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

Why a finite alphabet of bases is not enough.

Kebis, Luca, Ouaknine, Scoones and Worrell [27] work with fixed-base series whose coefficients lie in a finite algebraic alphabet. Their echoing condition requires long near-repetitions, separated intervals containing the mismatches, and nonzero weighted mismatch sums on at least two of those intervals. Claim 7 supplies the non-cancellation step in the proof of the value theorem. Merely having finitely many letters does not give these properties. The four-letter radix sequence is not the numerator sequence: the actual $m_{a}$ is unbounded. A recoding would have to preserve the scalar value, identify its coefficient alphabet and verify the echoing conditions. No fixed-base recoding with verified echoing properties is proved here.

Two natural bases expose the tradeoff. The least integer base whose powers clear every $P_{n}$ is $30$ , while base $8$ matches the growth of $P_{n} = H (2^{n})$ . The former keeps integrality; the latter keeps polynomial size.

Proposition 12.4 (what direct fixed-base recoding preserves). The following identities converge absolutely:

α = \sum_{a \geq 0} \frac{e_{a}}{30^{a + 1}} = \sum_{a \geq 0} \frac{v_{a}}{8^{a + 1}}, e_{a} = \frac{m_{a} 30^{a + 1}}{P_{a + 1}}, v_{a} = \frac{m_{a} 8^{a + 1}}{P_{a + 1}} .

Here $e_{a} \in Z_{> 0}$ and $e_{a} \geq (15 / 4)^{a + 1}$ , whereas $v_{a} \in Z [1 / 15]$ and $0 < v_{a} < 225 (a + 1)^{2}$ . For an integer $q \geq 2$ , the termwise divisibility $P_{n} ∣ q^{n}$ for every $n \geq 1$ holds exactly when $30 ∣ q$ ; that direct recoding then has coefficients at least $(q / 8)^{a + 1}$ .

Proof. Each exponent in $P_{n}$ is at most $n$ , so $P_{n} ∣ 30^{n}$ . The estimate $8^{n} / 15 < P_{n} \leq 8^{n}$ was proved in Section 8. These facts, $m_{a} \geq 1$ and $m_{a} \leq 15 (a + 1)^{2}$ , give all the coefficient bounds and identities. Necessity of $30 ∣ q$ follows because each of $2, 3, 5$ divides some $P_{n}$ ; sufficiency follows from $P_{n} ∣ 30^{n}$ . The general lower bound follows again from $P_{n} \leq 8^{n}$ . ◻

The base- $30$ coefficients are integral but grow exponentially, so the cited polynomial-growth criterion does not apply. The base- $8$ coefficients have polynomial size but are not all integral: already $v_{1} = 4 \cdot 8^{2} / 12 = 64 / 3$ . In fact, their reduced denominators grow exponentially even after cancellation. Since $P_{a + 1} / 2^{a + 1}$ is odd, the expression

v_{a} = \frac{m_{a} 4^{a + 1}}{P_{a + 1} / 2^{a + 1}}

can cancel an odd factor only through $m_{a}$ . Consequently,

den (v_{a}) \geq \frac{P_{a + 1}}{2^{a + 1} m_{a}} > \frac{4^{a + 1}}{225 (a + 1)^{2}} .

The last inequality uses $P_{a + 1} > 8^{a + 1} / 15$ and $m_{a} \leq 15 (a + 1)^{2}$ . Thus the bound on absolute value does not control Diophantine height even for these particular coefficients; fixed denominator-prime support is not enough.

More generally, no fixed integer base gives this termwise recoding both eventually integral coefficients and polynomial growth. If $30 ∣ q$ , the proposition gives exponential growth. Otherwise choose $p \in {2, 3, 5}$ not dividing $q$ . Cancellation by $m_{a}$ removes at most a factor $m_{a}$ , so

den (\frac{m_{a} q^{a + 1}}{P_{a + 1}}) \geq \frac{p^{⌊ (a + 1) \log_{p} 2 ⌋}}{m_{a}} > \frac{2^{a + 1}}{15 p (a + 1)^{2}} .

This denominator tends to infinity. Thus matching the denominator growth by base $8$ keeps the coefficients small but introduces large rational denominators; making the coefficients integral by base $30$ makes them exponentially large. This obstruction concerns termwise rescaling only, not regrouping or carrying.

Carrying resolves a different problem: it bounds the digits but leaves the radices variable. It also requires the integer parts of the actual tails, not just the known block numerators. Set

δ_{a} = m_{a} + ⌊ X_{a + 1} ⌋ - b_{a} ⌊ X_{a} ⌋ = ⌊ b_{a} {X_{a}} ⌋ .

