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

Integer recurrences in their general form

Consider an arbitrary integer recurrence with positive integer radices. Divisibility and the existence of segments satisfying the residue inequality must now be supplied as hypotheses. For the actual three-prime radices, the criterion below shows that each integer solution is eventually divisible by the smooth part of a denominator. It still leaves the residue-inequality hypothesis to be proved. The comparison with Cantor-series criteria is in Section 9.

Let $D$ and $B$ be positive integers with $D = D_{sm} B$ , where $D_{sm} = 2^{u} 3^{v} 5^{w}$ , $u, v, w \in N$ and $gcd (B, 30) = 1$ . Let $(c_{n})$ be an integer sequence satisfying $c_{n + 1} = b_{n} c_{n} - D m_{n}$ for integer sequences $(b_{n})$ and $(m_{n})$ . For these general sequences, form $W$ and $F$ by the recursions in Section 10; a segment satisfying the residue inequality must have $W \neq 0$ , and its modulus is $| W |$ . In the factorisation below the quotients $d_{n}$ are required to be integers; writing $c_{n} = D_{sm} d_{n}$ asserts divisibility, not just an identity in $Q$ .

Proposition 13.1 (conditional denominator reduction). If $c_{n} = D_{sm} d_{n}$ for every $n$ , with $D_{sm} > 0$ , then the recurrence and iterated recurrence for $(d_{n})$ have multiplier $B$ in place of $D$ . Moreover, for every $n$ and every real $t$ ,

0 < c_{n} \leq D t ⟺ 0 < d_{n} \leq B t .

Proof. Substitute $c_{n} = D_{sm} d_{n}$ and $D = D_{sm} B$ in the recurrence and divide by $D_{sm}$ . Dividing $0 < c_{n} \leq D_{sm} B t$ by this positive factor gives the stated bound equivalence. Dividing the iterated recurrence $c_{ℓ + h} = W_{ℓ, h} c_{ℓ} - D F_{ℓ, h}$ likewise gives $d_{ℓ + h} = W_{ℓ, h} d_{ℓ} - B F_{ℓ, h}$ . ◻

For a general bound $c_{n} \leq G (D, n)$ , division and integrality give only

d_{n} \leq ⌊ \frac{G (D, n)}{D_{sm}} ⌋,

not automatically $d_{n} \leq G (B, n)$ . The bounds used for the actual tails are linear in the multiplier before rounding. For any real $t$ ,

⌊ \frac{⌊ D t ⌋}{D_{sm}} ⌋ = ⌊ B t ⌋ .

Indeed, an integer $z$ satisfies $D_{sm} z \leq ⌊ D t ⌋$ exactly when $D_{sm} z \leq D t$ , or $z \leq B t$ . Thus rounding does not change the reduced bound in this case.

Growth of the radix product is not enough: $b_{n} = 3$ , $m_{n} = 1$ , $D = 2$ and $c_{n} = 1$ satisfy the recurrence, but $2$ never divides $c_{n}$ . The precise condition follows by reducing the recurrence modulo $D_{sm}$ . For any starting index $A$ and every $n \geq A$ ,

\begin{aligned} c_{n} & \equiv (\prod_{j = A}^{n - 1} b_{j}) c_{A} (\mod D_{sm}), \\ D_{sm} ∣ c_{n} & ⟺ \frac{D_{sm}}{gcd (D_{sm}, c_{A})} ∣ \prod_{j = A}^{n - 1} b_{j} . \end{aligned}

The congruence is an induction, since $D_{sm} ∣ D$ . For the equivalence, cancel the positive greatest common divisor; the remaining factor of $c_{A}$ is coprime to the remaining modulus. Thus it is growth of the required prime valuations, not growth of the product as a real number, that ensures divisibility.

