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

Recovering the prime-power jumps

As in long division, periodicity requires both finitely many remainders and a deterministic rule for the next step. The varying bases make the second requirement nontrivial. We obtain it by separating the five possible block images, using a bound on the actual arithmetic tails.

Proof of Theorem 1.1. Finitely many tails under rationality. Write $H = H_{{2, 3, 5}}$ in this proof and put $P_{a} = H (2^{a})$ . Every running LCM at or before $2^{a}$ divides $P_{a}$ . We therefore normalise the tail after $2^{a}$ by

\begin{matrix} (1) & Y_{a} = P_{a} \sum_{\begin{matrix} t > 2^{a} \\ t = 2^{n}, 3^{n} or 5^{n}, n \geq 1 \end{matrix}} \frac{1}{H (t)} . \end{matrix}

Each subsequent jump multiplies the running LCM by at least $2$ , and a jump by $3$ eventually occurs. Comparison with $\sum_{j \geq 1} 2^{- j}$ gives $0 < Y_{a} < 1$ .

Suppose that $D_{{2, 3, 5}} = N / K$ , where $N$ is an integer and $K \geq 1$ is an integer. Clearing the finite prefix gives

K Y_{a} = P_{a} N - K (P_{a} + \sum_{\begin{matrix} t \leq 2^{a} \\ t = 2^{n}, 3^{n} or 5^{n}, n \geq 1 \end{matrix}} \frac{P_{a}}{H (t)}) \in Z .

Thus the $Y_{a}$ lie in the finite set $(K^{- 1} Z) \cap (0, 1)$ . It remains to show why equality of two tails propagates.

Recovering a block from its tail. Consider the jumps between $16$ and $32$ , including the last endpoint. They occur at $25, 27, 32$ , multiplying the running LCM successively by $5, 3, 2$ . After normalisation at $16$ , their contribution is $1 / 5 + 1 / 15 + 1 / 30$ . The remaining tail is normalised at a running LCM $30$ times as large, so

Y_{4} = \frac{1}{5} + \frac{1}{15} + \frac{1}{30} + \frac{Y_{5}}{30} = \frac{9 + Y_{5}}{30} .

In the block $(64, 128]$ , the interior jumps occur in the opposite order, at $81, 125$ . Its constant term is $1 / 3 + 1 / 15 + 1 / 30 = 13 / 30$ . The multiplier $30$ records the primes, but not their order; the numerators $9$ and $13$ retain the order. The remaining question is whether the unknown following tail can conceal this difference.

In general, $(2^{a}, 2^{a + 1}]$ contains its terminal power of $2$ , at most one power of $3$ and at most one power of $5$ . Let $τ_{a}$ denote the sequence of interior prime labels, omitting the terminal $2$ . Splitting off the block gives

\begin{matrix} (2) & Y_{a} = G_{τ_{a}} (Y_{a + 1}), G_{τ} (y) = \frac{μ_{τ} + y}{b_{τ}}, \begin{array}{cccccc} τ & \emptyset & 3 & 5 & 35 & 53 \\ b_{τ} & 2 & 6 & 10 & 30 & 30 \\ μ_{τ} & 1 & 3 & 3 & 13 & 9 \end{array} \end{matrix}

Here $b_{τ}$ is the factor by which the running LCM increases across the block, and $μ_{τ} / b_{τ}$ is the block’s normalised contribution.

The five formulae alone do not recover $τ_{a}$ from $Y_{a}$ . The coarse images $G_{53} ([0, 1]) = [3 / 10, 1 / 3]$ and $G_{5} ([0, 1]) = [3 / 10, 2 / 5]$ overlap. We need a smaller interval known to contain the next arithmetic tail. The table gives the lower bound $Y_{a} \geq 3 / 10$ . With that lower endpoint, separating these two images requires an upper endpoint $U$ satisfying

G_{53} (U) < G_{5} (3 / 10), \frac{9 + U}{30} < \frac{33}{100}, U < \frac{9}{10} .

This specifies the improvement needed from the arithmetic sequence.

A power $5^{f}$ belongs to block $⌊ f \log_{2} 5 ⌋$ . These indices start at $2$ and their successive differences are $2$ or $3$ . Every three consecutive blocks therefore contain a power of $5$ . Given $a$ , let $j \in {a, a + 1, a + 2}$ be the first such block. The maps for blocks containing $5$ give $Y_{j} \leq 7 / 15$ , since $Y_{j + 1} < 1$ and $G_{35} (1) = 7 / 15$ bounds $G_{5} (1)$ and $G_{53} (1)$ as well. We now work backwards from $j$ to $a$ . Each intervening block has map $G_{\emptyset}$ or $G_{3}$ , and $G_{3} (y) \leq G_{\emptyset} (y) = (1 + y) / 2$ for $y \geq 0$ . At most two backward steps give the successive bounds $7 / 15$ , $11 / 15$ , $13 / 15$ . Hence

\begin{matrix} (3) & Y_{a} \in I = [\frac{3}{10}, \frac{13}{15}], G_{\emptyset} (G_{\emptyset} (7 / 15)) = \frac{13}{15} < \frac{9}{10} . \end{matrix}

The bound uses the occurrence of a power of $5$ within three blocks. It does not assert invariance under arbitrary words in the five maps: indeed, $G_{\emptyset} (13 / 15) = 14 / 15$ lies outside $I$ .

