# Superpolynomial source mixing at prescribed Collatz run durations

Sylve Recursions. 6 November 2026.

## 1. Setup and results

Let $C(x)=x/2$ for even $x$ and $C(x)=(3x+1)/2$ for odd $x$. An odd Haar source $X\in\mathbb Z_2^\times$ has first parity bit 1 and independent fair subsequent bits, by the parity conjugacy [1]. A parity word of length $M$ determines exactly one source residue modulo $2^M$, with uniform lifts. Denote odd Haar probability by $\mathbb P_0$.

Write the parity sequence as consecutive blocks $1^{A_i}0^{B_i}$, each confirmed by the first 1 of the next block. The positive geometric lengths $A_i,B_i$ are independent under $\mathbb P_0$ and each has mass $2^{-j}$ at $j\ge1$. Write $\lambda=\log_2 3$, $\beta=\lambda-1$, and define the block duration and height by
$$
Q_k=\sum_{i\le k}(A_i+B_i),\quad S_k=\sum_{i\le k}(\beta A_i-B_i),
\quad \tau_n=Q_n+1,\quad \mathcal A_n=\{S_i>0,\ 1\le i\le n\}.
$$
At bit time $j$ the height is $H_j=\lambda a_j-j$, where $a_j$ counts ones in the first $j$ bits. Block positivity is equivalent to positivity at every positive bit time through $\tau_n$. The stopping bit is the 1 of the $n$-th $01$. Let $\mathbb P_{n,q}$ be odd Haar conditioned on $\mathcal A_n,Q_n=q$.

Use the marked record $\mathcal R_n$ of Paper 107 [2]. Cut the word through $\tau_n$ into aligned four-bit chunks and a residual word. Mark either $1101$ or $1011$ by the same symbol when the entering height exceeds $2-\lambda$ and at least three confirmations remain. Keep all other chunks, the residual word and the duration. Both marked completions agree in their endpoint height, preceding bit and count of $01$ transitions. Each fiber with $k$ marks consists of $2^k$ equally likely orientations under $\mathbb P_{n,q}$. Put $e(u)=e^{2\pi i u}$ and
$$
T_{n,q,m}(b)=
\mathbb E_{n,q}\big|\mathbb E_{n,q}[e(bX/2^m)\mid\mathcal R_n]\big|,
\qquad F(m)=\frac{(\log m)^2}{\log\log m}.
$$
Unmarked logarithms are natural.

**Theorem 1.** Let $I$ be a compact interval contained in $(\lambda/\beta,\infty)$ and fix $0<\alpha<1$. There are $C,c>0$ such that for sufficiently large $n,m$, all integers $q$ with $q/n\in I$, and odd $b$ with $0<|b|\le\exp(m^\alpha)$,
$$
T_{n,q,m}(b)\le C e^{-cF(m)}.
\tag{1}
$$
For $m>q+1$ the norm is zero. The constants may depend on $I$ and $\alpha$. The proof also gives superpolynomial decay for any prescribed subexponential numerator budget $\log |b|\le B_0(m)=o(m)$, with an onset depending on the budget.

**Corollary 2.** Fix $I$ as in Theorem 1. Set $M=q+1$, $L=M-d$, and let $d$ be a nonnegative integer with $d\le c_0 F(M)$ for a sufficiently small fixed $c_0>0$. Among the $2^{L-2}$ odd integers in $[2^{L-1},2^L)$, let $N(n,q,L)$ count those with $\tau_n=M$ and $C^j(x)>x$ for every $1\le j\le M$. Uniformly for $q/n\in I$,
$$
N(n,q,L)=2^{L-2}\mathbb P_0(\mathcal A_n,Q_n=q)
       [1+O_I(e^{-c_1F(M)})]
\tag{2}
$$
for some $c_1>0$. The marked record law for these integers is within $O_I(e^{-c_1F(M)})$ in total variation of its $\mathbb P_{n,q}$ law. A complete period $M\le L-1$ gives exact agreement after the first-crossing guard. The same relative estimate and marked-law comparison hold for a union of durations in a fixed shell if every duration in the union is covered by one of these two conclusions with uniform constants. Theorem 4 below extends this comparison to the shortest feasible duration. Actual survival after a central cutoff still leaves an unresolved tail.

The following local asymptotic identifies the reference mass even at atypical linear durations. For $\ell=q/n\in I$ set
$$
u=1-\lambda/\ell,\qquad v=1-\lambda/(\beta\ell),\qquad
\eta=\lambda^{-1}\log(u/v),\qquad t=\log(2v)+\eta,\qquad
K=\frac{(1-u)(1-v)}{uv}.
\tag{3}
$$
Use $\widetilde{\mathbb P}_\ell$ for independent positive geometric lengths with tails $u,v$. Put $v_A=u/(1-u)^2$ and $v_B=v/(1-v)^2$. On the nonnegative integers define the finite measures
$$
D_\ell(r)=\sum_{k\ge1}\frac1k
 \widetilde{\mathbb E}_\ell[e^{-\eta S_k}; S_k>0,Q_k=r],
\qquad \mathcal C_\ell=\sum_{j\ge0} D_\ell^{*j}/j!,\quad D_\ell^{*0}=\delta_0,
$$
and the bounded periodic function
$$
J_\ell(r)=\frac{\exp[-\eta\{\lambda(\lfloor r/\lambda\rfloor+1)-r\}]}
 {1-e^{-\eta\lambda}}.
$$
Here the braces in the numerator group the expression, and the convention at a jump selects the next strictly positive height.

