egyptian fraction expansions for rational numbers …nidia/writing/summer2011.pdf · 2 elvia nidia...

EGYPTIAN FRACTION EXPANSIONS FOR RATIONAL NUMBERS BETWEEN 0

AND 1 OBTAINED WITH ENGEL SERIES

ELVIA NIDIA GONZALEZ AND JULIA BERGNER, PHDDEPARTMENT OF MATHEMATICS

Abstract. The ancient Egyptians expressed rational numbers as the finite sum of distinct unit fractions.Were the Egyptians limited by this notation? In fact, they were not as every rational number can bewritten as a finite sum of distinct unit fractions. Moreover, these expansions are not unique. There existseveral di↵erent algorithms for computing Egyptian fraction expansions, all of which produce di↵erentrepresentations of the same rational number.

One such algorithm, called an Engel series, produces a finite increasing sequence of integers for everyrational number. This sequence is then used to obtain an Egyptian fraction expansion. Motivated by thework of M. Mays, this project aims to investigate properties of natural number denominators n that producelength x Egyptian fraction expansions using Engel series for x

n

between 0 and 1.While computing Engel expansions using Mathematica, a helpful pattern emerged: rational numbers x

n

which produce length x Egyptian fraction expansions were those whose denominator minus one was divisibleby every natural number between (and including) x and 2. We conjecture and prove that this will alwayshold for length x Engel series for x

n

between 0 and 1 where n = k · lcm(x, x � 1, . . . , 3, 2) + 1 for k 2 N.Furthermore, we conjecture that n = lcm(x, x � 1, . . . , 3, 2) + 1 is the smallest n such that x

n

produces alength x Egyptian fraction expansion. A proof that n = k · lcm(x, x�1, . . . , 3, 2)+1 works is included alongwith 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 ancientEgyptians. It is clear from this document that the Egyptians had an unintuitive way of expressing rationalnumbers. Unlike the current precedent, in which one integer is written over another integer, they would writeout rational numbers as a sum of distinct unit fractions. For example, 2

7 would be written as 14 +

128 instead.

Additionally, the Rhind papyrus included extensive tables, one of which included di↵erent representationsof 2

n

for odd n between 5 and 101 using the following identity [7]

1

n

=1

n+ 1+

1

n · (n+ 1).

However impractical these expansions may seem, they make certain real-life problems much easier tosolve. 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 eightpeople? The solution can be written as the proper fraction, 5

8 . This means each person gets five eighths ofa loaf. Now, applying this is not as easy as writing out the solution as a fraction. How would someone cutfive-eighths from eight loaves? Each loaf could be cut in half, then those halves into quarters, then thosequarters into eighths. From here, each of the eight persons would receive five small pieces of bread. Thissolution, although correct, is not very practical.

If we represent 58 as 1

2 + 18 , 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, Number Theory.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 withbeing 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 thatEgyptian fraction expansions are not unique. Di↵erent algorithms may produce di↵erent representationsof the same fraction. Our example of 5

8 can also be expressed as 12 + 1

10 + 140 . Table 1 lists several more

examples.

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

x

n

Engel Series Greedy Algorithm Continued Fraction

57

12 + 1

6 + 124 + 1

16812 + 1

5 + 170

12 + 1

6 + 121

413

14 + 1

20 + 1140 + 1

182014 + 1

18 + 1468

14 + 1

28 + 170 + 1

130

718

13 + 1

1813 + 1

1813 + 1

24 + 1104 + 1

234

Another question that arose from Egyptian fraction expansions inquired whether or not the Egyptianswere limited by the use of this notation. That is, could they express any rational number as a sum of distinctunit fraction? The Egyptians, as it turns out, were not at all limited. Any rational number can be expressedas the finite sum of distinct unity fractions. Furthermore, even irrational numbers can be expressed as aninfinite sum of distinct unit fractions.

The focus of this project is Egyptian fraction expansions for rational numbers between 0 and 1 (but notincluding 0 and 1) obtained using an Engel series. Some examples of these expansions can be seen on thesecond column of Table 1.

