Some applications of combinatorics on words in Number Theory

Amy Glen

School of Engineering & IT Murdoch University, Perth, Australia [email protected] http://amyglen.wordpress.com

Number Theory Down Under 4 @ The University of Newcastle September 23–26, 2016

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 1 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 001001001001001001001001001001 ···

I 1100111100011011101111001101110010111111101 ···

I 100102110122220102110021111102212222201112012 ···

I 0123456789101112131415 ···

I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... ↑ I 1100111100011011101111001101110010111111101 ···

I 100102110122220102110021111102212222201112012 ···

I 0123456789101112131415 ···

I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2

I 1100111100011011101111001101110010111111101 ···

I 100102110122220102110021111102212222201112012 ···

I 0123456789101112131415 ···

I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2

I 1.100111100011011101111001101110010 ... ↑ I 100102110122220102110021111102212222201112012 ···

I 0123456789101112131415 ···

I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2 √ I 1.100111100011011101111001101110010 ... = ((1 + 5)/2)2

I 100102110122220102110021111102212222201112012 ···

I 0123456789101112131415 ···

I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2 √ I 1.100111100011011101111001101110010 ... = ((1 + 5)/2)2

I 10.0102110122220102110021111102212222201112012 ... ↑ I 0123456789101112131415 ···

I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2 √ I 1.100111100011011101111001101110010 ... = ((1 + 5)/2)2

I 10.0102110122220102110021111102212222201112012 ... =( π)3

I 0123456789101112131415 ···

I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2 √ I 1.100111100011011101111001101110010 ... = ((1 + 5)/2)2

I 10.0102110122220102110021111102212222201112012 ... =( π)3

I 0.123456789101112131415 ... ↑ I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2 √ I 1.100111100011011101111001101110010 ... = ((1 + 5)/2)2

I 10.0102110122220102110021111102212222201112012 ... =( π)3

I 0.123456789101112131415 ... = Champernowne’s number (C10)

I 1121212121212 ···

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2 √ I 1.100111100011011101111001101110010 ... = ((1 + 5)/2)2

I 10.0102110122220102110021111102212222201112012 ... =( π)3

I 0.123456789101112131415 ... = Champernowne’s number (C10) √ I [1; 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2,...] = 3

I 212114116118 ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2 √ I 1.100111100011011101111001101110010 ... = ((1 + 5)/2)2

I 10.0102110122220102110021111102212222201112012 ... =( π)3

I 0.123456789101112131415 ... = Champernowne’s number (C10) √ I [1; 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2,...] = 3

I [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8,...]

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 Finite & Infinite Words Words By a word, I mean a finite or infinite sequence of symbols (letters) taken from a non-empty countable set A (alphabet). Examples

I 001

∞ I (001) = 0.01001001001001001001001001001 ... = (2/7)2 √ I 1.100111100011011101111001101110010 ... = ((1 + 5)/2)2

I 10.0102110122220102110021111102212222201112012 ... =( π)3

I 0.123456789101112131415 ... = Champernowne’s number (C10) √ I [1; 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2,...] = 3

I [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1,...] = e

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 2 For instance:

I Expansions of real numbers in integer bases (e.g., binary and decimal expansions) or continued fraction expansions allow us to associate with every real number a finite or infinite sequence of digits.

I Combinatorial group theory involves the study of words that represent group elements.

Finite & Infinite Words Words...

In mathematics, words naturally arise when one wants to represent elements from some set in a systematic way.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 3 I Combinatorial group theory involves the study of words that represent group elements.

Finite & Infinite Words Words...

In mathematics, words naturally arise when one wants to represent elements from some set in a systematic way. For instance:

I Expansions of real numbers in integer bases (e.g., binary and decimal expansions) or continued fraction expansions allow us to associate with every real number a finite or infinite sequence of digits.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 3 Finite & Infinite Words Words...

In mathematics, words naturally arise when one wants to represent elements from some set in a systematic way. For instance:

I Expansions of real numbers in integer bases (e.g., binary and decimal expansions) or continued fraction expansions allow us to associate with every real number a finite or infinite sequence of digits.

I Combinatorial group theory involves the study of words that represent group elements.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 3 A palindrome is a word that reads the same backwards as forwards. Examples: eye, civic, radar

Basic measure: number of distinct blocks (factors) of each length occurring in the word.

I Most commonly studied words are those which satisfy one or more strong regularity properties; for instance, words containing many repetitions or palindromes.

I The extent to which a word exhibits strong regularity properties is generally inversely proportional to its “complexity”.

Finite & Infinite Words Words...

I Depending on the problem to be solved, it may be fruitful to study combinatorial and structural properties of the words representing the elements of a particular set or to impose certain combinatorial conditions on such words.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 4 Basic measure: number of distinct blocks (factors) of each length occurring in the word.

A palindrome is a word that reads the same backwards as forwards. Examples: eye, civic, radar

I The extent to which a word exhibits strong regularity properties is generally inversely proportional to its “complexity”.

Finite & Infinite Words Words...

I Depending on the problem to be solved, it may be fruitful to study combinatorial and structural properties of the words representing the elements of a particular set or to impose certain combinatorial conditions on such words.

I Most commonly studied words are those which satisfy one or more strong regularity properties; for instance, words containing many repetitions or palindromes.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 4 Basic measure: number of distinct blocks (factors) of each length occurring in the word.

I The extent to which a word exhibits strong regularity properties is generally inversely proportional to its “complexity”.

Finite & Infinite Words Words...

I Depending on the problem to be solved, it may be fruitful to study combinatorial and structural properties of the words representing the elements of a particular set or to impose certain combinatorial conditions on such words.

I Most commonly studied words are those which satisfy one or more strong regularity properties; for instance, words containing many repetitions or palindromes. A palindrome is a word that reads the same backwards as forwards. Examples: eye, civic, radar

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 4 Basic measure: number of distinct blocks (factors) of each length occurring in the word.

Finite & Infinite Words Words...

I Depending on the problem to be solved, it may be fruitful to study combinatorial and structural properties of the words representing the elements of a particular set or to impose certain combinatorial conditions on such words.

I Most commonly studied words are those which satisfy one or more strong regularity properties; for instance, words containing many repetitions or palindromes. A palindrome is a word that reads the same backwards as forwards. Examples: eye, civic, radar

I The extent to which a word exhibits strong regularity properties is generally inversely proportional to its “complexity”.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 4 Finite & Infinite Words Words...

I Depending on the problem to be solved, it may be fruitful to study combinatorial and structural properties of the words representing the elements of a particular set or to impose certain combinatorial conditions on such words.

I Most commonly studied words are those which satisfy one or more strong regularity properties; for instance, words containing many repetitions or palindromes. A palindrome is a word that reads the same backwards as forwards. Examples: eye, civic, radar

I The extent to which a word exhibits strong regularity properties is generally inversely proportional to its “complexity”. Basic measure: number of distinct blocks (factors) of each length occurring in the word.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 4 √ n Conjecture: Cx(n) = 2 for all n as it is believed 2 is normal in base 2.

F1(x) = {0, 1}, Cx(1) = 2

F2(x) = {00, 01, 10, 11}, Cx(2) = 4

F3(x) = {000, 001, 010, 100, 101, 110, 111}, Cx(3) = 8

Example √ x = ( 2)2 = 1.0110101000001001111 ...

Finite & Infinite Words Complexity Factor Complexity

The factor complexity function of a finite or infinite word w, denoted by Cw(n), counts the number of distinct factors of w of each length n.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 5 √ n Conjecture: Cx(n) = 2 for all n as it is believed 2 is normal in base 2.

F1(x) = {0, 1}, Cx(1) = 2

F2(x) = {00, 01, 10, 11}, Cx(2) = 4

F3(x) = {000, 001, 010, 100, 101, 110, 111}, Cx(3) = 8

Finite & Infinite Words Complexity Factor Complexity

The factor complexity function of a finite or infinite word w, denoted by Cw(n), counts the number of distinct factors of w of each length n.

Example √ x = ( 2)2 = 1.0110101000001001111 ...

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 5 √ n Conjecture: Cx(n) = 2 for all n as it is believed 2 is normal in base 2.

F2(x) = {00, 01, 10, 11}, Cx(2) = 4

F3(x) = {000, 001, 010, 100, 101, 110, 111}, Cx(3) = 8

Finite & Infinite Words Complexity Factor Complexity

The factor complexity function of a finite or infinite word w, denoted by Cw(n), counts the number of distinct factors of w of each length n.

Example √ x = ( 2)2 = 1.0110101000001001111 ...

F1(x) = {0, 1}, Cx(1) = 2

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 5 √ n Conjecture: Cx(n) = 2 for all n as it is believed 2 is normal in base 2.

F3(x) = {000, 001, 010, 100, 101, 110, 111}, Cx(3) = 8

Finite & Infinite Words Complexity Factor Complexity

The factor complexity function of a finite or infinite word w, denoted by Cw(n), counts the number of distinct factors of w of each length n.

Example √ x = ( 2)2 = 1.0110101000001001111 ...

F1(x) = {0, 1}, Cx(1) = 2

F2(x) = {00, 01, 10, 11}, Cx(2) = 4

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 5 √ n Conjecture: Cx(n) = 2 for all n as it is believed 2 is normal in base 2.

Finite & Infinite Words Complexity Factor Complexity

The factor complexity function of a finite or infinite word w, denoted by Cw(n), counts the number of distinct factors of w of each length n.

Example √ x = ( 2)2 = 1.0110101000001001111 ...

F1(x) = {0, 1}, Cx(1) = 2

F2(x) = {00, 01, 10, 11}, Cx(2) = 4

F3(x) = {000, 001, 010, 100, 101, 110, 111}, Cx(3) = 8

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 5 Finite & Infinite Words Complexity Factor Complexity

The factor complexity function of a finite or infinite word w, denoted by Cw(n), counts the number of distinct factors of w of each length n.

Example √ x = ( 2)2 = 1.0110101000001001111 ...

F1(x) = {0, 1}, Cx(1) = 2

F2(x) = {00, 01, 10, 11}, Cx(2) = 4 F (x) = {000, 001, 010, 100, 101, 110, 111}, C (3) = 8 3 x √ n Conjecture: Cx(n) = 2 for all n as it is believed 2 is normal in base 2.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 5 That is: w is aperiodic ⇔ Cw(n) ≥ n + 1 for all n ∈ N.

I An infinite word w is called Sturmian if and only if Cw(n) = n + 1 for each n.

I Sturmian words are the aperiodic infinite words of minimal complexity.

I Their low complexity accounts for many interesting features, as it induces certain regularities in such words without, however, making them periodic.

