EGYPTIAN EXPANSIONS FOR RATIONAL BETWEEN 0 AND 1 OBTAINED WITH ENGEL

ELVIA NIDIA GONZALEZ´ AND JULIA BERGNER, PHD DEPARTMENT OF MATHEMATICS

Abstract. The ancient Egyptians expressed rational numbers as the finite sum of distinct unit . Were the Egyptians limited by this notation? In fact, they were not as every rational can be written as a finite sum of distinct unit fractions. Moreover, these expansions are not unique. There exist several di↵erent algorithms for computing expansions, all of which produce di↵erent representations of the same . One such algorithm, called an Engel series, produces a finite increasing sequence of for every rational number. This sequence is then used to obtain an Egyptian fraction expansion. Motivated by the work of M. Mays, this project aims to investigate properties of natural number denominators n that produce x length x Egyptian fraction expansions using Engel series for n between 0 and 1. x While computing Engel expansions using Mathematica, a helpful pattern emerged: rational numbers n which produce length x Egyptian fraction expansions were those whose denominator minus one was divisible by every natural number between (and including) x and 2. We conjecture and prove that this will always x hold for length x Engel series for between 0 and 1 where n = k lcm(x, x 1,...,3, 2) + 1 for k N. n · 2 Furthermore, we conjecture that n = lcm(x, x 1,...,3, 2) + 1 is the smallest n such that x produces a n length x Egyptian fraction expansion. A proof that n = k lcm(x, x 1,...,3, 2) + 1 works is included along · with an investigation of whether or not n = lcm(x, x 1,...,3, 2) + 1 the least such n that does.

1. Introduction The Rhind papyrus is responsible for preserving the mathematical methods employed by the ancient Egyptians. It is clear from this document that the Egyptians had an unintuitive way of expressing rational numbers. Unlike the current precedent, in which one is written over another integer, they would write 2 1 1 out rational numbers as a sum of distinct unit fractions. For example, 7 would be written as 4 + 28 instead. Additionally, the Rhind papyrus included extensive tables, one of which included di↵erent representations 2 of n for odd n between 5 and 101 using the following identity [7] 1 1 1 = + . n n +1 n (n + 1) · However impractical these expansions may seem, they make certain real-life problems much easier to solve. One such real-life application is the sharing problem. This problem focuses on sharing whole items (like loaves of bread) with multiple people. For example, how can five loaves be split evenly amongst eight 5 people? The solution can be written as the proper fraction, 8 . This means each person gets five eighths of a loaf. Now, applying this is not as easy as writing out the solution as a fraction. How would someone cut five-eighths from eight loaves? Each loaf could be cut in half, then those halves into quarters, then those quarters into eighths. From here, each of the eight persons would receive five small pieces of bread. This solution, although correct, is not very practical. 5 1 1 If we represent 8 as 2 + 8 , this sharing problem becomes significantly easier to put into practice. Each person receives half a loaf first, which ensures that four loaves are distributed evenly amongst all eight people.

Key words and phrases. Egyptian fractions, Engel Series, . This work is funded through UC LEADS at the University of California, Berkeley. 1 2 ELVIA NIDIA GONZALEZ´ AND JULIA BERGNER, PHD DEPARTMENT OF MATHEMATICS

Next, the final loaf can be cut into eighths and each person would receive one of those pieces. Along with being easier to distribute the bread, this solution allows each person to have one larger piece (the half loaf) along with the smaller piece (the eighth loaf). These representations inspired many questions. Some have been answered. For example, we know that Egyptian fraction expansions are not unique. Di↵erent algorithms may produce di↵erent representations 5 1 1 1 of the same fraction. Our example of 8 can also be expressed as 2 + 10 + 40 . Table 1 lists several more examples.

