Erdős Problem #282 — open problem in combinatorics

Erdős Problem #282

Let A⊆ ℕ be an infinite set and consider the following greedy algorithm for a rational x∈ (0,1): choose the minimal n∈ A not used so far such that n≥ 1/x and repeat with x replaced by x-1/n.

From the catalogue. Imported from The Formal Conjectures Authors (Google DeepMind and contributors) (Apache-2.0) — original. Nobody has started on it here yet: tasks are created as soon as someone asks for one or submits a claim. A Lean proof is checked against the statement below by the Lean kernel; a curator confirms before the problem counts as resolved.

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Cite

@misc{cairn-erdos-282,
  title        = {Erdős Problem #282},
  author       = {{Cairn Commons contributors}},
  howpublished = {\url{https://cairn-commons.com/problems/erdos-282}},
  year         = {2026},
  note         = {Open problem on Cairn Commons, CC BY 4.0. Accessed 2026-09-28}
}

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Claims

Verified

Disputed

Refuted

On the literature board

The problem

The question

Let $A \subseteq N$ be an infinite set and consider the following greedy algorithm for a rational $x \in (0, 1)$ : choose the minimal $n \in A$ not used so far such that $n \geq 1/ x$ and repeat with $x$ replaced by $x - \frac{1}{n}$ . If this terminates after finitely many steps then this produces a representation of $x$ as the sum of distinct unit fractions with denominators from $A$ .

Does this process always terminate if $x$ has odd denominator and $A$ is the set of odd numbers?

Formal statement (Lean 4)

From Formal Conjectures, module FormalConjectures.ErdosProblems.«282».

theorem erdos_282 {x : ℚ} (hx : x ∈ Set.Ioo 0 1) (hx_den : Odd x.den) :
    greedyUnitFractionRem { n | Odd n } x =ᶠ[atTop] 0

What counts as progress

A Lean proof of the pinned statement (or of its negation, for a yes/no question) — checked by the Lean kernel against the upstream statement; a curator confirms before the problem is marked resolved.
Partial results: special cases, weaker bounds, reductions — as verified claims.
Computations and numerical evidence with published code (reproducible).
Literature: the problem may have been solved or partly solved already — check erdosproblems.com/282. Report it as a literature claim.
A precise flaw in the formal statement (a misformalisation) — report it upstream too.

Variants

erdos_282.variants.general — More generally, for which pairs x and A does this process terminate?
erdos_282.variants.graham — Graham has shown that m/n is the sum of distinct unit fractions with denominators ≡ apmodd if and only if (n/(n,a,d),d/(a,d))=1.
erdos_282.variants.sq — Graham has also shown that x is the sum of distinct unit fractions with square denominators if and only if x∈ [0,π^2/6-1)∪ [1,π^2/6).

References

erdosproblems.com/282

Source and licence

Imported from Formal Conjectures (Erdős problems), commit e6d1743831c2. Statements and descriptions © The Formal Conjectures Authors, Apache License 2.0; reformatted for this page.