I They have numerous equivalent definitions and characterisations . . .

Finite & Infinite Words Complexity Complexity & Periodicity

Theorem (Morse-Hedlund 1940)

An infinite word w is eventually periodic if and only if Cw(n) ≤ n for + some n ∈ N .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 6 I An infinite word w is called Sturmian if and only if Cw(n) = n + 1 for each n.

I Sturmian words are the aperiodic infinite words of minimal complexity.

I Their low complexity accounts for many interesting features, as it induces certain regularities in such words without, however, making them periodic.

I They have numerous equivalent definitions and characterisations . . .

Finite & Infinite Words Complexity Complexity & Periodicity

Theorem (Morse-Hedlund 1940)

An infinite word w is eventually periodic if and only if Cw(n) ≤ n for + some n ∈ N .

That is: w is aperiodic ⇔ Cw(n) ≥ n + 1 for all n ∈ N.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 6 I Sturmian words are the aperiodic infinite words of minimal complexity.

I Their low complexity accounts for many interesting features, as it induces certain regularities in such words without, however, making them periodic.

I They have numerous equivalent definitions and characterisations . . .

Finite & Infinite Words Complexity Complexity & Periodicity

Theorem (Morse-Hedlund 1940)

An infinite word w is eventually periodic if and only if Cw(n) ≤ n for + some n ∈ N .

That is: w is aperiodic ⇔ Cw(n) ≥ n + 1 for all n ∈ N.

I An infinite word w is called Sturmian if and only if Cw(n) = n + 1 for each n.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 6 I Their low complexity accounts for many interesting features, as it induces certain regularities in such words without, however, making them periodic.

I They have numerous equivalent definitions and characterisations . . .

Finite & Infinite Words Complexity Complexity & Periodicity

Theorem (Morse-Hedlund 1940)

An infinite word w is eventually periodic if and only if Cw(n) ≤ n for + some n ∈ N .

That is: w is aperiodic ⇔ Cw(n) ≥ n + 1 for all n ∈ N.

I An infinite word w is called Sturmian if and only if Cw(n) = n + 1 for each n.

I Sturmian words are the aperiodic infinite words of minimal complexity.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 6 I They have numerous equivalent definitions and characterisations . . .

Finite & Infinite Words Complexity Complexity & Periodicity

Theorem (Morse-Hedlund 1940)

An infinite word w is eventually periodic if and only if Cw(n) ≤ n for + some n ∈ N .

That is: w is aperiodic ⇔ Cw(n) ≥ n + 1 for all n ∈ N.

I An infinite word w is called Sturmian if and only if Cw(n) = n + 1 for each n.

I Sturmian words are the aperiodic infinite words of minimal complexity.

I Their low complexity accounts for many interesting features, as it induces certain regularities in such words without, however, making them periodic.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 6 Finite & Infinite Words Complexity Complexity & Periodicity

Theorem (Morse-Hedlund 1940)

An infinite word w is eventually periodic if and only if Cw(n) ≤ n for + some n ∈ N .

That is: w is aperiodic ⇔ Cw(n) ≥ n + 1 for all n ∈ N.

I An infinite word w is called Sturmian if and only if Cw(n) = n + 1 for each n.

I Sturmian words are the aperiodic infinite words of minimal complexity.

I Their low complexity accounts for many interesting features, as it induces certain regularities in such words without, however, making them periodic.

I They have numerous equivalent definitions and characterisations . . .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 6 I Words over a 2-letter alphabet {a, b} that are factors of (infinite) Sturmian words are called finite Sturmian words – they are the cyclic shifts of so-called Christoffel words which can be obtained via the following construction.

Finite & Infinite Words Sturmian words Constructing Sturmian words

I Let’s consider a nice geometric realisation, starting with a special class of finite words . . .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 7 Finite & Infinite Words Sturmian words Constructing Sturmian words

I Let’s consider a nice geometric realisation, starting with a special class of finite words . . .

I Words over a 2-letter alphabet {a, b} that are factors of (infinite) Sturmian words are called finite Sturmian words – they are the cyclic shifts of so-called Christoffel words which can be obtained via the following construction.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 7 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

L(3, 5) = a

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

L(3, 5) = aa

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

L(3, 5) = aab

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

L(3, 5) = aaba

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

L(3, 5) = aabaa

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

L(3, 5) = aabaab

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

L(3, 5) = aabaaba

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower Christoffel word of slope 5

L(3, 5) = aabaabab

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 8 Finite & Infinite Words Sturmian words Christoffel words: Construction by example

3 Lower & Upper Christoffel words of slope 5

L(3, 5) = aabaabab U(3, 5) = babaabaa

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 9 I f = abaababaabaababaaba ··· (note: disregard 1st a in construction) √ 5−1 I Characteristic Sturmian word of slope 2 , golden ratio conjugate

√ 5−1 Example: y = 2 x −→ Fibonacci word f

Finite & Infinite Words Sturmian words From Christoffel words to Sturmian words Sturmian words are obtained *similarly* by replacing the line segment by a half-line: y = αx + ρ with irrational α ∈ (0, 1), ρ ∈ R.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 10 I f = abaababaabaababaaba ··· (note: disregard 1st a in construction) √ 5−1 I Characteristic Sturmian word of slope 2 , golden ratio conjugate

Finite & Infinite Words Sturmian words From Christoffel words to Sturmian words Sturmian words are obtained *similarly* by replacing the line segment by a half-line: y = αx + ρ with irrational α ∈ (0, 1), ρ ∈ R.

√ 5−1 Example: y = 2 x −→ Fibonacci word f

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 10 √ 5−1 I Characteristic Sturmian word of slope 2 , golden ratio conjugate

Finite & Infinite Words Sturmian words From Christoffel words to Sturmian words Sturmian words are obtained *similarly* by replacing the line segment by a half-line: y = αx + ρ with irrational α ∈ (0, 1), ρ ∈ R.

√ 5−1 Example: y = 2 x −→ Fibonacci word f

I f = abaababaabaababaaba ··· (note: disregard 1st a in construction)

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 10 Finite & Infinite Words Sturmian words From Christoffel words to Sturmian words Sturmian words are obtained *similarly* by replacing the line segment by a half-line: y = αx + ρ with irrational α ∈ (0, 1), ρ ∈ R.

√ 5−1 Example: y = 2 x −→ Fibonacci word f

I f = abaababaabaababaaba ··· (note: disregard 1st a in construction) √ 5−1 I Characteristic Sturmian word of slope 2 , golden ratio conjugate

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 10 Properties

I L(p, q) = awb ⇐⇒ U(p, q) = bwa

I |L(p, q)|a = q, |L(p, q)|b = p =⇒ |L(p, q)| = p + q

I L(p, q) is the reversal of U(p, q)

I Christoffel words are of the form awb, bwa where w is a palindrome.

Finite & Infinite Words Sturmian words Christoffel words: Properties

Examples Slope p/q 3/4 4/3 4/7 5/7 L(p, q) aababab abababb aabaabaabab aababaababab U(p, q) bababaa bbababa babaabaabaa bababaababaa

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 11 I |L(p, q)|a = q, |L(p, q)|b = p =⇒ |L(p, q)| = p + q

I L(p, q) is the reversal of U(p, q)

I Christoffel words are of the form awb, bwa where w is a palindrome.

Finite & Infinite Words Sturmian words Christoffel words: Properties

Examples Slope p/q 3/4 4/3 4/7 5/7 L(p, q) aababab abababb aabaabaabab aababaababab U(p, q) bababaa bbababa babaabaabaa bababaababaa

Properties

I L(p, q) = awb ⇐⇒ U(p, q) = bwa

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 11 I L(p, q) is the reversal of U(p, q)

I Christoffel words are of the form awb, bwa where w is a palindrome.

Finite & Infinite Words Sturmian words Christoffel words: Properties

Examples Slope p/q 3/4 4/3 4/7 5/7 L(p, q) aababab abababb aabaabaabab aababaababab U(p, q) bababaa bbababa babaabaabaa bababaababaa

Properties

I L(p, q) = awb ⇐⇒ U(p, q) = bwa

I |L(p, q)|a = q, |L(p, q)|b = p =⇒ |L(p, q)| = p + q

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 11 I Christoffel words are of the form awb, bwa where w is a palindrome.

Finite & Infinite Words Sturmian words Christoffel words: Properties

Examples Slope p/q 3/4 4/3 4/7 5/7 L(p, q) aababab abababb aabaabaabab aababaababab U(p, q) bababaa bbababa babaabaabaa bababaababaa

Properties

I L(p, q) = awb ⇐⇒ U(p, q) = bwa

I |L(p, q)|a = q, |L(p, q)|b = p =⇒ |L(p, q)| = p + q

I L(p, q) is the reversal of U(p, q)

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 11 Finite & Infinite Words Sturmian words Christoffel words: Properties

Examples Slope p/q 3/4 4/3 4/7 5/7 L(p, q) aababab abababb aabaabaabab aababaababab U(p, q) bababaa bbababa babaabaabaa bababaababaa

Properties

I L(p, q) = awb ⇐⇒ U(p, q) = bwa

I |L(p, q)|a = q, |L(p, q)|b = p =⇒ |L(p, q)| = p + q

I L(p, q) is the reversal of U(p, q)

I Christoffel words are of the form awb, bwa where w is a palindrome.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 11 race car

, and for any word w and letter x, P al(wx) = (P al(w)x)+.

+ I (race) =

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

+ I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word)

Example

I P al is the iterated palindromic closure operator, defined as follows.

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 race car

+ I (race) =

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

, and for any word w and letter x, P al(wx) = (P al(w)x)+. Example

+ I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word)

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 race car

+ I (race) =

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

, and for any word w and letter x, P al(wx) = (P al(w)x)+. Example

I We define P al(ε) = ε (empty word)

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 race car

+ I (race) =

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

Example

, and for any word w and letter x, P al(wx) = (P al(w)x)+.

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word)

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 race car

+ I (race) =

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

Example

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word), and for any word w and letter x, P al(wx) = (P al(w)x)+.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 car

+ I (race) = race

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word), and for any word w and letter x, P al(wx) = (P al(w)x)+. Example

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 car

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word), and for any word w and letter x, P al(wx) = (P al(w)x)+. Example

+ I (race) = race

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word), and for any word w and letter x, P al(wx) = (P al(w)x)+. Example

+ I (race) = race car

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word), and for any word w and letter x, P al(wx) = (P al(w)x)+. Example

+ I (race) = race car

I P al(race) = rarcrarerarcrar

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 I L(7, 4) = aabaabaabab = aP al(abaa)b

Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word), and for any word w and letter x, P al(wx) = (P al(w)x)+. Example

+ I (race) = race car

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 Finite & Infinite Words Palindromic closure Christoffel words & palindromes Theorem (folklore) A finite word w is a Christoffel word if and only if w = aub or w = bua where u = P al(v) for some word v over {a, b}.