Table 1. Expansions of various rational numbers using di↵erent algorithms.

x n Engel Series Greedy Algorithm

5 1 1 1 1 1 1 1 1 1 1 7 2 + 6 + 24 + 168 2 + 5 + 70 2 + 6 + 21

4 1 1 1 1 1 1 1 1 1 1 1 13 4 + 20 + 140 + 1820 4 + 18 + 468 4 + 28 + 70 + 130

7 1 1 1 1 1 1 1 1 18 3 + 18 3 + 18 3 + 24 + 104 + 234

Another question that arose from Egyptian fraction expansions inquired whether or not the Egyptians were limited by the use of this notation. That is, could they express any rational number as a sum of distinct ? The Egyptians, as it turns out, were not at all limited. Any rational number can be expressed as the finite sum of distinct unity fractions. Furthermore, even irrational numbers can be expressed as an infinite sum of distinct unit fractions. The focus of this project is Egyptian fraction expansions for rational numbers between 0 and 1 (but not including 0 and 1) obtained using an Engel series. Some examples of these expansions can be seen on the second column of Table 1.

2. Engel Expansions An Engel series, for any , is sequence of increasing integers [6]. Any real number, say z, can be expressed as an Egyptian fraction expansion using an Engel series in a unique way. This representation for any real z can be computed by defining u = z then letting a = 1 .From here, each subsequent a 1 1 u1 i+1 d e 1 and ui+1 is obtained by first computing ui+1 = ui ai 1 then computing ai+1 = ,where is the · d ui e de ceiling function defined as r = s if and only if s 1

1 1 . a a a i=1 1 2 i X · ··· Any Egyptian fraction expansion derived using an Engel series will be referred to as an from now on. Below we prove that Engel expansions are finite for all rational numbers. Theorem 2.1. Engel expansions are finite for all rational numbers. EGYPTIANFRACTIONEXPANSIONSUSINGENGELSERIES 3

x Proof. Suppose m Q.Thenm = n for x, n Z and n = 0. Begin to compute the Engel expansion by first letting 2 2 6 x n u = and a = . 1 n 1 d x e Next, compute u2 using a1 and u1 to obtain x x a n u = u a 1=( a ) 1= · 1 . 2 1 · 1 n · 1 n 1 n n n By the definition of the ceiling function, = = a1 if and only if a1 1 < a1.Thusa1 1 < d u1 e d x e x  x if and only if x a1 x

3. Results The work within this project was inspired by a paper by Michael E. Mays titled, “A Worst Case of the -Sylvester Expansion” [4]. Mays explored Egyptian fraction expansions of rational numbers using the greedy algorithm. More specifically, his paper investigated the properties of fraction expansions which had lengths that matched the numerator using the greedy algorithm. In a similar manner, this paper investigates fraction expansions whose lengths match their numerators using Engel expansions. Initially, this project began by simply looking at tables of Engel expansions. Tables 2 and 3 o↵er a few examples of Engel expansions whose lengths match their numerators. Each table begins with a denominator that is one greater than the numerator and ends when an Engel expansion of length equal to the numerator is obtained. The first column is simply the rational number being investigated, the second column is the Engel series, and the third column is the Engel expansion 1. It was through tables such as these that a pattern first emerged. First, we noticed every integer between (and including) x and 2 had to divide one minus the denominator (that is, n 1). This observation led us to conjecture that as long as n = lcm(x, x 1,...,3, 2)+1, x the Engel series produced for n will always be of length x. Next, we noticed that n = lcm(x, x 1,...,3, 2) + 1 seems to be the least value for the denominator of x n that does produce the desired expansion. We do prove that n = lcm(x, x 1,...,3, 2) + 1 does indeed always produce a length x Engel expansion. We also conjecture (but do not prove) that this n is the least such n with this property. Instead, we logically demonstrate that n = lcm(x, x 1,...,3, 2) + 1 is the least 6 denominator that produces a six term Engel expansion for n . Finally, we observed that denominators other than n = lcm(x, x 1,...,3, 2) + 1 produced an expansion of length x. These denominators seemed to increase in some predictable fashion. Table 5 demonstrates that any denominator equal to 1 plus any multiple of lcm(x, x 1,...,3, 2) will produce a length x expansion for x n .

