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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An integral basis and the possible values of $M$

Fix $D \geq 2$ . We temporarily allow a coefficient at index one so that the adjacent-difference vectors form a triangular basis. After solving $V_{2} = \dots = V_{D} = 0$ , we impose one further equation to remove that coordinate. Only the coefficient space is enlarged; the series $S$ continues to start at $d = 2$ , with no term $1 / (1! - 1)$ .

Let $e_{n}$ be the unit vector at index $n$ . For $n \geq 2$ put

T_{n} = n e_{n - 1} - e_{n} .

We have $M (T_{n}) = 0$ . If $d ∤ n$ , the two floor exponents in its weighted sum agree, and $n W_{d, n - 1} = W_{d, n}$ . If $d ∣ n$ , the exponent increases by one, so $n W_{d, n - 1} = d! W_{d, n}$ . Hence

V_{d} (T_{n}) = (d! - 1) W_{d, n} 1_{d ∣ n} .

For example, $T_{4} = 4 e_{3} - e_{4}$ has weighted sums $V_{2} = 6$ and $V_{4} = 23$ , and all its other weighted sums vanish. Since $T_{2} = 2 e_{1} - e_{2}$ has only $V_{2} = 1$ nonzero, $T_{4} - 6 T_{2}$ isolates the fourth weighted sum.

We therefore define, in increasing order of $n$ ,

U_{n} = T_{n} - \sum_{\begin{matrix} d ∣ n \\ 2 \leq d < n \end{matrix}} W_{d, n} U_{d} .

Assume the construction has isolated the weighted sums below $n$ . Subtracting $W_{d, n} U_{d}$ removes the contribution at a proper divisor $d$ and changes no other weighted sum. At $d = n$ we have $W_{n, n} = 1$ . Induction therefore gives $M (U_{n}) = 0$ and $V_{d} (U_{n}) = (d! - 1) 1_{d = n}$ for every $d \geq 2$ .

For the basis claim, restrict to indices $1, \dots, N$ . The matrix with columns $e_{1}, T_{2}, \dots, T_{N}$ is triangular over $Z$ with diagonal $1, - 1, \dots, - 1$ . The change from $T_{n}$ to $U_{n}$ is triangular with diagonal one. Both determinants are units in $Z$ , so both inverses have integer entries. In back-substitution the only divisors are $1$ and $- 1$ ; no rational coefficients are needed. Thus $e_{1}, U_{2}, \dots, U_{N}$ represents every integer vector uniquely. If this expansion is $b e_{1} + \sum z_{d} U_{d}$ , then $M (c) = b$ and $V_{d} (c) = b + (d! - 1) z_{d}$ . These identities determine the coordinates:

c = M (c) e_{1} + \sum_{d \geq 2} \frac{V_{d} (c) - M (c)}{d! - 1} U_{d} .

The congruence (1) also verifies the integrality of the displayed coefficients. They vanish for $d > N$ , since then $V_{d} (c) = M (c)$ , so the expansion is finite. These same coefficients occur in the integer-difference identity at the start of §1: their sum is $R (c) - M (c) S$ . Thus the coordinates also recover the integer term in the linear form.

Cancellation through $D$ forces $L_{D} ∣ M (c)$ . To obtain $M (c) = L_{D}$ in the enlarged space, start with $L_{D} e_{1}$ , whose weighted sums all equal $L_{D}$ . The vector $U_{d}$ changes only the $d$ th sum, by $d! - 1$ . Hence define

K_{D} = L_{D} e_{1} - \sum_{d = 2}^{D} \frac{L_{D}}{d! - 1} U_{d} .

This vector satisfies $M = L_{D}$ and $V_{2} = \dots = V_{D} = 0$ . All solutions in the enlarged coefficient space consequently have the unique form

\begin{matrix} (2) & c = t K_{D} + \sum_{n > D} z_{n} U_{n}, M (c) = t L_{D}, \end{matrix}

where $t, z_{n}$ are integers and only finitely many $z_{n}$ are nonzero. The only forbidden coefficient is now the one at index $1$ . Write $a_{D} = (K_{D})_{1}$ and $u_{n} = (U_{n})_{1}$ . Removing this coefficient is equivalent to

\begin{matrix} (3) & t a_{D} + \sum_{n > D} z_{n} u_{n} = 0. \end{matrix}

The free higher basis coefficients can change the first coefficient by every multiple of $gcd {u_{n} : n > D}$ , and only by such a multiple. Thus the gcd gives the exact solvability condition. We next show how to compute it using a bounded list of coefficients. Here $u_{2} = 2$ , whereas for $n > 2$ the recurrence reads

u_{n} = - \sum_{\begin{matrix} d ∣ n \\ 2 \leq d < n \end{matrix}} W_{d, n} u_{d} .

It gives $u_{n} = 0$ for odd $n > 1$ . If $p$ is prime, the only proper divisor of $2 p$ with a nonzero contribution is $2$ , and therefore $u_{2 p} = - (2 p)! / 2^{p - 1}$ , including $p = 2$ .

