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

A bound using the order of the jumps

The bound in Theorem 9.1 uses only the fact that every new prime factor is at least $2$ . Successive powers of $2$ cannot continue indefinitely without a power of $3$ between them. Incorporating that restriction improves the geometric majorant.

Proposition 9.6 (a smaller quadratic bound). For every $a \geq 0$ ,

0 < X_{a} \leq \tilde{Q} (n_{a}) < Q (n_{a}), \tilde{Q} (n) = \frac{1210 n^{2} + 9130 n + 18847}{11979} .

Proof. For a point $x \geq 2^{a}$ , let $d_{p}$ be the increase in the exponent of $p$ in its running LCM from the boundary $2^{a}$ , and put $k = d_{2} + d_{3} + d_{5}$ . The definition of $d_{3}$ gives $x < 3^{⌊ \log_{3} 2^{a} ⌋ + d_{3} + 1} \leq 2^{a} 3^{d_{3} + 1} < 2^{a + 2 d_{3} + 2}$ , hence $d_{2} \leq 2 d_{3} + 1$ . It follows that $k \leq 3 (d_{3} + d_{5}) + 1$ , so at least $j = ⌊ (k + 1) / 3 ⌋$ of the $k$ new prime factors are odd. Each of these contributes at least $3$ , and each remaining factor at least $2$ . Hence

\frac{H (x)}{P_{a}} \geq 2^{k - j} 3^{j} =: d (k) .

There is at most one interval of constancy with each exponent sum, with at most $(n_{a} + k + 3)^{2} / 9$ supported points by the earlier projection bound. Consequently

X_{a} \leq \sum_{k \geq 0} \frac{(n_{a} + k + 3)^{2}}{18 d (k)} .

Now $d (3 m) = 12^{m}$ , $d (3 m + 1) = 2 \cdot 12^{m}$ and $d (3 m + 2) = 6 \cdot 12^{m}$ . Grouping in threes and summing the quadratic geometric series gives $\tilde{Q} (n_{a})$ . Finally $11979 (Q (n) - \tilde{Q} (n)) = 121 n^{2} + 1518 n + 5111 > 0$ for $n \geq 0$ . ◻

The estimate uses the original block multiplicities and the order of the prime-power jumps. It is not a bound for an arbitrary recurrence with the same bases. For an integral $B X_{a}$ , the bound can be rounded down to $⌊ B \tilde{Q} (n_{a}) ⌋$ . This improves the bound available in a rationality contradiction. Theorem 10.2 and the recorded finite tests use $K$ , not this smaller bound. No optimality claim is made for either quadratic bound.

Section 10 explains how the smaller bound can also replace $K$ in the residue argument.

Relation to Cantor-series criteria.

Since $h_{0} = 1 / 2$ , the value $S / 2 = X_{0}$ is the Cantor series $\sum_{a \geq 0} m_{a} / P_{a + 1}$ , with $P_{0} = 1$ and $P_{a + 1} = b_{a} P_{a}$ . Its normalised tails $X_{a}$ are the usual objects in the rationality criteria for Cantor series. Under a small-numerator hypothesis, Erdős and Straus characterise rationality by the existence of a positive integer $B$ and integers $c_{a}$ satisfying, eventually,

B m_{a} = b_{a} c_{a} - c_{a + 1}, | c_{a + 1} | < b_{a} / 2

[18]. The second condition is part of the criterion, not a consequence of the recurrence alone. Their proof already chooses nearest integers. Hančl and Tijdeman make that choice part of the criterion: they prescribe $c_{a}$ as a nearest integer to $B m_{a} / b_{a}$ and require the recurrence [19]. Koutsoukou-Argyraki and Li formalised the Erdős–Straus criteria in Isabelle/HOL [20].

Both cited criteria require $m_{a} / (b_{a - 1} b_{a}) \to 0$ in our notation. Here the weighted-triangle count gives $m_{a} = Θ ((a + 1)^{2})$ , and

\frac{m_{a}}{900} \leq \frac{m_{a}}{b_{a - 1} b_{a}} \leq \frac{m_{a}}{4} (a \geq 1) .