1The appendix contains similar tables with more details and for larger numerators. Since the denominators can get unruly fairly quickly (see the last row of Table 4 for a good demonstration of this) the majority of the tables will include the Engel series rather than the Engel expansion for rational numbers. 4 ELVIA NIDIA GONZALEZ´ AND JULIA BERGNER, PHD DEPARTMENT OF MATHEMATICS

3 Table 2. Engel expansions for n

3 n Engel Series Engel Expansion

3 2, 2 1 + 1 4 { } 2 4 3 2, 5 1 + 1 5 { } 2 10 3 = 1 2 1 6 2 { } 2 3 3, 4, 7 1 + 1 + 1 7 { } 3 12 84 4 Table 3. Engel expansions for n

4 n Engel Series Engel Expansion

4 2, 2, 5 1 + 1 + 1 5 { } 2 4 20 2 2, 3 1 + 1 3 { } 2 6 4 2, 7 1 + 1 7 { } 2 14 4 = 1 2 1 8 2 { } 2 4 3, 3 1 + 1 9 { } 3 9 2 3, 5 1 + 1 5 { } 3 15 4 3, 11 1 + 1 11 { } 3 33 4 = 1 3 1 12 3 { } 3 4 4, 5, 7, 13 1 + 1 + 1 + 1 13 { } 4 20 140 1820 x Table 4. Length x Engel series and Engel expansions for n .

x x n Engel Series Engel Expansion 2 2 2, 3 1 + 1 3 { } 2 6 3 3 3, 4, 7 1 + 1 + 1 7 { } 3 12 84 4 4 4, 5, 7, 13 1 + 1 + 1 + 1 13 { } 4 20 140 1820 5 5 13, 16, 21, 31, 61 1 + 1 + 1 + 1 + 1 61 { } 13 208 4,368 135,408 8,259,888 EGYPTIANFRACTIONEXPANSIONSUSINGENGELSERIES 5

x Table 5. Other possible denominators that produce a length x Engel Expansion for n when x =3

3 n n Engel Series

3 7=1 lcm(3, 2) + 1 3, 4, 7 7 · { } 3 13 = 2 lcm(3, 2) + 1 5, 7, 13 13 · { } 3 19 = 3 lcm(3, 2) + 1 7, 10, 19 19 · { } 3 25 = 4 lcm(3, 2) + 1 9, 13, 25 25 · { } 3 31 = 5 lcm(3, 2) + 1 11, 16, 31 31 · { } 3 37 = 6 lcm(3, 2) + 1 13 19 37 37 · { · · }

4. Theorems and Proofs This section will include proofs of conjectures made within the previous section. First we prove that n = k lcm(x, x 1,...,3, 2) + 1 will produce a length x Engel expansion. · x Theorem 4.1. Suppose x is a positive integer and n = k lcm(x, x 1,...,3, 2) + 1. Then n will have a length x Engel expansion. · In order to prove Theorem 3.1, we need a lemma first. Lemma 4.2. [2, Page 69, (3.6)]

r + s = r + sor r + s = r + sfors Z. d e d e b c b c 2 x Proof Of Theorem 4.1. Suppose n = k lcm(x, x 1,...,3, 2) + 1 and n (0, 1) where x N. To obtain the x · 2 2 Engel expansion of n begin by letting x u = . 1 n By assumption, n = k lcm(x, x 1,...,3, 2) + 1 hence · x u = . 1 k lcm(x, x 1,...,3, 2) + 1 · 1 Next, a1 = which can be rewritten as d u1 e k lcm(x, x 1,...,3, 2) + 1 k lcm(x, x 1,...,3, 2) 1 a = · = · + . 1 d x e d x xe

The result of dividing k lcm(x, x 1,...,3, 2) by x will be a natural number so by the lemma, a1 can be rewritten again as · k lcm(x, x 1,...,3, 2) 1 a = · + . 1 x dxe 1 1 Since x N, x will always be a positive unit fraction and x will always equal 1. Again then, a1 can be rewritten2 as d e k lcm(x, x 1,...,3, 2) k lcm(x, x 1,...,3, 2) + x a = · +1= · . 1 x x 6 ELVIA NIDIA GONZALEZ´ AND JULIA BERGNER, PHD DEPARTMENT OF MATHEMATICS

