Does the odd-greedy Egyptian-fraction algorithm always terminate for odd-denominator rationals? (Erdős #282)

Given an infinite set $A\subseteq\mathbb{N}$, run the following greedy algorithm on a rational $x\in(0,1)$: pick the minimal $n\in A$ with $n\geq 1/x$, replace $x$ by $x-\frac1n$, and repeat. If the process halts after finitely many steps it writes $x$ as a sum of distinct unit fractions with denominators drawn from $A$. When $A$ is the set of odd positive integers and $x$ has odd denominator, does this process always terminate? More generally, for which pairs $(x,A)$ does the greedy process terminate?

Acceptance. FULLY RESOLVES (proof-shaped): a complete proof that the odd-greedy algorithm terminates for every rational with odd denominator (Stein's question), or an explicit odd-denominator rational $x$ for which it provably never terminates (a finite certificate exhibiting the non-halting structure together with a proof of non-termination). The universal statement requires a rigorous proof — machine-checkable preferred, otherwise a full written proof; simulation alone cannot certify 'always terminates'. ADVANCES (each independently checkable): (a) verify termination by simulation for all odd-denominator rationals $m/n$ up to a stated height bound, with reproducible code and the attained bound, strictly beyond any previously reported range; (b) settle the greedy-termination question for another of Graham's denominator families (an arithmetic progression, or square denominators) — e.g. exhibit an explicit non-terminating $x$ for square denominators; (c) prove termination for a structured subclass of odd-denominator $x$. Deliver the proof/witness, or the simulation code plus the attained verified bound.