For the actual radices the product is $P_{n} / P_{A}$ . Its valuations at $2, 3, 5$ all tend to infinity, so every fixed $D_{sm}$ eventually divides $c_{n}$ , for any integral initial value $c_{A}$ . Under rationality $S = N / D$ in lowest terms, one may start at $A = 1$ : $c_{1} = D X_{1} = N - D$ is integral and coprime to $D_{sm}$ . Since $P_{1} = 2$ , the criterion becomes $D_{sm} ∣ P_{n} / 2 = h_{n}$ , exactly the onset already obtained from the finite-prefix identity in Proposition 9.4. This is an alternative proof of the divisibility step, not an additional arithmetic hypothesis left to verify for $S$ . The finite-prefix proof in Theorem 9.3 remains valid.

Proposition 13.2 (segments satisfying the residue inequality exclude a positive bounded integer solution). Let $(b_{n})$ and $(m_{n})$ be sequences of nonnegative integers, let $G : N_{> 0} \times N \to N$ , and assume the residue condition (20) for these sequences and $G$ , using $| W_{ℓ, h} | > 0$ as the modulus. Fix $B > 0$ coprime to $30$ . There is no integral sequence $(d_{n})$ satisfying simultaneously $d_{n + 1} = b_{n} d_{n} - B m_{n}$ , $d_{n} > 0$ and $| d_{n} | \leq G (B, n)$ for every $n \geq 0$ .

Proof. Choose one segment satisfying the residue inequality $(ℓ, h)$ . The iterated recurrence gives $d_{ℓ + h} \equiv - B F_{ℓ, h}$ modulo $| W_{ℓ, h} |$ . The endpoint value is positive and at most $G (B, ℓ + h)$ , whereas the least positive residue of the right-hand side exceeds that bound, so Proposition 10.1 applies. ◻

Product growth gives room for a large residue; it does not put the actual residue there. For the constant data $b_{n} = 2$ , $m_{n} = 1$ , the positive solution is $d_{n} = B$ . Every segment has $W = 2^{h}$ , $F = 2^{h} - 1$ and $- B F \equiv B (\mod W)$ . Its least positive residue is therefore at most $B$ , however large $W$ becomes, so no bound $G (B, n) \geq B$ can pass the test. For the actual three-prime coefficients and $G = K$ , Theorem 10.2 makes the residue-inequality hypothesis equivalent to irrationality of $S$ . That equivalence does not hold for an arbitrary choice of $G$ .

Coprimality with $30$ selects the denominators covered by the hypothesis; once a segment is fixed, the finite contradiction does not use it. The conventions cover the edge cases: a zero product of the bases is excluded, a zero residue is represented by the full modulus, and positivity prevents the endpoint value from vanishing. The version stated with a nonzero smooth factor assumes the exact factorisation $c_{n} = D_{sm} d_{n}$ , the corresponding recurrence for $(c_{n})$ and the positive upper bound for $(d_{n})$ , and gives the same contradiction.

Even retaining the actual radix sequence and quadratic growth does not force irrationality if the numerators are changed. For the same $b_{a}$ , prescribe the positive integer carries $(a + 3)^{2}$ and set

{\hat{m}}_{a} = b_{a} (a + 3)^{2} - (a + 4)^{2} .

The shift by three ensures positivity even at $a = 0$ : since $b_{a} \geq 2$ ,

{\hat{m}}_{a} \geq 2 (a + 3)^{2} - (a + 4)^{2} = a^{2} + 4 a + 2 > 0.

Both these integer coefficients and the prescribed carries have quadratic order, because $2 \leq b_{a} \leq 30$ . Nevertheless, finite telescoping gives

\sum_{a = 0}^{N - 1} \frac{{\hat{m}}_{a}}{P_{a + 1}} = 9 - \frac{(N + 3)^{2}}{P_{N}} ⟶ 9.

This example keeps the exact radices, not the actual block multiplicities or their numerical upper bound $Q$ . It shows why the arithmetic of the specific $m_{a}$ , rather than the radix alphabet and growth orders alone, must enter a proof for $S$ .

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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