I P al is the iterated palindromic closure operator, defined as follows. + I Let w denote the unique shortest palindrome beginning with w. I We define P al(ε) = ε (empty word), and for any word w and letter x, P al(wx) = (P al(w)x)+. Example

+ I (race) = race car

I P al(race) = rarcrarerarcrar

I L(3, 5) = aabaabab = aP al(aba)b

I L(7, 4) = aabaabaabab = aP al(abaa)b

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 12 (i.e., iff ∆ is of the form ua∞ or ub∞ for some finite word u).

I The remaining (aperiodic) infinite words P al(∆) are precisely the characteristic (or standard) Sturmian words. [de Luca 1997]

∞ I The well-known Fibonacci word, for example, is P al(∆) where ∆ = (ab) .

I Then P al(∆) := lim P al(x1x2 ··· xn) n→∞ is an infinite word having infinitely many palindromic prefixes.

I All such words are palindromically rich in the sense that they contain the maximum possible number of palindromic factors (i.e., a new palindrome is introduced at each position).

I Note that P al(∆) is periodic if and only if ∆ degenerates to an infinitely repeated single letter

Finite & Infinite Words Palindromic closure Palindromic richness

I Now suppose ∆ := x1x2x3x4 ··· is an infinite word over the 2-letter alphabet {a, b}.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 13 (i.e., iff ∆ is of the form ua∞ or ub∞ for some finite word u).

I The remaining (aperiodic) infinite words P al(∆) are precisely the characteristic (or standard) Sturmian words. [de Luca 1997]

∞ I The well-known Fibonacci word, for example, is P al(∆) where ∆ = (ab) .

I All such words are palindromically rich in the sense that they contain the maximum possible number of palindromic factors (i.e., a new palindrome is introduced at each position).

I Note that P al(∆) is periodic if and only if ∆ degenerates to an infinitely repeated single letter

Finite & Infinite Words Palindromic closure Palindromic richness

I Now suppose ∆ := x1x2x3x4 ··· is an infinite word over the 2-letter alphabet {a, b}.

I Then P al(∆) := lim P al(x1x2 ··· xn) n→∞ is an infinite word having infinitely many palindromic prefixes.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 13 (i.e., iff ∆ is of the form ua∞ or ub∞ for some finite word u).

I The remaining (aperiodic) infinite words P al(∆) are precisely the characteristic (or standard) Sturmian words. [de Luca 1997]

∞ I The well-known Fibonacci word, for example, is P al(∆) where ∆ = (ab) .

I Note that P al(∆) is periodic if and only if ∆ degenerates to an infinitely repeated single letter

Finite & Infinite Words Palindromic closure Palindromic richness

I Now suppose ∆ := x1x2x3x4 ··· is an infinite word over the 2-letter alphabet {a, b}.

I Then P al(∆) := lim P al(x1x2 ··· xn) n→∞ is an infinite word having infinitely many palindromic prefixes.

I All such words are palindromically rich in the sense that they contain the maximum possible number of palindromic factors (i.e., a new palindrome is introduced at each position).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 13 I The remaining (aperiodic) infinite words P al(∆) are precisely the characteristic (or standard) Sturmian words. [de Luca 1997]

∞ I The well-known Fibonacci word, for example, is P al(∆) where ∆ = (ab) .

(i.e., iff ∆ is of the form ua∞ or ub∞ for some finite word u).

Finite & Infinite Words Palindromic closure Palindromic richness

I Now suppose ∆ := x1x2x3x4 ··· is an infinite word over the 2-letter alphabet {a, b}.

I Then P al(∆) := lim P al(x1x2 ··· xn) n→∞ is an infinite word having infinitely many palindromic prefixes.

I All such words are palindromically rich in the sense that they contain the maximum possible number of palindromic factors (i.e., a new palindrome is introduced at each position).

I Note that P al(∆) is periodic if and only if ∆ degenerates to an infinitely repeated single letter

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 13 I The remaining (aperiodic) infinite words P al(∆) are precisely the characteristic (or standard) Sturmian words. [de Luca 1997]

∞ I The well-known Fibonacci word, for example, is P al(∆) where ∆ = (ab) .

Finite & Infinite Words Palindromic closure Palindromic richness

I Now suppose ∆ := x1x2x3x4 ··· is an infinite word over the 2-letter alphabet {a, b}.

I Then P al(∆) := lim P al(x1x2 ··· xn) n→∞ is an infinite word having infinitely many palindromic prefixes.

I All such words are palindromically rich in the sense that they contain the maximum possible number of palindromic factors (i.e., a new palindrome is introduced at each position).

I Note that P al(∆) is periodic if and only if ∆ degenerates to an infinitely repeated single letter (i.e., iff ∆ is of the form ua∞ or ub∞ for some finite word u).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 13 ∞ I The well-known Fibonacci word, for example, is P al(∆) where ∆ = (ab) .

Finite & Infinite Words Palindromic closure Palindromic richness

I Now suppose ∆ := x1x2x3x4 ··· is an infinite word over the 2-letter alphabet {a, b}.

I Then P al(∆) := lim P al(x1x2 ··· xn) n→∞ is an infinite word having infinitely many palindromic prefixes.

I All such words are palindromically rich in the sense that they contain the maximum possible number of palindromic factors (i.e., a new palindrome is introduced at each position).

I Note that P al(∆) is periodic if and only if ∆ degenerates to an infinitely repeated single letter (i.e., iff ∆ is of the form ua∞ or ub∞ for some finite word u).

I The remaining (aperiodic) infinite words P al(∆) are precisely the characteristic (or standard) Sturmian words. [de Luca 1997]

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 13 Finite & Infinite Words Palindromic closure Palindromic richness

I Now suppose ∆ := x1x2x3x4 ··· is an infinite word over the 2-letter alphabet {a, b}.

I Then P al(∆) := lim P al(x1x2 ··· xn) n→∞ is an infinite word having infinitely many palindromic prefixes.

I All such words are palindromically rich in the sense that they contain the maximum possible number of palindromic factors (i.e., a new palindrome is introduced at each position).

I Note that P al(∆) is periodic if and only if ∆ degenerates to an infinitely repeated single letter (i.e., iff ∆ is of the form ua∞ or ub∞ for some finite word u).

I The remaining (aperiodic) infinite words P al(∆) are precisely the characteristic (or standard) Sturmian words. [de Luca 1997]

∞ I The well-known Fibonacci word, for example, is P al(∆) where ∆ = (ab) .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 13 baababaaba ···

Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

f = P al(∆) = a

Finite & Infinite Words Palindromic closure Recall: Fibonacci word

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

baababaaba ···

Finite & Infinite Words Palindromic closure Recall: Fibonacci word

∆ = (ab)(ab)(ab) · · · −→ f = P al(∆) = a

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

aababaaba ···

Finite & Infinite Words Palindromic closure Recall: Fibonacci word

∆ = (ab)(ab)(ab) · · · −→ f = P al(∆) = ab

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

ababaaba ···

Finite & Infinite Words Palindromic closure Recall: Fibonacci word

∆ = (ab)(ab)(ab) · · · −→ f = P al(∆) = aba

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

babaaba ···

Finite & Infinite Words Palindromic closure Recall: Fibonacci word

∆ = (ab)(ab)(ab) · · · −→ f = P al(∆) = abaa

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

baaba ···

Finite & Infinite Words Palindromic closure Recall: Fibonacci word

∆ = (ab)(ab)(ab) · · · −→ f = P al(∆) = abaaba

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

aaba ···

Finite & Infinite Words Palindromic closure Recall: Fibonacci word

∆ = (ab)(ab)(ab) · · · −→ f = P al(∆) = abaabab

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

Finite & Infinite Words Palindromic closure Recall: Fibonacci word

∆ = (ab)(ab)(ab) · · · −→ f = P al(∆) = abaababaaba ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Finite & Infinite Words Palindromic closure Recall: Fibonacci word

∆ = (ab)(ab)(ab) · · · −→ f = P al(∆) = abaababaaba ···

Note: Palindromic prefixes have lengths (Fn+1 − 2)n≥1 = 0, 1, 3, 6, 11, 19,... where (Fn)n≥0 is the sequence of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21,..., defined by: F0 = F1 = 1,Fn = Fn−1 + Fn−2 for n ≥ 2.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 14 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

r = P al(∆) = abacabaabacababacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

r = P al(∆) = abacabaabacababacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word:

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

bacabaabacababacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = a

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

acabaabacababacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = ab

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

cabaabacababacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = aba

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

abaabacababacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abac

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

abacababacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abacaba

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

bacababacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abacabaa

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

bacabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abacabaabacaba

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

acabaabacabacabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abacabaabacabab

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

cabaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abacabaabacababacabaabacaba

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

abaabaca ···

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abacabaabacababacabaabacabac

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3.

Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abacabaabacababacabaabacabacabaabaca ···

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 Finite & Infinite Words Palindromic closure A natural generalisation: Episturmian words {a, b} −→ A (finite alphabet) gives standard episturmian words.

Theorem (Droubay-Justin-Pirillo 2001) An infinite word s over A is a standard episturmian word if and only if there exists an infinite word ∆ = x1x2x3 ··· over A such that

s = lim P al(x1x2 ··· xn) = P al(∆). n→∞

Example: ∆ = (abc)∞ “directs” the so-called Tribonacci word: r = P al(∆) = abacabaabacababacabaabacabacabaabaca ···

Note: Palindromic prefixes have lengths ((Tn+2 + Tn + 1)/2 − 2)n≥1 = 0, 1, 3, 7, 14, 27, 36 ... where (Tn)n≥0 is the sequence of Tribonacci numbers 1, 1, 2, 4, 7, 13, 24, 44, . . . , defined by:

T0 = T1 = 1,T2 = 2,Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3. Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 15 For instance:

I Transcendental numbers

I Distribution modulo 1

Finite & Infinite Words Palindromic closure Remarks

(Epi)Sturmian words, Christoffel words, and the P al operator have all played important roles in the solutions to some number-theoretical problems . . .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 16 Finite & Infinite Words Palindromic closure Remarks

(Epi)Sturmian words, Christoffel words, and the P al operator have all played important roles in the solutions to some number-theoretical problems . . .

