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Beatty sequences and integers in golden mean base Michel Dekking CIRM November 6, 2019 11:40-12:20 CIRM November 5, 2019 17:50-18:30 Integers in golden mean base := (1 + 5) / 2 Base phi representations / beta-expansions of the natural

  1. Beatty sequences and integers in golden mean base Michel Dekking CIRM November 6, 2019 11:40-12:20 CIRM November 5, 2019 17:50-18:30

  2. Integers in golden mean base √ ϕ := (1 + 5) / 2 Base phi representations / beta-expansions of the natural numbers (beta = ϕ ) George Bergman in 1957.

  3. Integers in golden mean base √ ϕ := (1 + 5) / 2 Base phi representations / beta-expansions of the natural numbers (beta = ϕ ) George Bergman in 1957. A natural number N is written in base phi if N has the form ∞ � d i ϕ i , N = i = −∞ with digits d i = 0 or 1, and where d i d i +1 = 11 is not allowed. We write these representations as β ( N ) = d L d L − 1 . . . d 1 d 0 · d − 1 d − 2 . . . d R +1 d R .

  4. Integers in golden mean base √ ϕ := (1 + 5) / 2 Base phi representations / beta-expansions of the natural numbers (beta = ϕ ) George Bergman in 1957. A natural number N is written in base phi if N has the form ∞ � d i ϕ i , N = i = −∞ with digits d i = 0 or 1, and where d i d i +1 = 11 is not allowed. We write these representations as β ( N ) = d L d L − 1 . . . d 1 d 0 · d − 1 d − 2 . . . d R +1 d R . “Significant digits”: d L = d R = 1

  5. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman).

  6. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). Examples.

  7. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1.

  8. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01

  9. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2

  10. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ.

  11. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 =

  12. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 = 10.01 + 1.0 =

  13. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 =

  14. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 = 100.01

  15. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 = 100.01 4 = 3+1 = 100.01 + 1.0 = 101.01

  16. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 = 100.01 4 = 3+1 = 100.01 + 1.0 = 101.01 5 = 4+1 = 101.01 + 1.0 =

  17. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 = 100.01 4 = 3+1 = 100.01 + 1.0 = 101.01 5 = 4+1 = 101.01 + 1.0 = 102.01 =

  18. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 = 100.01 4 = 3+1 = 100.01 + 1.0 = 101.01 5 = 4+1 = 101.01 + 1.0 = 102.01 = 110.02 =

  19. Integers in golden mean base, examples The base phi representation of a number N is unique (Bergman). ϕ 2 = 1 + ϕ Examples. 1 = 1. 2 = 10.01 ϕ + ϕ − 2 = 2 ⇔ ϕ + 1 / (1 + ϕ ) = 2 ⇔ (1+ ϕ ) ϕ +1 = 2+2 ϕ ⇔ ϕ + 1+ ϕ = 1+2 ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 = 100.01 4 = 3+1 = 100.01 + 1.0 = 101.01 5 = 4+1 = 101.01 + 1.0 = 102.01 = 110.02 = 1000.1001

  20. Integers in golden mean base, a list N β ( N ) N N in base 2 1 1 1 1 2 10 · 01 2 10 3 100 · 01 3 11 4 101 · 01 4 100 5 1000 · 1001 5 101 6 110 6 1010 · 0001 7 10000 · 0001 7 111 8 1000 8 10001 · 0001 9 10010 · 0101 9 1001 10 10100 · 0101 10 1010 11 10101 · 0101 11 1011 12 100000 · 101001 12 1100 13 100010 · 001001 13 1101 14 100100 · 001001 14 1110 15 1111 15 100101 · 001001

  21. Distribution of k th digit d k = d k ( N ) ? Distribution of d k = d k ( N ) over the natural numbers N ∈ N , where k is an integer.

  22. Distribution of k th digit d k = d k ( N ) ? Distribution of d k = d k ( N ) over the natural numbers N ∈ N , where k is an integer. Distribution in the frequency sense: The case k = 0 conjectured by Bergman, and proved in 1999 by E. Hart and L. Sanchis, On the occurrence of F n in the Zeckendorf decomposition of nF n : THEOREM The frequency of 1’s in ( d 0 ( N )) exists, and √ N 1 ϕ + 2 = 5 − 1 5 � lim d 0 ( M ) = . N 10 N →∞ M =1

  23. Distribution of digit d 0 = d 0 ( N ) and d 1 = d 1 ( N ) Stronger property for d 0 (Baruchel 2018): CONJECTURE 1 Digit d 0 ( N ) = 1 if and only if N = ⌊ n ϕ ⌋ + 2 n + 1 for some natural number n , or N = 1.

