# 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 $r\ge1$, $$\mathbb P\{M_n^1(Y_n)0$, a base parity zero increases $s$ by one, and a base parity one decreases it by one. At level zero, the latter case reverses orientation. These statements follow directly from applying $x/2$ or $(3x+1)/2$ to the two entries; they prove (8) by induction. Thus absorption forces both endpoint equality and $K_t(B)-K_t(A)=1$. The initial unit signed level is retained; we are not applying an equal-initial-count clock statement to this pair. Putting $\lambda=1/n$ in (7) and using $1-e^{-\tau/n}\ge1-e^{-1}$ on $\{\tau>n\}$ gives, with the constant enlarged for $n=0$, $$\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}$$ Also $\mathbb P\{\mathcal M_k^1(\mathcal Y_k)k\mid A\text{ odd}\}\le C/\sqrt{k}.\tag{A7}$$ 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