**Theorem 3.** With $\mathcal I(\ell)=\log K+t\ell$,
$$
\mathbb P_0(\mathcal A_n,Q_n=q)
=e^{-n\mathcal I(\ell)}n^{-2}
 \left[
 \frac{\sum_{r\ge0}\mathcal C_\ell(r)J_\ell(q-r)}
 {2\pi\sqrt{v_Av_B}}+O_I(n^{-1/4})
 \right].
\tag{4}
$$
The leading factor inside the brackets is bounded above and away from zero uniformly on $I$. For each fixed tilt parameter $\ell$, its periodic function of the phase $q\bmod\lambda$ is nonconstant. Thus (2) and (4) give a relative integer count including the oscillation at individual durations and the exponential cost of atypical durations. The estimate concerns ensembles of sources.

## 2. A change of measure matching the duration

The tilted means are $\mu_A=\ell/\lambda$ and $\mu_B=\beta\ell/\lambda$. Thus the height has zero mean increment and the duration has mean $\ell$. The compact interval $I$ keeps the two geometric tails in a compact subset of $(0,1)$ and $\eta>0$ bounded away from zero. The one-block likelihood ratio is
$$
\frac{\widetilde{\mathbb P}_\ell(A=a,B=b)}{\mathbb P_0(A=a,B=b)}
   =K\exp[\eta(\beta a-b)+t(a+b)].
$$
Consequently on $\{\mathcal A_n,Q_n=q\}$ the probability law is tilted by the terminal factor $e^{-\eta S_n}$ alone, with normalizer
$$
Z_{n,q}=\widetilde{\mathbb E}_\ell[e^{-\eta S_n};\mathcal A_n,Q_n=q].
\tag{5}
$$
In particular the original event mass equals $e^{-n\mathcal I(\ell)}Z_{n,q}$. The exponent is the relative entropy per block of the two geometric laws, since the tilted mean of the height is zero.

The estimates below are uniform over the matched tilts. For $r$ tilted complete blocks started at a real height $x\ge0$, the killed probability of prescribed integer totals $a,b$ at height $y=x+\beta a-b>0$ is at most $C_I(1+x)(1+y)r^{-2}$. The killed probability of height in $[y,y+1)$ is at most $C_I(1+x)(1+y)r^{-3/2}$. These are the arguments of [2, Lemma 3.1], uniformly here. For details, the two negative binomial sums have maximal atoms $O_I(r^{-1/2})$ by Stirling's formula at their modes. A unit height interval allows a bounded number of $A$-totals for any fixed $B$-total. The centered increment and its negative have bounded first and second overshoot moments at every nonnegative threshold, uniformly over $I$, by conditioning on one geometric variable and using memorylessness in the other. Optional stopping on exiting $(0,R)$ gives upper-exit probability $O_I((1+x)/R)$ and mean exit time $O_I(R(1+x))$. Taking $R=\sqrt r$ gives half-line survival $O_I((1+x)r^{-1/2})$ in both directions. In the three-part decomposition, the outside pieces have forward and reverse survival bounds, and the middle has the unrestricted joint atom or interval bound. All variances are bounded away from zero and all overshoot and mode constants are uniform on $I$. The argument needs no integer lattice for the height.

For a prescribed duration $v_1$ the endpoints lie on a translate of $\lambda\mathbb Z$. Summing the joint estimate against the terminal weight yields
$$
\widetilde{\mathbb E}_{\ell,x}[e^{-\eta S_r};
 S_i>0\ (1\le i\le r),Q_r=v_1]\le C_I(1+x)r^{-2}.
\tag{6}
$$
In particular $Z_{n,q}\ll_I n^{-2}$. For its lower bound choose the totals $a=\lfloor q/\lambda\rfloor+1$, $b=q-a$. Both are within a bounded distance of their tilted means, and $0<\beta a-b<\lambda$. Their joint negative binomial atom is at least $c_I/n$. At least one cyclic rotation of any word at these totals stays strictly positive, by starting just after its minimum among partial sums through time $n-1$. The sums are distinct by irrationality of $\beta$, and a positive total preserves strict positivity when wrapping. Conditional rotation invariance therefore supplies a surviving fraction at least $1/n$. The endpoint weight is at least $e^{-\eta\lambda}$. Thus $Z_{n,q}\asymp_I n^{-2}$ for sufficiently large $n$.

Generate the tilted bits after the forced first 1 by a two-state Markov chain with stay probabilities $u$ in 1 and $v$ in 0. At a prefix of length $j\ge1$ with last bit 1, the continuation through the next valley has a nonnegative geometric number of extra ones and a positive geometric number of zeros. At last bit 0 there are just a nonnegative geometric number of extra zeros. If this valley is at bit time $j+\ell_0$ and height $y$, its confirming 1 starts the next independent full block and belongs to that block. For $r$ confirmations remaining, the required duration of the remaining $r-1$ complete blocks is $q-j-\ell_0$. The valley is imposed positive, and its expected positive increment over the current height is uniformly bounded. Thus (6) gives the weighted continuation bound $C_I(1+x)n^{-2}$ whenever $r\ge n/2$, where $x$ is the current height. This argument holds at adapted bit stopping times bounded by a deterministic horizon with at least $n/2$ confirmations remaining. In particular for any such prefix event its $\mathbb P_{n,q}$ probability is at most $C_I$ times its tilted expectation weighted by $(1+x)$ and killed on nonpositive height, by (5).

## 3. Unweighted complete clocks under the Markov law

The arithmetic mechanism is the ternary proxy and valuation bound of Paper 90 [3] and the complete-clock geometry and count potential of Paper 91 [4]. The reference bits here are Markov. We first work with unweighted probabilities and pay for the selected law after an adapted stopping time. This suffices because the final rate is faster than any fixed power of the conductor.