2. Engel Expansions

An Engel series, for any real number, is sequence of increasing integers [6]. Any real number, say z, canbe expressed as an Egyptian fraction expansion using an Engel series in a unique way. This representationfor any real z can be computed by defining u1 = z then letting a1 = d 1

u1e.From here, each subsequent a

i+1

and u

i+1 is obtained by first computing u

i+1 = u

i

· ai

� 1 then computing a

i+1 = d 1u

i

e, where d e is theceiling function defined as dre = s if and only if s� 1 < r s [2, Page 69, (3.5)].

If z is a rational number, this process will halt whenever uk

= 0 and the Engel series is {a1, a2, . . . , ak�1}.The Egyptian fraction expansion for a rational number is derived from the series as follows:

z =k�1X

i=1

1

a1 · a2 · · · ai=

1

a1+

1

a1a2+ · · ·+ 1

a1a2 · · · ak�1.

Otherwise, if z is irrational, this algorithm continues and we end up with an infinite Engel series {a1, a2, a3, . . . }.This infinite Engel series can be used to obtain an Egyptian fraction expansion with the following sum [3]:

1X

i=1

1

a1 · a2 · · · ai.

Any Egyptian fraction expansion derived using an Engel series will be referred to as an Engel expansionfrom 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.

EGYPTIAN FRACTION EXPANSIONS USING ENGEL SERIES 3

Proof. Suppose m 2 Q. Then m = x

n

for x, n 2 Z and n 6= 0. Begin to compute the Engel expansion by firstletting

u1 =x

n

and a1 = dnx

e.

Next, compute u2 using a1 and u1 to obtain

u2 = u1 · a1 � 1 = (x

n

· a1)� 1 =x · a1 � n

n

.

By the definition of the ceiling function, d 1u1e = dn

x

e = a1 if and only if a1�1 <

n

x

a1. Thus a1�1 <

n

x

if and only if x · a1 � x < n. Subtracting n from both sides of the inequality and adding x to both leaves uswith x · a1 � n < x. The numerator for u2 is x · a1 � n. From the inequalities above, x · a1 � n < x which isthe numerator for u1.

We can now find another positive integer a2 such that a2(x · a1 � n)� n < x · a1 � n. The left hand sideof this inequality is precisely the numerator for u3 and the right hand side is precisely the numerator for u2.So no matter which u

i

we start with, the numerator for u

i+1 will be less than the numerator for u

i

. Sincethe numerators for the u

i

’s are strictly decreasing, eventually we will end up with zero in the numerator forsome u

i

(say u

k

has a numerator of 0) and this process halts with

k�1X

i=1

1

a1 · a2 · · · ai

which is a finite Engel expansion as desired. ⇤

3. Results

The work within this project was inspired by a paper by Michael E. Mays titled, “A Worst Case ofthe Fibonacci-Sylvester Expansion” [4]. Mays explored Egyptian fraction expansions of rational numbersusing the greedy algorithm. More specifically, his paper investigated the properties of fraction expansionswhich had lengths that matched the numerator using the greedy algorithm. In a similar manner, this paperinvestigates 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 fewexamples of Engel expansions whose lengths match their numerators. Each table begins with a denominatorthat is one greater than the numerator and ends when an Engel expansion of length equal to the numerator isobtained. The first column is simply the rational number being investigated, the second column is the Engelseries, and the third column is the Engel expansion 1. It was through tables such as these that a patternfirst emerged. First, we noticed every integer between (and including) x and 2 had to divide one minus thedenominator (that is, n�1). This observation led us to conjecture that as long as n = lcm(x, x�1, . . . , 3, 2)+1,the Engel series produced for x

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 indeedalways produce a length x Engel expansion. We also conjecture (but do not prove) that this n is the leastsuch n with this property. Instead, we logically demonstrate that n = lcm(x, x� 1, . . . , 3, 2) + 1 is the leastdenominator that produces a six term Engel expansion for 6

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 thatany denominator equal to 1 plus any multiple of lcm(x, x� 1, . . . , 3, 2) will produce a length x expansion forx

n

.

