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

Separation of the three-prime kernel

We test an exact kernel identity: can finitely many products separate one exponent from the other two? A single binary carry rules this out for three primes. Diagonal rescaling leaves threshold columns; independent row and column choices produce staircase minors of every order. This says nothing yet about identities after summation or approximation, which we treat separately. The initial two-generator identity is valid for real generators; distinct primes enter the later density argument.

Proposition 7.1 (two generators separate). For real $p, q > 1$ , with $L_{p, q} (t) = p^{⌊ \log_{p} t ⌋} q^{⌊ \log_{q} t ⌋}$ as above, and all integers $i, j \geq 0$ , the two-prime kernel $K_{2} (i, j) = 1 / L_{p, q} (p^{i} q^{j})$ is the outer product

K_{2} (i, j) = (p^{i} q^{⌊ \log_{q} p^{i} ⌋})^{- 1} (p^{⌊ \log_{p} q^{j} ⌋} q^{j})^{- 1},

so every two-by-two minor of $K_{2}$ vanishes.

Proof. Since $i$ and $j$ are integers, $⌊ \log_{p} (p^{i} q^{j}) ⌋ = i + ⌊ \log_{p} q^{j} ⌋$ and $⌊ \log_{q} (p^{i} q^{j}) ⌋ = j + ⌊ \log_{q} p^{i} ⌋$ . Hence $L_{p, q} (p^{i} q^{j})$ is the product of $p^{i} q^{⌊ \log_{q} p^{i} ⌋}$ , which depends on $i$ alone, and $p^{⌊ \log_{p} q^{j} ⌋} q^{j}$ , which depends on $j$ alone. A matrix whose entries are a product of a row function and a column function has vanishing two-by-two minors. For distinct primes $p, q$ the product $L_{p, q}$ is the running least common multiple by the argument of Theorem 3.1. ◻

At three primes the smallest rectangle already disproves a single product. A factorisation $f (i) g (j) h (k)$ would force $K (0, 0, 0) K (1, 1, 0) = K (1, 0, 0) K (0, 1, 0)$ .

Proposition 7.2 (non-separability at ${2, 3, 5}$ ). With $(p, q, r) = (2, 3, 5)$ ,

det (\begin{matrix} K (0, 0, 0) & K (0, 1, 0) \\ K (1, 0, 0) & K (1, 1, 0) \end{matrix}) = det (\begin{matrix} 1 & 1 / 6 \\ 1 / 2 & 1 / 60 \end{matrix}) = - \frac{1}{15} \neq 0 .

Proof. The four values are computed from $H (1) = 1$ , $H (2) = 2$ , $H (3) = 2 \cdot 3 = 6$ and $H (6) = 4 \cdot 3 \cdot 5 = 60$ , so the determinant is $1 / 60 - 1 / 12 = - 1 / 15$ . ◻

Theorem 7.3 (no finite separation of the kernel). Let $p, q, r$ be primes with $p \neq q$ , $p \neq r$ and $q \neq r$ . For every $n \geq 1$ there are injective maps $I, J : {0, \dots, n - 1} \to N$ such that, for every $k \geq 0$ ,

det (K (I (a), J (b), k))_{0 \leq a, b < n} \neq 0 .

Consequently, for no finite $d$ do there exist rational-valued functions $f_{ℓ} : N \to Q$ and $G_{ℓ} : N^{2} \to Q$ , $0 \leq ℓ < d$ , satisfying $K (i, j, k) = \sum_{ℓ < d} f_{ℓ} (i) G_{ℓ} (j, k)$ for all $i, j, k$ .

Proof. For fixed $k$ , the $p$ - and $q$ -exponents still split into row and column terms. The $r$ -exponent also includes the carry from adding two fractional parts. To isolate it, put $α = \log_{r} p$ , $β = \log_{r} q$ , $x_{i} = {i α}$ and $y_{j} = {j β}$ . Write $v_{p} (N)$ for the exponent of the prime $p$ in the positive integer $N$ . The three prime exponents are

\begin{aligned} v_{p} (H (p^{i} q^{j} r^{k})) & = i + ⌊ j \log_{p} q + k \log_{p} r ⌋, \\ v_{q} (H (p^{i} q^{j} r^{k})) & = j + ⌊ i \log_{q} p + k \log_{q} r ⌋, \\ v_{r} (H (p^{i} q^{j} r^{k})) & = k + ⌊ i α ⌋ + ⌊ j β ⌋ + ⌊ x_{i} + y_{j} ⌋ . \end{aligned}

Every term except the last floor depends on $i$ and $k$ alone or on $j$ and $k$ alone, so with positive rational $R_{i} (k)$ and $C_{j} (k)$ ,

\begin{matrix} (10) & K (i, j, k) = R_{i} (k) C_{j} (k) t^{⌊ x_{i} + y_{j} ⌋}, t = r^{- 1} . \end{matrix}

The remaining matrix is independent of $k$ . Since $x_{i} + y_{j} \in [0, 2)$ , its entry is $1$ when there is no carry and $t$ when there is one.

Each of $α$ and $β$ is irrational, since a rational value would give an equality of positive powers of distinct primes, so each fractional-part orbit is dense in $[0, 1]$ and each is injective. The row and column indices are chosen independently, so no density theorem for a single orbit of pairs is required. First choose indices with $0 < x_{I (0)} < \dots < x_{I (n - 1)} < 1$ . We want the columns to cross their thresholds at successive selected rows, so that consecutive row differences will leave a triangular matrix. Write $s_{b} = 1 - y_{J (b)}$ , and use density of the second orbit to choose

s_{0} \in (0, x_{I (0)}), s_{b} \in (x_{I (b - 1)}, x_{I (b)}) (1 \leq b < n) .

