egyptian fraction expansions

Egyptian Fraction Expansions

Elvia Nidia Gonzalez & Julia Bergner PhDDepartment of Mathematics

University of California, Riverside

What is an Egyptian Fraction Expansion?

• The Egyptians expressed their rational numbers as sums of unique unit fractions.

• A unit fraction has the number one in the numerator.


• The Egyptians expressed their rational numbers as sums of unique unit fractions.

• A unit fraction has the number one in the numerator.


= 110= 1

3

58 = 1

2 + 18

Were the Egyptians limited?


⇡ = 18 + 1

61 + 15020 + 1

128541455 + . . .


⇡ = 18 + 1

61 + 15020 + 1

128541455 + . . .

e = 12 + 1

5 + 155 + 1

9999 + 13620211523 . . .

Are Egyptian fraction expansions unique?


58 = 1

2 + 18


58 = 1

2 + 18

= 12 + 1

10 + 140


58 = 1

2 + 18

= 12 + 1

10 + 140

= 13 + 1

5 + 111 + 1

1320

Engel Expansion

Engel Expansion

u1 = x

n

a1 = d 1u1e

dre

Ceiling Function

The ceiling function is the least integer no less than than r

dred5.32e = 6

Ceiling Function


dre

d 12e = 1

d5.32e = 6

Ceiling Function


dre

d 12e = 1

d5.32e = 6

Ceiling Function


d� 12e = 0

dre

d 12e = 1

d5.32e = 6

Ceiling Function


d� 12e = 0

d3e = 3

Engel Expansion

u1 = x

n

a1 = d 1u1e

Engel Expansion

ui+1 = ui · ai � 1

ai+1 = d 1ui+1

e

Engel Expansion

uk = 0

Engel Expansion

uk = 0

{a1, a2, . . . , ak�1}

Engel Expansion

uk = 0x

n

=1

a1+

1

a1a2+ . . .

+1

a1a2 · · · ak�1

Examples

Let u1 = 38

Examples

then a1 = d 83e

Examples

then a1 = d 83e

= 3

Examples

u2 = u1 · a1 � 1

Examples

u2 = u1 · a1 � 1

= 38 · 3� 1

Examples

u2 = u1 · a1 � 1

= 38 · 3� 1

= 18

Examples

a2 = d 81e

Examples

a2 = d 81e

= 8

Examples

a2 = d 81e

= 8u3 = 1

8 · 8� 1

Examples

a2 = d 81e

= 8u3 = 1

8 · 8� 1

= 0

Examples

38 = 1

3 + 13·8

Examples

38 = 1

3 + 13·8

= 13 + 1

24

Engel Series expansions with lengths equal to their numerators

x

x

n

{a1, a2, . . . , ax}

2 23 {2, 3}

3 37 {3, 4, 7}

4 413 {4, 5, 7, 13}

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

6 661 {11, 13, 16, 21, 31, 61}

The Pattern

The Pattern

We conjectured and proved that

n = lcm(x, x� 1, . . . , 3, 2) + 1 will

produce a length x Engel expansion.

The Pattern

Is this the only “n” that works?


37 = 1

3 + 13·4 + 1

3·4·7


37 = 1

3 + 13·4 + 1

3·4·7313 = 1

5 + 15·7 + 1

5·7·13


37 = 1

3 + 13·4 + 1

3·4·7313 = 1

5 + 15·7 + 1

5·7·13319 = 1

7 + 17·10 + 1

7·10·19

The Second Pattern

x 2 N where N = {1, 2, 3, 4, . . . }

The Second Pattern

x 2 N where N = {1, 2, 3, 4, . . . }

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

k 2 N

The Second Pattern

We conjectured and proved that

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

produce an entire family of length x

Engel expansions for

x

n

2 (0, 1).

Families of expansions whose lengths equal their numerators

3n n Engel Expansion

37 7 = 1 · lcm(3, 2) + 1 1

3 + 13·4 + 1

3·4·7

313 13 = 2 · lcm(3, 2) + 1 1

5 + 15·7 + 1

5·7·13

319 19 = 3 · lcm(3, 2) + 1 1

7 + 17·10 + 1

7·10·19

325 25 = 4 · lcm(3, 2) + 1 1

9 + 19·13 + 1

9·13·25

331 31 = 5 · lcm(3, 2) + 1 1

11 + 111·16 + 1

11·16·31

337 37 = 6 · lcm(3, 2) + 1 1

13 + 113·19 + 1

13·19·37

Conjectures to Prove

• We would like to prove

n = lcm(x, x� 1, . . . , 3, 2) + 1 is the

least n that will produce a length x

Engel expansion for

x

n

2 (0, 1).

Related Questions

• What conditions produce

“best case” scenarios. That is,

expansions of length 2?

• Does n = k · lcm(x, x� 1, . . . , 3, 2) + 1produce a length x expansion for

x

n

< 0? Or x

n

> 1?

Acknowledgements

• Professor Julia Bergner, Department of Mathematics

• Maria Franco-Aguilar and Maggie Gover, MSRIP

• Diana Lizarraga and Chris Noble, UCLEADS

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