Let $f(s)=\lambda s-1$ and $P$ be the bit transition matrix. The chain is reversible with stationary odd probability $p=1/\lambda$. The identity $1-u=\beta(1-v)$ gives both assertions. If $g(s)=(\mu_B-\lambda)s$, then $g-Pg=Pf$. In the forward direction $H_j+g(s_j)$ is a martingale. In the reverse direction the physical increment along a reversed edge is $-f(s_{j-1})$, using the state from which the reverse transition departs. A corrector is $g^-(s)=-g(s)-f(s)$, since $g^--Pg^-=-f$. Reversing a path with specified endpoint states changes its probability by a ratio of their stationary probabilities, uniformly bounded. Both corrected increments are bounded, and their conditional variances are bounded below by a positive constant. Indeed the two possible next corrected values differ in absolute value by $\lambda+g(1)-g(0)=\mu_B$, and both transition probabilities are bounded below. Azuma's inequality gives a uniform Gaussian tail for physical displacement, since the endpoint correction is bounded.

For either direction the probability of half-line survival through $r$ bits from height $x\ge0$ is at most $C_I(1+x)/\sqrt{r+1}$. Exit $(0,R)$ from $x<R$ and take the first exit after time zero. A sufficiently long string of states with one sign forces exit with positive probability uniformly over the current state, so the exit has finite mean. The corrected exit heights are within a constant of 0 or $R$. Optional first moment bounds the upper-exit probability by $C_I(1+x)/R$, and the stopped square identity then bounds the mean exit time by $C_I R(1+x)$ using the conditional variance lower bound. The passage from bounded stopping follows by the martingale square isometry and finite mean exit time. Survival is contained in upper exit or an exit time greater than $r$. Take $R=\sqrt r$, enlarging constants for small $r$ or $x\ge R$.

Also every unrestricted height atom at time $r$ is at most $C_I/\sqrt{r+1}$, including when a terminal state is specified. Use a uniform mixture $P(s,\cdot)=\epsilon\nu+(1-\epsilon)P'(s,\cdot)$ with $\nu$ uniform on the two states. The refresh decisions and their $\nu$ draws can be made independently. For disjoint consecutive pairs of transitions strictly inside the interval, whenever both transitions refresh, the intermediate bit is independent of all bits outside such selected sites after conditioning on the refresh decisions. The second refresh separates this bit from the subsequent path. There are a binomial number of such sites with expectation at least a fixed multiple of $r$ for large $r$. Except for exponentially small probability, at least a fixed multiple of $r$ independent fair bits remain in the count. Their maximum binomial atom is $O_I(r^{-1/2})$. This works in both directions, with the same interior bits and opposite signs.

We use these two estimates to extend the soft band bound [4, Lemma 3]. Sample every four bits and put $I_0=(-5,0)$. Let $N_j$ count the visits to $I_0$ from time 0 through $j$ at those sampled times. For $\omega>0$, if $z\ge0$, $r\ge0$ and $y\le r$, then
$$
\mathbb E_{z,s}[e^{-\omega N_j}; H_{4j}=y]\le C_{I,\omega}(r+1)/(j+1).
\tag{7}
$$
Here $z$ and $y$ are real heights relative to the band and there is no positivity restriction. To prove it, a path started inside the band and avoiding its first positive return for $j$ blocks stays on one side after the first step, because a four-bit jump has magnitude at most four. It starts that avoidance within four of the edge. The half-line bound applies after shifting the bit path by four in the appropriate direction. The first-return tail is therefore $O_I((j+1)^{-1/2})$, in either direction, uniformly in the state. If at most $k$ returns occur by $j$, one of the first $k+1$ waiting times exceeds $j/(k+1)$. The strong Markov property and a union bound give $O_I((k+1)^{3/2}/\sqrt{j+1})$. Summing with geometric weights gives $O_{I,\omega}((j+1)^{-1/2})$ discounted mass from within the band. Splitting into two halves, using the discount on the first half and the free atom on the second, gives a discounted endpoint atom $O_{I,\omega}((j+1)^{-1})$ from within the band. From $z\ge0$, paths with no visit stay above the band. Reverse the final half from $y\in[0,r]$ and apply the half-line bound, then bound the first-half atom. For a first visit in the first half, use the bound from within the band. If all visits are in the second half, reverse the complete path and use the first-visit argument in that direction. The sampled visit count is invariant under reversal and the stationary ratio is bounded. Summing over the two possible endpoint states proves (7).

Work at old conductor $\mathfrak m=m+1$ and numerator $b'=3b$. At four-bit time $i$ let $a(i)=a_{4i}$ be the running odd-bit count and put $d_i=\mathfrak m-2-4i$. The centered representative $r_i=b'3^{-(a(i)+3)}\bmod 2^{d_i}$ is called good if $98|r_i|>2^{d_i}$ and black otherwise. Only the nonterminal columns $i<\lfloor(\mathfrak m-3)/4\rfloor$ will be used. At a black entrance $i_0$, write its label $3^{a(i_0)+3}r_{i_0}=3^\nu w$ with $3\nmid w$. Its count gap is $s=\nu-3-a(i_0)\ge0$. Write $x$ for its physical height, $h_0=s+1$, and $L_0=\log_2(98|3^\nu w|)-(\mathfrak m-2)-3\lambda$ for the lower height of its triangle. Define
$$
T=\min\{j\ge1:\#\text{ ones in the next }4j\text{ bits}>s\},\quad
k_0=(\#\text{ ones through }T)-s,\quad
Q=\lfloor\tfrac12\log_3 h_0\rfloor+1,\quad
P_0=\max(0,\lfloor(Q-4)/8\rfloor).
$$
The overshoot satisfies $1\le k_0\le4$. The original full top interval ends $E=\lfloor(x+\lambda s-L_0)/4\rfloor$ columns after the entrance. Follow the count clock even after any earlier departure from its triangle. The complete episode ends at $\rho=T+P_0$ if $T\le E-P_0$, and at $\rho=T$ otherwise. This is a stopping time. Let $G$ count the good vertices from the entrance through $\rho$. For every fixed $\omega>0$ and any available horizon $R$ wholly in the nonterminal columns, we claim
$$
\mathbb E[e^{-\omega G}; \rho\le R,\ \text{all heights survive to the endpoint}]
   \le C_{I,\omega}(s+1)^{-c_{I,\omega}}
\tag{8}
$$
for some positive power. The initial Markov state is arbitrary and the estimate is uniform in the other entrance data.

