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

Bounding the tails and clearing a rational denominator

We sharpen the quadratic estimate by counting points of equal running LCM. As the cutoff increases, one prime exponent increases at each jump, so the sum of the exponents indexes the nonempty cells. At $2^{a}$ it is

n_{a} = a + ⌊ \log_{3} (2^{a}) ⌋ + ⌊ \log_{5} (2^{a}) ⌋ .

Counting the points in each cell and summing their geometric weights gives the polynomial

Q (n) = \frac{n^{2} + 8 n + 18}{9} .

Theorem 9.1 (a quadratic upper bound). For every $a \geq 0$ , $0 < X_{a} \leq Q (n_{a})$ .

Proof. Partition the smooth integers $x \geq 2^{a}$ by the exponent vector of their running LCM

(A, B, C) = (⌊ \log_{2} x ⌋, ⌊ \log_{3} x ⌋, ⌊ \log_{5} x ⌋) .

The cell of a vector is the interval $[λ, η)$ , where

λ = max (2^{A}, 3^{B}, 5^{C}), η = min (2^{A + 1}, 3^{B + 1}, 5^{C + 1}) .

Thus $η \leq 2^{A + 1} \leq 2 λ$ . By Lemma 3.5, fixing the exponents of $3$ and $5$ leaves at most one exponent of $2$ . Since $A \geq B \geq C$ , the cell contains at most

(B + 1) (C + 1) \leq \frac{(A + B + C + 3)^{2}}{9}

smooth points: writing $u = B + 1 \geq v = C + 1$ and using $A + 1 \geq u$ , the difference $(2 u + v)^{2} - 9 u v = (u - v) (4 u - v)$ is nonnegative. Unique factorisation identifies these exponent triples with distinct smooth integers.

As the cutoff increases all three prime exponents are nondecreasing, so two nonempty cells with the same exponent sum are the same cell, and there is at most one nonempty cell with each exponent sum. Every cell above $2^{a}$ has exponent sum at least $n_{a}$ , and a cell with sum $n_{a} + k$ has running LCM at least $P_{a} 2^{k}$ , since each of the $k$ extra prime factors is at least $2$ . Nonnegative summation over exponent sums, allowing empty cells, gives

X_{a} \leq \frac{1}{18} \sum_{k \geq 0} \frac{(n_{a} + k + 3)^{2}}{2^{k}} = \frac{n_{a}^{2} + 8 n_{a} + 18}{9},

the evaluation using the geometric moments $\sum 2^{- k} = 2$ , $\sum k 2^{- k} = 2$ and $\sum k^{2} 2^{- k} = 6$ . Positivity follows from the block at $2^{a}$ . ◻

Each extra prime factor costs at least two in the denominator; each cell contains at most quadratically many points. This is a geometric sum of elementary projection counts, not an asymptotic smooth-number estimate or a Hecke–Mahler argument.

The quadratic order is also necessary. The positive next tail and the recurrence give

X_{a} = \frac{m_{a} + X_{a + 1}}{b_{a}} > \frac{m_{a}}{b_{a}} \geq \frac{m_{a}}{30} .

The lower bound of Lemma 8.3, together with $n_{a} \leq 3 a$ , therefore gives $X_{a} = Θ ((a + 1)^{2})$ . In particular, the normalised tails are not uniformly bounded. This order estimate does not assert an asymptotic constant or optimality of $Q$ .

The bound is used at the endpoint of a segment, where the natural index is the positive prime-power jump count strictly below the cutoff. Put

\begin{matrix} (17) & j_{a} = # {p^{e} < 2^{a} : p \in {2, 3, 5}, e \geq 1} = n_{a} - 1 (a \geq 1), \end{matrix}

the equality holding for $a \geq 1$ because the powers of $2$ below $2^{a}$ number $a - 1$ while the powers of $3$ and of $5$ below $2^{a}$ number $⌊ \log_{3} 2^{a} ⌋$ and $⌊ \log_{5} 2^{a} ⌋$ ; at $a = 0$ the count is $j_{0} = 0$ and $n_{0} - 1 = - 1$ . Substituting $n_{a} = j_{a} + 1$ into $Q$ gives the integer bound used throughout the rest of the note:

\begin{matrix} (18) & K (B, a) = ⌊ B Q (n_{a}) ⌋, K (B, a) = ⌊ \frac{B (j_{a}^{2} + 10 j_{a} + 27)}{9} ⌋ (a \geq 1) . \end{matrix}

