Theorem 4.2. https://rstation.ai/problems/collatz/papers/11/v1