For each specified crossing column $j$, overshoot $k_0$ and last bit, the probability is at most $C_I/\sqrt{h_0}$. Indeed the total count $s+k_0$ fixes the height, and $4j\ge h_0$, so drop the crossing constraint and use the unrestricted atom. By the displacement concentration and monotonicity of the count, the probability outside $|j-h_0/(4p)|\le C_D\sqrt{h_0\log(h_0+1)}$ is $O_{I,D}(h_0^{-D})$ for any fixed $D>0$. One applies concentration at the two deterministic window edges. This gives the stated tail without a sum over crossing columns.

On the branch $T\le E-P_0$, all observations from $T$ through $T+P_0$ are inside the original full top interval. Its seed at a column $i$ is $-w2^{-d_i}$ modulo $3^Q$. Multiplication by $16$ advances one column and has order $3^{Q-1}$ modulo $3^Q$, by the valuation of $16^{3^a}-1$. Each possible seed in its fixed nonzero class modulo 3 therefore selects one clock progression. Since $3^{Q-1}\le\sqrt{h_0}$, the probability of any central seed, together with a specified overshoot and specified last bit, is at most
$$
C_I\sqrt{\log(h_0+1)}\,3^{-(Q-1)}.
\tag{9}
$$
The tail is charged just once. After conditioning on the crossing's last bit, the next word has its usual Markov law from that bit. We keep the joint submeasure in (9) and never divide by the probability of the bit or of the overshoot.

For a fixed future word, follow the first count crossings of precisions $16,32,\ldots,16\lfloor P_0/16\rfloor$ above the old top. If the future word has at least $P_0$ ones in $4P_0$ bits, each such crossing occurs, with precision in $[16j,16j+3]$ and at most $k_0+4P_0\le Q$. The failed count event has probability $O_I(e^{-c_I P_0})$ under each starting state by the displacement bound. At any observation with precision $k$, blackness implies that the centered seed modulo $3^k$ lies in the interval of radius $(3^k+1)/98$ [3, equation (16)]. Indeed if the observed residue is $r$, then $3^k r\equiv w\bmod 2^{d_i}$, and $(3^k r-w)/2^{d_i}$ represents that centered seed. The full top interval and blackness give its asserted magnitude.

For a uniform reference seed in the fixed unit class modulo 3, condition on any preceding exposed lower trits. Precision increases by at least 13 and its first value is at least 16. Multiplying by the known power of 16 preserves the residue filtration. In the interval the proportion of fresh lifts is at most $2/98+2/3^8\le3/98$, by counting an arithmetic progression in a centered interval. Thus for any specified set of $k$ observations their joint proxy-black probability is at most $(3/98)^k$, by successive conditioning. If there are $a_0=\lfloor P_0/16\rfloor$ observations, the probability of at least half proxy black is at most $(2\sqrt{3/98})^{a_0}$. Multiply this reference bound by the domination in (9) and integrate over the future word separately for each last bit and overshoot. On the other event the discount is at most $e^{-\omega a_0/2}$. Adding the total clock tail and failed count mass gives a negative power of $h_0$, since $P_0$ and $a_0$ grow as fixed multiples of $\log h_0$. The restriction that the endpoint be within $R$ can be dropped after these pointwise implications.

On the late branch $T>E-P_0$, for fixed central $T=j$ and overshoot the endpoint height is $y=x+\lambda(s+k_0)-4j<L_0+4P_0+4\lambda$. Before crossing the old top, each visit to $(L_0-5,L_0)$ is good by [4, Lemma 4]. This holds even after previous departures from the original triangle. Hence the discount in (8) is bounded by the band discount including the endpoint, times $e^\omega$. With the endpoint height already specified, drop the clock constraint and apply (7) from $x-L_0\ge0$. Sum its bound $O_{I,\omega}((P_0+1)/(j+1))$ over four overshoots and $O_I(\sqrt{h_0\log(h_0+1)})$ central columns. Since $4j\ge h_0$, the result is a negative power of $h_0$. Outside the window use the total clock tail. Small $s$ is absorbed into the constant. This proves (8).

## 4. Occupation and transfer to integer sources

Use $V=\lfloor\varepsilon\mathfrak m\rfloor$ four-bit steps, with $\varepsilon>0$ fixed small enough in terms of $\sup I$ that $4(V+1)<n/2$ for large $m\le q+1$. This is inside the nonterminal conductor. To supply a state before the first bit, start the unconditioned Markov chain with state 1 at height zero. Its first bit is 1 with probability $u$. Every positive-length surviving prefix starts with 1, so all its probabilities under the forced-first-bit law are obtained by dividing by $u$, uniformly bounded away from zero. Subsequent starting-state assertions are unaffected.

Beginning at zero, search for a black entrance, follow its complete clock, then search strictly after its endpoint. Stop by column $V$ and count good vertices through $V$ inclusive. Denote this count by $W_V$. The completed episodes have disjoint vertex sets. Every completed episode contains a good vertex because consecutive black vertices keep their label by [3, Section 3], and at its original top crossing the count has exceeded that label's allowed count. The search and completion indices are stopping times.

