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Statement
For a positive rational and a positive integer , let
so that holds the integers above that never stand in position when the denominators of a representation of are listed in decreasing order (p. 3). An integer has , so it can occur only in the one-term representation ; the paper leaves such out.
Theorem 3 (p. 3). "Let be a positive rational number. The set is finite for any integer , and there exists an integer such that is empty for all ."
The paper's own consequences (p. 3): is finite, and "possibly is a complete list of integers (greater than ) with this property"; is empty once is large, and "possibly this holds for every ". The paper does not compute or .
Source. G. Martin, Denser Egyptian fractions, arXiv:math/9811112v1 (18 November 1998), Theorem 3 on p. 3, read on the page image and in the text layer of that preprint; the proof is Section 6 (pp. 20--24). The journal version, Acta Arith. 95 (2000), no. 3, 231--260, was not compared.
Read depth. Claims checked: the statement and the definition of were read clause by clause on the page image of p. 3. Of the proof, pp. 20--21 (Lemmas 18 and 19, the construction for and the passage to ) were read for structure on the page images; the proof of Lemma 19, through Lemma 20 (pp. 21--24), was not checked.
Proof pointer
Section 6 (p. 20) restates Proposition 5 as Lemma 18: for a closed interval there is such that for all and all with there is a set of positive integers not exceeding with . Lemma 19 (deduced on p. 22 from Lemma 20, which is stated on p. 21 and proved on pp. 22--24) gives, for every integer , a positive integer with . The proof of Theorem 3 begins (p. 20) by showing that is finite: for , and a large integer , take from Lemma 19. The remainder lies in , and since with an integer, its denominator has no prime-power divisor above ; so Lemma 18 with represents by integers at most , and adding and gives a representation of with largest denominator and second-largest denominator (pp. 20--21). The higher positions use a different argument (p. 21): splitting the term with the largest denominator by identity (4) gives , and for each a representation of by denominators exceeding shows for some , so the sets are eventually empty.
Dependencies
Same-paper Proposition 5 (through Lemma 18) and Lemmas 19--20.
Bears on
- Problem 292: the site's "Explore " invites finer questions. The analogue for the second-largest and later denominators is not posed in the 1980 monograph; the paper says it is mentioned in Guy's Unsolved problems in number theory (p. 3). Theorem 3 answers it for representations of up to finitely many exceptions, while the density question itself is Theorem 4.