For instance:

I Transcendental numbers

I Distribution modulo 1

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 16 a1 a2 a3 a4 a5 I Then cα = P al(a b a b a ··· ). [Arnoux-Rauzy 1991]

Example

Recall the Fibonacci√ word is f := cα with α = 1/τ = ( 5 − 1)/2 = [0; 1, 1, 1, 1,...] and combinatorially:

f = P al(abababab ··· ) = abaababaaba ··· .

I cα can also be constructed as the limit of an infinite sequence of finite words, defined with respect to the CF of α.

I Many nice combinatorial properties of cα are related to the CF of α.

Some Applications in Number Theory Continued Fractions & Sturmian Words Characteristic Sturmian words via CF expansions

I Let cα denote the characteristic Sturmian word corresponding to the line of irrational slope α ∈ (0, 1) through (0, 0) where

α = [0; a1, a2, a3,...].

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 17 Example

Recall the Fibonacci√ word is f := cα with α = 1/τ = ( 5 − 1)/2 = [0; 1, 1, 1, 1,...] and combinatorially:

f = P al(abababab ··· ) = abaababaaba ··· .

I cα can also be constructed as the limit of an infinite sequence of finite words, defined with respect to the CF of α.

I Many nice combinatorial properties of cα are related to the CF of α.

Some Applications in Number Theory Continued Fractions & Sturmian Words Characteristic Sturmian words via CF expansions

I Let cα denote the characteristic Sturmian word corresponding to the line of irrational slope α ∈ (0, 1) through (0, 0) where

α = [0; a1, a2, a3,...].

a1 a2 a3 a4 a5 I Then cα = P al(a b a b a ··· ). [Arnoux-Rauzy 1991]

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 17 I cα can also be constructed as the limit of an infinite sequence of finite words, defined with respect to the CF of α.

I Many nice combinatorial properties of cα are related to the CF of α.

Some Applications in Number Theory Continued Fractions & Sturmian Words Characteristic Sturmian words via CF expansions

I Let cα denote the characteristic Sturmian word corresponding to the line of irrational slope α ∈ (0, 1) through (0, 0) where

α = [0; a1, a2, a3,...].

a1 a2 a3 a4 a5 I Then cα = P al(a b a b a ··· ). [Arnoux-Rauzy 1991]

Example

Recall the Fibonacci√ word is f := cα with α = 1/τ = ( 5 − 1)/2 = [0; 1, 1, 1, 1,...] and combinatorially:

f = P al(abababab ··· ) = abaababaaba ··· .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 17 I Many nice combinatorial properties of cα are related to the CF of α.

Some Applications in Number Theory Continued Fractions & Sturmian Words Characteristic Sturmian words via CF expansions

I Let cα denote the characteristic Sturmian word corresponding to the line of irrational slope α ∈ (0, 1) through (0, 0) where

α = [0; a1, a2, a3,...].

a1 a2 a3 a4 a5 I Then cα = P al(a b a b a ··· ). [Arnoux-Rauzy 1991]

Example

Recall the Fibonacci√ word is f := cα with α = 1/τ = ( 5 − 1)/2 = [0; 1, 1, 1, 1,...] and combinatorially:

f = P al(abababab ··· ) = abaababaaba ··· .

I cα can also be constructed as the limit of an infinite sequence of finite words, defined with respect to the CF of α.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 17 Some Applications in Number Theory Continued Fractions & Sturmian Words Characteristic Sturmian words via CF expansions

I Let cα denote the characteristic Sturmian word corresponding to the line of irrational slope α ∈ (0, 1) through (0, 0) where

α = [0; a1, a2, a3,...].

a1 a2 a3 a4 a5 I Then cα = P al(a b a b a ··· ). [Arnoux-Rauzy 1991]

Example

Recall the Fibonacci√ word is f := cα with α = 1/τ = ( 5 − 1)/2 = [0; 1, 1, 1, 1,...] and combinatorially:

f = P al(abababab ··· ) = abaababaaba ··· .

I cα can also be constructed as the limit of an infinite sequence of finite words, defined with respect to the CF of α.

I Many nice combinatorial properties of cα are related to the CF of α.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 17 That is to say, an irrational number whose CF expansion has bounded partial quotients is either quadratic or transcendental.

I Liouville (1844): Transcendental CFs whose sequences of partial quotients grow very fast (too fast to be algebraic) I Transcendental CFs with bounded partial quotients: Maillet (1906)  Baker (1962, 1964)   Shallit (1979)  Davison (1989) transcendence criteria from DA Queffélec (1998)   Allouche, Davison, Queffélec, Zamboni (2001)  Adamczewski-Bugeaud (2005–2007) 

Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions Long-standing Conjecture (Khintchine 1949) The CF expansion of an irrational algebraic real number α is either eventually periodic (iff α is a quadratic irrational) or it contains arbitrarily large partial quotients.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 18 I Liouville (1844): Transcendental CFs whose sequences of partial quotients grow very fast (too fast to be algebraic) I Transcendental CFs with bounded partial quotients: Maillet (1906)  Baker (1962, 1964)   Shallit (1979)  Davison (1989) transcendence criteria from DA Queffélec (1998)   Allouche, Davison, Queffélec, Zamboni (2001)  Adamczewski-Bugeaud (2005–2007) 

Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions Long-standing Conjecture (Khintchine 1949) The CF expansion of an irrational algebraic real number α is either eventually periodic (iff α is a quadratic irrational) or it contains arbitrarily large partial quotients. That is to say, an irrational number whose CF expansion has bounded partial quotients is either quadratic or transcendental.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 18 I Transcendental CFs with bounded partial quotients: Maillet (1906)  Baker (1962, 1964)   Shallit (1979)  Davison (1989) transcendence criteria from DA Queffélec (1998)   Allouche, Davison, Queffélec, Zamboni (2001)  Adamczewski-Bugeaud (2005–2007) 

Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions Long-standing Conjecture (Khintchine 1949) The CF expansion of an irrational algebraic real number α is either eventually periodic (iff α is a quadratic irrational) or it contains arbitrarily large partial quotients. That is to say, an irrational number whose CF expansion has bounded partial quotients is either quadratic or transcendental.

I Liouville (1844): Transcendental CFs whose sequences of partial quotients grow very fast (too fast to be algebraic)

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 18 Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions Long-standing Conjecture (Khintchine 1949) The CF expansion of an irrational algebraic real number α is either eventually periodic (iff α is a quadratic irrational) or it contains arbitrarily large partial quotients. That is to say, an irrational number whose CF expansion has bounded partial quotients is either quadratic or transcendental.

I Liouville (1844): Transcendental CFs whose sequences of partial quotients grow very fast (too fast to be algebraic) I Transcendental CFs with bounded partial quotients: Maillet (1906)  Baker (1962, 1964)   Shallit (1979)  Davison (1989) transcendence criteria from DA Queffélec (1998)   Allouche, Davison, Queffélec, Zamboni (2001)  Adamczewski-Bugeaud (2005–2007) 

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 18 Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f =

I All such numbers are transcendental. More generally:

Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f =

More generally:

Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f =

Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental. More generally: Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f =

Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental. More generally: Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 Example: f =

Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental. More generally: Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental. More generally: Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f = Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental. More generally: Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f = aba · ababaabaababaababaabaababaaba ··· Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental. More generally: Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f = abaab · abaabaababaababaabaababaaba ··· Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental. More generally: Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f = abaababa · abaababaababaabaababaaba ··· Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 Some Applications in Number Theory Transcendental Numbers Transcendental continued fractions: Examples

I Let ξa,b := [0; a, b, a, a, b, a, b, a, a, b, a, . . .] where the sequence of partial quotients is the Fibonacci word on positive integers a, b.

I All such numbers are transcendental. More generally: Theorem (Allouche-Davison-Queffélec-Zamboni 2001) Any real number in (0, 1) that has a Sturmian continued fraction – of which the sequence of partial quotients forms a Sturmian word on two distinct positive integers – is transcendental.

I Proved by showing that Sturmian continued fractions admit very good approximations by quadratic real numbers.

I In particular, they used the fact that any Sturmian word begins with arbitrarily long squares (words of the form XX = X2).

Example: f = abaababaabaab · abaababaabaababaaba ··· Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 19 I And even more generally, aperiodic infinite words beginning with infinitely many palindromes (including Sturmian and episturmian words) also give rise to transcendental numbers . . . Theorem (Adamczewski-Bugeaud 2007)

Let (an)n≥1 be an infinite sequence of positive integers. If (an)n≥1 begins with infinitely many palindromes, then the real number ξ in (0, 1) with continued fraction expansion [0; a1, a2, . . . , an,...] is either quadratic irrational or transcendental.

I Adamczewksi and Bugeaud have also proved that if the sequence of partial quotients in the CF expansion of a non-quadratic irrational ξ ∈ (0, 1) begins with arbitrarily long blocks that are almost squares, then ξ is transcendental.

I Analogous results hold for expansions of real numbers in integer bases.

Some Applications in Number Theory Transcendental Numbers Generalisations

I Any real number in (0, 1) that has a strict standard episturmian continued fraction is transcendental. [Glen 2005]

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 20 Theorem (Adamczewski-Bugeaud 2007)

Let (an)n≥1 be an infinite sequence of positive integers. If (an)n≥1 begins with infinitely many palindromes, then the real number ξ in (0, 1) with continued fraction expansion [0; a1, a2, . . . , an,...] is either quadratic irrational or transcendental.

I Adamczewksi and Bugeaud have also proved that if the sequence of partial quotients in the CF expansion of a non-quadratic irrational ξ ∈ (0, 1) begins with arbitrarily long blocks that are almost squares, then ξ is transcendental.

I Analogous results hold for expansions of real numbers in integer bases.

Some Applications in Number Theory Transcendental Numbers Generalisations

I Any real number in (0, 1) that has a strict standard episturmian continued fraction is transcendental. [Glen 2005]

I And even more generally, aperiodic infinite words beginning with infinitely many palindromes (including Sturmian and episturmian words) also give rise to transcendental numbers . . .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 20 I Adamczewksi and Bugeaud have also proved that if the sequence of partial quotients in the CF expansion of a non-quadratic irrational ξ ∈ (0, 1) begins with arbitrarily long blocks that are almost squares, then ξ is transcendental.

I Analogous results hold for expansions of real numbers in integer bases.

Some Applications in Number Theory Transcendental Numbers Generalisations

I Any real number in (0, 1) that has a strict standard episturmian continued fraction is transcendental. [Glen 2005]

I And even more generally, aperiodic infinite words beginning with infinitely many palindromes (including Sturmian and episturmian words) also give rise to transcendental numbers . . . Theorem (Adamczewski-Bugeaud 2007)

Let (an)n≥1 be an infinite sequence of positive integers. If (an)n≥1 begins with infinitely many palindromes, then the real number ξ in (0, 1) with continued fraction expansion [0; a1, a2, . . . , an,...] is either quadratic irrational or transcendental.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 20 I Analogous results hold for expansions of real numbers in integer bases.