Write $s_i,G_i$ for the gap and good count of the $i$-th completed episode. By (8) at $\omega=1$, the product of factors
$$
(s_i+1)^{c_*}e^{-G_i}/C_*,
$$
killed on failure of positivity or completion, is a nonnegative supermartingale in the episode filtration for some fixed $C_*\ge1,c_*>0$. The search between episodes costs at most one in conditional probability. Assign value zero to all stages following a failed search or incomplete episode. Optional stopping bounds the probability that this product ever reaches $e^u$ by $e^{-u}$. A successful first crossing has an adapted bit endpoint at most $4V$. The conditional prefix comparison following (6) multiplies this bound by at most $C_I(1+m)$, since the height at such a prefix is $O(m)$. The factor for the forced first bit is included in $C_I$. Consequently
$$
\mathbb P_{n,q}\!\left(\exists k:\ 
 c_*\sum_{i\le k}\log(s_i+1)-k\log C_*-\sum_{i\le k}G_i\ge u\right)
 \le C_I(1+m)e^{-u}.
\tag{10}
$$

For precision, the deterministic count argument here is the one in [4, Section 5]. Let $B\ge\max((\log\mathfrak m)^4,\log|b'|+5)$ with $B=o(\mathfrak m)$, and put
$$
D=100[\log(1+\mathfrak m/B)+11]^2,\quad
Y=V/(D+1),\quad
\mathcal B=\frac{(\log Y)^2-(\log B)^2}{\log(D+1)}.
$$
At every supported entrance with count $A$, the valuation estimate of [3, Appendix A] gives $s\le D(A+B)$. Its lift condition holds because the lift $(3^{A+3}r_i-b')/2^{d_i}$ has magnitude less than $14\,3^A$ on a positive prefix. This follows from $3^A\ge2^{4i}$ and $|b'|<2^{\mathfrak m-3}$. The function $v\mapsto v[\log(1+\mathfrak m/v)+11]^2$ is increasing, so the estimate with $A+\log|b'|+5$ implies the one with $A+B$. Every observation budget $P_0$ is $O(\log m)$ since $A\le4V$ and $s\ll m(\log m)^2$.

If the final episode is unfinished, let $a_*^{\rm cnt}$ be its entrance count. Otherwise let $a_*^{\rm cnt}$ be the count at $V$. An unfinished episode has gained at most $s$ before its old top, or at most $s+4+4P_0$ after it. Therefore
$$
B+a(V)\le(D+1)(B+a_*^{\rm cnt})+O(\log m).
$$
A surviving prefix has $a(V)>4V/\lambda>2V$, so $B+a_*^{\rm cnt}\ge Y$ for large $m$. If $W_V\le g\le(\log\mathfrak m)^2$, at most $g$ episodes completed. Apart from their first $s_i$ count units, the count gain before $a_*^{\rm cnt}$ is $O((g+1)\log m)$. The uncovered search edges and boundaries number $O(W_V+g+1)$, each adding at most four, and the remaining overshoots and observations cost $O(\log m)$ per complete episode. Apply the increasing potential $\Phi(y)=(\log y)^2/(2\log(D+1))$ for $y\ge B$. For $0\le s\le Dy$ it satisfies $\Phi(y+s)-\Phi(y)\le2\log(s+1)$ [4, Lemma 7]. Its derivative decreases, so the uncovered units contribute at most $O((g+1)\log m)\Phi'(B)=o(1)$ by $B\ge(\log\mathfrak m)^4$. Telescoping gives $\sum_i\log(s_i+1)\ge\mathcal B/4-o(1)$. Since $\sum_iG_i\le g$, choosing $g=\lfloor c_2\mathcal B\rfloor$ with sufficiently small $c_2>0$ makes the accumulated payment in (10) at least $c_3\mathcal B-O(1)$ on $\{W_V\le g\}$. This proves
$$
\mathbb P_{n,q}(W_V\le g)\le C_I(1+m)e^{-c_3\mathcal B}.
\tag{11}
$$
For $r=\log(\mathfrak m/B)\to\infty$, $\log(Y/B)=r-O_I(\log(2+r))$, $\log Y+\log B\asymp\log m$, and $\log(D+1)\asymp\log(2+r)$. Thus $\mathcal B/\log m\asymp r/\log(2+r)\to\infty$. For $|b|\le e^{m^\alpha}$ choose $B=\mathfrak m^{(1+\alpha)/2}$ above a fixed onset. Then $\mathcal B\asymp_\alpha F(m)$.

The quota is at least $n/2$ throughout the first $V+1$ chunks, so only the height test can remove a good vertex from the Fourier product there. A positive height at most $a_*=2-\lambda$ on a surviving word can only follow a 0 since $\beta>a_*$. The next bit must be 1 to stay positive, so that height is a block valley. Apart from the origin, early visits to the strip are therefore among the first $\lfloor n/2\rfloor$ block valleys. For a proposed ordered tuple $0<i_1<\dots<i_k\le\lfloor n/2\rfloor$ of visits, the killed interval bound of Section 2 gives a product $\prod_{j=1}^k C_I(i_j-i_{j-1})^{-3/2}$ with $i_0=0$. At the last visit the exact-duration continuation (6) is at most $C_I n^{-2}$. Division by $Z_{n,q}$ and summation over the gaps give factorial moments $\mathbb E_{n,q}\binom{V_{\rm low}}k\le C_I C_4^k$ for the valley count. A fixed small binomial generating parameter then gives an exponential tail, also after adding the origin.

