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

An exact weighted-shift identity for the repeated series

Set $α = S / 2$ and retain the boundary running LCMs $P_{a}$ . To combine shifted tails, we first express them with the same denominators. After multiplication by $P_{t}$ , the term with index $a + t$ becomes

\frac{P_{t} m_{a + t}}{P_{a + t + 1}} = \frac{P_{t} P_{a + 1}}{P_{a + t + 1}} \frac{m_{a + t}}{P_{a + 1}} .

The first factor on the right is the weight $γ_{a, t}$ below. It is forced by the change of denominator, rather than chosen to produce cancellation. The removed finite prefix supplies the rational correction. The identity concerns the repeated sum $S$ only. In this subsection and the next, $r$ is a positive integer shift; the prime-generator notation is no longer in use.

Lemma 12.1 (weighted shifts preserve the actual value). Fix integers $c_{0}, \dots, c_{σ}$ , not all zero, independently of the positive integer shift $r$ . Put

γ_{a, t} = \frac{P_{t} P_{a + 1}}{P_{a + t + 1}}, D_{r, a} = 15 \sum_{j = 0}^{σ} c_{j} γ_{a, j r} m_{a + j r} .

Then $γ_{a, t} \in {1, 1 / 3, 1 / 5, 1 / 15}$ and $D_{r, a} \in Z$ . Furthermore, with

A_{r} = 15 \sum_{j = 0}^{σ} c_{j} P_{j r}, Z_{r} = 15 \sum_{j = 0}^{σ} c_{j} P_{j r} \sum_{k < j r} \frac{m_{k}}{P_{k + 1}} \in Z,

we have the absolutely convergent identity

\begin{matrix} (21) & A_{r} α - Z_{r} = \sum_{a \geq 0} \frac{D_{r, a}}{P_{a + 1}} . \end{matrix}

One may take $C = 225 \sum_{j = 0}^{σ} | c_{j} | max (1, j)^{2}$ in $| D_{r, a} | \leq C (a + r + 1)^{2}$ . If $J$ is the largest index with $c_{J} \neq 0$ , then $A_{r} \neq 0$ whenever $\sum_{j < J} | c_{j} | 2^{- (J - j) r} < | c_{J} |$ ; an empty sum is zero.

Proof. The exponent of $2$ in $γ_{a, t}$ is $t + (a + 1) - (a + t + 1) = 0$ . For $p = 3, 5$ , the exponent is $⌊ t θ_{p} ⌋ + ⌊ (a + 1) θ_{p} ⌋ - ⌊ (a + t + 1) θ_{p} ⌋$ , which is $0$ or $- 1$ by floor addition. Thus the only possible denominator factors are one $3$ and one $5$ ; the prefactor $15$ in $D_{r, a}$ clears both. For $k < j r$ , $P_{k + 1}$ divides $P_{j r}$ , so $Z_{r}$ is an integer. For each $t$ , absolute convergence gives

\sum_{a \geq 0} \frac{γ_{a, t} m_{a + t}}{P_{a + 1}} = P_{t} \sum_{k \geq t} \frac{m_{k}}{P_{k + 1}} = P_{t} (α - \sum_{k < t} \frac{m_{k}}{P_{k + 1}}) .

The finite linear combination proves the identity. Lemma 8.3, $0 < γ \leq 1$ and $a + j r + 1 \leq max (1, j) (a + r + 1)$ give the stated constant. For $j < J$ , $P_{j r} / P_{J r} \leq 2^{- (J - j) r}$ . Dividing $A_{r}$ by $15 P_{J r}$ and bounding the remaining terms proves the nonvanishing condition, which holds for all sufficiently large $r$ . ◻

For any fixed nonzero coefficient vector, including $(1, - 3, 3, - 1)$ , the nonvanishing condition is automatic for sufficiently large $r$ : each term on its left tends to zero. No arithmetic assumption on $S$ is needed. The identity still represents $S / 2$ , but gives no sparsity or cancellation of $D_{r, a}$ .

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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