The intervals are disjoint, so the $J (b)$ are distinct. Their strict endpoints ensure that no selected entry lies on a threshold, and $x_{I (a)} + y_{J (b)} \geq 1$ holds exactly when $b \leq a$ . Dividing row $a$ by $R_{I (a)} (k)$ and column $b$ by $C_{J (b)} (k)$ therefore leaves

C_{n} (t) = (\begin{matrix} t & 1 & 1 & \dots & 1 \\ t & t & 1 & \dots & 1 \\ ⋮ & ⋮ & ⋱ & ⋱ & ⋮ \\ t & t & \dots & t & 1 \\ t & t & \dots & t & t \end{matrix}), det C_{n} (t) = t (t - 1)^{n - 1} \neq 0,

the determinant following by subtracting from each row its predecessor, working upwards from the last. Before the division, the determinant equals

det C_{n} (t) \prod_{a < n} R_{I (a)} (k) \prod_{b < n} C_{J (b)} (k),

which is nonzero for every $k$ ; this also explains why the same indices work in every layer.

For the last assertion, suppose that a representation with $d$ summands exists and fix $k$ . On the selected rows $I (0), \dots, I (d)$ and columns $J (0), \dots, J (d)$ the resulting matrix factors as

(f_{ℓ} (I (a)))_{0 \leq a \leq d, ℓ < d} (G_{ℓ} (J (b), k))_{ℓ < d, 0 \leq b \leq d} .

It has rank at most $d$ and hence zero determinant, contradicting the nonzero minor of order $d + 1$ . ◻

Index selection is essential. In the $k = 0$ layer of the ${2, 3, 5}$ kernel, the leading $4 \times 4$ minor is singular because $K (3, j, 0) = K (0, j, 0) / 120$ for $0 \leq j \leq 3$ . This is not a proportionality of the full rows: at the next column,

K (3, 4, 0) - \frac{K (0, 4, 0)}{120} = - \frac{1}{19440000} .

A singular leading minor therefore does not settle the rank.

The selected minors put thresholds at distinct cuts among the row phases. A general sample need not do so: several columns can have the same cut, and two can be constant. Counting the distinct cuts gives its rank.

Proposition 7.4 (rank of threshold columns). Let $m \geq 1$ and let $c$ lie in a field with $c \neq 0, 1$ . For $0 \leq h \leq m$ , let $v_{h}$ be the length- $m$ column whose first $h$ entries are $1$ and whose remaining entries are $c$ . If the distinct columns of a matrix are the $v_{h}$ with $h$ in a nonempty set $E \subseteq {0, \dots, m}$ , then its rank is

| E | - 1_{{0, m} \subseteq E} .

Proof. Write $E = {h_{1} < \dots < h_{s}}$ . The $s - 1$ consecutive differences are

v_{h_{a + 1}} - v_{h_{a}} = (1 - c) 1_{{h_{a}, \dots, h_{a + 1} - 1}},

so they are linearly independent because their nonempty supports are disjoint. If $h_{1} > 0$ or $h_{s} < m$ , those supports miss a coordinate on which $v_{h_{1}}$ is nonzero, and adjoining $v_{h_{1}}$ gives rank $s$ . If $h_{1} = 0$ and $h_{s} = m$ , the differences sum to $(1 - c) 1$ while $v_{0} = c 1$ , so $v_{0}$ is already in their span and the rank is $s - 1$ . ◻

The restrictions $c \neq 0, 1$ exclude the zero column at $c = 0$ and the collapse of all columns at $c = 1$ . They hold automatically at $c = 1 / r$ over $Q$ . To apply the proposition, fix $k$ and sort the chosen row phases $x_{i}$ . Each normalised column is a threshold column $v_{h}$ , where $h$ is the number of sampled phases strictly below $1 - y_{j}$ . Repeated threshold positions give proportional columns before column normalisation. The row and column factors in (10) are nonzero, so the formula gives the exact rank of every nonempty rectangular sample within this layer, independently of $k$ . Repeated rows or columns do not change rank.

This also gives an exact algorithm. Order the sampled rows by the rational numbers $p^{i} / r^{⌊ \log_{r} p^{i} ⌋}$ . For column $j$ , count those strictly below $r^{⌊ \log_{r} q^{j} ⌋ + 1} / q^{j}$ , then apply the endpoint correction to the set of resulting counts. Indeed, raising $x_{i} < 1 - y_{j}$ to base $r$ gives exactly this rational comparison, and $x_{i} + y_{j} = 1$ belongs to the $c$ side of the threshold. The integer logarithms are found by comparing powers, so the whole calculation uses integer arithmetic rather than numerical logarithms.

The fixed-layer restriction is essential. With $(p, q, r) = (2, 3, 5)$ , rows $i = 0, 1$ and columns indexed by $(j, k) = (0, 0), (0, 1)$ give

(\begin{matrix} K (0, 0, 0) & K (0, 0, 1) \\ K (1, 0, 0) & K (1, 0, 1) \end{matrix}) = (\begin{matrix} 1 & 1 / 60 \\ 1 / 2 & 1 / 360 \end{matrix}), det = - \frac{1}{180} .

Both columns have the same phase $y_{0} = 0$ , yet their rank is $2$ , not $1$ . When $k$ varies with the column, the row factor also varies with that column and cannot be removed by one common diagonal rescaling. The threshold formula is not a rank formula for such mixed-layer samples.

Scope of the rank theorem.

Smith factors of the selected staircase

Uniform approximation of the rescaled kernel

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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