1The appendix contains similar tables with more details and for larger numerators. Since the denominators can get unrulyfairly quickly (see the last row of Table 4 for a good demonstration of this) the majority of the tables will include the Engelseries rather than the Engel expansion for rational numbers.


Table 2. Engel expansions for 3n

3n

Engel Series Engel Expansion

34 {2, 2} 1

2 + 14

35 {2, 5} 1

2 + 110

36 = 1

2 {2} 12

37 {3, 4, 7} 1

3 + 112 + 1

84

Table 3. Engel expansions for 4n

4n


45 {2, 2, 5} 1

2 + 14 + 1

20

23 {2, 3} 1

2 + 16

47 {2, 7} 1

2 + 114

48 = 1

2 {2} 12

49 {3, 3} 1

3 + 19

25 {3, 5} 1

3 + 115

411 {3, 11} 1

3 + 133

412 = 1

3 {3} 13

413 {4, 5, 7, 13} 1

4 + 120 + 1

140 + 11820

Table 4. Length x Engel series and Engel expansions for x

n

.

x

x

n


2 23 {2, 3} 1

2 + 16

3 37 {3, 4, 7} 1

3 + 112 + 1

84

4 413 {4, 5, 7, 13} 1

4 + 120 + 1

140 + 11820

5 561 {13, 16, 21, 31, 61} 1

13 + 1208 + 1

4,368 + 1135,408 + 1

8,259,888


Table 5. Other possible denominators that produce a length x Engel Expansion for x

n

when x = 3

3n

n Engel Series

37 7 = 1 · lcm(3, 2) + 1 {3, 4, 7}

313 13 = 2 · lcm(3, 2) + 1 {5, 7, 13}

319 19 = 3 · lcm(3, 2) + 1 {7, 10, 19}

325 25 = 4 · lcm(3, 2) + 1 {9, 13, 25}

331 31 = 5 · lcm(3, 2) + 1 {11, 16, 31}

337 37 = 6 · lcm(3, 2) + 1 {13 · 19 · 37}

4. Theorems and Proofs

This section will include proofs of conjectures made within the previous section. First we prove thatn = k · lcm(x, x� 1, . . . , 3, 2) + 1 will produce a length x Engel expansion.

Theorem 4.1. Suppose x is a positive integer and n = k · lcm(x, x � 1, . . . , 3, 2) + 1. Then x

n

will have alength x Engel expansion.

In order to prove Theorem 3.1, we need a lemma first.

Lemma 4.2. [2, Page 69, (3.6)]

dr + se = dre+ s or br + sc = brc+ s for s 2 Z.

Proof Of Theorem 4.1. Suppose n = k · lcm(x, x� 1, . . . , 3, 2)+1 and x

n

2 (0, 1) where x 2 N. To obtain theEngel expansion of x

n

begin by letting

u1 =x

n

.

By assumption, n = k · lcm(x, x� 1, . . . , 3, 2) + 1 hence

u1 =x

k · lcm(x, x� 1, . . . , 3, 2) + 1.

Next, a1 = d 1u1e which can be rewritten as

a1 = dk · lcm(x, x� 1, . . . , 3, 2) + 1

x

e = dk · lcm(x, x� 1, . . . , 3, 2)

x

+1

x

e.

The result of dividing k · lcm(x, x � 1, . . . , 3, 2) by x will be a natural number so by the lemma, a1 can berewritten again as

a1 =k · lcm(x, x� 1, . . . , 3, 2)

x

+ d 1x

e.

Since x 2 N, 1x

will always be a positive unit fraction and d 1x

e will always equal 1. Again then, a1 can berewritten as

a1 =k · lcm(x, x� 1, . . . , 3, 2)

x

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

x

.


