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

The recurrence for tails between powers of two

Return to $S = R_{{2, 3, 5}}$ and $P = {2, 3, 5}$ . In place of the three jumps in $(16, 32]$ , we now count all seven smooth integers in $[16, 32)$ : $16, 18, 20, 24, 25, 27, 30$ . Their running LCMs are $720, 720, 720, 720, 3600, 10800, 10800$ . Two changes need keeping separate: we retain multiplicities, and we include the left rather than the right endpoint. The first changes the coefficients; the second changes the normalisation.

We group the terms of $S$ into $[2^{a}, 2^{a + 1})$ , including the left endpoint. The prefix to be cleared is therefore strictly before $2^{a}$ : it excludes the jump by $2$ at that boundary. Put $P_{a} = H (2^{a})$ and $h_{a} = P_{a} / 2$ . Every running LCM in that prefix divides $h_{a}$ . Here $h_{0} = 1 / 2$ , and $h_{a}$ is an integer for $a \geq 1$ . The powers of $3$ and $5$ selected at $2^{a}$ exceed $2^{a} / 3$ and $2^{a} / 5$ , respectively, while the power of $2$ is $2^{a}$ itself. Thus

\frac{8^{a}}{15} < P_{a} \leq 8^{a} (a \geq 0) .

With this normalisation, define the block masses and tails by

\begin{matrix} (13) & s_{a} = \sum_{\begin{matrix} i, j, k \geq 0 \\ 2^{a} \leq 2^{i} 3^{j} 5^{k} < 2^{a + 1} \end{matrix}} \frac{1}{H (2^{i} 3^{j} 5^{k})}, T_{a} = \sum_{j \geq a} s_{j}, X_{a} = h_{a} T_{a} . \end{matrix}

For $a \geq 1$ , the prefix has exponent of $2$ at most $a - 1$ and other exponents at most their boundary values. This proves the divisibility by $P_{a} / 2$ needed in Lemma 9.2.

The jumps after $2^{a}$ and up to $2^{a + 1}$ consist of any powers of $3$ or $5$ strictly inside that interval, followed by the factor $2$ at its right endpoint. There is at most one power of each odd prime: successive powers have ratio greater than two. Let $I_{a}$ list the pairs $(p, e)$ with $p \in {3, 5}$ and $2^{a} < p^{e} < 2^{a + 1}$ , ordered by the value $p^{e}$ .

Proposition 8.1 (four possible bases). For every $a$ ,

\begin{matrix} (14) & b_{a} = \frac{P_{a + 1}}{P_{a}} = 2 \prod_{(p, e) \in I_{a}} p \in {2, 6, 10, 30}, so 2 \leq b_{a} \leq 30 . \end{matrix}

Proof. A power of $3$ may occur and a power of $5$ may occur, each at most once. The resulting factor is $2$ , $2 \cdot 3$ , $2 \cdot 5$ or $2 \cdot 3 \cdot 5$ . ◻

The coefficient subtracted at step $a$ is the block mass with its denominators cleared:

m_{a} = h_{a + 1} s_{a} = \sum_{\begin{matrix} x smooth \\ 2^{a} \leq x < 2^{a + 1} \end{matrix}} \frac{P_{a + 1}}{2 H (x)} .

Each summand is an integer, by the same strict-cutoff argument. For example, $[2, 4)$ contains just $2$ and $3$ , so $m_{1} = 6 (1 / 2 + 1 / 6) = 4$ and $X_{2} = 6 X_{1} - 4$ . The first four pairs $(b_{a}, m_{a})$ , starting at $a = 1$ , are $(6, 4)$ , $(10, 7)$ , $(6, 7)$ and $(30, 65)$ . In particular $m_{3} > b_{3}$ : these coefficients are not positional digits.

In the block $[16, 32)$ , the internal jumps are $25$ and $27$ , in that order. Multiplication by $h_{5} = 10800$ gives the weights for the seven points listed above. The four smooth numbers $16, 18, 20, 24$ precede both jumps and have weight $15$ ; $25$ precedes only the jump at $27$ and has weight $3$ ; and $27, 30$ have weight $1$ . Thus

m_{4} = 4 \cdot 15 + 1 \cdot 3 + 2 \cdot 1 = 7 + (5 - 1) \cdot 4 \cdot 3 + (3 - 1) \cdot 5 = 65.

The second expression starts with the seven points and adds the corrections before $25$ and $27$ . It is this form that extends to every block.

