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Source. Section 8, Open problems, item (4), arXiv:2202.00191v2, PDF p. 18. Published as J. Number Theory 242 (2023), 208--234; not compared.

Statement

Let $u_n(\theta)$ be the best $n$-term Egyptian underapproximation of $\theta\in(0,1]$ in the paper's convention (denominators $2\le x_1\le\cdots\le x_n$, repetitions allowed; it is attained and rational by Theorem 3, p. 6).

Open problem (4), as posed on p. 18 (reference [3] is Erdős and Graham, 1980):

Let $\theta\in(0,1]$. Erdős and Graham [3, p.31] asserted (without proof or reference to any publication) that for every rational number $\theta$ there exists an integer $n_0=n_0(\theta)$ such that, for all $n\ge n_0+1$,

$$> u_n(\theta)=u_{n_0}(\theta)+u_{n-n_0}\left(\theta-u_{n_0}(\theta)\right) >$$

and the best $(n-n_0)$-term underapproximation $u_{n-n_0}\left(\theta-u_{n_0}(\theta)\right)$ is always constructed by the greedy algorithm. They also wrote, "It is not difficult to construct irrationals for which the result fails." Prove or disprove these statements.

Standing on 2026-09-17

• The rational assertion: claimed proved for every positive rational, in this convention and in the distinct-denominator convention, by Kovač and Tang, Theorem 1 (arXiv:2607.28387v2, 2026), whose formulation of the problem is this item; an author preprint, recorded as a claim.
• The irrational assertion: proved, non-constructively and in the strongest measure sense, by Kovač, Theorem 1 (2025) and its Corollary 2; no explicit example is known.
• Item (1) of the same list asks whether irrational $\theta$ exist whose greedy $n$-term sequence is the unique best one for every $n$; Kovač and Tang's Example 2 gives one.

Read depth

Claims checked (the item read clause by clause on PDF p. 18); there is no proof to read.

Bears on. #206: the formulation of the rational companion question and of the irrational claim.