trureturingGitHub MATHEMATICAL ATLAS

RESEARCH / Focused target

The least self-divisor exponent in OEIS A092028

Literature status: not recheckedRepository record: proved

RELEASED FOUNDATIONS / PROPOSED CONNECTION

Research connections

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Problem

The question

OEIS A092028, NAME (%N, verbatim):

a(n) is the smallest m > 1 such that m divides n^m-1

COMMENTS (%C, verbatim; Farideh Firoozbakht, Mar 26 2004):

Conjecture 1: All terms of this sequence are primes.

Conjecture 2: a(n) is the smallest prime factor of n-1 or for n>2, A092028(n) = A020639(n-1).

The literal formal claim is ∀ n > 2, a(n) = minFac(n − 1), where a(n) = sInf {m ∈ ℕ | 1 < m ∧ m ∣ n^m − 1}. In the natural numbers, sInf ∅ = 0; for every n > 2, the defining set is nonempty. Conjecture 1 is a corollary because minFac(n − 1) is prime, and is not formalized as a separate public theorem.

Motivation

Our foothold

The equality identifies the least solution of a self-referential divisibility condition for every natural input above two. It proves the unbounded OEIS conjecture rather than extending the entry's finite data.

Gap

Missing bridges

The searches recorded in preregistration issue #7526 and its probe on 2026-09-13 found that OEIS history entries #1 through #16, including revision #16, still carry both conjectures. The sibling plus-sign entry A092067 was proved by Neder in 2019; it supplied a search pattern only and is not imported. Exact searches returned MathOverflow 0, Crossref 0, and GitHub issues 0. arXiv, Semantic Scholar, and OpenAlex were rate-limited, while GitHub code search was unauthenticated. Those unavailable surfaces are ASSUMED-UNVERIFIED. The completed searches are bounded and establish no priority claim.

Route

Proposed approach

For the upper bound, let p = minFac(n − 1). The number p is prime and p ∣ n − 1, hence n^p ≡ 1 (mod p) and p belongs to the defining set. For the lower bound, take any m > 1 with m ∣ n^m − 1 and let q = minFac(m). The multiplicative order of (n : ZMod q) divides both m and q − 1. Minimality of q gives gcd(m, q − 1) = 1, so the order is one. Thus q ∣ n − 1, and minFac(n − 1) ≤ q ≤ m. Together the two bounds give the claimed equality.

Falsifier

What would falsify this route

One natural number n > 2 for which the least m > 1 dividing n^m − 1 differs from minFac(n − 1) would contradict the theorem.

Evidence

Evidence to collect

  • Lean module: D5/S3/Arith/Congruence/FiroozbakhtLeastSelfDivisorPowerMinusOne.lean.
  • Main theorem: firoozbakht_a092028.
  • Axioms: std3 (propext, Classical.choice, Quot.sound).
  • The orchestrator checked every n with 3 ≤ n ≤ 3000 and found zero mismatches.
  • The probe checked all 9998 cases with 3 ≤ n ≤ 10000 and found zero mismatches. For n = 3..12, the values are 2, 3, 2, 5, 2, 7, 2, 3, 2, 11.

Triage

Scope assessment

theorem. The formal result proves the equality for every natural n > 2.

ASSUMED-UNVERIFIED

Unverified assumptions

arXiv, Semantic Scholar, and OpenAlex were rate-limited, and GitHub code search was unauthenticated. The successful literature searches were bounded. No claim of exhaustive literature coverage or first-publication priority is made.

Dossier & anchor history

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