Running least common multiples: distinct values and multiplicities

Will Cook

Long paper

Running least common multiples: distinct values and multiplicities

Erdős #269 By Will Cook July 2026; revised 5 October 2026 Authorship, AI use and citation

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Summary. For the prime set {2,3,5}, a five-map argument proves irrationality of the sum over distinct running LCMs. Two-prime sums reduce to a transcendental Hecke–Mahler value. For the repeated three-prime sum, an equivalent residue criterion leaves an arithmetic condition open. Complete ordinary arguments for general distinct-running-LCM sums and prime-power subseries are retained as remarks awaiting formalisation.

This paper covers distinct-height irrationality, all-rank kernel structure and the repeated-sum tail criteria, preserving the multiplicity distinction and the remaining open residue-window obligation.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #269, which remains open.

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In this paper

The two-prime Hecke–Mahler reduction

Let $P = {p, q}$ with $p < q$ , and put $L_{p, q} (t) = p^{⌊ \log_{p} t ⌋} q^{⌊ \log_{q} t ⌋}$ . The proof of Theorem 3.1, with one coordinate omitted, identifies this with the running LCM. The two sums are therefore

D_{{p, q}} = 1 + \sum_{t \in {p, p^{2}, \dots} \cup {q, q^{2}, \dots}} \frac{1}{L_{p, q} (t)}, R_{{p, q}} = \sum_{i, j \geq 0} \frac{1}{L_{p, q} (p^{i} q^{j})} .

We first express the contribution from powers of $q$ in terms of that from powers of $p$ . The repeated kernel then factors, and this relation puts both factors in a single transcendental parameter. This is the factorisation and Hecke–Mahler reduction credited to Fan in the introduction.

Proof of Theorem 1.2. Set

θ = \frac{\log p}{\log q}, x = \frac{1}{p}, y = \frac{1}{q}, m_{n} = ⌊ n θ ⌋, δ_{n} = m_{n + 1} - m_{n} .

Here $0 < θ < 1$ , and $θ$ is irrational, since a rational value would give $p^{b} = q^{a}$ for positive integers $a, b$ . Consequently $δ_{n} \in {0, 1}$ . Put

A = \sum_{n \geq 0} x^{n} y^{m_{n}}, B_{*} = \sum_{n \geq 0} δ_{n} x^{n} y^{m_{n} + 1} .

The initial value and the positive powers of $p$ contribute $A$ , since $L_{p, q} (p^{n}) = p^{n} q^{m_{n}}$ . A power of $q$ lies strictly between $p^{n}$ and $p^{n + 1}$ exactly when $δ_{n} = 1$ , it is then $q^{m_{n} + 1}$ , and its post-jump reciprocal is $x^{n} y^{m_{n} + 1}$ ; so the positive powers of $q$ contribute $B_{*}$ and $D_{{p, q}} = A + B_{*}$ . All these series converge absolutely.

Since $y^{m_{n + 1}} - y^{m_{n}} = δ_{n} y^{m_{n}} (y - 1)$ , an index shift gives $A - 1 - x A = x (y - 1) B_{*} / y$ , so $B_{*} = ((p - 1) A - p) / (1 - q)$ and

\begin{matrix} (4) & D_{{p, q}} = \frac{(q - p) A + p}{q - 1} . \end{matrix}

At a smooth point, $\log_{p} (p^{i} q^{j}) = i + j / θ$ and $\log_{q} (p^{i} q^{j}) = j + i θ$ , so $L_{p, q} (p^{i} q^{j}) = p^{i + ⌊ j / θ ⌋} q^{j + m_{i}}$ and absolute convergence permits the factorisation

R_{{p, q}} = A \sum_{j \geq 0} y^{j} x^{⌊ j / θ ⌋} .

For $j \geq 1$ , the index $n = ⌊ j / θ ⌋$ is precisely the one for which $p^{n} < q^{j} < p^{n + 1}$ ; the inequalities are strict because distinct primes have no common positive power. This interval contains at most one power of $q$ , since $p < q$ . Thus $δ_{n} = 1$ and $j = m_{n} + 1$ . Conversely, every $n$ with $δ_{n} = 1$ contains that unique power of $q$ . The second factor is therefore $1 + B_{*}$ , giving

\begin{matrix} (5) & R_{{p, q}} = \frac{(p + q - 1) A - (p - 1) A^{2}}{q - 1} . \end{matrix}

It remains to prove that $A$ is transcendental. For the Hecke–Mahler series

F_{θ} (x, y) = \sum_{n \geq 1} \sum_{k = 1}^{⌊ n θ ⌋} x^{n} y^{k}

a finite geometric sum gives $((1 - y) / y) F_{θ} (x, y) = x / (1 - x) - (A - 1)$ , that is

\begin{matrix} (6) & A = \frac{1}{1 - x} - \frac{1 - y}{y} F_{θ} (x, y) . \end{matrix}

Bugeaud and Laurent’s Theorem 1.1 states, in particular, that $F_{θ} (β, α)$ is transcendental when $θ \in (0, 1)$ is irrational, $α$ and $β$ are nonzero algebraic numbers, $| β | < 1$ and $| β α^{θ} | < 1$ [15]; the $ρ = 0$ case used here goes back to Loxton and van der Poorten [16]. Take $(β, α) = (x, y)$ : then $| β | = 1 / p < 1$ and

| x y^{θ} | = \frac{1}{p} {(\frac{1}{q})}^{\log p / \log q} = \frac{1}{p^{2}} < 1 .

So $F_{θ} (x, y)$ is transcendental, and (6) makes $A$ transcendental. The coefficient of $A$ in (4) is $(q - p) / (q - 1) \neq 0$ and the coefficient of $A^{2}$ in (5) is $- (p - 1) / (q - 1) \neq 0$ , both rational. If either value were algebraic, its identity would exhibit $A$ as a root of a nonzero polynomial over the algebraic numbers. ◻

A product of two transcendental numbers need not be transcendental. What proves the repeated sum transcendental is its nonconstant quadratic expression in the single value $A$ , not the factorisation by itself. The value theorem adds no unverified hypothesis in this two-prime case: distinct primes give an irrational slope and the displayed reciprocal arguments satisfy its size conditions. Primality is stronger than the calculation needs. For coprime integers $1 < p < q$ , enumerate the monoid ${p^{i} q^{j} : i, j \geq 0}$ , not all integers supported on the prime factors of $p q$ . Unique exponent pairs, the running-LCM product and the irrationality of $\log p / \log q$ still hold, so both identities and transcendence conclusions remain valid. For example, this applies to generators $4, 9$ . By contrast, the monoid generated by $4, 8$ has running LCM $8$ at the cutoff $8$ , not $4^{⌊ \log_{4} 8 ⌋} 8^{⌊ \log_{8} 8 ⌋} = 32$ ; its slope is rational as well. The coprime extension was already noted in the forum discussion (Section 13.1).

A third prime introduces an additional floor term that cannot be separated in this way; Section 7 makes that obstruction precise. For the sums over distinct running LCMs, the next section proves irrationality for every finite set of primes by an elementary argument that uses no value theorem.

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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