Continuing to find the Engel expansion requires computing u = u a 1 which becomes 2 1 · 1 x k lcm(x, x 1,...,3, 2) + x u = · 1 2 k lcm(x, x 1,...,3, 2) + 1 · x ·

k lcm(x, x 1,...,3, 2) + x k lcm(x, x 1,...,3, 2) 1 = · · k lcm(x, x 1,...,3, 2) + 1 ·

x 1 = . k lcm(x, x 1,...,3, 2) + 1 · Notice that the numerator of u1 = x and the numerator of u2 = x 1. The numerator of u2 is one less than the numerator of u .Indeed,u will have a numerator of x 2, u will have a numerator of x 3, and so 1 3 4 on. Eventually, The ux+1 term will have a numerator of 0 and this process halts with the Egyptian fraction x 1 1 1 expansion n = a + a a + + a a a as desired. ⇤ 1 1· 2 ··· 1 2··· x

5. Conjecture Conjecture 5.1. The n from Theorem 4.1 is the least such n which produces a length x Engel expansion.

The tables included in the appendix do make a strong case for n = lcm(x, x 1,...,3, 2) + 1 being the x least denominator that produces the desired length expansion for n . Below is an example that proves it, logically, for a small fixed value of x. First we need these additional lemmas along with Lemma 4.2.

Lemma 5.2. [2, Page 96, Exercise 12]

r r + s 1 = dse b s c Lemma 5.3. [2, Page 68]

r = r r Z r = r d e , 2 ,b c Example. Let x = 6. Then by Theorem 4.1 n = lcm(6, 5, 4, 3, 2) + 1 = 61. We will show that this is the smallest possible value for n that will produce a length 6 Engel expansion.

6 6 1 5.1. Case 1: n =6p. Suppose n =6p.Thenn = 6p = p . This is a one-term Engel expansion. Hence the following numbers are eliminated from consideration for n are 12, 18, 24, 30, 36, 42, 48, 54, 60.

6 6 2 5.2. Case 2: n =3p. Suppose n =3p.Thenn = 3p = p .

5.2.1. Subcase 1: p is even. If p is even, then 2 p then p can be written as p =2 p0 for p N.Thus | · 2 6 6 6 6 1 = = = = n 3p 3 2p 6 p p · 0 · 0 as in case 1. EGYPTIANFRACTIONEXPANSIONSUSINGENGELSERIES 7

5.2.2. Subcase 2: p is odd. If p is odd then p =2p0 + 1. Now apply Engel’s algorithm. 6 6 2 2 u1 = = = = n 3p p 2p0 +1 2p +1 a = 0 1 d 2 e By Lemma 5.2, a1 can be rewritten using the floor function:

(2p0 + 1) + 2 1 2p0 +2 a = = = p0 +2 . 1 b 2 c b 2 c b c Next, use Lemma 4.2 to separate the 2 from the floor function:

a = p0 +2. 1 b c Finally, since p0 N Z by Lemma 5.3, p0 = p0 and a1 = p0 + 2. Find u2 by continuing the algorithm for Engel series: 2 ⇢ b c 2 1 u2 = a1 u1 1=(p0 + 1) 1= · · 2p0 +1 2p0 +1 To find a2 take the ceiling of the reciprocal of u1:

a = 2p0 +1 = 2p0 +1=2p0 +1. 2 d e d e 1 Repeat the algorithm once more to obtain u3 = a2 u2 1= (2p0 + 1) 1 = 0. Since u3 = 0, we halt · p0+1 · and the Engel expansion is 6 6 6 2 1 1 = = = = + . n 3p 3(2p0 + 1) 2p0 +1 p0 +1 (p0 + 1)(2p0 + 1) This is still not a six-term expansion. Now eliminated as candidates for n are 9, 15, 21, 27, 33, 39, 45, 51, and 57. 6 6 3 5.3. Case 3: n =2p. Suppose n =2p.Thenn = 2p = p . 6 6 6 5.3.1. Subcase 1: p is a multiple of 3. If p is a multiple of 3, I can write p =3p0 for p0 N and = = 2 n 2 3 p0 6p0 which has already been covered by case 1. · ·

