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RESEARCH / Focused target

Oudra's divisor-count formula for the Stechkin function

Literature status: not recheckedRepository record: proved

RELEASED FOUNDATIONS / PROPOSED CONNECTION

Research connections

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Problem

The question

OEIS A274010, NAME (%N, verbatim):

Boris Stechkin function: a(n) is the number of m with 2 <= m <= n and floor(n(m-1)/m) divisible by m-1.

Ridouane Oudra's FORMULA conjecture (%F, verbatim):

Conjecture: a(n) = tau(n) + tau(n-1) - 2, for n>=2. - Ridouane Oudra, Feb 28 2020

AUTHOR (%A, verbatim):

Robert Israel, Jun 06 2016

The formal definition counts the natural numbers m in the closed interval from two through n for which m-1 divides the natural quotient n*(m-1)/m. Natural division is the floor of the corresponding nonnegative rational quotient. For every n >= 2, the formal theorem states stechkinFunction n + 2 = n.divisors.card + (n-1).divisors.card. Both divisor finsets contain one, so their cardinalities sum to at least two; the subtraction-free equality is therefore equivalent to the quoted formula.

Motivation

Our foothold

The formula identifies a filtered divisibility count with the sum of two adjacent divisor counts. It gives a closed arithmetic description of every term of the sequence rather than a finite-prefix verification.

Gap

Missing bridges

The OEIS entry labels the formula as a conjecture and carries no settlement line. Its cited bibliography predates the formula, and the linked table gives values rather than a proof. Exact repository and pinned-library searches found no existing declaration of the definition, formula, source predicate, or divisibility equivalence used below.

Route

Proposed approach

Reindex the defining interval by k=m-1. For 1 <= k < n, Euclidean division shows that k divides n*k/(k+1) exactly when k+1 divides n or n-1. This identifies the filtered interval with the disjoint union of the divisors greater than one of those adjacent numbers. Taking cardinalities and restoring the omitted divisor one on both sides gives the formula.

Falsifier

What would falsify this route

Any natural number n >= 2 for which the filtered count plus two differs from the sum of the two divisor counts would refute the result. Agreement on any finite initial range would not establish the universal statement.

Evidence

Evidence to collect

  • Module: D5/S3/Arith/StechkinFunctionDivisorCount.lean.
  • stechkinFunction formalizes the quoted interval, quotient, divisibility predicate, and cardinality.
  • result proves the universal adjacent divisor-count equality.
  • The theorem's axiom closure is [propext, Classical.choice, Quot.sound].

Triage

Scope assessment

theorem; resolution proved for every natural number n >= 2.

ASSUMED-UNVERIFIED

Unverified assumptions

The source quotations, contributor attribution, absence of an OEIS settlement line, bibliography date, hyperlink description, and external-search readings are attributed evidence without independent verification. Historical openness beyond the checked sources remains unverified; no exhaustive literature or priority claim is made.

Dossier & anchor history

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