OPEN
This is open, and cannot be resolved with a finite computation.
Let $A\subseteq \mathbb{N}$ be an infinite set and consider the following greedy algorithm for a rational $x\in (0,1)$: choose the minimal $n\in A$ such that $n\geq 1/x$ and repeat with $x$ replaced by $x-\frac{1}{n}$. If this terminates after finitely many steps then this produces a representation of $x$ as the sum of distinct unit fractions with denominators from $A$.
Does this process always terminate if $x$ has odd denominator and $A$ is the set of odd numbers? More generally, for which pairs $x$ and $A$ does this process terminate?
In 1202 Fibonacci observed that this process terminates for any $x$ when $A=\mathbb{N}$. The problem when $A$ is the set of odd numbers is due to Stein.
Graham
[Gr64b] has shown that $\frac{m}{n}$ is the sum of distinct unit fractions with denominators $\equiv a\pmod{d}$ if and only if\[\left(\frac{n}{(n,(a,d))},\frac{d}{(a,d)}\right)=1.\]Does the greedy algorithm always terminate in such cases?
Graham
[Gr64c] has also shown that $x$ is the sum of distinct unit fractions with square denominators if and only if $x\in [0,\pi^2/6-1)\cup [1,\pi^2/6)$. Does the greedy algorithm for this always terminate? Erdős and Graham believe not - indeed, perhaps it fails to terminate almost always.
See also
[206].
View the LaTeX source
Additional thanks to: Zach Hunter
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #282, https://www.erdosproblems.com/282, accessed 2026-02-13
