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 logarithmic flow.

Consider

T = \prod_{q \in P} R / (\log q) Z, Θ (s) = (s mod \log q)_{q \in P} .

For $θ \in T$ , a $q$ -crossing is a time $t$ at which $θ_{q} + t \in (\log q) Z$ . At $θ_{k} = Θ (\log t_{k})$ the positive crossings occur at $\log (q^{j} / t_{k})$ with $q^{j} > t_{k}$ , so their labels, read in time order, are $q_{k + 1} q_{k + 2} \dots$ . Crossings at this phase never coincide.

Let $T_{0}$ be the closure of $Θ (R)$ . Every forward orbit in $T_{0}$ is dense there. Indeed, the pigeonhole principle gives arbitrarily large $d > 0$ with $Θ (d)$ arbitrarily close to $0$ , so negative times are limits of positive ones. Translation then proves the same assertion for every starting point and every later starting time.

In the normalised coordinates $φ_{q} = θ_{q} / \log q$ , Kronecker’s theorem identifies $T_{0}$ with the image of the smallest rational subspace $L \subseteq R^{P}$ containing $ω = (1 / \log q)_{q \in P}$ . Any two coordinate forms $φ_{a}, φ_{b}$ are linearly independent on $L$ . To see this, their dependence would put a nonzero vector supported on ${a, b}$ in the rational annihilator of $L$ . That intersection is rational, so it would contain a nonzero integer vector. Its pairing with $ω$ would force $\log a / \log b \in Q$ , contrary to unique factorisation. In particular, $(\log b) φ_{b} - (\log a) φ_{a}$ is a nonzero form on $L$ . We write it as $θ_{b} - θ_{a}$ when using real representatives.

To interchange just two jumps, we seek a phase at which exactly those two crossings coincide. The next argument finds such a phase inside $T_{0}$ ; it does not require this orbit closure to be the whole torus $T$ .

Remark 5.5 (two walls). If $| P | \geq 4$ , there is $x \in T_{0}$ with $θ_{q} (x) = 0$ for exactly two primes $q \in P$ .

Proof. Choose a rational plane in $L$ on which every pair of coordinate forms remains independent. Such a plane exists: for each pair, the determinant on $L \times L$ is a nonzero polynomial, and finitely many of these polynomials cannot vanish on all rational pairs. An integer basis of the plane identifies its image with a two-dimensional subtorus of $T_{0}$ . On this subtorus each wall $θ_{q} = 0$ has the equation $⟨ γ_{q}, s ⟩ \in Z$ , where $γ_{q} \in Z^{2} ∖ {0}$ , and the vectors $γ_{q}$ are pairwise nonparallel. A wall may have several components if $γ_{q}$ is not primitive.

The walls form a finite arrangement of closed geodesics. Let $k_{v}$ count the wall families through a vertex $v$ . Each such family contributes two incident half-edges, so $E = \sum_{v} k_{v}$ . Two families alone divide the torus into open parallelograms: their two integer linear forms give a finite covering of the standard torus. Adding the remaining straight lines subdivides these parallelograms into convex polygons, each with at least three sides. Thus $2 E \geq 3 F$ , while Euler’s formula gives $V - E + F = 0$ . Together these imply

\sum_{v} (k_{v} - 3) = E - 3 V \leq 0.

Every vertex belongs to at least two wall families, and the last inequality says that the average number of families through a vertex is at most three. The origin belongs to all $| P | \geq 4$ families. Another vertex must therefore belong to exactly two, as required. ◻

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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