Which integer sequences form denominators of Egyptian fractions?
I have a new paper out: “Egyptian Fractions with Denominators from Sequences Closed Under Doubling”, in the Journal of Integer Sequences. (There should also be an arXiv version soon but despite my long association with arXiv they made me get an endorser before I could upload to the number theory category, slowing down my submission there.)
Anyway, it’s basically a journal version of an old blog post here, “Egyptian fractions with practical denominators”. That post concerned the practical numbers, positive integers \(n\) such that all other positive integers up to \(n\) can be represented as sums of distinct divisors of \(n\). This definition gives the practical numbers a natural connection to Egyptian fractions, representations of rational numbers as sums of distinct unit fractions: if you represent a number \(k<n\) by a sum of distinct divisors of \(n\), and then divide everything by \(n\), you get an Egyptian fraction for \(k/n\). ZhiWei Sun asked whether the practical numbers and Egyptian fractions were connected in a different way, with every positive integer having an Egyptian fraction representation in which all denominators are practical, and my blog post provides a positive answer to Sun’s question.
The new paper simplifies the presentation of the solution, compared to the blog post, by providing direct formulas for the representation rather than an iterative computational method for finding it. But beyond that, it also shows that the same method (based on dividing by a large power of two and dealing separately with the quotient and remainder) works for many other integer sequences beyond the practical numbers. All it needs from an integer sequence is two simple properties:

The sequence should include a multiple of every integer. In order to represent \(k/p\) as an Egyptian fraction, when \(p\) is prime, the denominators must include at least one multiple of \(p\), so the requirement of including multiples is pretty natural in this context.

For every number \(n\) that belongs to the sequence, \(2n\) should also belong to the sequence. This is the “closed under doubling” of the new article’s title, and is closely connected to the method used by the article involving division by powers of two.
Whenever a sequence \(S\) of positive integers has both properties, we can find Egyptian fractions for all positive rationals up to \(\sum_{x\in S}1/x\), the natural limiting value for such representations. When \(\sum_{x\in S}1/x\) diverges, we get all positive rationals. As well as the practical numbers, this works for some other nice sequences including the odious and evil numbers, the orders of isomorphism groups of trees, and the fibbinary numbers, numbers whose binary representation avoids consecutive ones. Because they’re based on binary representations, the doubling property of odious, evil, and fibbinary numbers follows from their definitions; the existence of multiples of other integers in these sequences is less obvious but was more or less already known. Isomorphism groups of trees have orders that are the products of factorials, from which (because 2 is a factorial and \(n!\) is a multiple of \(n\)) both properties follow immediately.
Although these two properties are sufficient for a sequence to form Egyptian fractions for rationals up to its natural limit, they are not necessary. Ron Graham’s PhD dissertation was on the same topic, and showed that the sequence of squares greater than one has the same property. (The sequence of all squares, including one, is a little more complicated: its sums of distinct reciprocals can represent all rationals in the intervals \([0,\pi^2/61)\) and \([1,\pi^2/6)\) but can’t cover the gap between these two intervals.) Characterizing which sequences do or do not form representations of this type more generally seems like an interesting question, but one that I don’t have much idea how to attack at this point.