The two inverse maps for $1101$ and $1011$ are $(16y-23)/27$ and $(16y-29)/27$ [2]. At chunk entrance $i$ their exchange translates the source by $-2^{4i+1}/3^{a(i)+2}$. The prefix slope and the resulting translation are unchanged by other orientations. Therefore the inner modulus in (1) is the product over marked chunks of
$$
\left|\cos\left(\pi\,\frac{b\,3^{-(a(i)+2)}}{2^{m-4i-1}}\right)\right|,
$$
where the inverse and phase are interpreted modulo the displayed positive power of two. A nonpositive exponent gives an integral phase. The shifted conductor $\mathfrak m=m+1$ and numerator $b'=3b$ used above give exactly this phase, so each good marked vertex contributes at most $\cos(\pi/98)<1$.

Discard all factors after the first $V+1$ chunks. Under the Markov prefix law, both possible chunks have probability at least a fixed constant at every eligible entrance because all bit transitions have a uniform positive lower bound. They are both legal from height $>a_*$. Use factor 1 for an unmarked chunk. The conditional mean is at most some fixed $\vartheta<1$ at any predictably good eligible entrance, and at most one elsewhere. Dividing by $\vartheta$ for each such entrance gives a nonnegative supermartingale. This argument can start with the auxiliary state and kill on nonpositive height. On paths with at least $\lfloor g/2\rfloor$ good eligible entrances, the unweighted product expectation is at most $\vartheta^{\lfloor g/2\rfloor}$. Transfer it by (6), costing at most $C_I(1+m)$ as in (10). The two exceptions are $W_V\le g$ and more than $g/2$ strip visits. Using (11) and the strip tail proves (1), and the subexponential-budget assertion follows because $\mathcal B/\log m\to\infty$. For $m>q+1$, the conditional residue is uniform on higher lifts and the primitive character sums to zero. This completes the proof of Theorem 1.

For Corollary 2 apply the Erdős–Turán interval inequality conditional on the marked record, then average. Use a harmonic cutoff $\exp(c_5 F(M))$ with $c_5$ sufficiently small. The primitive conductor of any such character is $m=M-O(F(M))$ and its numerator lies in the range of (1), for example with $\alpha=1/2$. The expected interval discrepancy is $O_I(e^{-c_6F(M)})$ after decreasing $c_6>0$. The shell interval divided by $2^M$ has length $2^{-d-1}$. The selected residue mass in it is its length with relative error $O_I(2^{d+1}e^{-c_6F(M)})$. Conditioning the record on that interval biases its law by the conditional interval mass divided by the unconditional one. The same averaged discrepancy divided by the interval length bounds the resulting total variation, up to a constant. Choose $c_0$ small enough. The standard first-crossing guard [2, Section 5 and references there] applies to all sources in the shell for large $M$, since they exceed $900(M+1)^{14.3}$. Thus coefficient and actual survival agree through the specified stop. For $M\le L-1$ each residue occurs equally often. Relative errors over a union of durations add weighted by the reference masses. Normalizing both mixtures proves the claimed marked-law transfer.

This transfer has a precise distinction from the full ordered word. When $M\ge L$, a full stopped word determines one residue modulo $2^M$, so its actual conditional law is the odd Haar selected word law restricted to the shell interval. Its total variation from the unconditioned selected word law is one minus the selected residue mass of the interval. In the range of (2) this is $1-2^{-d-1}(1+o(1))$. The marked record and the relative count remain accurate even when that full-word distance is close to one.

## 5. Uniform local asymptotic at the matched tilt

For a fixed parameter $\ell\in I$, denote the $k$-th summand of $D_\ell(r)$ by $b_k(r)$ and its mass by $a_k=\sum_r b_k(r)$. Three bounds are uniform over $I$:
$$
a_k\le C_I k^{-3/2},\qquad
\sup_r b_k(r)\le C_I k^{-2},\qquad
\sum_r r b_k(r)\le C_I k^{-1/2}.
\tag{12}
$$
For a fixed total of the $A$'s, summing positive endpoint heights against $e^{-\eta y}$ costs at most $C_I/\sqrt k$ in the other coordinate, by its negative binomial maximum and the integer spacing of the heights as the $B$-total varies. This gives the first bound after division by $k$. Also $Q_k<\lambda\sum A_i$ on positive height, so the first moment of the $A$-total gives the third bound. At fixed duration the two-coordinate maximum is $C_I/k$ and the heights have spacing $\lambda$. This gives the second bound.

For $|r-\ell k|\le C_0\sqrt k$ set $x=(r-\ell k)/\sqrt k$ and $\sigma_Q^2=\lambda^2 v_A v_B/(\beta^2 v_A+v_B)$. The uniform estimate is
$$
k^2 b_k(r)=
 \frac{e^{-x^2/(2\sigma_Q^2)}}{2\pi\sqrt{v_Av_B}}\,
 J_\ell(r)+O_{I,C_0}((1+\log k)/\sqrt k).
\tag{13}
$$
To see this, at fixed $r$ the two totals are $a$ and $r-a$, and the positive heights are $y=\lambda a-r>0$. The first is $\lambda(\lfloor r/\lambda\rfloor+1)-r$ and the spacing is $\lambda$. For $y\le C\log k$, the two standardized deviations from their means are $x/\lambda+O(y/\sqrt k)$ and $\beta x/\lambda+O(y/\sqrt k)$. Stirling's expansion of the two independent negative binomial masses gives their product
$$
(2\pi k\sqrt{v_Av_B})^{-1}
 \big[e^{-x^2/(2\sigma_Q^2)}+O_{I,C_0}((1+y)/\sqrt k)\big].
$$
The expansion is uniform on the indicated windows since the geometric parameters stay in a compact interior interval. The sum of exponential terminal weights over the full positive height progression is $J_\ell(r)$. Heights above $C\log k$ have scaled contribution $O_I(k^{-1})$ by the two maxima and exponential damping for sufficiently large $C$. This proves (13).

