Factorial Linear Forms and Denominators: Detailed Proofs and Rationality Criteria

Will Cook

Long paper

Factorial Linear Forms and Denominators: Detailed Proofs and Rationality Criteria

Erdős #68 By Will Cook 1 September 2026; revised 7 October 2026 Authorship, AI use and citation

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Summary. The record proves liminf (log L_N)/(N^3/2 log N) $\geq 2 \sqrt{2} / 3$ for L_N=lcm(2!−1,…,N!−1), by comparing a terminal block of denominators with pairwise gcds rather than an external multiplicity theorem. Specialising a Louwsma–Martino valuation, it exhibits primes 139 and 2593 absent from all later reduced denominators. Coefficient constructions give integer forms MS+k; at fixed moment the remainder changes only by an integer. Rationality criteria use factorial digits and the next integer above a scaled partial sum; the digit argument also recovers the irrationality of e. Two exact calculations give q∤299999! and $q \geq 2^{39990} > 10^{12038}$ . These exclusions do not prove irrationality.

This paper covers the complete problem-specific reasoning surface for Erdős #68, including all registered result families and their boundaries.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #68, which remains open.

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In this paper

An elementary criterion for any real number

The fractional-part condition in (20) is an instance of Cantor’s termination criterion. For any real $x$ and positive sequence $c_{m}$ satisfying $c_{m} \leq 1 / m$ eventually, we have

x \notin Q ⟺ {(m - 1)! x} \geq c_{m} for arbitrarily large m .

For rational $x$ the fractional parts vanish eventually. Conversely, if ${(m - 1)! x} < c_{m} \leq 1 / m$ for all large $m$ , then $m {(m - 1)! x} < 1$ , so every sufficiently late canonical digit is zero. Cantor’s criterion then makes $x$ rational. Equality $c_{m} = 1 / m$ is allowed: it is failure of the displayed weak inequality that makes $m {(m - 1)! x}$ strictly less than one. For $x = S$ , the choice $c_{m} = E_{m} / m$ gives exactly (20); indeed $0 < E_{m} < 1$ ensures the required threshold bound. The finite test (21) is a separate sufficient condition at one index. Its cofinal converse uses the carry theorem, as shown in §4, not substitution into this general criterion. Neither argument proves that $S$ meets the test at arbitrarily large indices.

The threshold $1 / m$ cannot be replaced by $λ / m$ for a fixed $λ > 1$ . For $x = e$ and $m \geq 3$ , the factorial tail gives

m {(m - 1)! e} = 1 + m! \sum_{n > m} \frac{1}{n!} < 1 + \frac{1}{m} < λ eventually .

Thus the irrational number $e$ would never meet the enlarged threshold at sufficiently large indices. Positivity is also essential: with $c_{m} = 0$ , rational numbers would meet the threshold eventually. These examples explain the two assumptions without imposing any equidistribution hypothesis.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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