Choose a prime $p$ with $D / 2 < p \leq D$ . Such a prime exists by Bertrand’s postulate when $D \geq 3$ , and $p = 2$ works when $D = 2$ . The coefficient $u_{2 p}$ lies in the tail and divides $(2 p)!$ . Once a finite initial interval includes that coefficient, its gcd therefore divides this factorial. To control the small-divisor weights beyond the interval, we shall require $n / d \geq 2 p$ whenever $d \leq D$ . This leads to the cutoff $H = D (2 p - 1)$ . We claim that

\begin{matrix} (4) & gcd {u_{n} : n > D} = gcd (u_{D + 1}, \dots, u_{H}) > 0, H < 2 D^{2} . \end{matrix}

Let $g$ be the finite gcd on the right. Its value could in principle decrease if we included later coefficients. To rule this out, we prove $g ∣ u_{n}$ for every $n > D$ by strong induction. The index $2 p$ lies in the finite interval and has a nonzero coefficient, so $g > 0$ and $g ∣ (2 p)!$ . The definition proves the induction claim through $H$ .

For $n > H$ , each term of the recurrence is $W_{d, n} u_{d}$ with $d < n$ . If $d > D$ , divisibility comes from $u_{d}$ by induction. If $d \leq D$ , we cannot use that hypothesis and instead use the weight. Put $k = n / d$ . The cutoff was chosen so that $n > D (2 p - 1)$ forces $k \geq 2 p$ for every such divisor. Also,

\frac{W_{d, n}}{k!} = \frac{n!}{(d!)^{k} k!} \in Z,

because the quotient counts partitions of $n$ labelled objects into $k$ unordered blocks of size $d$ . Hence $g ∣ (2 p)! ∣ k! ∣ W_{d, n}$ . Thus every product in the recurrence is divisible by $g$ , completing the induction. The finite gcd divides the entire tail, and inclusion of the finite interval gives the reverse divisibility. Finally, $H \leq D (2 D - 1) < 2 D^{2}$ .

Theorem 1.2 (the set of attainable values of $M$ ). Fix $D \geq 2$ and a prime $p$ with $D / 2 < p \leq D$ . Put

H = D (2 p - 1), G_{D} = gcd {| u_{n} | : D < n \leq H}, μ_{D} = L_{D} \frac{G_{D}}{gcd (G_{D}, a_{D})} .

For finite integer vectors supported on $n \geq 2$ and cancelling all weighted sums $2, \dots, D$ , the possible values of $M$ are exactly $μ_{D} Z$ . The integer $μ_{D}$ is positive, is independent of the eligible prime $p$ , and is attained by a vector of coefficients with gcd one, that is, by a primitive vector.

Proof. Equation (4) identifies $G_{D}$ with the gcd of all $u_{n}$ for $n > D$ . Their finite integer combinations are exactly $G_{D} Z$ . Hence (3) has a solution if and only if $G_{D} ∣ t a_{D}$ , or equivalently $G_{D} / gcd (G_{D}, a_{D}) ∣ t$ . Substitution in (2) gives the claimed set of values of $M$ . Bézout’s identity applied to the finite list through $H$ supplies a vector attaining its least positive element. A common coefficient divisor larger than one could be divided out, giving a smaller positive attainable value of $M$ . The attaining vector is therefore primitive. Finally, the set of values of $M$ is intrinsic to the cancellation equations, so its positive generator is independent of the eligible prime $p$ . ◻

Fix $M \in μ_{D} Z$ and put $t = M / L_{D}$ . The support equation (3) determines which basis coefficients are admissible. The following formula records their integer contribution and shows why changing the vector at that value of $M$ cannot alter nonintegrality.

Theorem 1.3 (how the coefficient choices change the remainder). For the vector in (2),

R (t K_{D} + \sum_{n > D} z_{n} U_{n}) = t L_{D} (S - H_{D}) + \sum_{n > D} z_{n} .

The remainder series converges for every finite vector supported away from index zero. A vector with $M = 0$ has integral remainder, and any two finite vectors with the same value of $M$ have remainders differing by an integer.

Proof. For $d > D$ , we have $V_{d} (K_{D}) = L_{D}$ and $V_{d} (U_{n}) / (d! - 1) = 1_{d = n}$ . Summing these identities gives $t L_{D} (S - H_{D}) + \sum_{n > D} z_{n}$ , as claimed. The first series converges and the second sum is finite.

Absolute convergence and the general integer-difference identity were proved at the start of §1, including for the auxiliary index $1$ . Taking $M = 0$ proves integrality; subtracting the identities for two vectors with equal $M$ proves the last assertion. ◻

Here $M = t L_{D}$ , so every denominator in $H_{D}$ divides $M$ . Thus $M H_{D} \in Z$ , and the integer in the original form $M S + k$ is $k = - M H_{D} + \sum_{n > D} z_{n}$ .

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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An integral basis and the possible values of 𝑀M

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An integral basis and the possible values of $M$