The following table gives the exact images of $I$ . They are pairwise disjoint, with successive gaps $1 / 900, 17 / 300, 79 / 900, 1 / 180$ . Writing $L = 3 / 10$ and $U = 13 / 15$ , the two narrow gaps are

G_{5} (L) - G_{53} (U) = \frac{3 L - U}{30} = \frac{1}{900}, G_{\emptyset} (L) - G_{3} (U) = \frac{3 L - U}{6} = \frac{1}{180} .

Both are positive because $U < 3 L = 9 / 10$ .

\begin{array}{cc} τ & G_{τ} (I) \\ 53 & [31 / 100, 74 / 225] \\ 5 & [33 / 100, 29 / 75] \\ 35 & [133 / 300, 104 / 225] \\ 3 & [11 / 20, 29 / 45] \\ \emptyset & [13 / 20, 14 / 15] \end{array}

Since $Y_{a + 1} \in I$ , the actual tail $Y_{a}$ belongs to exactly one of these five image intervals, identifying $τ_{a}$ . For instance, $Y_{a} \in [31 / 100, 74 / 225]$ forces $τ_{a} = 53$ and gives $Y_{a + 1} = 30 Y_{a} - 9$ . In general we calculate $Y_{a + 1} = b_{τ_{a}} Y_{a} - μ_{τ_{a}}$ , identify its image interval, and continue. Each recovered value is the next arithmetic tail, so the same bound $I$ remains available at every step.

Repetition would give a rational frequency. The finitely many tail values obtained above include a repeated value $Y_{u} = Y_{v}$ with $u < v$ . Since the image intervals are disjoint, this forces $τ_{u} = τ_{v}$ . Applying the corresponding inverse map gives $Y_{u + 1} = Y_{v + 1}$ . Induction yields $τ_{u + j} = τ_{v + j}$ for every $j \geq 0$ . Thus the entire later block word has period $v - u$ .

To exclude this possibility, let $α = \log 2 / \log 3$ . The number of powers of $3$ in block $a$ is $⌊ (a + 1) α ⌋ - ⌊ a α ⌋$ , and is determined by $τ_{a}$ . If the block word has eventual period $ℓ \geq 1$ , let $c$ be the number of $3$ -powers in one period. For every sufficiently large fixed $a$ and every $n \geq 1$ we then have

⌊ (a + n ℓ) α ⌋ = ⌊ a α ⌋ + n c .

Dividing by $n$ and taking the limit gives $ℓ α = c$ , contrary to unique factorisation, since it would imply $2^{ℓ} = 3^{c}$ . ◻

The Lean proof follows this route. It identifies $H (t)$ with the running least common multiple, shows that the sum counts each distinct value once, identifies the contribution of each dyadic block with one of the five pairs $(b_{τ}, μ_{τ})$ , and proves that every three consecutive blocks contain a power of $5$ and that the blocks containing a power of $3$ are not eventually periodic. It then applies an irrationality criterion for block series with these five pairs, whose proof contains the bounds, the separation and the return above.

The first twelve block types are $\emptyset, 3, 5, 3, 53, \emptyset, 35, 3, \emptyset, 53, \emptyset, 35$ . The saved numerical calculation over the first $3000$ blocks reports tails in $[0.3186, 0.7799]$ . These approximations are separate from the rational bounds used in the proof.

Grouping the jumps into dyadic blocks matters. One jump at a time, the tails $x_{k} = (q_{1} \dots q_{k}) \sum_{j > k} (q_{1} \dots q_{j})^{- 1}$ satisfy $x_{k} = (1 + x_{k + 1}) / q_{k + 1}$ and $1 / 4 \leq x_{k} \leq 1$ , and the images of $[1 / 4, 1]$ under $x \mapsto (1 + x) / 2$ and $x \mapsto (1 + x) / 3$ overlap on $[5 / 8, 2 / 3]$ . A tail before a jump by $2$ cannot be told from a tail before a jump by $3$ from these bounds alone.

Listing the jumps one at a time writes $D_{{2, 3, 5}}$ as a Cantor series with every digit equal to $1$ and every base in ${2, 3, 5}$ . The criteria of Erdős–Straus [18] and Hančl–Tijdeman [19], recalled in Section 9, assume that each digit divided by the product of the last two bases in its denominator tends to $0$ . Here that quotient is at least $1 / 25$ , so neither criterion applies.

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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