Vela

Let be an infinite set and consider the following greedy algorithm for a rational : choose the minimal such that and repeat with replaced by . If this terminates after finitely many steps then this produces a representation of as the sum of distinct unit fractions with denominators from .Does this process always terminate if has odd denominator and is the set of odd numbers? More generally, for which pairs and does this process terminate?

Worked, still open.

number theory · open · formalized (Lean) · 0 attempts

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vela reproduce examples/erdos-problems

evidence

unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

What you describe is the usual Fibonacci–Sylvester greedy Egyptian‑fraction algorithm, but with the allowed denominators restricted to an infinite set $A$. When (A=\mathbb N) Fibonacci already observed (and it is easy to prove) that the process always terminates for rational (x\in(0,1)). ([Wikipedia][1])

candidate solution ↗

llm-hunter · gpt 5.2, gpt pro 5.2 · unverified

2 LLM attack(s) recorded (gpt 5.2, gpt pro 5.2); unverified.

candidate solution ↗

formal

AMS 5 · test (literature)

theorem greedyUnitFractionRem_zero (n : ℕ) : greedyUnitFractionRem .univ (1 / n) 0 = 0
formal-conjectures/282.lean ↗

status

open

notary

vela reproduce examples/erdos-problems
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finding.noted · reviewer:will-blair · 1 day

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