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Erdős ProblemsCombinatorics

Erdős Problem 282: Sq

Graham has also shown that xx is the sum of distinct unit fractions with square denominators if and only if x∈[0,π2/6−1)∪[1,π2/6)x\in [0,\pi^2/6-1)\cup [1,\pi^2/6). Does the greedy algorithm for this always terminate? Erdős and Graham believe not - indeed, perhaps it fails t...

CombinatoricsAMS 5

Mathematical statement

Graham has also shown that xx is the sum of distinct unit fractions with square denominators if and only if x∈[0,π2/6−1)∪[1,π2/6)x\in [0,\pi^2/6-1)\cup [1,\pi^2/6). Does the greedy algorithm for this always terminate? Erdős and Graham believe not - indeed, perhaps it fails to terminate almost always.

Statement source: Erdős Problems statement material

Statement terms: Source-specific

Source-specific terms. Therefore does not assert reuse rights beyond attributed display.

Statement artifacts, not proofs

These records expose exact Lean propositions and statement-only wrappers. Defining a proposition does not supply a proof of it. A placeholder-bearing target also contains no proof. Elaboration checks syntax and types; it does not certify that a formalization perfectly captures every nuance of the informal problem.

Pinned Lean formulation 1

erdos_282.variants.sq

Canonical source
Complete statement target, proof intentionally absentLean 4
theorem erdos_282.variants.sq :    answer(sorry) ↔ ∀ x : ℚ, (x : ℝ) ∈ Set.Ico 0 (π ^ 2 / 6 - 1) ∪ Set.Ico 1 (π ^ 2 / 6) →      greedyUnitFractionRem { n | IsSquare n } x =ᶠ[atTop] 0 := by  sorry
Statement source
Formal Conjectures
Lean version
v4.27.0
Placeholder
Present; no proof artifact
Source evidence
Pinned source index
Fidelity review
Community formulation

References

Pinned Formal Conjectures sourceformalizationerdosproblems.com/282background