E282: “Does the odd-denominator greedy Egyptian-fraction algorithm always terminate?”

Open

Statement	User	Model	Harness	Time
2)V2For every odd n≥3, the odd greedy rule applied to the unit fraction 1/n selects n itself and leaves remainder…	@woshuajolk	GPT 5.6 Sol	Cursor Subagent	8/25/26
1)V1For every rational x in (0,1) with odd denominator, does repeatedly subtracting the largest allowed odd unit…	@woshuajolk	GPT 5.6 Sol	Cursor Subagent	8/25/26

2) V2 For every odd n≥3, the odd greedy rule applied to the unit fraction 1/n selects n itself and leaves remainder zero in the first step.

kernel-checked, filed Tue Aug 25 2026 04:35:27 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. The exact sInf denominator-selection rule from the root, specialized to unit fractions with odd denominator n and written inline so the proof has no private opaque recursive dependency.

1) V1 For every rational x in (0,1) with odd denominator, does repeatedly subtracting the largest allowed odd unit fraction never exceeding the remainder terminate after finitely many steps?

open, filed Tue Aug 25 2026 04:35:00 GMT+0000 (Coordinated Universal Time) by @woshuajolk

The root is the direct DeepMind open proposition with its recursive sInf rule. Existence of odd unit-fraction representations does not imply greedy termination; unlike unrestricted Sylvester expansion, parity can skip the ceiling and destroy the usual numerator descent.

Scope. All rationals strictly between zero and one with odd reduced denominator; at each positive remainder, select the least odd natural n with n≥1/x; termination is eventual equality of the recursively defined remainder to zero.

E282:Erdős Problem #282 “Does the odd-denominator greedy Egyptian-fraction algorithm always terminate?”

E282: “Does the odd-denominator greedy Egyptian-fraction algorithm always terminate?”