  24. Distribution of digit d 0 = d 0 ( N ) and d 1 = d 1 ( N ) Stronger property for d 0 (Baruchel 2018): CONJECTURE 1 Digit d 0 ( N ) = 1 if and only if N = ⌊ n ϕ ⌋ + 2 n + 1 for some natural number n , or N = 1. ( ⌊ n ϕ ⌋ ) = 1 , 3 , 4 , 6 , 8 , 9 , 11 , 12 , 14 , 16 , 17 , 19 , 21 , 22 , 24 , 25 , 27 , ... is the well known lower Wythoff sequence.

  25. Distribution of digit d 0 = d 0 ( N ) and d 1 = d 1 ( N ) Stronger property for d 0 (Baruchel 2018): CONJECTURE 1 Digit d 0 ( N ) = 1 if and only if N = ⌊ n ϕ ⌋ + 2 n + 1 for some natural number n , or N = 1. ( ⌊ n ϕ ⌋ ) = 1 , 3 , 4 , 6 , 8 , 9 , 11 , 12 , 14 , 16 , 17 , 19 , 21 , 22 , 24 , 25 , 27 , ... is the well known lower Wythoff sequence. Corresponding property for digit d 1 (Kimberling 2012): CONJECTURE 2 Digit d 1 ( N ) = 1 if and only if N = ⌊ n ϕ ⌋ + 2 n − 1 for some natural number n .

  26. Generalized Beatty sequences In the conjectures occur sequences V of the type V ( n ) = p ⌊ n α ⌋ + q n + r , n ≥ 1, where α is a real number, and p , q , and r are integers.

  27. Generalized Beatty sequences In the conjectures occur sequences V of the type V ( n ) = p ⌊ n α ⌋ + q n + r , n ≥ 1, where α is a real number, and p , q , and r are integers. If S is a sequence then ∆ S ( n ) = S ( n + 1) − S ( n ) , for n = 1 , 2 . . . .

  28. Generalized Beatty sequences In the conjectures occur sequences V of the type V ( n ) = p ⌊ n α ⌋ + q n + r , n ≥ 1, where α is a real number, and p , q , and r are integers. If S is a sequence then ∆ S ( n ) = S ( n + 1) − S ( n ) , for n = 1 , 2 . . . . The sequence ∆( ⌊ n ϕ ⌋ ) is equal to the Fibonacci word x 2 , 1 = 21221212 . . . on the alphabet { 2 , 1 } .

  29. Generalized Beatty sequences In the conjectures occur sequences V of the type V ( n ) = p ⌊ n α ⌋ + q n + r , n ≥ 1, where α is a real number, and p , q , and r are integers. If S is a sequence then ∆ S ( n ) = S ( n + 1) − S ( n ) , for n = 1 , 2 . . . . The sequence ∆( ⌊ n ϕ ⌋ ) is equal to the Fibonacci word x 2 , 1 = 21221212 . . . on the alphabet { 2 , 1 } . LEMMA (JPA & FMD, 2019) Let V ( n ) = p ⌊ n ϕ ⌋ + qn + r be a golden mean GBS. Then ∆ V = x 2 p + q , p + q . Conversely, if x a , b is the Fibonacci word on the alphabet { a , b } , then any V with ∆ V = x a , b is a GBS V = (( a − b ) ⌊ n ϕ ⌋ ) + (2 b − a ) n + r ) for some integer r .

  30. Morphisms Morphism: is a map from the set of infinite words over an alphabet to itself, respecting the concatenation operation. Example: Fibonacci morphism σ on { 0 , 1 } σ (0) = 01 , σ (1) = 0 .

  31. Morphisms Morphism: is a map from the set of infinite words over an alphabet to itself, respecting the concatenation operation. Example: Fibonacci morphism σ on { 0 , 1 } σ (0) = 01 , σ (1) = 0 . A central role in this talk: γ on the alphabet { A , B , C , D } γ ( A ) = AB , γ ( B ) = C , γ ( C ) = D , γ ( D ) = ABC .

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