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

Boundary terms in weighted finite differences

Follow one lattice point through four successive triangles. If it lies in all four and its weight is unchanged, the coefficients $(1, - 3, 3, - 1)$ sum to zero. If it first enters the second triangle, its coefficient is $- 3 + 3 - 1 = - 1$ ; entry in the third gives $3 - 1 = 2$ , and entry in the fourth gives $- 1$ . Thus a finite difference depends both on which triangles contain the point and on changes of its weight. We separate these two effects below. Avoiding floor crossings controls the weights, but does not remove the newly added strips.

Recall that $w_{j, k} = j \log_{2} 3 + k \log_{2} 5$ , $t_{j, k} = {w_{j, k}}$ and $θ_{p} = 1 / \log_{2} p$ for $p = 3, 5$ . To distinguish the shared points from the new ones, let $T_{a} = {(j, k) \in N^{2} : w_{j, k} < a + 1}$ and, for $0 \leq t < 1$ , set

ω_{a} (t) = \prod_{p \in {3, 5}} p^{⌊ (a + 1) θ_{p} ⌋ - ⌊ (a + t) θ_{p} ⌋} .

This weight is defined even for pairs outside $T_{a}$ . For $ν \geq 0$ , put

\begin{aligned} κ_{p} (a, t, ν r) & = ⌊ (a + t + ν r) θ_{p} ⌋ - ⌊ (a + t) θ_{p} ⌋ - ⌊ ν r θ_{p} ⌋, \\ χ_{ν} (a, t) & = 3^{- κ_{3} (a, t, ν r)} 5^{- κ_{5} (a, t, ν r)} . \end{aligned}

Each $κ_{p}$ is $0$ or $1$ . A value $κ_{p} = 1$ records a floor-addition carry. This is different from a new lattice point entering $T_{a}$ as $a$ increases.

Proposition 12.2 (an exact interior-and-strip decomposition). For a fixed operator $c_{0}, \dots, c_{σ}$ and $r \geq 1$ , let $E_{0} = T_{a}$ and $E_{s} = T_{a + s r} ∖ T_{a + (s - 1) r}$ for $1 \leq s \leq σ$ . Then

\frac{D_{r, a}}{15} = \sum_{s = 0}^{σ} \sum_{(j, k) \in E_{s}} ω_{a} (t_{j, k}) \sum_{ν = s}^{σ} c_{ν} χ_{ν} (a, t_{j, k}) .

For the cubic operator $(1, - 3, 3, - 1)$ , if all these crossing bits vanish, then

\frac{D_{r, a}}{15} = - M_{0} + 2 M_{1} - M_{2}, M_{s} = \sum_{(j, k) \in T_{a + (s + 1) r} ∖ T_{a + s r}} ω_{a} (t_{j, k}) .

Thus absence of floor crossings cancels the common interior, but not necessarily the three boundary strips.

Proof. For each $p = 3, 5$ , the exponent of $p$ on either side of

γ_{a, ν r} ω_{a + ν r} (t) = ω_{a} (t) χ_{ν} (a, t)

is

⌊ (a + 1) θ_{p} ⌋ + ⌊ ν r θ_{p} ⌋ - ⌊ (a + t + ν r) θ_{p} ⌋ .

On the left the upper-boundary exponents cancel; on the right the $⌊ (a + t) θ_{p} ⌋$ terms cancel. This proves the pointwise identity. Insert the triangle formula for each $m_{a + ν r}$ and group by the first triangle containing a pair. A pair in $E_{s}$ occurs exactly in the terms $ν \geq s$ , proving the first formula. Under the no-crossing hypothesis, $χ_{ν} = 1$ . The full cubic sum is zero, while its three successive suffix sums are $- 1, 2, - 1$ , proving the second formula. ◻

Example 12.3 (cubic cancellation can fail with no floor crossings). Take $a = 0$ , $r = 35$ and $(c_{0}, c_{1}, c_{2}, c_{3}) = (1, - 3, 3, - 1)$ . Exact integer comparisons give

\begin{array}{crrrr} u & 0 & 35 & 70 & 105 \\ ⌊ \log_{3} 2^{u} ⌋ = ⌊ \log_{3} 2^{u + 1} ⌋ & 0 & 22 & 44 & 66 \\ ⌊ \log_{5} 2^{u} ⌋ = ⌊ \log_{5} 2^{u + 1} ⌋ & 0 & 15 & 30 & 45 \\ m_{u} & 1 & 195 & 723 & 1582 \end{array}

The first two rows imply $b_{u} = 2$ and $⌊ (u + t) θ_{p} ⌋ = ⌊ u θ_{p} ⌋$ for every $0 \leq t < 1$ , $p = 3, 5$ . Hence every relevant $κ_{p} (0, t, u)$ vanishes, $γ_{0, u} = 1$ , and all triangle weights at these four indices are $1$ . The last row is therefore the count of pairs satisfying $3^{j} 5^{k} < 2^{u + 1}$ , not a floating-point estimate. The strips have sizes $194, 528, 859$ , so

D_{35, 0} = 15 (- 194 + 2 \cdot 528 - 859) = 45 \neq 0.

Exact enumeration reproduces all four counts and checks the endpoint power inequalities using integers. This example disproves the claim that avoiding floor crossings alone forces the third difference to vanish. It does not exclude a different fixed operator, different shifts, a boundary correction or a sparse-defect statement restricted to a suitable infinite subsequence.

What the Hecke–Mahler comparison does and does not supply.

Why a finite alphabet of bases is not enough.

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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