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Claim. The answer to Problem 315 is yes, as the case $\lambda=1$ of Kovač and Tang's Corollary 3 and Theorem 4 in their preprint Eventually greedy best Egyptian underapproximations of rational numbers via optimal control (arXiv:2607.28387, v1 30 July 2026, v2 4 August 2026). For a rational $\lambda>0$ their Theorem 1 gives an eventually greedy sequence $(b_n)$ of best $n$-term underapproximations of $\lambda$ by unit fractions. Corollary 3 says that if the best $n$-term tuple with repeated denominators allowed is unique for every $n$, every other nondecreasing sequence $(a_n)$ of integers $a_n\ge2$ with $\sum1/a_n=\lambda$ has $\liminf a_n^{2^{-n}}<\lim b_n^{2^{-n}}$; Theorem 4, added in v2 after a suggestion of van Doorn in the thread, drops the uniqueness hypothesis and says that every such sequence either satisfies the inequality or agrees with $(b_n)$ from some index on. For $\lambda=1$ the sequence $(b_n)$ is Sylvester's $2,3,7,43,\ldots$, its best $n$-term tuples are unique (Curtiss and Takenouchi, as the paper cites) and the limit is the Vardi constant $c_0=1.264085\ldots$, so the statement follows for nondecreasing, hence for strictly increasing, sequences; Tang's thread comment of 31 July 2026 says the result recovers this problem. The route is non-constructive: it feeds Theorem 1 into Li and Tang's conditional theorem (Theorem 1.6 of their Acta Math. Hungar. paper, Theorem 1.9 of arXiv:2503.12277v4), which derives the extremality from the eventually-greedy property. The paper presents the case $\lambda=1$ as already proved independently by Li and Tang and by Kamio, not as a new result.

Submission note. Posted to the site's forum by Q. Tang on 31 July 2026:

It may also be worth mentioning that Corollary 3 of a recent paper by Vjeko and me gives the following generalization of this problem.

Let $0<\lambda\leq 1$ be rational, and suppose that, for every positive integer $n$, the maximizing $n$-term denominator tuple in the nondecreasing-denominator convention is unique. Let $(b_n)_{n\geq 1}$ be the eventually greedy extremal sequence supplied by our main theorem. Then, for every strictly increasing sequence of integers

$$2\leq a_1<a_2<\cdots,\qquad > \sum_{n=1}^{\infty}\frac1{a_n}=\lambda,$$

different from $(b_n)_{n\geq1}$, we

have

$$\liminf_{n\to\infty}a_n^{1/2^{n}}<\lim_{n\to\infty}b_n^{1/2^{n}}.$$

Known

uniqueness results make this applicable to several explicit classes of rational numbers described immediately after Corollary 3. For $\lambda=1$, the sequence $(b_n)$ is the Sylvester sequence, so the result recovers [315].

Depends on.

• Theorem 1 of the paper, which supplies the eventually-greedy property.

The route also rests on Li and Tang's conditional theorem, Theorem 1.9 of arXiv:2503.12277v4 (Theorem 1.6 of the Acta Math. Hungar. paper), which derives the extremality from the eventually-greedy property.

AI assistance. The paper's declaration of AI usage says that key steps of the proof of Theorem 1 (the payoff function, the existence of optimal terminal decompositions and the treatment of competing non-greedy underapproximations) were provided by OpenAI's GPT-5.6 Sol and rewritten by the authors, who take responsibility for correctness; the chat transcripts are public in a repository the paper links.

Standing. The paper is an author preprint: the arXiv listing carries no journal reference, a Crossref bibliographic query of 2026-09-17 found no record, and no independent review is located; the site's commentary credits Kamio and Li and Tang and does not name this paper. The library card records Theorem 1 at claims-checked depth and its proof as read for its scheme only; the corpus has not verified the proof. The claim stays claimed, and the problem's standing rests on the accepted claims of Kamio and Li and Tang.