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

Certificates for three and four primes

The saved computations apply the interval criterion in exact rational arithmetic to the following finite family. They accompany the ordinary proof of Remark 5.7 and have not been formalised. The exclusions in part (ii) concern these interval calculations; the ordinary torus argument also treats those sets.

Remark 5.8 (three and four primes up to $31$ ). Let $P$ be a set of primes, each at most $31$ .

If $| P | = 3$ , then $D_{P}$ is irrational.
If $| P | = 4$ and $P$ is not one of the $38$ sets listed below, then $D_{P}$ is irrational.

Proof. For each set, the program certificate.py of the programs for sums over distinct values builds the automaton, computes intervals $J (s)$ by $80$ rounds of the operation above from $[0, 1 / (p - 1)]$ , with endpoints rounded outwards to multiples of $2^{- 96}$ , and tests the disjointness of the sets $U_{τ}$ for $k = 1, 2, \dots$ . Every comparison is made in exact rational arithmetic, and each $U_{τ}$ is computed as a union of intervals merged where they meet. The test passes for each of the $165$ sets of three primes with some $k \leq 4$ , and for each set of four primes outside the list with some $k \leq 5$ . Remark 5.7 then gives irrationality. ◻

For three primes the least passing depth is $k = 1$ for $141$ sets and $k = 2$ for the $17$ sets ${2, 3, 11}$ , ${2, 3, 13}$ , ${2, 3, 17}$ , ${2, 3, 19}$ , ${2, 3, 23}$ , ${2, 3, 29}$ , ${2, 3, 31}$ , ${2, 5, 7}$ , ${2, 11, 17}$ , ${2, 13, 19}$ , ${2, 17, 19}$ , ${2, 19, 29}$ , ${2, 19, 31}$ , ${3, 13, 17}$ , ${3, 17, 23}$ , ${3, 23, 31}$ and ${3, 29, 31}$ . It is $k = 3$ for ${2, 3, 7}$ , ${2, 11, 13}$ , ${2, 17, 29}$ and ${2, 19, 23}$ , and $k = 4$ for ${2, 17, 31}$ , ${2, 23, 29}$ and ${2, 29, 31}$ . The smallest gap between sets $U_{τ}$ of different types, among all passing triples, is about $9.4 \cdot 10^{- 6}$ , for ${2, 17, 31}$ ; for ${2, 3, 5}$ it is $27983 / 963915 \approx 0.029$ at depth $1$ , much larger than the gap $1 / 900$ of the single interval used above. For four primes the least passing depth is $1, 2, 3, 4, 5$ for $184$ , $34$ , $28$ , $42$ and $4$ sets respectively, and the smallest gap is about $1.8 \cdot 10^{- 8}$ , for ${3, 7, 17, 19}$ . All these gaps are far above the rounding scale $2^{- 96}$ . The $38$ four-prime sets with no passing depth $k \leq 5$ are ${2, 3, 5, 7}$ , ${2, 3, 5, 11}$ , ${2, 3, 5, 13}$ , ${2, 3, 5, 17}$ , ${2, 3, 5, 19}$ , ${2, 3, 5, 23}$ , ${2, 3, 7, 11}$ , ${2, 3, 7, 13}$ , ${2, 3, 7, 17}$ , ${2, 3, 7, 23}$ , ${2, 3, 11, 13}$ , ${2, 3, 11, 17}$ , ${2, 3, 11, 19}$ , ${2, 3, 11, 29}$ , ${2, 3, 11, 31}$ , ${2, 3, 13, 19}$ , ${2, 3, 13, 29}$ , ${2, 3, 17, 19}$ , ${2, 3, 17, 29}$ , ${2, 3, 29, 31}$ , ${2, 5, 7, 11}$ , ${2, 5, 17, 29}$ , ${2, 5, 23, 29}$ , ${2, 5, 23, 31}$ , ${2, 11, 13, 19}$ , ${2, 11, 17, 23}$ , ${2, 11, 17, 29}$ , ${2, 11, 23, 31}$ , ${2, 13, 19, 31}$ , ${2, 19, 23, 31}$ , ${3, 5, 13, 17}$ , ${3, 5, 17, 19}$ , ${3, 5, 23, 31}$ , ${3, 7, 11, 13}$ , ${3, 19, 23, 29}$ , ${3, 19, 23, 31}$ , ${3, 19, 29, 31}$ and ${7, 13, 29, 31}$ . Twenty of them contain both $2$ and $3$ .

A second computation replaces the rounded intervals by the exact extremes $[L (s), U (s)]$ , found by policy iteration and checked by substitution into their defining equations. It returns the same least depth for all $495$ sets. A third check compares the automaton with the first $400$ actual blocks of every set: every actual word is a path, and every exact enclosure of an actual tail lies in the interval of its state. This comparison is a consistency check only. The whole triple computation takes about three seconds, and the four-prime computation about seventy seconds, on one laptop; the saved logs record every depth and gap.

For ${2, 3, 5, 7}$ the automaton has $18$ states and $16$ types, and the test fails through depth $7$ . Its tail intervals range from about $0.188$ to $0.785$ , so the largest exceeds three times the smallest, and the images of the empty block and of the block with one power of $3$ overlap. The largest overlap between two first types is always of this pair. For each depth $k$ the table gives the number of merged intervals in the sets $U_{τ}$ and the length of that largest overlap.

$k$	intervals	largest overlap
$1$	$16$	$2.4 \cdot 10^{- 2}$
$2$	$58$	$1.8 \cdot 10^{- 3}$
$3$	$386$	$1.0 \cdot 10^{- 4}$
$4$	$1413$	$4.6 \cdot 10^{- 5}$
$5$	$8395$	$7.6 \cdot 10^{- 7}$
$6$	$38494$	$2.4 \cdot 10^{- 7}$
$7$	$193905$	$1.8 \cdot 10^{- 8}$

At depth $7$ two further pairs of types still overlap, $(5, 7, 3)$ with $(7)$ by about $1.5 \cdot 10^{- 9}$ and $(3, 5, 7)$ with $(3, 7)$ by about $5.8 \cdot 10^{- 10}$ .

The persistent overlap is not a rounding error or merely slow convergence. An exact collision explains why no depth succeeds. The two-block words $(5, 7, 3), \emptyset$ and $(7), (3, 5)$ have the same map,

G_{(5, 7, 3)} \circ G_{\emptyset} (y) = G_{(7)} \circ G_{(3, 5)} (y) = \frac{103 + y}{420},

which is the collision $f (5, 7, 3, 2) = f (7, 2, 3, 5) = 51$ of Section 5.1 read in blocks: the letters are $5, 7, 3, 2, 2$ and $7, 2, 3, 5, 2$ , and $f = 103$ for both. From the states $(1, 2, 2)$ and $(1, 2, 3)$ , written as $(s_{3}, s_{5}, s_{7})$ , both words continue by the blocks $(3), (5, 7)$ and end in the state $(2, 1, 1)$ , so the four-block words have the same map and the same end state. Hence the sets $U_{(5, 7, 3)}$ and $U_{(7)}$ of Remark 5.7 meet at every depth, for any valid choice of intervals, and the criterion cannot decide ${2, 3, 5, 7}$ with this automaton. Both four-block words occur in the actual block sequence, $12$ and $40$ times among the first $20000$ blocks. Within this automaton, a tail value therefore need not determine its first block. The identity does not exhibit equal tails at two positions of the actual prime-power sequence: those tails have different later phases. It proves failure of this interval criterion. The proof of Remark 5.1 uses the isolated reversal of two actual jumps instead.

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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