The equality follows from the tail recurrence and $m_{a} \in Z$ . Thus $0 \leq δ_{a} < b_{a}$ and ${X_{a}} = (δ_{a} + {X_{a + 1}}) / b_{a}$ . Iterating gives the finite identity

{X_{0}} = \sum_{a = 0}^{N - 1} \frac{δ_{a}}{P_{a + 1}} + \frac{{X_{N}}}{P_{N}} .

The last term is nonnegative, is less than $1 / P_{N} \leq 2^{- N}$ , and tends to zero. This is the greedy mixed-radix expansion of the fractional part of $X_{0} = S / 2$ , with digits in ${0, \dots, 29}$ . For the actual series its first eight digits are

(δ_{0}, \dots, δ_{7}) = (1, 4, 8, 3, 10, 1, 8, 5) .

To certify them, use the twelve-block enclosure from Section 11, either propagated by the recurrence or evaluated directly as

P_{a} \sum_{j = a}^{11} \frac{m_{j}}{P_{j + 1}} < X_{a} \leq P_{a} \sum_{j = a}^{11} \frac{m_{j}}{P_{j + 1}} + \frac{P_{a}}{P_{12}} Q (n_{12}) (0 \leq a < 12) .

Both descriptions give the same rational endpoints. At indices $0$ through $8$ , each interval lies strictly between consecutive integers, certifying all nine floors needed for the eight digits. Width alone would not suffice: an interval crossing an integer leaves its floor undetermined. For example, $⌊ X_{4} ⌋ = 2$ and $⌊ X_{5} ⌋ = 5$ , giving $δ_{4} = 65 + 5 - 30 \cdot 2 = 10$ , not the uncarried numerator $m_{4} = 65$ . This finite calculation makes no claim about the later digit complexity.

A rational number need not have eventually periodic digits when the radices vary. For example, take radix $5$ at the indices $1, 2, 4, 8, \dots$ and radix $3$ elsewhere. The constant tail fraction $1 / 2$ gives digit $2$ at those indices and $1$ elsewhere. These digits are not eventually periodic, but their mixed-radix expansion sums to $1 / 2$ . This example uses different radices from our four possible bases; it shows why fixed-base eventual periodicity cannot be transferred to a variable-radix expansion. Our digits also depend on the actual tail fractions, not just on the four-letter radix sequence. The construction supplies neither a fixed-base expansion with verified echoing properties nor a bound on its digit complexity.

Adamczewski–Bugeaud’s complexity theorem [30] concerns the actual digits of an integer-base expansion: for an algebraic irrational their length- $n$ block complexity divided by $n$ tends to infinity. Neither $(b_{a})$ nor the uncarried $(e_{a})$ is such a digit expansion of $α$ .

Problem 12.5 (the repeated three-prime target). Prove irrationality of $S = R_{{2, 3, 5}}$ . Equivalent forms are $E (K)$ and

\begin{matrix} (22) & B X_{a} \notin Z for every B \geq 1 with gcd (B, 30) = 1 and every a \geq 1. \end{matrix}

The residue formulation uses finite integer computations, but asks for a success for every eligible denominator and beyond every prescribed start. The tail formulation requires that no positive integer multiplier coprime to $30$ make any $X_{a}$ , $a \geq 1$ , integral. These are reformulations of the same irrationality assertion, not extra assumptions on the series, and neither follows from the rank obstruction. The next proposition proves the tail reformulation, complementing Theorem 10.2.

Proposition 12.6 (irrationality is equivalent to nonintegrality of every reduced tail). Statement (22), quantified over every $B \geq 1$ coprime to $30$ and every $a \geq 1$ , is equivalent to irrationality of $S$ .

Proof. If $B X_{a} \in Z$ for some such $B$ and $a$ , then $X_{a} \in Q$ , and since $S = \sum_{j < a} s_{j} + X_{a} / h_{a}$ with $h_{a}$ a positive rational and the prefix a finite sum of rationals, $S \in Q$ . Conversely if $S \in Q$ , then Theorem 9.3 produces $B$ coprime to $30$ with $B X_{a} \in Z$ for every $a \geq a_{D}$ . ◻

Integrality at one index propagates forwards, but nonintegrality at one early index does not propagate forever: a rational denominator can clear later. Proposition 9.4 describes that onset exactly. The dichotomy in Proposition 8.4 also permits integral tails, so it does not settle this question.

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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