The ratio therefore has quadratic order and tends to infinity. There is also a direct lower count that does not use the real-logarithm triangle estimate. For $a \geq 1$ , each pair $0 \leq j, k \leq ⌊ a / 5 ⌋$ has $3^{j} 5^{k} < 2^{a}$ , since $15 < 2^{5}$ , and hence a unique power-of-two multiple in $[2^{a}, 2^{a + 1})$ . Each such point contributes at least one to $m_{a}$ . Thus $m_{a} \geq (⌊ a / 5 ⌋ + 1)^{2}$ , an elementary lower bound sufficient to see directly that the small-numerator condition fails.

The actual scaled tails also satisfy

B X_{a} - \frac{B m_{a}}{b_{a}} = \frac{B X_{a + 1}}{b_{a}} = Θ ((a + 1)^{2})

for each fixed $B \geq 1$ , by the recurrence, the quadratic tail bounds and $2 \leq b_{a} \leq 30$ . Thus any eventual integer tails supplied by a hypothetical rational value could not be the prescribed nearest integers. In fact, those prescribed integers fail the recurrence at every sufficiently large index, regardless of how ties are resolved. If $c_{a}$ is a nearest integer to $B m_{a} / b_{a}$ , then

| b_{a} c_{a} - B m_{a} | \leq 15, c_{a + 1} \geq \frac{B m_{a + 1}}{30} - \frac{1}{2} ⟶ \infty .

Consequently $c_{a + 1} = b_{a} c_{a} - B m_{a}$ is impossible for all sufficiently large $a$ . This is a failure of the proposed construction of integer coefficients, not an irrationality contradiction: the criterion’s small-numerator hypothesis also fails.

The denominator clearing itself is standard; the calculations specific to this series retain its multiplicities, use the strict endpoint to divide the normalising factor by two, and bound the resulting positive tails.

For polynomial Cantor data, Hančl–Tijdeman [25] give a polynomial cancellation criterion and a division mechanism. Their nonconstant polynomial radix is not our bounded $(b_{a})$ . Their separate Theorem 4.2 uses a finite-product rearrangement. We do not use an unrestricted infinite reindexing from that argument: for the integer decompositions considered here, the terminal terms must be checked separately. In the first-order case $m_{a} = b_{a} c_{a} - c_{a + 1}$ , finite summation gives

\sum_{a = 0}^{N - 1} \frac{m_{a}}{P_{a + 1}} = c_{0} - \frac{c_{N}}{P_{N}} .

Every integer $c_{0}$ generates an integer solution; integrality alone places no restriction on the initial value. The terminal condition is the restriction: $c_{N} / P_{N} \to c_{0} - S / 2$ , so decay to zero selects the actual initial tail $c_{0} = S / 2$ . This is the same uniqueness mechanism as for the small solutions above, not a consequence of solving the recurrence.

For these actual numerators the terminal term cannot vanish for any integer $c_{0}$ . Indeed, $P_{3} = 120$ , $n_{3} = 5$ and $(m_{0}, m_{1}, m_{2}) = (1, 4, 7)$ , so the tail bound gives

0 < \frac{S}{2} = \frac{1}{2} + \frac{4}{12} + \frac{7}{120} + \frac{X_{3}}{120} \leq \frac{107}{120} + \frac{83}{1080} = \frac{523}{540} < 1.

Thus $c_{0} - S / 2 \neq 0$ for every integer $c_{0}$ . Since $8^{N} / 15 < P_{N} \leq 8^{N}$ , every such unscaled integer solution satisfies $| c_{N} | = Θ (8^{N})$ . This is a concrete reason that quadratic growth of $m_{a}$ does not imply polynomial growth of recursively defined carries. The starting index is essential: this does not exclude the actual tails becoming integral at a later index. After replacing $m_{a}$ by $D m_{a}$ for an integer $D \geq 1$ , however, a vanishing terminal term would require $D S / 2 = c_{0}$ . Excluding that possibility for every $D$ would require irrationality itself; the unscaled calculation does not do so.

In the cited rearrangement the transformed integer numerator must be $o (b_{a})$ , which for bounded radices means eventual zero. Only after the terminal terms have been controlled does this cancellation determine the infinite sum. A polynomial bound on each individual decomposition coefficient would suffice for a fixed number of shifts: the corresponding series are absolutely convergent, since their denominators are products of at least $2$ at each step. A polynomial bound on the combined numerators is not a substitute, as the exponentially growing integer solutions above demonstrate. Nor does quadratic growth force an integer solution to be a polynomial in the floor coordinates.

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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