The strict cutoff in (17) is $2^{a}$ . There are two distinct uses of the bound: $B X_{a} \leq B Q (n_{a})$ holds for the actual real tails, whereas $B X_{a} \leq K (B, a)$ follows only when $B X_{a}$ is integral. For example, $Q (n_{3}) = 83 / 9$ gives the integer bound $K (1, 3) = 9$ , not an upper bound of $9$ for an arbitrary real quantity below $83 / 9$ . The short paper instead uses $K_{s h o r t} (B, a) = 90 B (a + 1)^{2}$ , valid without integrality. Here $K$ always means (18); either bound may be used in the residue contradiction.

Lemma 9.2 (finite denominator clearing). For all integers $0 \leq u \leq b$ the sum over the segment $h_{b} \sum_{a = u}^{b - 1} s_{a}$ is a natural number. If $S = N / D$ with $N \in Z$ and $D \in N_{> 0}$ , then $D X_{a} \in Z$ for every $a \geq 1$ , and there are indices $1 \leq i < j \leq D + 1$ for which $X_{i} - X_{j} \in Z$ .

Proof. An empty segment has mass zero. Otherwise $b \geq 1$ , and every integer $x < 2^{b}$ has exponent of $2$ at most $b - 1$ while its other prime exponents are at most those at $2^{b}$ , so $H (x) ∣ P_{b} / 2 = h_{b}$ and every term of the finite segment clears at $h_{b}$ . Since $h_{a} \sum_{u < a} s_{u}$ is an integer,

D X_{a} = h_{a} N - D h_{a} \sum_{u < a} s_{u} \in Z .

Among $D + 1$ of these integers two share a residue modulo $D$ , and the corresponding tails differ by an integer. ◻

The strict upper endpoint is what permits division by two. A cutoff including $2^{b}$ would not clear its term at the normalising factor $h_{b} = P_{b} / 2$ . The pigeonhole argument identifies two tails modulo $Z$ , not two equal tails. Since $X_{a}$ is unbounded, the lattice condition does not force the exact repetition used in the distinct-value proof.

Theorem 9.3 (rationality gives positive integer tails). Suppose $S = N / D$ with $N \in Z$ , $D \in N_{> 0}$ , and write

D = 2^{u} 3^{v} 5^{w} B, u, v, w \in N, B \in N_{> 0}, gcd (B, 30) = 1, a_{D} = u + 1 + 2 v + 3 w .

Then for every $a \geq a_{D}$ the number $d_{a} = B X_{a}$ is a positive integer and

d_{a + 1} = b_{a} d_{a} - B m_{a}, 1 \leq d_{a} \leq K (B, a) \leq 90 B (a + 1)^{2} .

Proof. Let $M = 2^{u} 3^{v} 5^{w}$ . For $a \geq a_{D}$ we have $2^{a} \geq 2^{u + 1}$ , $2^{a} \geq 3^{v}$ and $2^{a} \geq 5^{w}$ , using $3 < 2^{2}$ and $5 < 2^{3}$ ; hence $M ∣ h_{a}$ . By Lemma 9.2, $X_{a}$ differs from $h_{a} N / D$ by an integer. Since $M ∣ h_{a}$ and $D = M B$ , multiplying by $B$ shows that $B X_{a}$ is an integer. Positivity and the recurrence come from (16), and the upper bound is Theorem 9.1 with the floor taken, since $d_{a}$ is an integer at most $B Q (n_{a})$ . Finally $n_{a} \leq 3 a$ , so $Q (n_{a}) \leq a^{2} + \frac{8}{3} a + 2 \leq 90 (a + 1)^{2}$ . ◻

The factorisation restricts no possible denominator: remove its powers of $2$ , $3$ and $5$ , leaving $B$ coprime to $30$ . These parts have different roles. The removed part controls the onset of integrality; $B$ scales the integer tails and their bound. The sufficient $a_{D}$ need not be the first clearing index. For the exact answer, write $den (x)$ for the positive reduced denominator of a rational $x$ .

Proposition 9.4 (exact denominators and minimal clearing). Suppose $S = N / (M B)$ is in lowest terms, with $M = 2^{u} 3^{v} 5^{w}$ and $gcd (B, 30) = 1$ . For every $a \geq 1$ ,

den (X_{a}) = \frac{M B}{gcd (M, h_{a})}, den (B X_{a}) = \frac{M}{gcd (M, h_{a})} .

