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.

Start here, by the editorial route

In this paper

A residue criterion and the bounds it allows

Two iterations already show how the recurrence determines an endpoint residue:

X_{2} = 6 X_{1} - 4, X_{3} = 10 X_{2} - 7 = 60 X_{1} - 47.

Whenever $B X_{1}$ is integral, $B X_{3} \equiv - 47 B (\mod 60)$ . An integral starting tail therefore puts the terminal tail in an arithmetic progression. Positivity and the quadratic bound restrict its possible integer values. If the least positive representative exceeds the bound, the starting tail cannot have been integral. For the zero class, that representative is the modulus, not zero. To exclude every rational denominator, the test must succeed for every eligible $B$ beyond every prescribed clearing onset.

A segment starts at $ℓ \geq 0$ and consists of $h \geq 0$ steps. Its product of bases and accumulated numerator are integers, even when the tails are not. Set

W_{ℓ, 0} = 1, F_{ℓ, 0} = 0, W_{ℓ, h + 1} = b_{ℓ + h} W_{ℓ, h}, F_{ℓ, h + 1} = b_{ℓ + h} F_{ℓ, h} + m_{ℓ + h} .

Induction gives

\begin{matrix} (19) & X_{ℓ + h} = W_{ℓ, h} X_{ℓ} - F_{ℓ, h}, d_{ℓ + h} = W_{ℓ, h} d_{ℓ} - B F_{ℓ, h} \end{matrix}

for any sequence satisfying $d_{n + 1} = b_{n} d_{n} - B m_{n}$ . Since $b_{a} = P_{a + 1} / P_{a}$ , the product telescopes to $W_{ℓ, h} = P_{ℓ + h} / P_{ℓ}$ . For the actual block masses, $m_{a} = h_{a + 1} s_{a}$ and $h_{a} = P_{a} / 2$ identify the two terms after division by $W_{ℓ, h}$ :

\frac{F_{ℓ, h}}{W_{ℓ, h}} = h_{ℓ} \sum_{a = ℓ}^{ℓ + h - 1} s_{a}, \frac{X_{ℓ + h}}{W_{ℓ, h}} = h_{ℓ} T_{ℓ + h} .

Thus the accumulated numerator records a finite part of the original reciprocal sum, in the same normalisation as the starting tail. No assumption of rationality or integrality is used in these identities. The previously proved bounds $8^{a} / 15 < P_{a} \leq 8^{a}$ also give $W_{ℓ, h} > 8^{h} / 15$ . For a positive integer $C$ and an integer $N$ , use ${lpr}_{C} (N) = 1 + ((N - 1) mod C) \in {1, \dots, C}$ . In particular ${lpr}_{C} (0) = C$ . Positivity of this representative is what allows comparison with a positive integer tail.

Proposition 10.1 (least positive residues). Let $C > 0$ and let $c$ be an integer with $0 < c$ and $| c | \leq K$ . If $c \equiv N (\mod C)$ and $K < {lpr}_{C} (N)$ , then the hypotheses are contradictory.

Proof. Every positive integer congruent to $N$ is at least ${lpr}_{C} (N)$ . Thus ${lpr}_{C} (N) \leq c = | c | \leq K$ , contrary to the strict inequality. ◻

For a bound $G : N_{> 0} \times N \to N$ let $E (G)$ be the statement

\begin{matrix} (20) & \begin{matrix} for every B \geq 1 with gcd (B, 30) = 1 and every a_{0} \geq 1, \\ there are ℓ \geq a_{0} and h \geq 1 with {lpr}_{W_{ℓ, h}} (- B F_{ℓ, h}) > G (B, ℓ + h) . \end{matrix} \end{matrix}

The segment may depend on both $B$ and $a_{0}$ . All quantities in the inequality are finite integers. The quantifiers over denominators and prescribed onsets are nevertheless unbounded; a search over a finite rectangle of $(B, ℓ)$ does not verify them. Every $b_{a} > 0$ , so $W_{ℓ, h} > 0$ is automatic.

Here $K$ is the integer bound obtained above from the jump count. Under the hypothesis that $B X_{ℓ}$ is integral, the recurrence makes $B X_{ℓ + h}$ integral as well, so $0 < B X_{ℓ + h} \leq K (B, ℓ + h)$ . Equivalently, the starting integer must lie in

(\frac{B F_{ℓ, h}}{W_{ℓ, h}}, \frac{B F_{ℓ, h} + K (B, ℓ + h)}{W_{ℓ, h}}] .

A residue exceeding $K$ says exactly that this interval contains no integer. The floor in the definition of $K$ is legitimate here because integrality is being assumed. The short paper uses the coarser real-tail bound $90 B (a + 1)^{2}$ instead. The argument needs a valid bound on the possible integer tail and the stated growth condition, rather than a particular numerical choice.

The two hypotheses on $G$ in the next theorem serve different purposes. Domination of $K$ lets a residue above $G$ exclude a positive integer tail. The limit $G (B, a) / 8^{a} \to 0$ lets every sufficiently long segment from a fixed nonintegral start overtake $G$ . It is a sufficient growth condition, not a necessary one: if only existence of a segment is required, a subsequence of small endpoint bounds can suffice, as we prove below. The stated theorem already covers every polynomial upper bound that dominates $K$ , and also $max {K (B, a), ⌈ B ρ^{a} ⌉}$ for $1 < ρ < 8$ . The zero bound fails to control the possible integer tail.

Theorem 10.2 (a residue criterion for every dominating bound of size $o (8^{a})$ ). Let $G : N_{> 0} \times N \to N$ satisfy $K (B, a) \leq G (B, a)$ for all $B$ and $a$ , and $G (B, a) / 8^{a} \to 0$ as $a \to \infty$ for each fixed $B$ . Then

E (G) ⟺ S \notin Q .