5.3.2. Subcase 2: p 2 mod 3. If p 2 mod 3 then p =3 p0 + 2 and ⌘ ⌘ · 6 6 6 3 = = = = u n 2 p 2(3 p + 2) 3 p +2 1 · · 0 · 0 so 3p0 +2 (3p0 + 2) + 3 1 3p0 +4 4 a = = = = p0 + . 1 d 3 e b 3 c b 3 c b 3c Since p N Z, by Lemma 4.2, ai can be rewritten as: 2 ⇢ 4 a = + p0 = p0 +1. 1 b3c Next, continue by finding u2 and a2: 3 1 u2 = u1 a1 1= (p0 + 1) 1= · 3p0 +2· 3p0 +2 and a2 = 3p0 +2 = 3p0 +2=3p0 +2. 1 d e d e Finally, u3 = u2 a2 1= (3p0 + 2) 1 = 0 and the algorithm halts. The Engel expansion for this · 3p0+2 · case becomes: 6 6 1 1 = = + , n 2(3p + 2) p +1 (p + 1) (3p + 2) 0 0 0 · 0 8 ELVIA NIDIA GONZALEZ´ AND JULIA BERGNER, PHD DEPARTMENT OF MATHEMATICS which is less than six terms long. Now eliminated as possibilities for n are 10, 16, 22, 28, 34, 40, 46, 52, and 58.

5.3.3. Subcase 3: p 1 mod 3. Notice that if p 1 mod 3 then p =3 p0 + 1. The following two cases will explore the possible⌘ Engel expansion produced for⌘ even p or odd p. ·

5.3.4. Sub-Subcase A: p0 is odd. If p is odd then p =3p0 + 1 for some p0 N and 2 6 6 3 3 u1 = = = = . n 2p p 3p0 +1

Next, use u1 to find a1:

3p0 +1 (3p0 + 1) + 3 1 3p0 +3 a = = = = p0 +1 = p0 +1=p0 +1. 1 d 3 e b 3 c b 3 c b c b c Continue by using a1 and u1 to find: 3 2 u2 = (p0 + 1) 1= 3p0 +1· 3p0 +1 and 3p +1 a = 0 2 d 2 e Because p0 is odd and positive it can be rewritten as p0 =2p00 + 1 (again p00 N). I can now continue to 2 rewrite a2 as follows:

3(2p00 + 1) + 1 6p00 +4 a = = = 3p00 +2 = 3p00 +2=3p00 +2. 2 d 2 e d 2 e d e d e

p0 1 Since p0 =2p00 + 1, this can be rewritten as p00 = 2 . This allows a2 to be rewritten one last time in terms of p0: p0 +1 3p0 +1 a =3p00 + 2 = 3( )+2= . 2 2 2 2 3p0+1 The next iteration will produce u3 = 1 = 0 and the algorithm halts. For this sub case, the 3p0+1 · 2 Engel expansion is: 6 6 3 1 1 = = = + . n 2p p p +1 (p + 1) (3p + 2) 0 0 · 00 This is still not a six term Engel expansion so 8, 20, 32, 44, and 56 have been eliminated as possible options 6 for the denominator of n .

5.3.5. Sub-Subcase B: p0 is even. Simillarly to 5.3.4, both u1 and a1 remain the same, so begin by computing:

3 3p0 +3 3p0 1 2 u2 = (p0 + 1) 1= = 3p0 +1· 3p0 +1 3p0 +1 and 3p +1 a = 0 . 2 d 2 e In order to simplify a2 further, use the fact that p0 is even so it can be rewritten as p0 =2p00 for some p00 N. Thus we get: 2