Some Applications in Number Theory Transcendental Numbers Generalisations

I Any real number in (0, 1) that has a strict standard episturmian continued fraction is transcendental. [Glen 2005]

I And even more generally, aperiodic infinite words beginning with infinitely many palindromes (including Sturmian and episturmian words) also give rise to transcendental numbers . . . Theorem (Adamczewski-Bugeaud 2007)

Let (an)n≥1 be an infinite sequence of positive integers. If (an)n≥1 begins with infinitely many palindromes, then the real number ξ in (0, 1) with continued fraction expansion [0; a1, a2, . . . , an,...] is either quadratic irrational or transcendental.

I Adamczewksi and Bugeaud have also proved that if the sequence of partial quotients in the CF expansion of a non-quadratic irrational ξ ∈ (0, 1) begins with arbitrarily long blocks that are almost squares, then ξ is transcendental.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 20 Some Applications in Number Theory Transcendental Numbers Generalisations

I Any real number in (0, 1) that has a strict standard episturmian continued fraction is transcendental. [Glen 2005]

I And even more generally, aperiodic infinite words beginning with infinitely many palindromes (including Sturmian and episturmian words) also give rise to transcendental numbers . . . Theorem (Adamczewski-Bugeaud 2007)

Let (an)n≥1 be an infinite sequence of positive integers. If (an)n≥1 begins with infinitely many palindromes, then the real number ξ in (0, 1) with continued fraction expansion [0; a1, a2, . . . , an,...] is either quadratic irrational or transcendental.

I Adamczewksi and Bugeaud have also proved that if the sequence of partial quotients in the CF expansion of a non-quadratic irrational ξ ∈ (0, 1) begins with arbitrarily long blocks that are almost squares, then ξ is transcendental.

I Analogous results hold for expansions of real numbers in integer bases.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 20 I Mahler proved that Z is at most countable.

I It is still an open problem to prove that Z is in fact empty!

I A more general question:

Given a real number α > 1 and an interval (x, y) ⊂ (0, 1), does there exists ξ > 0 such that all fractional parts {ξαn}, n ≥ 0, lie in the interval [x, y) or [x, y] ?

Some Applications in Number Theory Distribution of real numbers modulo 1 Distribution of real numbers modulo 1

I Mahler (1968): defined the set of Z-numbers by

  3n 1 Z := ξ ∈ + ∀n ≥ 0, 0 ≤ ξ < R 2 2 where {z} denotes the fractional part of z.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 21 I It is still an open problem to prove that Z is in fact empty!

I A more general question:

Given a real number α > 1 and an interval (x, y) ⊂ (0, 1), does there exists ξ > 0 such that all fractional parts {ξαn}, n ≥ 0, lie in the interval [x, y) or [x, y] ?

Some Applications in Number Theory Distribution of real numbers modulo 1 Distribution of real numbers modulo 1

I Mahler (1968): defined the set of Z-numbers by

  3n 1 Z := ξ ∈ + ∀n ≥ 0, 0 ≤ ξ < R 2 2 where {z} denotes the fractional part of z.

I Mahler proved that Z is at most countable.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 21 I A more general question:

Given a real number α > 1 and an interval (x, y) ⊂ (0, 1), does there exists ξ > 0 such that all fractional parts {ξαn}, n ≥ 0, lie in the interval [x, y) or [x, y] ?

Some Applications in Number Theory Distribution of real numbers modulo 1 Distribution of real numbers modulo 1

I Mahler (1968): defined the set of Z-numbers by

  3n 1 Z := ξ ∈ + ∀n ≥ 0, 0 ≤ ξ < R 2 2 where {z} denotes the fractional part of z.

I Mahler proved that Z is at most countable.

I It is still an open problem to prove that Z is in fact empty!

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 21 Some Applications in Number Theory Distribution of real numbers modulo 1 Distribution of real numbers modulo 1

I Mahler (1968): defined the set of Z-numbers by

  3n 1 Z := ξ ∈ + ∀n ≥ 0, 0 ≤ ξ < R 2 2 where {z} denotes the fractional part of z.

I Mahler proved that Z is at most countable.

I It is still an open problem to prove that Z is in fact empty!

I A more general question:

Given a real number α > 1 and an interval (x, y) ⊂ (0, 1), does there exists ξ > 0 such that all fractional parts {ξαn}, n ≥ 0, lie in the interval [x, y) or [x, y] ?

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 21 Bugeaud-Dubickas (2005) described all irrational numbers ξ > 0 such that for a fixed integer b ≥ 2 the fractional parts {ξbn}, n ≥ 0, all belong to an interval of length 1/b. Theorem (Bugeaud-Dubickas 2005) Let b ≥ 2 be an integer and ξ > 0 be an irrational number. Then: n I the frac. parts {ξb } cannot all lie in an interval of length < 1/b;

I there exists a closed interval of length 1/b containing the frac. parts {ξbn} for all n ≥ 0 iff the base b expansion of the fractional part of ξ is a Sturmian sequence on {k, k + 1} for some k ∈ {0, 1, . . . , b − 2}.

Some Applications in Number Theory Distribution of real numbers modulo 1

Theorem (Flatto-Lagarias-Pollington 1995) If p, q are coprime integers with p > q ≥ 2, then any interval (x, y) n + containing all fractional parts {ξ(p/q) }, n ≥ 0, for some ξ ∈ R must satisfy y − x ≥ 1/p.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 22 Theorem (Bugeaud-Dubickas 2005) Let b ≥ 2 be an integer and ξ > 0 be an irrational number. Then: n I the frac. parts {ξb } cannot all lie in an interval of length < 1/b;

I there exists a closed interval of length 1/b containing the frac. parts {ξbn} for all n ≥ 0 iff the base b expansion of the fractional part of ξ is a Sturmian sequence on {k, k + 1} for some k ∈ {0, 1, . . . , b − 2}.

Some Applications in Number Theory Distribution of real numbers modulo 1

Theorem (Flatto-Lagarias-Pollington 1995) If p, q are coprime integers with p > q ≥ 2, then any interval (x, y) n + containing all fractional parts {ξ(p/q) }, n ≥ 0, for some ξ ∈ R must satisfy y − x ≥ 1/p.

Bugeaud-Dubickas (2005) described all irrational numbers ξ > 0 such that for a fixed integer b ≥ 2 the fractional parts {ξbn}, n ≥ 0, all belong to an interval of length 1/b.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 22 Some Applications in Number Theory Distribution of real numbers modulo 1

Theorem (Flatto-Lagarias-Pollington 1995) If p, q are coprime integers with p > q ≥ 2, then any interval (x, y) n + containing all fractional parts {ξ(p/q) }, n ≥ 0, for some ξ ∈ R must satisfy y − x ≥ 1/p.

Bugeaud-Dubickas (2005) described all irrational numbers ξ > 0 such that for a fixed integer b ≥ 2 the fractional parts {ξbn}, n ≥ 0, all belong to an interval of length 1/b. Theorem (Bugeaud-Dubickas 2005) Let b ≥ 2 be an integer and ξ > 0 be an irrational number. Then: n I the frac. parts {ξb } cannot all lie in an interval of length < 1/b;

I there exists a closed interval of length 1/b containing the frac. parts {ξbn} for all n ≥ 0 iff the base b expansion of the fractional part of ξ is a Sturmian sequence on {k, k + 1} for some k ∈ {0, 1, . . . , b − 2}.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 22 Notation

I Let T denote the shift map on sequences: T ((sn)n≥0) := (sn+1)n≥0. k k-th shift: T ((sn)n≥0) := (sn+k)n≥0.

I Let  denote the lexicographic order on {0, 1}N induced by 0 < 1.

Theorem If s is a Sturmian word of (irrational) slope α > 0 over the alphabet {0, 1}, then k 0cα  T (s)  1cα for all k ≥ 0,

where cα is the (unique) standard Sturmian word of slope α.

That is: all shifts of a Sturmian sequence s ∈ {0, 1}N of slope α are lexicographically ≥ 0cα and lexicographically ≤ 1cα.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words... The core of Bugeaud & Dubickas’ result is the following fact that has been rediscovered several times in different contexts (since mid-late 80’s) . . .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 23 I Let  denote the lexicographic order on {0, 1}N induced by 0 < 1.

Theorem If s is a Sturmian word of (irrational) slope α > 0 over the alphabet {0, 1}, then k 0cα  T (s)  1cα for all k ≥ 0,

where cα is the (unique) standard Sturmian word of slope α.

That is: all shifts of a Sturmian sequence s ∈ {0, 1}N of slope α are lexicographically ≥ 0cα and lexicographically ≤ 1cα.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words... The core of Bugeaud & Dubickas’ result is the following fact that has been rediscovered several times in different contexts (since mid-late 80’s) . . . Notation

I Let T denote the shift map on sequences: T ((sn)n≥0) := (sn+1)n≥0. k k-th shift: T ((sn)n≥0) := (sn+k)n≥0.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 23 Theorem If s is a Sturmian word of (irrational) slope α > 0 over the alphabet {0, 1}, then k 0cα  T (s)  1cα for all k ≥ 0,

where cα is the (unique) standard Sturmian word of slope α.

That is: all shifts of a Sturmian sequence s ∈ {0, 1}N of slope α are lexicographically ≥ 0cα and lexicographically ≤ 1cα.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words... The core of Bugeaud & Dubickas’ result is the following fact that has been rediscovered several times in different contexts (since mid-late 80’s) . . . Notation

I Let T denote the shift map on sequences: T ((sn)n≥0) := (sn+1)n≥0. k k-th shift: T ((sn)n≥0) := (sn+k)n≥0.

I Let  denote the lexicographic order on {0, 1}N induced by 0 < 1.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 23 That is: all shifts of a Sturmian sequence s ∈ {0, 1}N of slope α are lexicographically ≥ 0cα and lexicographically ≤ 1cα.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words... The core of Bugeaud & Dubickas’ result is the following fact that has been rediscovered several times in different contexts (since mid-late 80’s) . . . Notation

I Let T denote the shift map on sequences: T ((sn)n≥0) := (sn+1)n≥0. k k-th shift: T ((sn)n≥0) := (sn+k)n≥0.

I Let  denote the lexicographic order on {0, 1}N induced by 0 < 1.

