TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 351, Number 9, Pages 3641–3657 S 0002-9947(99)02327-2 Article electronically published on March 22, 1999
DENSE EGYPTIAN FRACTIONS
GREG MARTIN
Abstract. Every positive rational number has representations as Egyptian fractions (sums of reciprocals of distinct positive integers) with arbitrarily many terms and with arbitrarily large denominators. However, such repre- sentations normally use a very sparse subset of the positive integers up to the largest denominator. We show that for every positive rational there exist representations as Egyptian fractions whose largest denominator is at most N and whose denominators form a positive proportion of the integers up to N, for sufficiently large N; furthermore, the proportion is within a small factor of best possible.
1. Introduction The ancient Egyptians wrote positive rational numbers as sums of distinct re- ciprocals of positive integers, or unit fractions. In 1202, Fibonacci published an algorithm (subsequently rediscovered by Sylvester in 1880, among others) for con- structing such representations, which have come to be called Egyptian fractions, for any positive rational number. Since that time, number theorists have been interested in some quantitative aspects of Egyptian fraction representations. For instance, there are algorithms that improve upon the Fibonacci–Sylvester algo- rithm in various ways, bounding the size of the largest denominator or limiting the number of terms. Bleicher [1] has a thorough survey of, and references to, such developments. One line of questions concentrates on the number of terms in Egyptian fraction representations. A positive rational m/n canalwaysbeexpandedintoanEgyptian fraction with at most m terms, for instance by the Farey series algorithm (see Golomb [3]). Erd˝os and Straus conjectured that, for all integers n 2, the fraction 4/n can actually be written as the sum of three unit fractions rather≥ than four. (In their conjecture, distinctness of the terms is not required.) Sierpi´nski [6] made the same conjecture for fractions of the form 5/n, and mentioned that Schinzel conjectured that, for any fixed numerator m, the fraction m/n could be written as the sum of three unit fractions for sufficiently large n. In this vein, Vaughan [7] showed that almost all positive fractions with numerator m can be written as the sum of three unit fractions. One can also investigate the behavior of the number of terms in Egyptian frac- tions at the other extreme. Any unit fraction can be split into two using the identity 1/n =1/(n+1)+1/(n(n+ 1)); consequently, given a particular Egyptian fraction representation, one can repeatedly use this splitting algorithm on the term with
Received by the editors July 7, 1997. 1991 Mathematics Subject Classification. Primary 11D68.
c 1999 American Mathematical Society
3641 3642 GREG MARTIN largest denominator to construct such representations with arbitrarily many terms. However, this process results in a tremendously thin set of denominators in the sense that if the largest denominator is x, then the number of denominators is log log x. One can try to use the splitting algorithm on the intermediate terms, but then the danger arises that the resulting unit fractions will no longer be distinct. It is therefore interesting to ask how many terms can be used to represent a rational number as an Egyptian fraction, given a bound on the size of the largest denominator. The purpose of this paper is to demonstrate that a positive pro- portion of the integers up to the bound can in fact be assembled to form such a representation. We will establish the following theorem: Theorem 1. Let r be a positive rational number and η>0a real number. For any real number x that is sufficiently large in terms of r and η, there is a set of S positive integers not exceeding x such that r = n 1/n and > (C(r) η)x, where ∈S |S| − P r C(r)=(1 log 2) 1 exp − . − − 1 log 2 −
That one can always find such a dense set of denominators is already surprising (though the Egyptians might not have been pleased to do their arithmetic this way!), but it turns out that the proportion given in Theorem 1 is comparable to the best possible result. Given a positive rational number r,anyset of positive r S integers not exceeding x with cardinality (1 e− )x satisfies b − c 1 (1) 1/n 1/n = r + O (x− ). ≥ r n e rx+1
We define log1 x =max log x, 1 and logj x =log1(logj 1 x) for any integer j k { k } k − k ≥ 2, and we write log x and logj x as shorthand for (log x) and (logj x) respectively. We use the notations P (n)andp(n) to denote the greatest and least prime factors of n respectively, and make the conventions that P (1) = 1 and p(1) = .Wesay that a prime power pl exactly divides an integer n,orthatnis exactly∞ divisible by pl,ifpl divides n but pl+1 does not. The constants implicit in the and O- notations in this paper may depend on any Greek variable (α, δ, ε, η,and λ)and also on k and the rational number r or a/b where appropriate, but they will never depend on p, q, t, v, w, x,ory; this remains the case when any of these variables is adorned with primes or subscripts. When the phrase “x is sufficiently large” is used, the size of x may depend on the Greek variables and k, r,anda/b as well. The author would like to thank Hugh Montgomery and Trevor Wooley for their guidance and support and the referee for many valuable comments. This mate- rial is based upon work supported under a National Science Foundation Graduate Research Fellowship.