3 2p00 +1 (6p00 + 1) + 2 1 6p00 +2 a = · = = = 3p00 +1 = 3p00 +1=3p00 +1. 2 d 2 e b 2 c b 2 c b c b c

p0 Furthermore, since p0 =2p00 we can rewrite this as p00 = 2 . This allows us to rewrite a2 as: 3p 3p +2 a = 0 +1= 0 . 2 2 2 EGYPTIANFRACTIONEXPANSIONSUSINGENGELSERIES 9

Then continue with the algorithm to find:

2 3p0 +2 1 u3 = 1= 3p0 +1· 2 3p0 +1 and 1 a3 = =3p0 +1. du3 e 1 3p0+1 Finally u4 = 1 = 0 and we halt. This produces the following three-term Engel expansion: 3p0+1 · 1 6 6 3 1 1 1 = = = + + . n 2p p p +1 (p + 1) (3p + 1) (p + 1) (3p + 1) (3p + 1) 0 0 · 00 0 · 00 · 0 Since this is still not a length 6 Engel expansion, 14,26, 38, and 50 have also been eliminated. 5.4. The remaining n<60. We now only have to eliminate 7, 11, 13, 17, 19, 23, 25, 29, 31, 25, 37, 41, 43, 47, 49, 53, 55 and 59 as possible choices for n . 5.4.1. n 5 mod 6. Since n 1 mod 6, n =6p + 5 for some p N. First compute: ⌘ ⌘ 2 6 6 u = = 1 n 6p +5 and 6p +5 (6p + 5) + 6 1 10 10 a = = = p + = + p = p +1. 1 d 6 e b 6 c b 6 c b 6 c Next, 6 1 u = (p + 1) 1= 2 6p +5· 6p +5 and 6p +5 a = = 6p +5=6p +5. 2 d 1 e d e Finally, u = 1 (6p + 5) 1 = 0. This produces the following Engel expansion: 3 6p+5 · 6 1 1 = + 5 p +1 (p + 1)(6p + 5) which is still not the desired length. Thus 11,17, 23, 29, 41, 47, 53, and 59 have been eliminated as possible 6 denominators for n .

6 Table 6. Values of n that do not yield length 6 Engel expansions for n .

n Engel Series 7 2, 2, 3, 4, 7 { } 13 3, 3, 7, 13 { } 19 4, 4, 19 { } 25 5, 5 { } 31 6, 7, 8, 31 { } 37 7, 8, 13, 19, 37 { } 43 8, 9, 22, 43 { } 49 9, 10, 49 { } 55 10, 11 { } 10 ELVIA NIDIA GONZALEZ´ AND JULIA BERGNER, PHD DEPARTMENT OF MATHEMATICS

5.4.2. n 1 mod 6. The remaining nine possible denominators are eliminated by brute force in Table 6. ⌘ 6 Finally, the smallest possible n such that n produces a length 6 Engel expansion is 61 = lcm(6, 5, 4, 3, 2) + 1 as desired. ⇤

6. Conclusion We have shown a method for choosing a denominator that will always produce Engel expansion whose length is equal to its denominator. This method is the relationship between the numerator x and the denominator n given by n = k lcm(x, x 1,...,3, 2) + 1. With this, we can obtain entire families of length x Engel expansions for fixed· x. We have also conjectured that n = lcm(x, x 1,...,3, 2) + 1 is the least denominator to produce our desired length expansion. In the future, we would like to prove that this is indeed the smallest denominator with this property. In , we are interested in the “best case” scenarios (Engel expansions of length two). We would like to employ similar methods used in this project to obtain a pattern for these length 2 expansions then prove this pattern holds. EGYPTIANFRACTIONEXPANSIONSUSINGENGELSERIES 11

7. Appendix

3 3 3 n Engel Series n Engel Series n Engel Series

3 2, 2 3 5, 7, 13 3 8, 11 4 { } 13 { } 22 { } 3 2, 5 3 5, 14 3 8, 23 5 { } 14 { } 23 { } 3 = 1 2 3 = 1 5 3 = 1 8 6 2 { } 15 5 { } 24 8 { } 3 3, 4, 7 3 6, 8 3 9, 13, 25 7 { } 16 { } 25 { } 3 3, 8 3 6, 17 3 9, 26 8 { } 17 { } 26 { } 3 = 1 3 3 = 1 6 3 = 1 9 9 3 { } 18 6 { } 27 9 { } 3 4, 5 3 7, 10, 19 3 10, 14 10 { } 19 { } 28 { } 3 4, 11 3 7, 20 3 10, 29 11 { } 20 { } 29 { } 3 = 1 4 3 = 1 7 3 = 1 10 12 4 { } 21 7 { } 30 10 { }