Both $K$ of (18) and $K_{0} (B, a) = 90 B (a + 1)^{2}$ satisfy these hypotheses, so $E (K)$ , $E (K_{0})$ and irrationality of $S$ are mutually equivalent. By contrast, $E (0)$ holds automatically, since every least positive residue is at least $1$ ; its truth alone therefore provides no contradiction to an integral tail.

Proof. Suppose $E (G)$ and suppose $S = N / D$ were rational. Theorem 9.3 supplies $B \geq 1$ coprime to $30$ , an onset $a_{D}$ , and positive integers $d_{a} = B X_{a} \leq K (B, a) \leq G (B, a)$ for $a \geq a_{D}$ satisfying the cleared recurrence. Apply (20) with $a_{0} = a_{D}$ to obtain a segment $(ℓ, h)$ with $ℓ \geq a_{D}$ . By (19), $d_{ℓ + h} \equiv - B F_{ℓ, h}$ modulo $W_{ℓ, h}$ , while $0 < d_{ℓ + h} \leq G (B, ℓ + h) < {lpr}_{W_{ℓ, h}} (- B F_{ℓ, h})$ . Proposition 10.1 is the contradiction, so $S$ is irrational.

Conversely suppose $S \notin Q$ , and fix $B \geq 1$ coprime to $30$ and $a_{0} \geq 1$ . Set $ℓ = a_{0}$ and $δ = ⌈ B X_{ℓ} ⌉ - B X_{ℓ} \in (0, 1)$ ; the prefix identity makes $B X_{ℓ}$ irrational. For all sufficiently large $h$ , $0 < δ + B X_{ℓ + h} / W_{ℓ, h} < 1$ , because $X_{ℓ + h} = O ((ℓ + h + 1)^{2})$ and $W_{ℓ, h} > 8^{h} / 15$ as shown above. The integer $⌈ B X_{ℓ} ⌉ W_{ℓ, h} - B F_{ℓ, h}$ is therefore exactly

{lpr}_{W_{ℓ, h}} (- B F_{ℓ, h}) = δ W_{ℓ, h} + B X_{ℓ + h} .

Its first term eventually exceeds $G (B, ℓ + h)$ , since $G (B, a) = o (8^{a})$ . Thus every sufficiently long segment from this fixed start satisfies the residue inequality, which is stronger than the required existence.

For the two stated bounds, $n_{a} \leq 3 a$ gives $K (B, a) \leq B Q (3 a) \leq 90 B (a + 1)^{2} = K_{0} (B, a)$ . Both are $O ((a + 1)^{2})$ for fixed $B$ , so both satisfy the required limit. Finally, ${lpr}_{C} (N) \geq 1 > 0$ always, independently of whether any tail is integral; a residue greater than zero cannot exclude a positive integer. ◻

The domination assumption is sufficient, not asserted to be necessary for equivalence. In particular, Proposition 9.6 already supplies the smaller valid bound $⌊ B \tilde{Q} (n_{a}) ⌋$ . Replacing $K$ by that bound in the forward implication leaves the argument unchanged; the converse uses only $G (B, a) = o (8^{a})$ . The zero-bound example explains why a size comparison with the possible integer tail is needed in the contradiction, not why this particular $K$ is indispensable.

At a fixed start, the same identity distinguishes two behaviours. If $B X_{ℓ}$ is nonintegral, the least positive residue eventually occupies a fixed positive fraction of the modulus. If $B X_{ℓ}$ is integral, the residue eventually equals the normalised tail $B X_{ℓ + h}$ itself. The next proposition gives both conclusions without assuming that $S$ is irrational.

Proposition 10.3 (the fixed-start residue limit). For fixed integers $B, ℓ \geq 1$ , write

R_{h} = {lpr}_{W_{ℓ, h}} (- B F_{ℓ, h}) .

For all sufficiently large $h$ ,

R_{h} = (⌈ B X_{ℓ} ⌉ - B X_{ℓ}) W_{ℓ, h} + B X_{ℓ + h},

and consequently

lim_{h \to \infty} \frac{R_{h}}{W_{ℓ, h}} = ⌈ B X_{ℓ} ⌉ - B X_{ℓ} .

In particular, if $B X_{ℓ}$ is integral, then $R_{h} = B X_{ℓ + h}$ for all sufficiently large $h$ .

Proof. The least positive residue is the modulus times the gap to the next strictly larger integer. Division with remainder gives

\frac{R_{h}}{W_{ℓ, h}} = ⌊ \frac{B F_{ℓ, h}}{W_{ℓ, h}} ⌋ + 1 - \frac{B F_{ℓ, h}}{W_{ℓ, h}} .

By the iterated recurrence, $B F_{ℓ, h} / W_{ℓ, h} = B X_{ℓ} - B X_{ℓ + h} / W_{ℓ, h}$ . These finite sums approach $B X_{ℓ}$ strictly from below, since $B X_{ℓ + h} > 0$ and $B X_{ℓ + h} / W_{ℓ, h} \to 0$ . Their floors therefore eventually equal $⌈ B X_{ℓ} ⌉ - 1$ , including when $B X_{ℓ}$ is an integer. Substitution proves both claims. The strict approach from below matters at an integer limit; the residue convention alone would assign a zero congruence class the value $W_{ℓ, h}$ , not zero. ◻

Thus a nonintegral $B X_{ℓ}$ , even for a rational $S$ , gives arbitrarily long successful segments against any bound $G (B, a) = o (8^{a})$ for that fixed $B$ . The rationality contradiction instead uses the denominator’s specific multiplier and a start beyond its clearing onset. Success for one pair $(B, ℓ)$ does not replace either quantifier.

The exact growth restriction on a dominating bound.

How fast the product of the bases grows

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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