Continuing to find the Engel expansion requires computing u2 = u1 · a1 � 1 which becomes

u2 =x

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

x

� 1

=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 thanthe numerator of u1. Indeed, u3 will have a numerator of x � 2, u4 will have a numerator of x � 3, and soon. Eventually, The u

x+1 term will have a numerator of 0 and this process halts with the Egyptian fractionexpansion x

n

= 1a1

+ 1a1·a2

+ · · ·+ 1a1a2···ax

as desired. ⇤

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 theleast denominator that produces the desired length expansion for x

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]

drs

e = br + s� 1

s

c

Lemma 5.3. [2, 8]

dre = r , r 2 Z , brc = r

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

5.1. Case 1: n = 6p. Suppose n = 6p. Then 6n

= 66p = 1

p

. This is a one-term Engel expansion. Hence thefollowing numbers are eliminated from consideration for n are 12, 18, 24, 30, 36, 42, 48, 54, 60.


= 63p = 2

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 2 N. Thus

6

n

=6

3p=

6

3 · 2p0 =6

6 · p0 =1

p

as in case 1.


5.2.2. Subcase 2: p is odd. If p is odd then p = 2p0 + 1. Now apply Engel’s algorithm.

u1 =6

n

=6

3p=

2

p

=2

2p0 + 1

a1 = d2p0 + 1

2e

By Lemma 5.2, a1 can be rewritten using the floor function:

a1 = b (2p0 + 1) + 2� 1

2c = b2p

0 + 2

2c = bp0 + 2c.

Next, use Lemma 4.2 to separate the 2 from the floor function:

a1 = bp0c+ 2.

Finally, since p

0 2 N ⇢ Z by Lemma 5.3, bp0c = p

0 and a1 = p

0 + 2. Find u2 by continuing the algorithm forEngel series:

u2 = a1 · u1 � 1 = (p0 + 1) · 2

2p0 + 1� 1 =

1

2p0 + 1To find a2 take the ceiling of the reciprocal of u1:

a2 = d2p0 + 1e = d2p0e+ 1 = 2p0 + 1.

Repeat the algorithm once more to obtain u3 = a2 · u2 � 1 = 1p

0+1 · (2p0 + 1)� 1 = 0. Since u3 = 0, we haltand the Engel expansion is

6

n

=6

3p=

6

3(2p0 + 1)=

2

2p0 + 1=

1

p

0 + 1+

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.


= 62p = 3

p

.

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 2 N and 6n

= 62·3·p0 =

66p0

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

n

=6

2 · p =6

2(3 · p0 + 2)=

3

3 · p0 + 2= u1

so

a1 = d3p0 + 2

3e = b (3p

0 + 2) + 3� 1

3c = b3p

0 + 4

3c = bp0 + 4

3c.

Since p 2 N ⇢ Z, by Lemma 4.2, ai

can be rewritten as:

a1 = b43c+ p

0 = p

0 + 1.

Next, continue by finding u2 and a2:

u2 = u1 · a1 � 1 =3

3p0 + 2· (p0 + 1)� 1 =

1

3p0 + 2

anda2 = d3p0 + 2e = d3p0e+ 2 = 3p0 + 2.

Finally, u3 = u2 · a2 � 1 = 13p0+2 · (3p0 + 2) � 1 = 0 and the algorithm halts. The Engel expansion for this

case becomes:6

n

=6

2(3p0 + 2)=

1

p

0 + 1+

1

(p0 + 1) · (3p0 + 2),


which is less than six terms long. Now eliminated as possibilities for n are 10, 16, 22, 28, 34, 40, 46, 52, and58.

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 willexplore 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 p

0 2 N and

u1 =6

n

=6

2p=

3

p

=3

3p0 + 1.

Next, use u1 to find a1:

a1 = d3p0 + 1

3e = b (3p

0 + 1) + 3� 1

3c = b3p

0 + 3

3c = bp0 + 1c = bp0c+ 1 = p

0 + 1.

Continue by using a1 and u1 to find:

u2 =3

3p0 + 1· (p0 + 1)� 1 =

2

3p0 + 1

and

a2 = d3p0 + 1

2e

Because p

0 is odd and positive it can be rewritten as p

0 = 2p00 + 1 (again p

00 2 N). I can now continue torewrite a2 as follows:

a2 = d3(2p00 + 1) + 1

2e = d6p

00 + 4

2e = d3p00 + 2e = d3p00e+ 2 = 3p00 + 2.