2. Elementary lemmas The following lemma is an easy consequence of the Cauchy–Davenport Theorem (see [8, Lemma 2.14], for instance), but we provide a direct proof.
Lemma 2. Let t be a nonnegative integer, and let x1,...,xt be nonzero elements of Zp, not necessarily distinct. Then the number of elements of Zp that can be written as the sum of some subset (possibly empty) of the xi is at least min p, t +1 .In particular, if t p 1, then every element of Z can be so written. { } ≥ − p We remark that the conclusion of the lemma is best possible, since we could take x1 x 0(modp), in which case the set of elements of Z that can ≡ ··· ≡ t 6≡ p be written as the sum of a subset of the xi is precisely 0,x1,...,tx1 , which has cardinality t +1ift
Suppose that t We now cite an existing Egyptian fraction algorithm, which we will use near the end of the proof of Theorem 1. Lemma 4. Let c/d be a positive rational number with d odd. Suppose that c/d < 1/P (d). Then there exists a set of distinct odd positive integers, with C max n d p, { ∈C} p P(d) ≤Y such that c/d = n 1/n. ∈C Proof. Breusch hasP shown that any positive rational number with odd denominator can be written as the sum of reciprocals of distinct odd positive integers. When DENSE EGYPTIAN FRACTIONS 3645 one examines his construction [2, Lemmas 1–3], one finds that, when c/d < 1/P (d), the integers involved in fact do not exceed 5lcm d, 32 p , { } 3 3. Smooth number sets For real numbers x, y>1, we recall the definition of (x, y), the set of y-smooth numbers up to x: A (x, y)= n x:P(n) y . A { ≤ ≤ } Let w>1and0 λ<1 be real numbers and k 2 be an integer. We will need to work with the following≤ sets of smooth numbers≥ with various specified properties: (3) (x, y; w, λ)= n:λx < n x; P (n) y; n is k-free; d2 n P (d) w , A { ≤ ≤ | ⇒ ≤ } (x, y; w, λ; pl)= n (x, y; w, λ):n=mpl with P (m) (x, y; w, λ)= (x, y0; w, λ) (x, y; w, λ; p) A A ∪ A y0 w[≤ k 1 − (x, y; w, λ; pl) , ∪ A y0 0 to be the continuous solution to the differential-difference equation ρ(u)=1, 01. − − The following lemma, due to Hildebrand, describes this asymptotic formula more specifically. We have not made any effort to optimize the error term or the range of y in the hypothesis, as it will suffice for our purposes as stated. 