Let $Z_{k,r}^{(\ell)}$ denote the terminally weighted killed mass at duration $r$ for any $k$ under this fixed $\ell$. The strict Sparre Andersen fluctuation identity [5,6] with duration marked gives
$$
\delta_0+\sum_{k\ge1} z^k Z_{k,\cdot}^{(\ell)}
  =\exp_*\!\left(\sum_{k\ge1}z^k b_k\right).
\tag{14}
$$
Coefficients are measures on the duration integers and multiplication convolves that mark. To justify the mark, apply the scalar strict identity to the independent increments under the probability measure reweighted in each block by $e^{-\eta(\beta A-B)+w(A+B)}$, divided by its one-step mass. Here $w$ is real near zero. Positivity is unchanged as a test on the words. Replace $z$ by $z$ times the one-step mass. The resulting Laplace identities on an open interval determine the coefficient measures. All required one-step transforms are finite near zero by the geometric tails. No nonempty block word has total height zero, by irrationality of $\beta$.

For completeness we give the uniform remainder in the convolution of (14). Let $A=\sum_k a_k$, uniformly bounded by (12), and consider the $j$-factor contribution at time $n$ and duration $q$. Its time lengths are positive integers summing to $n$, and a longest is at least $n/j$. When the others and their durations are fixed, the longest factor's duration is fixed. By (12) its maximum times $n^2$ is at most $C_I j^2$. Summing over the $j$ choices of longest position gives an upper bound $C_I j^3 A^{j-1}$ for the scaled contribution. For $j\ge2$, terms with a second length greater than $\sqrt n$ have scaled total at most
$$
C_I j^3(j-1)A^{j-2}\sum_{k>\sqrt n}a_k
 \ll_I j^3(j-1)A^{j-2}n^{-1/4}.
$$
For the other terms the sum of the time and duration marks of the small factors has truncated first moment at most
$$
C_I(j-1)A^{j-2}\sum_{k\le\sqrt n} k^{-1/2}
 \ll_I(j-1)A^{j-2}n^{1/4}.
$$
Here (12) controls both marks. Thus the contribution where that sum exceeds $\sqrt n$ has the same scaled error after using the largest-factor bound and Markov's inequality.

On the remaining terms the longest index is unique for large $n$. Write its time as $n-l$ and its duration as $q-r$. Both small-mark sums $l,r$ are nonnegative and $l+r\le\sqrt n$. At the matched tilt $q=\ell n$ the standardized deviation is $(\ell l-r)/\sqrt{n-l}$. It is bounded uniformly, and the Gaussian envelope in (13) differs from 1 by at most $C_I(l+r)/\sqrt n$. The normalization ratio $(n/(n-l))^2$ differs from 1 by $O(l/n)$. Keep $J_\ell(q-r)$ exactly. The errors from (13), from the Gaussian envelope and from this ratio sum to $O_I(n^{-1/4})$ after division by $j!$ and summation over $j$, using the truncated first moment and the bound on $A$. Removing the small-mark truncations in the leading term costs the same error by (12). For $j=1$ use (13) directly with zero deviation. The factor $j$ for the long position changes $1/j!$ to $1/(j-1)!$. We obtain
$$
n^2 Z_{n,q}^{(\ell)}
 =\frac{1}{2\pi\sqrt{v_Av_B}}\sum_{r\ge0}\mathcal C_\ell(r)J_\ell(q-r)
  +O_I(n^{-1/4}).
$$
Use (5) to get (4).

The bounds on the leading factor follow from (12), the positive uniform lower bound on $J_\ell$, and the unit atom at zero of $\mathcal C_\ell$. At fixed $\ell$, the Fourier coefficient of $J_\ell$ in the basis $e^{2\pi i h r/\lambda}$ is $1/[\lambda(\eta-2\pi i h/\lambda)]$ by integration over one period. Translation convolution multiplies it by $\exp(\sum_r D_\ell(r)e^{-2\pi i h r/\lambda})$, which is nonzero. Thus the phase factor is nonconstant. Formula (4) retains the jump convention even for terms where $q=r$. This proves Theorem 3.

## 6. The short-duration edge

**Theorem 4.** Fix $L_0>\lambda/\beta$, $\delta>0$ and $0<\alpha<1$. For sufficiently large $n$, Theorem 1's bound (1) holds uniformly for
$$
\lambda n/\beta<q\le L_0 n,\qquad \delta n\le m\le q+1,\qquad
0<|b|\le e^{m^\alpha},\quad b\ {\rm odd}.
$$
Thus Corollary 2 holds uniformly all the way down to the shortest feasible duration, for $q\le L_0n$. The local asymptotic (4) keeps its compact-interior hypothesis.

For any selected word the total number of zeros is at least $n$. Positive endpoint height implies $\beta q>\lambda\sum_i B_i$, so $q>\lambda n/\beta$ is necessary. Conversely at each integer $q$ above this threshold, $a=\lfloor q/\lambda\rfloor+1$ and $b=q-a=\lfloor\beta q/\lambda\rfloor$ both exceed or equal $n$, and the endpoint height is strictly positive. A rotation of a word at these totals survives. This proves feasibility.

Use the change of measure of Section 2, now taking
$$
v=\max(1-\lambda/(\beta(q/n)),\,1/n),\qquad
u=1-\beta(1-v).
$$
The two means have ratio $\beta$ as before. If the clipping changes $v$, it shifts the mean total duration through $n$ blocks by only a bounded amount, so the specified totals $a,b$ still lie within bounded distance of their respective means. The likelihood formula stays exact with $\eta,t,K$ computed from these $u,v$, and the duration exponential remains constant on $Q_n=q$. The new $u$ stays in a fixed compact interior interval and $0<v\le v_{\max}<1$, with $nv\ge1$.