Since p

0 = 2p00 +1, this can be rewritten as p00 = p

0�12 . This allows a2 to be rewritten one last time in terms

of p0:

a2 = 3p00 + 2 = 3(p

0 + 1

2) + 2 =

3p0 + 1

2.

The next iteration will produce u3 = 23p0+1 · 3p0+1

2 � 1 = 0 and the algorithm halts. For this sub case, theEngel expansion is:

6

n

=6

2p=

3

p

=1

p

0 + 1+

1

(p0 + 1) · (3p00 + 2).

This is still not a six term Engel expansion so 8, 20, 32, 44, and 56 have been eliminated as possible optionsfor the denominator of 6

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:

u2 =3

3p0 + 1· (p0 + 1)� 1 =

3p0 + 3� 3p0 � 1

3p0 + 1=

2

3p0 + 1

and

a2 = d3p0 + 1

2e.

In order to simplify a2 further, use the fact that p0 is even so it can be rewritten as p0 = 2p00 for some p00 2 N.Thus we get:

a2 = d3 · 2p00 + 1

2e = b (6p

00 + 1) + 2� 1

2c = b6p

00 + 2

2c = b3p00 + 1c = b3p00c+ 1 = 3p00 + 1.

Furthermore, since p

0 = 2p00 we can rewrite this as p00 = p

0

2 . This allows us to rewrite a2 as:

a2 =3p0

2+ 1 =

3p0 + 2

2.


Then continue with the algorithm to find:

u3 =2

3p0 + 1· 3p

0 + 2

2� 1 =

1

3p0 + 1

and

a3 = d 1

u3e = 3p0 + 1.

Finally u4 = 13p0+1 · 3p0+1

1 � 1 = 0 and we halt. This produces the following three-term Engel expansion:

6

n

=6

2p=

3

p

=1

p

0 + 1+

1

(p0 + 1) · (3p00 + 1)+

1

(p0 + 1) · (3p00 + 1) · (3p0 + 1).

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 2 N. First compute:

u1 =6

n

=6

6p+ 5

and

a1 = d6p+ 5

6e = b (6p+ 5) + 6� 1

6c = bp+ 10

6c = b10

6c+ p = p+ 1.

Next,

u2 =6

6p+ 5· (p+ 1)� 1 =

1

6p+ 5and

a2 = d6p+ 5

1e = d6pe+ 5 = 6p+ 5.

Finally, u3 = 16p+5 · (6p+ 5)� 1 = 0. This produces the following Engel expansion:

6

5=

1

p+ 1+

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 possibledenominators for 6

n

.

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

.

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}


5.4.2. n ⌘ 1 mod 6. The remaining nine possible denominators are eliminated by brute force in Table 6.Finally, the smallest possible n such that 6