To compute $m_{a}$ in general, count the smooth numbers before each prime-power jump. Let $N (t)$ be the number of positive ${2, 3, 5}$ -smooth integers strictly below $t$ . First give every point in $[2^{a}, 2^{a + 1})$ weight one. At an internal jump $p^{e}$ , the points before that jump need an additional weight $p - 1$ , multiplied by the prime factors at all later internal jumps. Thus, for $a \geq 1$ ,

\begin{matrix} (15) & \begin{aligned} m_{a} & = N (2^{a + 1}) - N (2^{a}) \\ + \sum_{(p, e) \in I_{a}} (p - 1) (N (p^{e}) - N (2^{a})) \prod_{\begin{matrix} (q, f) \in I_{a} \\ p^{e} < q^{f} \end{matrix}} q, \\ m_{0} & = 1. \end{aligned} \end{matrix}

The first difference counts the whole block; each later difference counts its points strictly before the indicated jump. The proof below justifies these weights, establishes convergence and derives the tail identities.

Theorem 8.2 (the tail recurrence). The block masses are summable and $S = \sum_{a \geq 0} s_{a}$ . For every $a \geq 0$ ,

\begin{matrix} (16) & m_{a} = h_{a + 1} s_{a} \in N_{> 0}, X_{a + 1} = b_{a} X_{a} - m_{a}, X_{a} = \sum_{j \geq a} \frac{m_{j}}{b_{a} b_{a + 1} \dots b_{j}} . \end{matrix}

Proof. Every exponent of $3$ or $5$ in the $a$ th block is at most $a$ , and for each such pair Lemma 3.5 leaves at most one exponent of $2$ placing the point in $[2^{a}, 2^{a + 1})$ . Each running LCM is at least $2^{a}$ and the block contains $2^{a}$ , so $0 < s_{a} \leq (a + 1)^{2} 2^{- a}$ . The majorant is summable, which justifies every tail splitting below, and unique prime factorisation identifies $\sum_{a} s_{a}$ with the original repeated series.

Write the internal jumps as $t_{ℓ} = p_{ℓ}^{e_{ℓ}}$ , $1 \leq ℓ \leq v$ , and set $t_{0} = 2^{a}$ , $t_{v + 1} = 2^{a + 1}$ and $n_{ℓ} = N (t_{ℓ})$ . On $[t_{ℓ}, t_{ℓ + 1})$ the running LCM is $P_{a} \prod_{u \leq ℓ} p_{u}$ , and the terminal jump has factor two, so

h_{a + 1} s_{a} = \sum_{ℓ = 0}^{v} (n_{ℓ + 1} - n_{ℓ}) \prod_{u > ℓ} p_{u} = n_{v + 1} - n_{0} + \sum_{ℓ = 1}^{v} (p_{ℓ} - 1) (\prod_{u > ℓ} p_{u}) (n_{ℓ} - n_{0}) .

The second equality is finite summation by parts and is exactly (15). The count itself is finite: for $p = 2, 3, 5$ , with $q, r$ the other primes, counting by the exponent of $p$ gives

N (p^{e}) = \sum_{u = 1}^{e} # {(i, j) \in N^{2} : q^{i} r^{j} < p^{u}} .

At $a = 0$ the block is the single point $1$ , giving $m_{0} = 1$ . Positivity follows from $s_{a} > 0$ . Splitting $T_{a} = s_{a} + T_{a + 1}$ and multiplying by $h_{a}$ gives the recurrence, and $b_{a} \dots b_{j} = P_{j + 1} / P_{a}$ with $m_{j} = h_{j + 1} s_{j}$ gives $m_{j} / (b_{a} \dots b_{j}) = h_{a} s_{j}$ , whose summation is the last identity. ◻

For estimates, it is useful to count by the odd part $3^{j} 5^{k}$ instead. Each such part below $2^{a + 1}$ has exactly one power-of-two multiple in $[2^{a}, 2^{a + 1})$ . This turns the same numerator into a two-dimensional count, with weights determined by the remaining odd-prime jumps.

Lemma 8.3 (the block numerator as a weighted lattice count). Put $λ_{3} = \log_{2} 3$ , $λ_{5} = \log_{2} 5$ and $θ_{p} = 1 / λ_{p}$ for $p = 3, 5$ . For $j, k \geq 0$ write $w_{j, k} = j λ_{3} + k λ_{5}$ and $t_{j, k} = {w_{j, k}}$ . The block numerator is

m_{a} = \sum_{\begin{matrix} j, k \geq 0 \\ w_{j, k} < a + 1 \end{matrix}} 3^{⌊ (a + 1) θ_{3} ⌋ - ⌊ (a + t_{j, k}) θ_{3} ⌋} 5^{⌊ (a + 1) θ_{5} ⌋ - ⌊ (a + t_{j, k}) θ_{5} ⌋} .