There is a uniform polynomial lower bound $Z_{n,q}\ge c n^{-3}$ under this tilted law. The sum of the $A$'s has mass $\gg n^{-1/2}$ at $a$ as before. For the $B$ sum, the excess over $n$ has variance $O(n)$, and Chebyshev's inequality puts a constant mass within $O(\sqrt n)$ of its mean. Its maximal atom is therefore $\gg n^{-1/2}$. A mode is within a bounded distance of the mean by the adjacent-mass ratio $v(n+j)/(j+1)$ at excess $j$. At nonnegative $j$ within any fixed distance of $nv/(1-v)$, these ratios are bounded above and away from zero uniformly because $nv\ge1$ and $v\le v_{\max}$. Thus the atom at $b$ is also $\gg n^{-1/2}$. The rotation loses at most a factor $n$. The terminal weight is at least $e^{-\eta\lambda}=v/u\ge1/n$. This proves the stated lower bound.

We check that the unweighted Markov inputs of Section 3 remain uniform as $v$ tends to zero. Reversibility and the two correctors remain unchanged and bounded. The corrected martingale still has bounded differences, hence the same uniform Azuma bound. For survival use two-step sampled martingales. From either state there is uniformly positive probability of state 1 at the next step, and from state 1 both outcomes have uniformly positive probability. The conditional variance accumulated in two steps is therefore bounded below uniformly in both directions. The first exit from a bounded interval has finite mean, since a sufficiently long run of 1's still has probability bounded below for any given length, uniformly in the current state. That run has a nonzero displacement of fixed sign in each direction. The exit proof of Section 3 now bounds survival at two-step times, hence also at all bit times.

For the free atom use a minorization with a deterministic refresh to 1. In disjoint groups of three transitions, condition on whether the first and third refresh. Where both do, the two surrounding states are fixed at 1, and the intermediate bit has its free distribution $P(1,\cdot)$. Its two values have probability bounded away from zero. After fixing the refresh flags for the first and third transitions in each group, the selected intermediate bits are independent of the other bits, because each is preceded and followed by a fixed refresh. The flags and unused randomness in the middle transitions are not conditioned. This supplies a binomial count of isolated Bernoulli variables, and its maximal atom is again $O(r^{-1/2})$. The terminal-state and reversal conventions are as in Section 3. Thus (7) and the unweighted clock proof (8) are uniform near the edge. Both marked words use only the transitions $11,10,01$ and an entering 1. Hence their local probabilities are also bounded below uniformly, even when $00$ is rare.

For the strip visits it suffices to retain the tilted killed interval estimate with exponent $3/2$. The $A$-sum alone gives the free unit-interval bound $O(r^{-1/2})$ after conditioning on the $B$ sum. The block increments remain centered with uniformly positive variance and bounded overshoot moments in both directions. Thus the forward and reverse half-line estimates and the three-part argument in Section 2 give the uniform killed interval bound. The ordered visit argument bounds the tilted probability of many low valleys on surviving paths by $Ce^{-cg}$, with no need to include the terminal exact-duration restriction in this upper bound. In detail the expectation of the binomial visit coefficient on survival is at most a sum of products of $C(i_j-i_{j-1})^{-3/2}$ by discarding survival after its last specified visit. Summing those products and using a binomial generating parameter gives the stated tail.

At an early adapted time the selected probability or expected nonnegative prefix factor is now at most $C n^3$ times its tilted bound, simply by dropping the terminal restriction and using $e^{-\eta S_n}\le1$ on selection. The same applies to the strip exception. Since $m\ge\delta n$, this cost is polynomial in $m$. It is absorbed by $\exp(-cF(m))$ in the payment, strip and cosine estimates of Section 4. This proves Theorem 4. In the interval discrepancy application the tested primitive conductors satisfy $m\sim q$ and $q>\lambda n/\beta$, so its lower conductor condition holds with a fixed $\delta$. This gives the extended integer consequence.


## 7. Scope and checks

Paper 107 supplies the run-preserving exchange, its phase, the geometric killed-kernel method and the actual-source guard. Its main theorem conditions on the run stop and then caps it. The public follow-up [7] handles a single central duration at rate $\exp[-c\log m/\log\log m]$, using the deterministic carry count of Paper 89 [8]. The present arithmetic part obtains the faster rate by adapting the actual complete clock of Papers 90–91 to the correlated bit law, paying for an adapted prefix only once. The choice of two means in (3) then covers a compact interval of atypical duration ratios. Paper 110 [9] concerns ternary concentration for critically tilted geometric words using stopped moments. Its atom certificate and exceptional-count bounds do not give this binary source-character estimate with a prescribed run duration. General conditioned local limits for finite-state Markov walks [10] provide background. The elementary refresh estimate here is for the exact count atoms needed at the clock crossing.

The bounded supplement compares literal word enumeration and aggregated Markov crossing tables in two rational proxy models. It checks the future split by the last bit, reverse rewards at specified endpoint states, the duration likelihood correction, ternary residue orders and lift counts. Separate exact controls check the first feasible duration, the two-step variance and refresh argument at the degenerate edge, and the marked fluctuation identity. A floating-point illustration compares finite leading phase sums to exact dyadic counts at twelve prescribed durations, and is separate from the proof. The omitted-state and wrong reverse-reward controls produce discrepancies, as expected. The exact-duration full-word obstruction is consistent with the information ceiling in Paper 94 [11]. The first small arithmetic survey produced no actual post-top black witness. Four explicitly constructed vertices give nonempty checks of that implication. The asymptotic statements use the proofs, with the imported arithmetic identified in Sections 3 and 4. The open arithmetic problem is to count actual survivors whose stopping duration extends beyond the covered range, in particular the uncapped tail at a central source length.

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