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

Consequences for the support parameter

The lcm estimates also constrain the size of a cancellation vector when its multiplier is bounded by a factorial. At depth $D = 2 t^{2}$ , cancellation gives $L_{D} ∣ M$ ; the next theorem additionally assumes $0 < M < (R + 1)! - 1$ . Its $R$ is an index, not the remainder or the modulus $R_{p}$ of §5. The conclusion bounds the largest supported index only when $R$ denotes that index. Taking $R$ to be a later cutoff gives a bound on the cutoff instead.

Theorem 3.5 (a lower bound for the support parameter). Let $t, M, R \in N$ satisfy

t \geq 2^{32}, M > 0, L_{2 t^{2}} ∣ M, M < (R + 1)! - 1 .

Then $3 t^{3} < 2 (R + 1)$ . Consequently no family satisfying these hypotheses for all sufficiently large $t$ has $(R (t) + 1) / t^{3} \leq 3 / 2$ eventually, and none has $R (t) = o (t^{3})$ .

Proof. Put $D = 2 t^{2}$ , $k = 2 t$ and $u = D - k + 1 = 2 t^{2} - 2 t + 1$ . Lemma 3.3 at $N = D$ , with $n! - 1 \geq n! / 2$ and $\log n! \geq n \log n - n$ , and with $L_{D} \leq M < (R + 1)!$ , gives

\begin{matrix} (11) & 2 t (u \log u - u - \log 2) < (R + 1) \log (R + 1) + (\binom{2 t + 1}{3}) \log (2 t^{2}) . \end{matrix}

Suppose $R + 1 \leq \frac{3}{2} t^{3}$ . Since $t \geq 16$ one has $\frac{15}{8} t^{2} \leq u \leq 2 t^{2}$ and $\log u \geq 2 \log t$ , so the left side of (11) is at least $\frac{15}{2} t^{3} \log t - 4 t^{3} - 2 t \log 2$ . Using $(\binom{2 t + 1}{3}) < \frac{4}{3} t^{3}$ and $\log \frac{3}{2} < \log 2$ , the right side is at most

\frac{3}{2} t^{3} (\log 2 + 3 \log t) + \frac{4}{3} t^{3} (\log 2 + 2 \log t) .

Dividing by $t^{3}$ and rearranging, (11) would require

\frac{1}{3} \log t < 4 + \frac{2 \log 2}{t^{2}} + \frac{17}{6} \log 2 .

But $\log t \geq 32 \log 2$ and $\frac{2}{3} < \log 2 < 1$ give $\frac{1}{3} \log t - \frac{17}{6} \log 2 \geq \frac{47}{6} \log 2 > 5$ , while $4 + 2 \log 2 / t^{2} < 5$ . This contradiction proves the finite inequality, and the two sequence conclusions follow. ◻

The hypotheses are compatible: take $M = L_{2 t^{2}}$ and then choose $R$ large enough. The conclusion rules out subcubic $R$ under the stated size condition; it neither constructs a coefficient vector nor proves irrationality.

Theorem 3.4 raises the asymptotic constant in the same estimate.

Corollary 3.6 (the asymptotic lower bound). Let $M (t), R (t)$ satisfy $M (t) > 0$ , $L_{2 t^{2}} ∣ M (t)$ and $M (t) < (R (t) + 1)! - 1$ for all sufficiently large $t$ . Then

\underset{t \to \infty}{lim inf} \frac{R (t) + 1}{t^{3}} \geq \frac{16}{9} .

Proof. Put $r = R (t) + 1$ , so $L_{2 t^{2}} \leq M (t) < r!$ and $\log L_{2 t^{2}} < r \log r$ . Since

\frac{(2 t^{2})^{3 / 2} \log (2 t^{2})}{t^{3} \log t} ⟶ 4 \sqrt{2},

Theorem 3.4 gives, for every $ε > 0$ and all sufficiently large $t$ ,

\log L_{2 t^{2}} \geq (\frac{16}{3} - ε) t^{3} \log t .

Suppose $r \leq c t^{3}$ for arbitrarily large $t$ , with $0 < c < 16 / 9$ fixed. On that subsequence $r \log r \leq c t^{3} (3 \log t + \log c) = (3 c + o (1)) t^{3} \log t$ . Choosing $0 < ε < 16 / 3 - 3 c$ contradicts the lower bound for sufficiently large $t$ . ◻

This corollary does not assert the strict inequality $R + 1 > \frac{16}{9} t^{3}$ eventually. Indeed, the finite logarithmic constraint (11) allows equality at that constant. To check this directly, set $R + 1 = 16 t^{3} / 9$ with $t \geq 4$ . Since $u = 2 t^{2} - 2 t + 1 \leq 2 t^{2}$ and $x \log x - x$ is increasing for $x \geq 1$ , the left side of (11) is at most $4 t^{3} (\log (2 t^{2}) - 1) - 2 t \log 2$ . Subtracting this bound from the right side gives

t^{3} (4 + \frac{16}{9} \log \frac{16}{9} - \frac{8}{3} \log 2) + t (\frac{5}{3} \log 2 - \frac{2}{3} \log t) > \frac{4}{3} t^{3} - \frac{2}{3} t^{2} > 0,

where $\log 2 < 1$ and $\log t \leq t$ suffice for the last comparison. Integer pairs with $9 (R + 1) = 16 t^{3}$ exist for every multiple $t \geq 6$ of $3$ . This verifies the numerical constraint, not the existence of a coefficient vector with that support. The asymptotic assertion in Corollary 3.6 instead follows from Theorem 3.4.

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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