3646 GREG MARTIN Lemma 5. Let x 3 and y be real numbers satisfying exp(log4 x) y x.Then ≥ 2 ≤ ≤ log x log x Ψ(x, y)=xρ 1+O 2 . log y log y Proof. Hildebrand shows [5, Theorem 1] that, for any ε>0, one has ulog u Ψ(x, x1/u)=xρ(u) 1+O 1 log x uniformly for x 3and1 u log x/(log x)5/3+ε. On setting ε =7/3and ≥ ≤ ≤ 2 u=logx/ log y (so that y = x1/u), the lemma follows immediately. From the definition (4) of ρ(u), one can easily derive that ρ is a positive, de- creasing function, and that ρ(u)=1 log u for 1 u 2, so that ρ(2) = 1 log 2 in particular. We will need the following− additional≤ properties≤ of ρ(u). − Lemma 6. Let x 3, v 1,andybe real numbers satisfying v x and exp(log4 x) y x.Then:≥ ≥ ≤ 2 ≤ ≤ log x/v log x log log (a) ρ ρ v 2 x/ y; log y log y log x 1 (b) for any ε>0, we have ρ − xε; log y (c) for any real number α>1, we have α α log xn− log x α+1 n− ρ ρ (log x)− ; log y log y log x log x α α log x/ log y log x α+α/ log3 x ρ n− (n ) 2 ρ n− 2 . log y log y log x Finally, for any real numbers 1 ρ(t)=ρ(s)(1 + O((t s)log2 s)). − 1 Letting s = log(x/v)/ log y and t =logx/ log y,weseethat log x log x/v log v log2 x ρ = ρ 1+O 2 ; log y log y log y 2 and under the hypothesis that log v = o(log y/ log2 x), the (1 + O( ))-term can be inverted and moved to the other side of the equation, which establishes· part (d). With Lemmas 5 and 6 at our disposal, we can establish the following lemmas concerning the distributions of the sets of smooth numbers defined in equation (3). 4 Lemma 7. Let x 3, y,andwbe real numbers satisfying exp(log2 x) y x and w log x.Then:≥ ≤ ≤ ≥ x log x log3 x (a) Ψ(x, y; w, 0) = ρ 1+O 2 ; ζ(k) log y log y 3 x log x log2 x k (b) Ψ0(x, y; w, 0) = ρ 1+O ,whereξ(k)=2(1 2− )ζ(k). ξ(k) log y log y − Proof. From the definition (3) of (x, y; w, λ), we have A 1 1 x− Ψ(x, y; w, 0) = x− µ(d) µ(f) . n x k f 2 n ≤ d n P (Xn) y P (Xd)| w p(Xf)|>w ≤ ≤ Interchanging the order of summation yields 1 1 x− Ψ(x, y; w, 0) = x− µ(d) µ(f) 1 d x1/k f √xd k n x ≤X ≤X − P (Xn≤) y P (d) min y,w P (f) y ≤ ≤ { } p(f)>w≤ dk n | f 2 n | 1 k 2 = µ(d) µ(f)(x− Ψ(xd− f − ,y)) 1/k k d x f √xd− P (d) ≤Xmin y,w P≤(Xf) y ≤ { } p(f)>w≤ k 2 k 2 log xd− f − log2 x = µ(d)d− µ(f)f − ρ 1+O , log y log y 1/k √ k d x f xd− P (d) ≤Xmin y,w P≤(Xf) y ≤ { } p(f)>w≤ where the final equality follows from Lemma 5. The primary contribution to the double sum will come from those terms with d small and f = 1 (notice that if 3648 GREG MARTIN f>1, then f>w log x), so we write it as ≥ k 1 k log xd− log2 x (7) x− Ψ(x, y; w, 0) = µ(d)d− ρ 1+O log y log y log d k k log xd− log x k+1 d− ρ ρ log− x, log y log y log x d x1/k ≤X≤ 2 ∞ k 2 log xf − log x 1 ∞ k d− f − ρ ρ log− x d− log y log y d=1 log x f √x d=1 X X≤ ≤ X log x 1 = ζ(k)ρ log− x. log y Also, for d log xd k log x (log(logkx)) log2 x ρ − = ρ 1+O 2 . log y log y log y Thus equation (7) becomes 1 x− Ψ(x, y; w, 0) 3 log x log2 x k k+1 1 = ρ 1+O µ(d)d− + O (log x)− +log− x . log y log y d