slug: divisor-doubling-multiple-divides-square bibkey: lowell2013a225004 doi: null url: https://oeis.org/A225004 triage: theorem motivation_gids:

• D5/S3/Factorization/DivisorDoublingMultipleDividesSquare.result

A Multiple with Fewer than Twice as Many Divisors Divides the Square

Problem

OEIS A225004, at revision #24 of Mar 17 2022, by J. Lowell, Apr 23 2013:

a(n) is the largest multiple of n with fewer than twice as many divisors as n.

Conjecture: a(n) is always a divisor of n^2.

The entry’s worked example: “a(6) = 18 because 6 has 4 divisors and term must have fewer than 8 divisors. Only 6, 12, and 18 are multiples of 6 with fewer than 8 divisors.”

Motivation

The frozen theorem D5/S3/Factorization/DivisorDoublingMultipleDividesSquare.result settles the comment, in the stronger form that every admissible multiple divides n^2.

Gap

Issue 9513 records the screen carried out before the work. A225004 appears nowhere under Problems/, D5/ or docs/ in this repository. It appears as one note-only row in the 2026-09-10 triage record in Library/Words/, where the same exponent argument was written out and the row was set aside for high bind risk. That disposition is reversed here: the risk named there is about the proof shape, which is reported as measured, and is not by itself a reason to leave a named external conjecture unjudged.

The entry’s only cross-reference is A000005, which carries no proof of this. Citation indices were not exhaustively reachable, so this is a bounded negative finding.

Route

Write f for the exponent in the prime factorisation, so that τ(k) = ∏ (f_k(p) + 1) over the primes of k. Divisibility by n^2 is equivalent to f_m ≤ 2 f_n pointwise, so the statement is a bound on one exponent at a time.

Let n ∣ m with both positive, giving f_n ≤ f_m pointwise and the primes of n among those of m. Suppose some prime q had f_m(q) > 2 f_n(q).

q is not a prime of n. Then q is a prime of m lying outside the primes of n. Split the product for τ(m) into the part over the primes of n and the rest. The first part is at least τ(n), factorwise. The second contains the factor at q, which is at least two, and every other factor in it is at least one.

q is a prime of n. Then f_m(q) + 1 ≥ 2 f_n(q) + 2 = 2 (f_n(q) + 1). Pull q out of the product over the primes of n: that one factor supplies the two, and each remaining factor is at least the corresponding factor of τ(n). The part of τ(m) outside the primes of n is at least one.

Either way τ(m) ≥ 2 τ(n), contradicting τ(m) < 2 τ(n).

Why the delivered statement is stronger than the sentence, deliberately. The entry’s a(n) is the largest admissible multiple. Writing a(n) down needs a bound on the search, and the only natural bound is n^2 — which is the conclusion, so defining it that way would presuppose what is to be proved. The statement delivered says every admissible multiple divides n^2. It contains the conjecture, since a(n) is one such multiple, and it establishes in addition that the admissible multiples are finitely many, so the maximum the entry names exists at all.

Falsifier

A reading in which the divisor count is compared with 2 τ(n) non-strictly would change the statement: with τ(m) ≤ 2 τ(n) allowed, m = p^{2a+1} for n = p^a has τ(m) = 2a + 2 = 2 τ(n) and does not divide n^2. The entry’s wording is “fewer than”, and its worked example at n = 6 lists exactly the multiples with fewer than eight divisors, so the strict reading is the entry’s.

Evidence

A divisor-count sieve to 200000 was used to test the delivered statement directly: for every n ≤ 2000, every multiple of n up to 200000 with fewer than 2 τ(n) divisors was checked to divide n^2. No violation. The largest admissible multiple was computed for each such n and divides n^2 in every case; those values reproduce the entry’s data line term by term over all 53 published terms, beginning 1, 4, 9, 16, 25, 18, 49, 64, 81, 50, 121, 48, and the worked example at n = 6 comes out as 18. Of the n ≤ 2000, exactly 106 have a(n) = n^2; the prime powers are among them, as the argument predicts.

Triage

theorem; Tier 1 named external open question, preregistered in issue 9513 before the work. The computational use is none: no declaration is a bounded enumeration, a checker, a numeric reduction or a certified instance, and the delivered statement is universally quantified over every n and every m.

ASSUMED-UNVERIFIED

The literature screen is bounded. The entry was read in full at revision #24 and its cross-reference was checked. Citation indices and printed sources were not exhaustively reachable, so no worldwide priority claim is made. The weight is stated plainly: the argument is elementary, and what is settled is that the sentence sat on the entry for thirteen years without being judged.