Hence the first integral reduced tail occurs at

a_{*} = min {a \geq 1 : 2^{a} \geq max (2^{u + 1}, 3^{v}, 5^{w})} \leq a_{D},

and $B X_{a}$ is integral exactly for $a \geq a_{*}$ . This onset is computable by integer powers; it is not an estimate obtained by rounding logarithms.

Proof. The finite prefix clears at $h_{a}$ , so Lemma 9.2 gives $X_{a} - h_{a} N / (M B) \in Z$ . Since $gcd (N, M B) = 1$ and every prime factor of $h_{a}$ is in ${2, 3, 5}$ , cancellation removes exactly $gcd (M, h_{a})$ from the denominator. Multiplication by $B$ gives the second formula. Now $M ∣ h_{a}$ means $a - 1 \geq u$ , $⌊ \log_{3} 2^{a} ⌋ \geq v$ and $⌊ \log_{5} 2^{a} ⌋ \geq w$ , precisely the three integer inequalities. All three persist when $a$ increases, giving the first and every later integral reduced tail. The earlier bounds $3 < 4$ and $5 < 8$ give $a_{*} \leq a_{D}$ . ◻

For example, a hypothetical reduced denominator $2^{3} 3^{2} 5 \cdot 7$ would require $2^{a} \geq max (16, 9, 5)$ , so $a_{*} = 4$ rather than the sufficient $a_{D} = 11$ . Lowest terms matter for this exact answer: unreduced factors could cancel against the numerator sooner.

The recurrence preserves integrality forward. Its homogeneous equation also determines how fast two distinct solutions separate.

Proposition 9.5 (propagation of integrality and uniqueness of a small solution). For every $a$ , $X_{a} = (m_{a} + X_{a + 1}) / b_{a} > 0$ , and if $X_{a} \in Z$ then $X_{n} \in Z$ for every $n \geq a$ . Moreover, fix $A$ , a positive width function $w$ with $w (A + k) / 8^{k} \to 0$ , and a real sequence $(y_{n})_{n \geq A}$ satisfying $y_{n + 1} = b_{n} y_{n} - m_{n}$ . If $y_{n}$ and $X_{n}$ both lie in $(m_{n} / b_{n}, m_{n} / b_{n} + w (n)]$ for every $n \geq A$ , then $y_{A} = X_{A}$ .

Proof. The identity is the recurrence solved for $X_{a}$ , and positivity holds because every block contains its dyadic left endpoint. Integer coefficients preserve integrality at every later step. For the last assertion, $y_{A + k} - X_{A + k} = (P_{A + k} / P_{A}) (y_{A} - X_{A})$ . The dyadic running LCM bounds from Section 8 give $P_{A + k} / P_{A} > 8^{k} / 15$ , whereas the common interval bounds the absolute difference by $w (A + k)$ . Thus $| y_{A} - X_{A} | < 15 w (A + k) / 8^{k} \to 0$ . ◻

The width condition allows every positive polynomial width and widths $ρ^{n}$ with $1 < ρ < 8$ , but not $8^{n}$ itself. Both solutions must lie in the stated intervals; the recurrence alone does not supply that condition. More generally, fix an integer $B \geq 1$ . Every real solution of $y_{a + 1} = b_{a} y_{a} - B m_{a}$ starting at an index $A$ satisfies

y_{a} = B X_{a} + \frac{P_{a}}{P_{A}} (y_{A} - B X_{A}) (a \geq A) .

Subtracting the recurrence for $B X_{a}$ proves the identity. Since $X_{a} = O ((a + 1)^{2})$ and $P_{a} = Θ (8^{a})$ , the unique solution with $y_{a} = o (8^{a})$ is $y_{a} = B X_{a}$ ; every other solution has $| y_{a} | = Θ (8^{a})$ . This classifies growth; integrality of $B X_{a}$ remains a separate question. The same formula explains a numerical precaution: a starting error $ε$ is multiplied by $P_{a} / P_{A}$ under forward iteration. A decimal approximation propagated forwards is therefore not a certificate of the tail floors. The digit calculation in Section 12 uses rational intervals with an explicit infinite-tail bound instead.

A bound using the order of the jumps

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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