Martin 1998 dense egyptian fractions

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theorem_1: Every positive rational r is, for large x, a sum of reciprocals of a set of more than (C(r) − η) x integers up to x, with C(r) within a factor 1 − log 2 of best possible.

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Greg Martin, Dense Egyptian fractions. arXiv preprint (1998). arXiv:math/9804045. The arXiv record carries no license field, so arXiv's assumed license applies (arXiv:math/9804045), every other right reserved.

The copy read for this card is arXiv:math/9804045v1 (8 April 1998), sixteen pages with a clean text layer, whose comments line says "to appear in Trans. Amer. Math. Soc"; the paper appeared as Trans. Amer. Math. Soc. 351 (1999), no. 9, 3641--3657, doi:10.1090/S0002-9947-99-02327-2. The published text was not compared with the preprint; locators refer to the preprint.

Theorem 1 states that for a positive rational r and eta > 0, once x is large enough in terms of r and eta, some set of more than (C(r) - eta)x positive integers up to x has reciprocals summing to r, where C(r) = (1 - log 2)(1 - exp(-r/(1 - log 2))). Martin also shows this density is nearly optimal: the trivial upper bound for such a set is (1 - e^{-r})x, and C(r)/(1 - e^{-r}) always exceeds 1 - log 2 = 0.30685, tending to 1 as r tends to 0. The construction subtracts from r the reciprocals of a suitable dense set A of integers up to x, then adds back a few multiples of each prime p dividing the resulting denominator to cancel that prime, which forces the elements of A to be roughly x^{1/2}-smooth; the density 1 - log 2 of x^{1/2}-smooth numbers is exactly the source of that factor in C(r). Lemmas 2 and 3 bound the number of multiples of p (up to p-1) needed for each per-prime cancellation, and a standard expansion algorithm mops up the small leftover rational. For problem 295 this is the maximum-count dual of the quantity k(N): where that problem asks how few distinct unit fractions with denominators at least N can sum to 1, Martin's theorem asks how many denominators below x can be used at once, and it says nothing about k(N). For problem 285, Theorem 1 with r = 1 bounds the least largest denominator f(k) by a constant multiple of k only for infinitely many k, as the result page derives.

Source: https://arxiv.org/abs/math/9804045.

Read status: claims checked. Theorem 1 and the optimality remark (p. 2) were read clause by clause in the text layer; the proof (Section 4, pp. 11--15) was read for structure only. Result page: theorem_1.

Bears on. #285, where Theorem 1 with r = 1 and 0 < eta < C(1) gives f(k) < k/(C(1) - eta) for infinitely many k (the sizes the construction produces), C(1) = 0.2950..., a deduction made on the result page and not in the paper, which leaves f(k) for other k and the asymptotic e/(e-1)k untouched; #295, an adjacent maximum-count result recorded for contrast, which says nothing about k(N) - (e-1)N

Results to transcribe.

• Theorem 1: Given a positive rational r and eta > 0, once x is large in terms of r and eta, some set of more than (C(r) - eta)x positive integers up to x has reciprocals summing to r, where C(r) = (1 - log 2)(1 - exp(-r/(1 - log 2))).
• Optimality remark: Any such set has at most about (1 - e^{-r})x elements, and C(r)/(1 - e^{-r}) > 1 - log 2 = 0.30685..., so C(r) is never less than 30 percent of best possible and is asymptotically optimal as r tends to 0.
• Lemmas 2 and 3 (pp. 3--4): If p^l exactly divides N, c/d is a rational with d dividing N, and S is a set of at least p - 1 divisors of N each exactly divisible by p^l, then adding the reciprocals of fewer than p elements of S to c/d gives a fraction whose reduced denominator divides N/p. Applied to each power of p in turn, this removes p from the denominator using at most p - 1 multiples of each power, which restricts the construction to roughly x^{1/2}-smooth denominators.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.