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Erdős problem 282

Let A⊆NA\subseteq \mathbb{N} be an infinite set and consider the following greedy algorithm for a rational x∈(0,1)x\in (0,1): choose the minimal n∈An\in A such that n≥1/xn\geq 1/x and repeat with xx replaced by x−1nx-\frac{1}{n}. If this terminates after finitely many steps then this produces a representation of xx as the sum of distinct unit fractions with denominators from AA.

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8 retained statements36c6d0fef71a

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FormalConjectures/ErdosProblems/

282.lean

Retained formal statement1 of 8

Let A⊆NA\subseteq \mathbb{N} be an infinite set and consider the following greedy algorithm for a rational x∈(0,1)x\in (0,1): choose the minimal n∈An\in A such that n≥1/xn\geq 1/x and repeat with xx replaced by x−1nx-\frac{1}{n}. If this terminates after finitely many steps then this produces a representation of xx as the sum of distinct unit fractions with denominators from AA.

Does this process always terminate if xx has odd denominator and AA is the set of odd numbers?

FormalConjectures/ErdosProblems/282.leanErdos282.erdos_2821 lineExact file
∀ {x : ℚ}, x ∈ Set.Ioo 0 1 → Odd x.den → Erdos282.greedyUnitFractionRem {n | Odd n} x =ᶠ[Filter.atTop] 0
OpenStatement only, no proof