Theorem If s is a Sturmian word of (irrational) slope α > 0 over the alphabet {0, 1}, then k 0cα  T (s)  1cα for all k ≥ 0,

where cα is the (unique) standard Sturmian word of slope α.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 23 Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words... The core of Bugeaud & Dubickas’ result is the following fact that has been rediscovered several times in different contexts (since mid-late 80’s) . . . Notation

I Let T denote the shift map on sequences: T ((sn)n≥0) := (sn+1)n≥0. k k-th shift: T ((sn)n≥0) := (sn+k)n≥0.

I Let  denote the lexicographic order on {0, 1}N induced by 0 < 1.

Theorem If s is a Sturmian word of (irrational) slope α > 0 over the alphabet {0, 1}, then k 0cα  T (s)  1cα for all k ≥ 0,

where cα is the (unique) standard Sturmian word of slope α.

That is: all shifts of a Sturmian sequence s ∈ {0, 1}N of slope α are lexicographically ≥ 0cα and lexicographically ≤ 1cα. Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 23 I The main tool used by Bugeaud and Dubickas was combinatorics on words: replace real numbers by their base b expansions, and transform inequalities between real numbers into (lexicographic) inequalities between infinite sequences representing their base b expansions.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words...

I Example: Consider the Fibonacci word on {0, 1} (a 7→ 0, b 7→ 1): f = 0100101001001010010 ··· , √ the characteristic Sturmian word cα of slope α = ( 5 − 1)/2.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 24 I The main tool used by Bugeaud and Dubickas was combinatorics on words: replace real numbers by their base b expansions, and transform inequalities between real numbers into (lexicographic) inequalities between infinite sequences representing their base b expansions.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words...

I Example: Consider the Fibonacci word on {0, 1} (a 7→ 0, b 7→ 1): f = 0100101001001010010 ··· , √ the characteristic Sturmian word cα of slope α = ( 5 − 1)/2.

f z }| { 001001010010010 ··· ≺ 01001010010010 · · · ≺ 101001010010010 ··· | {z } | {z } 0f 1f

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 24 I The main tool used by Bugeaud and Dubickas was combinatorics on words: replace real numbers by their base b expansions, and transform inequalities between real numbers into (lexicographic) inequalities between infinite sequences representing their base b expansions.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words...

I Example: Consider the Fibonacci word on {0, 1} (a 7→ 0, b 7→ 1): f = 0100101001001010010 ··· , √ the characteristic Sturmian word cα of slope α = ( 5 − 1)/2.

T (f) z }| { 001001010010010 ··· ≺ 1001010010010 · · · ≺ 101001010010010 ··· | {z } | {z } 0f 1f

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 24 I The main tool used by Bugeaud and Dubickas was combinatorics on words: replace real numbers by their base b expansions, and transform inequalities between real numbers into (lexicographic) inequalities between infinite sequences representing their base b expansions.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words...

I Example: Consider the Fibonacci word on {0, 1} (a 7→ 0, b 7→ 1): f = 0100101001001010010 ··· , √ the characteristic Sturmian word cα of slope α = ( 5 − 1)/2.

T 2(f) z }| { 001001010010010 ··· ≺ 001010010010 · · · ≺ 101001010010010 ··· | {z } | {z } 0f 1f

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 24 I The main tool used by Bugeaud and Dubickas was combinatorics on words: replace real numbers by their base b expansions, and transform inequalities between real numbers into (lexicographic) inequalities between infinite sequences representing their base b expansions.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words...

I Example: Consider the Fibonacci word on {0, 1} (a 7→ 0, b 7→ 1): f = 0100101001001010010 ··· , √ the characteristic Sturmian word cα of slope α = ( 5 − 1)/2.

T 3(f) z }| { 001001010010010 ··· ≺ 01010010010 · · · ≺ 101001010010010 ··· | {z } | {z } 0f 1f

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 24 I The main tool used by Bugeaud and Dubickas was combinatorics on words: replace real numbers by their base b expansions, and transform inequalities between real numbers into (lexicographic) inequalities between infinite sequences representing their base b expansions.

Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words...

I Example: Consider the Fibonacci word on {0, 1} (a 7→ 0, b 7→ 1): f = 0100101001001010010 ··· , √ the characteristic Sturmian word cα of slope α = ( 5 − 1)/2.

T 4(f) z }| { 001001010010010 ··· ≺ 1010010010 · · · ≺ 101001010010010 ··· | {z } | {z } 0f 1f and so on . . .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 24 Some Applications in Number Theory Distribution of real numbers modulo 1 Fractional parts of powers & Sturmian words...

I Example: Consider the Fibonacci word on {0, 1} (a 7→ 0, b 7→ 1): f = 0100101001001010010 ··· , √ the characteristic Sturmian word cα of slope α = ( 5 − 1)/2.

T 4(f) z }| { 001001010010010 ··· ≺ 1010010010 · · · ≺ 101001010010010 ··· | {z } | {z } 0f 1f and so on . . .

I The main tool used by Bugeaud and Dubickas was combinatorics on words: replace real numbers by their base b expansions, and transform inequalities between real numbers into (lexicographic) inequalities between infinite sequences representing their base b expansions.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 24 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 .

Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y).

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 .

∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y).

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example:

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 .

1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y).

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 .

I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y).

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 .

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 .

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y). Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

∞ 1 Example: Consider s = (01) . Then r(s) = 3 .

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y). Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y). Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y). Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 . 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s))

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . .

Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y). Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 . 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s))

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 Some Applications in Number Theory Distribution of real numbers modulo 1 I Given a sequence s ∈ {0, 1}N, let r(s) denote the real number in (0, 1) whose binary digits (after the binary point) are given by s. Example: ∞ 1 1 1 1 r((01) ) = 22 + 24 + 26 + ··· = 3 1 1 r(01000 ··· ) = 4 and r(0011111 ··· ) = 4 I Let x, y be sequences on {0, 1}. Then x  y ⇔ r(x) ≤ r(y). Example: 0100000 ··· ≺ 01010101 ··· | {z } | {z } (1/4)2 (1/3)2 Also note that, for any sequence s ∈ {0, 1}N, r(T k(s)) = {r(s)2k}.

∞ 1 Example: Consider s = (01) . Then r(s) = 3 . 1 2  1 1 1  ∞ { 3 · 2} = 3 = 2 × 22 + 24 + 26 + ··· = r((10) ) = r(T (s)) 1 2 1 ∞ 2 { 3 · 2 } = 3 = r((01) ) = r(T (s)) 1 3 2 ∞ 3 { 3 · 2 } = 3 = r((10) ) = r(T (s)) and so on . . . Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 25 For example: (01)∞  T k((10)∞)  (10)∞ for all k ≥ 0 1 2 k 2 3 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0

h 1 i Note: R = 2/3 is the smallest positive real number such that 3 ,R k + contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R . Recall: For any Sturmian word s ∈ {0, 1}N of slope α, k 0cα  T (s)  1cα for all k ≥ 0,

k 1 i.e., r(0cα) ≤ {r(s)2 } ≤ r(0cα) + 2 for all k ≥ 0.

1 Note: R = r(0cα) + 2 is the smallest positive real number such that k + [r(0cα),R] contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R (namely ξ = r(s) where s ∈ {0, 1}N is a Sturmian word of slope α).

Some Applications in Number Theory Distribution of real numbers modulo 1 So, for any given sequences x, y, s over {0, 1}, x  T k(s)  y ⇐⇒ r(x) ≤ {r(s)2k} ≤ r(y).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 26 h 1 i Note: R = 2/3 is the smallest positive real number such that 3 ,R k + contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R . Recall: For any Sturmian word s ∈ {0, 1}N of slope α, k 0cα  T (s)  1cα for all k ≥ 0,

k 1 i.e., r(0cα) ≤ {r(s)2 } ≤ r(0cα) + 2 for all k ≥ 0.

1 Note: R = r(0cα) + 2 is the smallest positive real number such that k + [r(0cα),R] contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R (namely ξ = r(s) where s ∈ {0, 1}N is a Sturmian word of slope α).

Some Applications in Number Theory Distribution of real numbers modulo 1 So, for any given sequences x, y, s over {0, 1}, x  T k(s)  y ⇐⇒ r(x) ≤ {r(s)2k} ≤ r(y). For example: (01)∞  T k((10)∞)  (10)∞ for all k ≥ 0 1 2 k 2 3 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 26 Recall: For any Sturmian word s ∈ {0, 1}N of slope α, k 0cα  T (s)  1cα for all k ≥ 0,

k 1 i.e., r(0cα) ≤ {r(s)2 } ≤ r(0cα) + 2 for all k ≥ 0.

1 Note: R = r(0cα) + 2 is the smallest positive real number such that k + [r(0cα),R] contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R (namely ξ = r(s) where s ∈ {0, 1}N is a Sturmian word of slope α).

Some Applications in Number Theory Distribution of real numbers modulo 1 So, for any given sequences x, y, s over {0, 1}, x  T k(s)  y ⇐⇒ r(x) ≤ {r(s)2k} ≤ r(y). For example: (01)∞  T k((10)∞)  (10)∞ for all k ≥ 0 1 2 k 2 3 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0

h 1 i Note: R = 2/3 is the smallest positive real number such that 3 ,R k + contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 26 1 Note: R = r(0cα) + 2 is the smallest positive real number such that k + [r(0cα),R] contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R (namely ξ = r(s) where s ∈ {0, 1}N is a Sturmian word of slope α).

Some Applications in Number Theory Distribution of real numbers modulo 1 So, for any given sequences x, y, s over {0, 1}, x  T k(s)  y ⇐⇒ r(x) ≤ {r(s)2k} ≤ r(y). For example: (01)∞  T k((10)∞)  (10)∞ for all k ≥ 0 1 2 k 2 3 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0

h 1 i Note: R = 2/3 is the smallest positive real number such that 3 ,R k + contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R . Recall: For any Sturmian word s ∈ {0, 1}N of slope α, k 0cα  T (s)  1cα for all k ≥ 0,

k 1 i.e., r(0cα) ≤ {r(s)2 } ≤ r(0cα) + 2 for all k ≥ 0.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 26 Some Applications in Number Theory Distribution of real numbers modulo 1 So, for any given sequences x, y, s over {0, 1}, x  T k(s)  y ⇐⇒ r(x) ≤ {r(s)2k} ≤ r(y). For example: (01)∞  T k((10)∞)  (10)∞ for all k ≥ 0 1 2 k 2 3 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0

h 1 i Note: R = 2/3 is the smallest positive real number such that 3 ,R k + contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R . Recall: For any Sturmian word s ∈ {0, 1}N of slope α, k 0cα  T (s)  1cα for all k ≥ 0,

k 1 i.e., r(0cα) ≤ {r(s)2 } ≤ r(0cα) + 2 for all k ≥ 0.