n

produces a length 6 Engel expansion is 61 = lcm(6, 5, 4, 3, 2)+ 1as desired. ⇤

6. Conclusion

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


7. Appendix

3n

Engel Series

3n

Engel Series

3n

Engel Series

34 {2, 2} 3

13 {5, 7, 13} 322 {8, 11}

35 {2, 5} 3

14 {5, 14} 323 {8, 23}

36 = 1

2 {2} 315 = 1

5 {5} 324 = 1

8 {8}

37 {3, 4, 7} 3

16 {6, 8} 325 {9, 13, 25}

38 {3, 8} 3

17 {6, 17} 326 {9, 26}

39 = 1

3 {3} 318 = 1

6 {6} 327 = 1

9 {9}

310 {4, 5} 3

19 {7, 10, 19} 328 {10, 14}

311 {4, 11} 3

20 {7, 20} 329 {10, 29}

312 = 1

4 {4} 321 = 1

7 {7} 330 = 1

10 {10}

4n

Engel Series

4n

Engel Series

4n

Engel Series

45 {2, 2, 5} 4

16 = 14 {4} 4

27 {7, 27}

46 = 2

3 {2, 3} 417 {5, 6, 17} 4

28 = 17 {7}

47 {2, 7} 4

18 = 29 {5, 9} 4

29 {8, 10, 29}

48 = 1

2 {2} 419 {5, 19} 4

30 = 215 {8, 15}

49 {3, 3} 4

20 = 15 {5} 4

31 {8, 31}

410 = 2

5 {3, 5} 421 {6, 7} 4

32 = 18 {9}

411 {3, 11} 4

22 = 211 {6, 11} 4

33 {9, 11}

412 = 1

3 {3} 423 {6, 23} 4

34 = 217 {9, 17}

413 {4, 5, 7, 13} 4

24 = 16 {6} 4

35 {9, 35}

414 = 2

7 {4, 7} 425 {7, 9, 13, 25} 4

36 = 19 {9}

415 {4, 15} 4

26 = 213 {7, 13} 4

37 {10, 13, 19, 37}


5n

Engel Series

5n

Engel Series

5n

Engel Series

56 {2, 2, 3} 5

25 = 15 {5} 5

44 {9, 44}

57 {2, 3, 4, 7} 5

26 {6, 7, 13} 545 = 1

9 {9}

58 {2, 4} 5

27 {6, 9} 546 {10, 12, 23}

59 {2, 9} 5

28 {6, 14} 547 {10, 16, 47}

510 = 1

2 {2} 529 {6, 29} 5

48 {10, 24}

511 {3, 3, 11} 5

30 = 16 {6} 5

49 {10, 49}

512 {3, 4} 5

31 {7, 8, 31} 550 = 1

10 {10}

513 {3, 7, 13} 5

32 {7, 11, 32} 551 {11, 13, 51}

514 {3, 14} 5

33 {7, 11, 33} 552 {11, 18, 26}

515 = 1

3 {3} 534 {7, 34} 5

53 {11, 27, 53}

516 {4, 4} 5

35 = 17 {7} 5

54 {11, 54}

517 {4, 6, 17} 5

36 {8, 9} 555 = 1

11 {11}

518 {4, 9} 5

37 {8, 13, 19, 37} 556 {12, 14}

519 {4, 19} 5

38 {8, 19} 557 {12, 19}

520 = 1

4 {4} 539 {8, 39} 5

58 {12, 29}

521 {5, 6, 7} 5

40 = 18 {8} 5

59 {12, 99}

522 {5, 8, 11} 5

41 {9, 11, 14, 41} 560 = 1

12 {12}

523 {5, 12, 23} 5

42 {9, 14} 561 {5, 16, 21, 31, 61}

524 {5, 24} 5

43 {9, 22, 43}


References

1. Erdos, Shallit. “New bounds on the length of finite Pierce and Engel Series.” Seminaire de Theorie des Nombres Bordeaux

3 (1991), 43-53.2. Graham, Knuth, Patashink. Concrete Mathematics. Addison Wesley, Massachusetts, 1st Edition (1989), Chapter 3.3. V. Laohakosol, T. Chaichana, J. Rattanamoong, and N.R. Kanasri. “Engel Series and Cohen-Egyptian Fraction Expan-

sions.” Hindawi Publishing Corporation. International Journal of Mathematics and Mathematical Sciences. Article ID

865705 (2009), 15 pages.4. Mays, Michael E. “A Worst Case of the Fibonacci-Sylvester Expansion.” The Journal of Combinatorial Mathematics and

Combinatorial Computing. 1 (1987), 141-148.5. O’Reilly, Declan “Creating Egyptian Fractions” Mathematics in School. Vol. 21, No. 5 (Nov., 1992), Page 41.6. Renyi, A. “A New Approach to the Theory of Engel’s Series.” Annales Universitatis Scientiarum Budapestinensis de

Rolando Eotvos Nominatae. Volume 5 (1962), 25-32.7. Weisstein, Eric W. “Egyptian Fraction.” From MathWorld–A Wolfram Web Resource.

http://mathworld.wolfram.com/EgyptianFraction.html

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