Every summand is in ${1, 3, 5, 15}$ . In particular, for $a \geq 0$ ,

(⌊ a / (2 λ_{3}) ⌋ + 1) (⌊ a / (2 λ_{5}) ⌋ + 1) \leq m_{a} \leq 15 (a + 1)^{2} .

Thus $m_{a} = Θ ((a + 1)^{2})$ and the numerator sequence is unbounded. These are integer numerators, not positional digits restricted to ${0, \dots, b_{a} - 1}$ .

Proof. For each pair with $w_{j, k} < a + 1$ , exactly one exponent $i = a - ⌊ w_{j, k} ⌋ \geq 0$ puts $2^{i} 3^{j} 5^{k} = 2^{a + t_{j, k}}$ in $[2^{a}, 2^{a + 1})$ . This pairs each point of the triangle with exactly one smooth integer in the block. Substituting in $P_{a + 1} / (2 H (2^{i} 3^{j} 5^{k}))$ cancels the power of $2$ and gives the displayed weight. Each remaining floor increment is zero or one since $0 < θ_{p} < 1$ . Finally $λ_{3}, λ_{5} > 1$ implies $j, k \leq a$ for every summation pair, proving the upper bound. For the lower bound, restrict to $0 \leq j \leq ⌊ a / (2 λ_{3}) ⌋$ and $0 \leq k \leq ⌊ a / (2 λ_{5}) ⌋$ . Then $w_{j, k} \leq a$ and each weight is at least one. ◻

The quadratic order comes from the size of the triangle, not from a polynomial formula for $m_{a}$ . As $a$ changes, new points enter and the weights of old points can change at floor crossings. A finite-difference argument must control both effects; polynomial growth alone cancels neither. Quadratic growth does not make this count a quadratic polynomial.

The recurrence also restricts how persistently a nonintegral tail can approach the integers. The following alternative uses only the integer coefficients and the bounds $2 \leq b_{a} \leq 30$ .

Proposition 8.4 (integer tails or repeated separation from the integers). For every integer $B \geq 1$ , either $B X_{a} \in Z$ for some $a \geq 0$ and every later $a$ , or for every $a_{0}$ there is $a \geq a_{0}$ with $| B X_{a} - z | \geq 1 / 31$ for every $z \in Z$ .

Proof. If the second alternative fails, there are $A$ and integers $z_{a}$ such that $e_{a} = B X_{a} - z_{a}$ satisfies $| e_{a} | < 1 / 31$ for all $a \geq A$ . The recurrence gives $z_{a + 1} - b_{a} z_{a} + B m_{a} = b_{a} e_{a} - e_{a + 1}$ . Since $B m_{a}$ is integral, the left side is an integer; the right side has absolute value less than $(30 + 1) / 31 = 1$ . Both sides therefore vanish, so $e_{a + 1} = b_{a} e_{a}$ . Hence $| e_{A + k} | \geq 2^{k} | e_{A} |$ for every $k$ , whereas $| e_{A + k} | < 1 / 31$ . Thus $e_{A} = 0$ , and the integer recurrence propagates integrality from $B X_{A}$ to every later $B X_{a}$ . ◻

The proposition does not determine which alternative holds for the tails of $S$ . For comparison, Erdős–Taylor [21] prove a countability statement for increasing integer sequences with bounded successive ratios; Fan [22] allows an unbounded, not necessarily increasing sequence, still with a uniform upper bound on successive ratios. For our divisibility chain the same error argument identifies the entire exceptional set, not just its cardinality. For each fixed integer $B \geq 1$ ,

{ξ \in R : dist (B h_{a} ξ, Z) ⟶ 0} = \frac{1}{B} Z [1 / 30], Z [1 / 30] = {m / 30^{t} : m \in Z, t \in N} .

Indeed, apply the proof above to $B h_{a} ξ$ , whose successive terms are related by multiplication by $b_{a}$ . Convergence of the distances forces $B h_{A} ξ \in Z$ for some $A \geq 1$ . Since $h_{A}$ has no prime factors outside $2, 3, 5$ , this gives membership in the right-hand side. Conversely, every fixed product of powers of $2, 3, 5$ eventually divides $h_{a}$ , so every member of the right-hand side gives integral values for all large $a$ .

Separation from the integers for one multiplier does not prove irrationality. For example, $ξ = 1 / 7$ satisfies $dist (h_{a} ξ, Z) \geq 1 / 7$ for every $a \geq 1$ , because $7 ∤ h_{a}$ ; with $B = 7$ the values $B h_{a} ξ$ are all integers. For $S$ , the denominator clearing in Theorem 9.3 below shows which multipliers matter: the integral alternative for the actual tails must be excluded for every $B \geq 1$ coprime to $30$ , not merely for $B = 1$ .

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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