1 Note: R = r(0cα) + 2 is the smallest positive real number such that k + [r(0cα),R] contains all the fractional parts {ξ2 }, k ≥ 0, for some ξ ∈ R (namely ξ = r(s) where s ∈ {0, 1}N is a Sturmian word of slope α). Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 26  1  For each x ∈ 0, 2 , the theorem gives the minimal interval beginning with x (namely [x, R(x)]) that contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

Definition Given a real number x ∈ (0, 1/2), let R(x) denote the smallest positive real number such that [x, R(x)] contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

Note: x ≤ R(x) < 1

The following theorem gives a complete description of the minimal intervals containing all fractional parts {ξ2k}, k ≥ 0, for some positive real number ξ.

Some Applications in Number Theory Distribution of real numbers modulo 1 h 1 i Note: For x ∈ 2 , 1 , there does not exist a real number ξ > 0 such that x ≤ {ξ2k} < 1 for all k ≥ 0 (Bugeaud-Dubickas with b = 2).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 27  1  For each x ∈ 0, 2 , the theorem gives the minimal interval beginning with x (namely [x, R(x)]) that contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

Note: x ≤ R(x) < 1

The following theorem gives a complete description of the minimal intervals containing all fractional parts {ξ2k}, k ≥ 0, for some positive real number ξ.

Some Applications in Number Theory Distribution of real numbers modulo 1 h 1 i Note: For x ∈ 2 , 1 , there does not exist a real number ξ > 0 such that x ≤ {ξ2k} < 1 for all k ≥ 0 (Bugeaud-Dubickas with b = 2).

Definition Given a real number x ∈ (0, 1/2), let R(x) denote the smallest positive real number such that [x, R(x)] contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 27  1  For each x ∈ 0, 2 , the theorem gives the minimal interval beginning with x (namely [x, R(x)]) that contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

The following theorem gives a complete description of the minimal intervals containing all fractional parts {ξ2k}, k ≥ 0, for some positive real number ξ.

Some Applications in Number Theory Distribution of real numbers modulo 1 h 1 i Note: For x ∈ 2 , 1 , there does not exist a real number ξ > 0 such that x ≤ {ξ2k} < 1 for all k ≥ 0 (Bugeaud-Dubickas with b = 2).

Definition Given a real number x ∈ (0, 1/2), let R(x) denote the smallest positive real number such that [x, R(x)] contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

Note: x ≤ R(x) < 1

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 27  1  For each x ∈ 0, 2 , the theorem gives the minimal interval beginning with x (namely [x, R(x)]) that contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

Some Applications in Number Theory Distribution of real numbers modulo 1 h 1 i Note: For x ∈ 2 , 1 , there does not exist a real number ξ > 0 such that x ≤ {ξ2k} < 1 for all k ≥ 0 (Bugeaud-Dubickas with b = 2).

Definition Given a real number x ∈ (0, 1/2), let R(x) denote the smallest positive real number such that [x, R(x)] contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

Note: x ≤ R(x) < 1

The following theorem gives a complete description of the minimal intervals containing all fractional parts {ξ2k}, k ≥ 0, for some positive real number ξ.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 27 Some Applications in Number Theory Distribution of real numbers modulo 1 h 1 i Note: For x ∈ 2 , 1 , there does not exist a real number ξ > 0 such that x ≤ {ξ2k} < 1 for all k ≥ 0 (Bugeaud-Dubickas with b = 2).

Definition Given a real number x ∈ (0, 1/2), let R(x) denote the smallest positive real number such that [x, R(x)] contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R .

Note: x ≤ R(x) < 1

The following theorem gives a complete description of the minimal intervals containing all fractional parts {ξ2k}, k ≥ 0, for some positive real number ξ.

 1  For each x ∈ 0, 2 , the theorem gives the minimal interval beginning with x (namely [x, R(x)]) that contains all fractional parts {ξ2k}, k ≥ 0, + for some ξ ∈ R . Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 27 ∞ ∞ (ii) If (2x)2 = (P al(v)01) or (2x)2 = (P al(v)10) for some word v over {0, 1}, then R(x) = r((1P al(v)0)∞). (iii) In all other cases, R(x) = r((1P al(v)0)∞) where v is the unique word over {0, 1} such that r((P al(v)01)∞) < 2x < r((P al(v)10)∞).

Furthermore, R(x) is the unique real number in (0, 1) that has a Sturmian binary expansion and satisfies x ≤ {R(x)2k} ≤ R(x) for all k ≥ 0.

Some Applications in Number Theory Distribution of real numbers modulo 1 Theorem  1  Let x be a real number in 0, 2 .

(i) If (2x)2 is a characteristic Sturmian sequence, then 1 R(x) = x + . 2

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 28 ∞ ∞ (ii) If (2x)2 = (P al(v)01) or (2x)2 = (P al(v)10) for some word v over {0, 1}, then R(x) = r((1P al(v)0)∞). (iii) In all other cases, R(x) = r((1P al(v)0)∞) where v is the unique word over {0, 1} such that r((P al(v)01)∞) < 2x < r((P al(v)10)∞).

Some Applications in Number Theory Distribution of real numbers modulo 1 Theorem  1  Let x be a real number in 0, 2 .

(i) If (2x)2 is a characteristic Sturmian sequence, then 1 R(x) = x + . 2 Furthermore, R(x) is the unique real number in (0, 1) that has a Sturmian binary expansion and satisfies x ≤ {R(x)2k} ≤ R(x) for all k ≥ 0.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 28 (iii) In all other cases, R(x) = r((1P al(v)0)∞) where v is the unique word over {0, 1} such that r((P al(v)01)∞) < 2x < r((P al(v)10)∞).

Some Applications in Number Theory Distribution of real numbers modulo 1 Theorem  1  Let x be a real number in 0, 2 .

(i) If (2x)2 is a characteristic Sturmian sequence, then 1 R(x) = x + . 2 Furthermore, R(x) is the unique real number in (0, 1) that has a Sturmian binary expansion and satisfies x ≤ {R(x)2k} ≤ R(x) for all k ≥ 0.

∞ ∞ (ii) If (2x)2 = (P al(v)01) or (2x)2 = (P al(v)10) for some word v over {0, 1}, then R(x) = r((1P al(v)0)∞).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 28 Some Applications in Number Theory Distribution of real numbers modulo 1 Theorem  1  Let x be a real number in 0, 2 .

(i) If (2x)2 is a characteristic Sturmian sequence, then 1 R(x) = x + . 2 Furthermore, R(x) is the unique real number in (0, 1) that has a Sturmian binary expansion and satisfies x ≤ {R(x)2k} ≤ R(x) for all k ≥ 0.

∞ ∞ (ii) If (2x)2 = (P al(v)01) or (2x)2 = (P al(v)10) for some word v over {0, 1}, then R(x) = r((1P al(v)0)∞). (iii) In all other cases, R(x) = r((1P al(v)0)∞) where v is the unique word over {0, 1} such that r((P al(v)01)∞) < 2x < r((P al(v)10)∞).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 28 What is R(1/4)?

I Observe that (2x)2 = (1/2)2 = 1000 ··· (or 0111 ··· ).

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem.

I We have ∞ ∞ (01) ≺ (2x)2 (= 1000 ··· ) ≺ (10) where (01)∞ = (P al(ε)01)∞ and (10)∞ = (P al(ε)10)∞.

I Hence, by Case (iii) of the Theorem, R(1/4) = r((10)∞) = 2/3.

1 2 k 2 I Moreover, 4 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples Take x = 1/4.

1 I The minimal interval beginning with x = 4 containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [1/4,R(1/4)].

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 29 I Observe that (2x)2 = (1/2)2 = 1000 ··· (or 0111 ··· ).

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem.

I We have ∞ ∞ (01) ≺ (2x)2 (= 1000 ··· ) ≺ (10) where (01)∞ = (P al(ε)01)∞ and (10)∞ = (P al(ε)10)∞.

I Hence, by Case (iii) of the Theorem, R(1/4) = r((10)∞) = 2/3.

1 2 k 2 I Moreover, 4 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples Take x = 1/4.

1 I The minimal interval beginning with x = 4 containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [1/4,R(1/4)]. What is R(1/4)?

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 29 I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem.

I We have ∞ ∞ (01) ≺ (2x)2 (= 1000 ··· ) ≺ (10) where (01)∞ = (P al(ε)01)∞ and (10)∞ = (P al(ε)10)∞.

I Hence, by Case (iii) of the Theorem, R(1/4) = r((10)∞) = 2/3.

1 2 k 2 I Moreover, 4 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples Take x = 1/4.

1 I The minimal interval beginning with x = 4 containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [1/4,R(1/4)]. What is R(1/4)?

I Observe that (2x)2 = (1/2)2 = 1000 ··· (or 0111 ··· ).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 29 I Thus, x falls into Case (iii) of the Theorem.

I We have ∞ ∞ (01) ≺ (2x)2 (= 1000 ··· ) ≺ (10) where (01)∞ = (P al(ε)01)∞ and (10)∞ = (P al(ε)10)∞.

I Hence, by Case (iii) of the Theorem, R(1/4) = r((10)∞) = 2/3.

1 2 k 2 I Moreover, 4 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples Take x = 1/4.

1 I The minimal interval beginning with x = 4 containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [1/4,R(1/4)]. What is R(1/4)?

I Observe that (2x)2 = (1/2)2 = 1000 ··· (or 0111 ··· ).

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 29 I We have ∞ ∞ (01) ≺ (2x)2 (= 1000 ··· ) ≺ (10) where (01)∞ = (P al(ε)01)∞ and (10)∞ = (P al(ε)10)∞.

I Hence, by Case (iii) of the Theorem, R(1/4) = r((10)∞) = 2/3.

1 2 k 2 I Moreover, 4 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples Take x = 1/4.

1 I The minimal interval beginning with x = 4 containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [1/4,R(1/4)]. What is R(1/4)?

I Observe that (2x)2 = (1/2)2 = 1000 ··· (or 0111 ··· ).

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 29 I Hence, by Case (iii) of the Theorem, R(1/4) = r((10)∞) = 2/3.

1 2 k 2 I Moreover, 4 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples Take x = 1/4.

1 I The minimal interval beginning with x = 4 containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [1/4,R(1/4)]. What is R(1/4)?

I Observe that (2x)2 = (1/2)2 = 1000 ··· (or 0111 ··· ).

