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

Support bounds and nonintegrality

There are three separate choices: the vector, its multiplier $M$ , and the cutoff used to estimate its remainder. The basis and support equations classify the first two. Once $M$ is fixed, $R (c) \equiv M (c) S (\mod Z)$ fixes the fractional part: integer translation may put a representative near zero but cannot change its distance from an integer.

To exclude every rational denominator, each $q > 0$ must divide the multiplier of some nonintegral form. Growth of the compulsory divisor $12 L_{D}$ alone is insufficient: its $3$ -adic valuation is exactly one for every $D$ , so it does not even force $9 ∣ M$ . By contrast, extending the cutoff changes neither $M$ nor the remainder, only the finite enclosure used to test it.

For $c = - 6 e_{2} + e_{4}$ , we have $M (c) = 12$ and $R (c) = 12 S - 17$ . At cutoff $4$ , its finite part is $- 239 / 115$ : the gap $9 / 115$ is smaller than the tail bound $24 / 119$ . At cutoff $5$ , the finite part is $- 27061 / 13685$ , whose gap $13376 / 13685$ exceeds the new bound $24 / 719$ . The finite part has crossed $- 2$ , so the next integer is now $- 1$ rather than $- 2$ . This explains the larger gap. The second cutoff certifies nonintegrality of the same remainder; failure at the first cutoff was inconclusive. The successful test excludes denominators dividing the unchanged value $M = 12$ , already covered by the finite exclusions.

For fixed $c$ with $M (c) > 0$ , the finite part increases to $R (c)$ with omitted tail $M (c) (S - H_{N})$ . If $R (c) \notin Z$ , its gap to the next integer is positive, whereas $2 M (c) / ((N + 1)! - 1)$ tends to zero; the test therefore succeeds at every sufficiently large cutoff. If $R (c) \in Z$ , the eventual gap equals the omitted tail and is smaller than this upper bound. Eventual success is consequently equivalent to nonintegrality of this fixed form, not an independent reason for it. The short paper’s remainder comparison gives the inequality explicitly. For each $q > 0$ we still need a vector with $q ∣ M (c)$ for which it succeeds; the successful cutoff need not be the largest supported index.

The support itself has a separate geometric restriction. Cancellation with $M \neq 0$ must cross a change in the relevant floor exponent; otherwise the weighted sum is proportional to $M$ .

Theorem 1.4 (constant values of the floor in the weights). Let $d \geq 2$ and $k \geq 0$ , and suppose every supported index $i$ satisfies $k d \leq i < (k + 1) d$ . Then $M (c) = (d!)^{k} V_{d} (c)$ . In particular, cancellation on the interval $d \leq i < 2 d$ forces $M (c) = 0$ ; and if every supported index is at least $d$ while $M (c) \neq 0$ and $V_{d} (c) = 0$ , then some supported index is at least $2 d$ .

Proof. On this interval the quotient $⌊ i / d ⌋$ is the constant $k$ , so $i! = (d!)^{k} W_{d, i}$ for every supported index and the factor $(d!)^{k}$ comes out of the sum. The two consequences follow by taking $k = 1$ and by contraposition. ◻

The constant-floor condition is local. A quantitative bound also needs control of the multiplier: cancellation alone does not give $0 < M < (R + 1)! - 1$ . After proving the common-denominator estimates, §3.1 derives the resulting bounds on $R$ .

At a prime index, the two-term vector $T_{p}$ already changes only one weighted sum; there are no proper divisors to eliminate.

Theorem 1.5 (changing just one weighted sum). Let $p \geq 3$ be prime and let $c_{p - 1} = p$ , $c_{p} = - 1$ , with every other coefficient zero. Then $M (c) = 0$ , $V_{p} (c) = p! - 1$ , and $V_{d} (c) = 0$ for every $d \geq 2$ with $d \neq p$ .

Proof. Since $p \geq 3$ , both indices $p - 1$ and $p$ lie in the prescribed coefficient domain $i \geq 2$ . Here $M = p (p - 1)! - p! = 0$ . For $2 \leq d < p$ the quotient identity $⌊ (p - 1) / d ⌋ = ⌊ p / d ⌋$ holds, since $d ∤ p$ , so the two factorial weights carry the same power of $d!$ and the weighted sum evaluates to $p (p - 1)! / (d!)^{⌊ p / d ⌋} - p! / (d!)^{⌊ p / d ⌋} = 0$ . For $d > p$ both indices lie below $d$ , so both weights are the plain factorials and the same cancellation occurs. At $d = p$ the weights are $(p - 1)!$ and $p! / p! = 1$ , giving $p (p - 1)! - 1 = p! - 1$ . ◻

For $p = 2$ , the same auxiliary identities would require $c_{1} = 2$ . They do not define an admissible vector on the domain $i \geq 2$ , since every vector supported on that domain has $c_{1} = 0$ .

Adding an integer multiple of this vector changes $V_{p}$ by that multiple of $p! - 1$ , without changing $M$ or any other $V_{d}$ . Suppose $M (c) \neq 0$ , $V_{2} (c) = \dots = V_{D} (c) = 0$ , and every supported index exceeds a prescribed integer $B$ . Choose a prime $p > max (D, B + 1, 2)$ and replace $c$ by

c - ⌊ R (c) + 1 / 2 ⌋ T_{p} .

Both new indices, $p - 1$ and $p$ , exceed $B$ . Since $p > D$ , the prescribed weighted sums remain zero, and $M$ is unchanged. The new remainder is $R (c) - ⌊ R (c) + 1 / 2 ⌋ \in [- 1 / 2, 1 / 2)$ . The progression construction below supplies such an initial vector: choose its first index $r > B$ .

An integral remainder rounds to zero; a nonintegral one rounds to a nonzero value. Thus rounding controls size but does not prove nonintegrality or exclude further denominators at fixed $M$ . Moreover, $⌊ R (c) + 1 / 2 ⌋$ depends on the unknown remainder: this is an existence formula for a representative, not an independent certificate of its nonvanishing.

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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