4 4 4 n Engel Series n Engel Series n Engel Series

4 2, 2, 5 4 = 1 4 4 7, 27 5 { } 16 4 { } 27 { } 4 = 2 2, 3 4 5, 6, 17 4 = 1 7 6 3 { } 17 { } 28 7 { } 4 2, 7 4 = 2 5, 9 4 8, 10, 29 7 { } 18 9 { } 29 { } 4 = 1 2 4 5, 19 4 = 2 8, 15 8 2 { } 19 { } 30 15 { } 4 3, 3 4 = 1 5 4 8, 31 9 { } 20 5 { } 31 { } 4 = 2 3, 5 4 6, 7 4 = 1 9 10 5 { } 21 { } 32 8 { } 4 3, 11 4 = 2 6, 11 4 9, 11 11 { } 22 11 { } 33 { } 4 = 1 3 4 6, 23 4 = 2 9, 17 12 3 { } 23 { } 34 17 { } 4 4, 5, 7, 13 4 = 1 6 4 9, 35 13 { } 24 6 { } 35 { } 4 = 2 4, 7 4 7, 9, 13, 25 4 = 1 9 14 7 { } 25 { } 36 9 { } 4 4, 15 4 = 2 7, 13 4 10, 13, 19, 37 15 { } 26 13 { } 37 { } 12 ELVIA NIDIA GONZALEZ´ AND JULIA BERGNER, PHD DEPARTMENT OF MATHEMATICS

5 5 5 n Engel Series n Engel Series n Engel Series

5 2, 2, 3 5 = 1 5 5 9, 44 6 { } 25 5 { } 44 { } 5 2, 3, 4, 7 5 6, 7, 13 5 = 1 9 7 { } 26 { } 45 9 { } 5 2, 4 5 6, 9 5 10, 12, 23 8 { } 27 { } 46 { } 5 2, 9 5 6, 14 5 10, 16, 47 9 { } 28 { } 47 { } 5 = 1 2 5 6, 29 5 10, 24 10 2 { } 29 { } 48 { } 5 3, 3, 11 5 = 1 6 5 10, 49 11 { } 30 6 { } 49 { } 5 3, 4 5 7, 8, 31 5 = 1 10 12 { } 31 { } 50 10 { } 5 3, 7, 13 5 7, 11, 32 5 11, 13, 51 13 { } 32 { } 51 { } 5 3, 14 5 7, 11, 33 5 11, 18, 26 14 { } 33 { } 52 { } 5 = 1 3 5 7, 34 5 11, 27, 53 15 3 { } 34 { } 53 { } 5 4, 4 5 = 1 7 5 11, 54 16 { } 35 7 { } 54 { } 5 4, 6, 17 5 8, 9 5 = 1 11 17 { } 36 { } 55 11 { } 5 4, 9 5 8, 13, 19, 37 5 12, 14 18 { } 37 { } 56 { } 5 4, 19 5 8, 19 5 12, 19 19 { } 38 { } 57 { } 5 = 1 4 5 8, 39 5 12, 29 20 4 { } 39 { } 58 { } 5 5, 6, 7 5 = 1 8 5 12, 99 21 { } 40 8 { } 59 { } 5 5, 8, 11 5 9, 11, 14, 41 5 = 1 12 22 { } 41 { } 60 12 { } 5 5, 12, 23 5 9, 14 5 5, 16, 21, 31, 61 23 { } 42 { } 61 { } 5 5, 24 5 9, 22, 43 24 { } 43 { } EGYPTIANFRACTIONEXPANSIONSUSINGENGELSERIES 13

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