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem.

I We have ∞ ∞ (01) ≺ (2x)2 (= 1000 ··· ) ≺ (10) where (01)∞ = (P al(ε)01)∞ and (10)∞ = (P al(ε)10)∞.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 29 1 2 k 2 I Moreover, 4 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples Take x = 1/4.

1 I The minimal interval beginning with x = 4 containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [1/4,R(1/4)]. What is R(1/4)?

I Observe that (2x)2 = (1/2)2 = 1000 ··· (or 0111 ··· ).

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem.

I We have ∞ ∞ (01) ≺ (2x)2 (= 1000 ··· ) ≺ (10) where (01)∞ = (P al(ε)01)∞ and (10)∞ = (P al(ε)10)∞.

I Hence, by Case (iii) of the Theorem, R(1/4) = r((10)∞) = 2/3.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 29 Some Applications in Number Theory Distribution of real numbers modulo 1 Examples Take x = 1/4.

1 I The minimal interval beginning with x = 4 containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [1/4,R(1/4)]. What is R(1/4)?

I Observe that (2x)2 = (1/2)2 = 1000 ··· (or 0111 ··· ).

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem.

I We have ∞ ∞ (01) ≺ (2x)2 (= 1000 ··· ) ≺ (10) where (01)∞ = (P al(ε)01)∞ and (10)∞ = (P al(ε)10)∞.

I Hence, by Case (iii) of the Theorem, R(1/4) = r((10)∞) = 2/3.

1 2 k 2 I Moreover, 4 ≤ { 3 · 2 } ≤ 3 for all k ≥ 0.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 29 What is R(x)?

I Observe that

(x)2 = 0 010010 0101001010100100011011010011000100001100 ··· | {z } P al(010)

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem. ∞ ∞ I We have (01001001) ≺ (2x)2 ≺ (01001010) where 010010 = P al(010).

I Hence, by Case (iii) of the Theorem, R(x) = r((1P al(010)0)∞) = 164/255.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples... √ Let x = 1/(π + 2e). √ I The minimal interval beginning with x = 1/(π + 2e) containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [x, R(x)].

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 30 I Observe that

(x)2 = 0 010010 0101001010100100011011010011000100001100 ··· | {z } P al(010)

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem. ∞ ∞ I We have (01001001) ≺ (2x)2 ≺ (01001010) where 010010 = P al(010).

I Hence, by Case (iii) of the Theorem, R(x) = r((1P al(010)0)∞) = 164/255.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples... √ Let x = 1/(π + 2e). √ I The minimal interval beginning with x = 1/(π + 2e) containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [x, R(x)]. What is R(x)?

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 30 I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem. ∞ ∞ I We have (01001001) ≺ (2x)2 ≺ (01001010) where 010010 = P al(010).

I Hence, by Case (iii) of the Theorem, R(x) = r((1P al(010)0)∞) = 164/255.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples... √ Let x = 1/(π + 2e). √ I The minimal interval beginning with x = 1/(π + 2e) containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [x, R(x)]. What is R(x)?

I Observe that

(x)2 = 0 010010 0101001010100100011011010011000100001100 ··· | {z } P al(010)

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 30 I Thus, x falls into Case (iii) of the Theorem. ∞ ∞ I We have (01001001) ≺ (2x)2 ≺ (01001010) where 010010 = P al(010).

I Hence, by Case (iii) of the Theorem, R(x) = r((1P al(010)0)∞) = 164/255.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples... √ Let x = 1/(π + 2e). √ I The minimal interval beginning with x = 1/(π + 2e) containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [x, R(x)]. What is R(x)?

I Observe that

(x)2 = 0 010010 0101001010100100011011010011000100001100 ··· | {z } P al(010)

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 30 ∞ ∞ I We have (01001001) ≺ (2x)2 ≺ (01001010) where 010010 = P al(010).

I Hence, by Case (iii) of the Theorem, R(x) = r((1P al(010)0)∞) = 164/255.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples... √ Let x = 1/(π + 2e). √ I The minimal interval beginning with x = 1/(π + 2e) containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [x, R(x)]. What is R(x)?

I Observe that

(x)2 = 0 010010 0101001010100100011011010011000100001100 ··· | {z } P al(010)

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 30 I Hence, by Case (iii) of the Theorem, R(x) = r((1P al(010)0)∞) = 164/255.

Some Applications in Number Theory Distribution of real numbers modulo 1 Examples... √ Let x = 1/(π + 2e). √ I The minimal interval beginning with x = 1/(π + 2e) containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [x, R(x)]. What is R(x)?

I Observe that

(x)2 = 0 010010 0101001010100100011011010011000100001100 ··· | {z } P al(010)

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem. ∞ ∞ I We have (01001001) ≺ (2x)2 ≺ (01001010) where 010010 = P al(010).

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 30 Some Applications in Number Theory Distribution of real numbers modulo 1 Examples... √ Let x = 1/(π + 2e). √ I The minimal interval beginning with x = 1/(π + 2e) containing all the fractional parts {ξ2k}, k ≥ 0, for some real number ξ > 0 is [x, R(x)]. What is R(x)?

I Observe that

(x)2 = 0 010010 0101001010100100011011010011000100001100 ··· | {z } P al(010)

I So (2x)2 is not a characteristic Sturmian sequence and it is not of the form (P al(v)01)∞ or (P al(v)10)∞ for some word v over {0, 1}.

I Thus, x falls into Case (iii) of the Theorem. ∞ ∞ I We have (01001001) ≺ (2x)2 ≺ (01001010) where 010010 = P al(010).

I Hence, by Case (iii) of the Theorem, R(x) = r((1P al(010)0)∞) = 164/255.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 30 The result of Bugeaud and Dubickas (2005) implies that Sturmian sequences on an alphabet {k, k + 1} for some k ∈ {0, 1, . . . , b − 2} will again play a fundamental role.

. . . episturmian words?

I One may ask what happens with base b expansions, where b ≥ 3, or what can be said about the intervals containing all {ξbn} for some ξ.

It is an open problem to consider what happens in the case of n fractional parts {ξ(p/q) }n≥0 for some positive real number ξ (which must all lie in an interval of length at least 1/p)

As a consequence of our theorem, we deduce: 1 If x is an algebraic real number in (0, 2 ), then R(x) is rational.

Some Applications in Number Theory Distribution of real numbers modulo 1 Remarks

I It is known (Ferenczi-Mauduit 1997) that any real number having a Sturmian binary expansion is transcendental.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 31 . . . episturmian words?

The result of Bugeaud and Dubickas (2005) implies that Sturmian sequences on an alphabet {k, k + 1} for some k ∈ {0, 1, . . . , b − 2} will again play a fundamental role. It is an open problem to consider what happens in the case of n fractional parts {ξ(p/q) }n≥0 for some positive real number ξ (which must all lie in an interval of length at least 1/p)

I One may ask what happens with base b expansions, where b ≥ 3, or what can be said about the intervals containing all {ξbn} for some ξ.

Some Applications in Number Theory Distribution of real numbers modulo 1 Remarks

I It is known (Ferenczi-Mauduit 1997) that any real number having a Sturmian binary expansion is transcendental. As a consequence of our theorem, we deduce: 1 If x is an algebraic real number in (0, 2 ), then R(x) is rational.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 31 . . . episturmian words?

The result of Bugeaud and Dubickas (2005) implies that Sturmian sequences on an alphabet {k, k + 1} for some k ∈ {0, 1, . . . , b − 2} will again play a fundamental role. It is an open problem to consider what happens in the case of n fractional parts {ξ(p/q) }n≥0 for some positive real number ξ (which must all lie in an interval of length at least 1/p)

Some Applications in Number Theory Distribution of real numbers modulo 1 Remarks

I It is known (Ferenczi-Mauduit 1997) that any real number having a Sturmian binary expansion is transcendental. As a consequence of our theorem, we deduce: 1 If x is an algebraic real number in (0, 2 ), then R(x) is rational.

I One may ask what happens with base b expansions, where b ≥ 3, or what can be said about the intervals containing all {ξbn} for some ξ.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 31 . . . episturmian words?

It is an open problem to consider what happens in the case of n fractional parts {ξ(p/q) }n≥0 for some positive real number ξ (which must all lie in an interval of length at least 1/p)

Some Applications in Number Theory Distribution of real numbers modulo 1 Remarks

I It is known (Ferenczi-Mauduit 1997) that any real number having a Sturmian binary expansion is transcendental. As a consequence of our theorem, we deduce: 1 If x is an algebraic real number in (0, 2 ), then R(x) is rational.

I One may ask what happens with base b expansions, where b ≥ 3, or what can be said about the intervals containing all {ξbn} for some ξ. The result of Bugeaud and Dubickas (2005) implies that Sturmian sequences on an alphabet {k, k + 1} for some k ∈ {0, 1, . . . , b − 2} will again play a fundamental role.

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 31 . . . episturmian words?

Some Applications in Number Theory Distribution of real numbers modulo 1 Remarks

I It is known (Ferenczi-Mauduit 1997) that any real number having a Sturmian binary expansion is transcendental. As a consequence of our theorem, we deduce: 1 If x is an algebraic real number in (0, 2 ), then R(x) is rational.

I One may ask what happens with base b expansions, where b ≥ 3, or what can be said about the intervals containing all {ξbn} for some ξ. The result of Bugeaud and Dubickas (2005) implies that Sturmian sequences on an alphabet {k, k + 1} for some k ∈ {0, 1, . . . , b − 2} will again play a fundamental role. It is an open problem to consider what happens in the case of n fractional parts {ξ(p/q) }n≥0 for some positive real number ξ (which must all lie in an interval of length at least 1/p)

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 31 Some Applications in Number Theory Distribution of real numbers modulo 1 Remarks

I It is known (Ferenczi-Mauduit 1997) that any real number having a Sturmian binary expansion is transcendental. As a consequence of our theorem, we deduce: 1 If x is an algebraic real number in (0, 2 ), then R(x) is rational.

I One may ask what happens with base b expansions, where b ≥ 3, or what can be said about the intervals containing all {ξbn} for some ξ. The result of Bugeaud and Dubickas (2005) implies that Sturmian sequences on an alphabet {k, k + 1} for some k ∈ {0, 1, . . . , b − 2} will again play a fundamental role. It is an open problem to consider what happens in the case of n fractional parts {ξ(p/q) }n≥0 for some positive real number ξ (which must all lie in an interval of length at least 1/p) . . . episturmian words?

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 31 Thank You!

Amy Glen (Murdoch University) Combinatorics on words & Number Theory Sept 25, 2016 32