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

Escape from a smaller interval

Write $θ_{m} = {m! S}$ . The next criterion compares a fractional part with the actual tail: $θ_{m - 1}$ must leave $[0, E_{m} / m)$ at arbitrarily large indices. The interval has width less than $2 / m^{2}$ . Since $E_{m}$ contains $S$ , this equivalence is not yet a finite test; the second assertion supplies a sufficient test using the finite quantities $Δ_{m}$ and $ε_{m}$ .

Proposition 4.3 (lower-interval criterion).

\begin{matrix} (20) & S \notin Q ⟺ \forall B \exists m > B : E_{m} \leq m θ_{m - 1} . \end{matrix}

For $m \geq 3$ the finite condition

\begin{matrix} (21) & m Δ_{m} \leq 1 + ε_{m} or 1 + ε_{m} + \frac{2}{m} \leq m Δ_{m} \end{matrix}

implies the escape inequality in (20). Cofinally many instances of (21) therefore imply $S \notin Q$ .

Proof. If $S$ is rational, the remainders $θ_{m - 1}$ vanish eventually while $E_{m} > 0$ , so the escape inequality fails eventually. Conversely, suppose $m θ_{m - 1} < E_{m}$ for every sufficiently large $m$ . By (13) this gives $m θ_{m - 1} < 1$ , so $a_{m} (S) = 0$ and $θ_{m} = m θ_{m - 1}$ . Fix $N$ after this recurrence begins. Then $θ_{N + k} = ((N + k)! / N!) θ_{N} < 1$ for every $k \geq 0$ , forcing $θ_{N} = 0$ . Hence $N! S$ is an integer and $S$ is rational.

For the finite implication, suppose $m θ_{m - 1} < E_{m}$ . The tail recurrence is $m E_{m - 1} = 1 + ε_{m} + E_{m}$ , so $E_{m - 1} > E_{m} / m > θ_{m - 1} \geq 0$ , and with $0 < E_{m - 1} < 1$ the strict-successor identity gives $Δ_{m} = E_{m - 1} - θ_{m - 1}$ . Hence

1 + ε_{m} < m Δ_{m} = 1 + ε_{m} + E_{m} - m θ_{m - 1} \leq 1 + ε_{m} + E_{m} < 1 + ε_{m} + \frac{2}{m},

which contradicts (21). ◻

There are two comparisons to keep separate: a finite sufficient test for escape at one index, and an equivalence when indices are unbounded. The finite test excludes only the open interval $1 + ε_{m} < m Δ_{m} < 1 + ε_{m} + 2 / m$ ; it accepts either endpoint and all values outside. The width $2 / m$ refers to $m Δ_{m}$ ; the interval for $Δ_{m}$ has width $2 / m^{2}$ . At one index this is a sufficient test for escape, obtained by replacing $E_{m}$ with the upper bound $2 / m$ .

At arbitrarily large integer indices, however, the finite test itself characterises irrationality:

S \notin Q ⟺ (21) holds for arbitrarily large integers m .

Only the forward implication remains to be shown. If $S$ is irrational, Theorem 4.1 gives $b_{m} \neq 1$ at arbitrarily large indices. By (14), each such index satisfies $m Δ_{m} \leq 1 + ε_{m}$ or $m Δ_{m} > 2 + ε_{m}$ . Since $2 / m < 1$ for $m \geq 3$ , either alternative implies (21). The converse is Proposition 4.3. This combines the two criteria; it does not prove either unbounded occurrence for $S$ . The equivalence is over all integer indices. A restriction to primes gives only a sufficient condition here.

For comparison at a single index, the exact classification is

\begin{matrix} (22) & E_{m} \leq m θ_{m - 1} ⟺ b_{m} \neq 1 or a_{m} (S) = m - 1 (m \geq 3) . \end{matrix}

To verify the classification, use $0 < E_{j} < 1$ to write

Z_{j} = ⌊ j! S ⌋ + 1_{{θ_{j} \geq E_{j}}} .

Substitution in the carry recurrence gives

b_{m} = 1 - a_{m} (S) + m 1_{{θ_{m - 1} \geq E_{m - 1}}} - 1_{{θ_{m} \geq E_{m}}} .

Since $0 \leq a_{m} (S) \leq m - 1$ , a unit carry has exactly two possibilities: both indicators are zero and $a_{m} (S) = 0$ , or both are one and $a_{m} (S) = m - 1$ . In the first case $m θ_{m - 1} = θ_{m} < E_{m}$ ; in the second, $m θ_{m - 1} \geq m - 1 > E_{m}$ . Conversely, $m θ_{m - 1} < E_{m} < 1$ forces $a_{m} (S) = 0$ and both indicators to be zero, using $m E_{m - 1} = 1 + ε_{m} + E_{m}$ . It therefore forces a unit carry with zero digit. This proves (22), including the equality case in its left-hand inequality.

Thus escape can occur even when $b_{m} = 1$ , provided the digit is $m - 1$ . This happens at $m = 52$ : exact rational arithmetic gives $b_{52} = 1$ together with $52 Δ_{52} > 1 + ε_{52} + 2 / 52$ , the margin being about $0.5689674908$ . This example separates the two events at a single index: the finite test can hold at a unit carry. Their occurrence at arbitrarily large indices nevertheless gives equivalent rationality criteria. Since $52$ is composite, the example supplies no prime-index instance.

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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