bibkey: lowell2013a225004 authors: J. Lowell year: 2013 title: “OEIS A225004, a(n) is the largest multiple of n with fewer than twice as many divisors as n” doi: null url: https://oeis.org/A225004 claim: “Conjecture: a(n) is always a divisor of n^2.” strata_touched:

• D5/S3/Factorization/DivisorDoublingMultipleDividesSquare license: citation-only triage: anchor

OEIS A225004

The entry records, for each n, the largest multiple of n whose divisor count is smaller than twice that of n. Its terms begin 1, 4, 9, 16, 25, 18, 49, 64, 81, 50, 121, 48, 169, ….

Verified locator

• URL: https://oeis.org/A225004
• Locator: NAME, “a(n) is the largest multiple of n with fewer than twice as many divisors as n”; COMMENTS, “Conjecture: a(n) is always a divisor of n^2.”
• Revision read: #24, Mar 17 2022. The comment stands and carries no answer.
• The entry’s only cross-reference is A000005, the divisor-counting function.

Reading of the statement

The entry’s own worked example fixes the reading: “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.”

So a multiple m of n is admissible when τ(m) < 2 τ(n), and a(n) is the largest admissible multiple. The conjecture asks whether a(n) always divides n^2.

Scope of the recorded answer

Every admissible multiple divides n^2, not only the largest, and one line of exponent bookkeeping is the whole reason.

Write f for the exponent in the prime factorisation, so τ(k) = ∏ (f_k(p) + 1) over the primes of k. Let n ∣ m with both positive, so f_n ≤ f_m pointwise and the primes of n are among those of m. Suppose some prime q had f_m(q) > 2 f_n(q).

• If q is not a prime of n, then q is a prime of m outside the primes of n. Splitting the product for τ(m) at that boundary, the part over the primes of n is at least τ(n) and the remaining part is at least 2, the factor at q alone being at least two.
• If q is a prime of n, then f_m(q) + 1 ≥ 2 f_n(q) + 2 = 2 (f_n(q) + 1). Pulling q out of the product over the primes of n, that single factor supplies the two and every other factor is at least the corresponding factor of τ(n).

Either way τ(m) ≥ 2 τ(n), contradicting admissibility. So f_m(p) ≤ 2 f_n(p) for every prime, which is exactly m ∣ n^2.

The same argument shows the admissible multiples are finitely many, all of them being divisors of n^2, so the largest one the entry names exists. The prime powers are the extreme case: for n = p^a admissibility reads b + 1 < 2(a+1), so b ≤ 2a and a(p^a) = p^{2a} = n^2.

Bounded prior-resolution evidence

The entry was read in full at revision #24 and records no proof and no reference to one; the comment still begins with the word Conjecture. Its only cross-reference, A000005, carries no proof of it. Citation indices were not exhaustively reachable, so this is a bounded negative finding and no worldwide priority claim is made.