Beatty sequences and integers in golden mean base Michel Dekking - - PowerPoint PPT Presentation

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

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

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 N =

∞

• i=−∞

diϕi, with digits di = 0 or 1, and where didi+1 = 11 is not allowed. We write these representations as β(N) = dLdL−1 . . . d1d0 · d−1d−2 . . . dR+1dR.

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 N =

∞

• i=−∞

diϕi, with digits di = 0 or 1, and where didi+1 = 11 is not allowed. We write these representations as β(N) = dLdL−1 . . . d1d0 · d−1d−2 . . . dR+1dR.

“Significant digits”: dL = dR = 1

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman).

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples.

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1.

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ.

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 3 = 2+1 =

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 3 = 2+1 = 10.01 + 1.0 =

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 =

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 = 100.01

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 3 = 2+1 = 10.01 + 1.0 = 11.01 = 100.01 4 = 3+1 = 100.01 + 1.0 = 101.01

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 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 =

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 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 =

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 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 =

Integers in golden mean base, examples

The base phi representation of a number N is unique (Bergman). Examples. ϕ2 = 1 + ϕ 1 = 1. 2 = 10.01 ϕ + ϕ−2 = 2 ⇔ ϕ + 1/(1 + ϕ) = 2 ⇔ (1+ϕ)ϕ+1 = 2+2ϕ ⇔ ϕ+ 1+ϕ = 1+2ϕ. 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

Integers in golden mean base, a list

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

Distribution of kth digit dk = dk(N)?

Distribution of dk = dk(N) over the natural numbers N ∈ N, where k is an integer.

Distribution of kth digit dk = dk(N)?

Distribution of dk = dk(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 Fn in the Zeckendorf decomposition of nFn: THEOREM The frequency of 1’s in (d0(N)) exists, and lim

N→∞

1 N

N

• M=1

d0(M) = 1 ϕ + 2 = 5 − √ 5 10 .

Distribution of digit d0 = d0(N) and d1 = d1(N)

Stronger property for d0 (Baruchel 2018): CONJECTURE 1 Digit d0(N) = 1 if and only if N = ⌊nϕ⌋ + 2n + 1 for some natural number n, or N = 1.

Distribution of digit d0 = d0(N) and d1 = d1(N)

Stronger property for d0 (Baruchel 2018): CONJECTURE 1 Digit d0(N) = 1 if and only if N = ⌊nϕ⌋ + 2n + 1 for some natural number n, or N = 1. (⌊nϕ⌋) = 1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, ... is the well known lower Wythoff sequence.

Distribution of digit d0 = d0(N) and d1 = d1(N)

Stronger property for d0 (Baruchel 2018): CONJECTURE 1 Digit d0(N) = 1 if and only if N = ⌊nϕ⌋ + 2n + 1 for some natural number n, or N = 1. (⌊nϕ⌋) = 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 d1 (Kimberling 2012): CONJECTURE 2 Digit d1(N) = 1 if and only if N = ⌊nϕ⌋ + 2n − 1 for some natural number n.

Generalized Beatty sequences

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

Generalized Beatty sequences

In the conjectures occur sequences V of the type V (n) = p⌊nα⌋ + 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 . . . .

Generalized Beatty sequences

In the conjectures occur sequences V of the type V (n) = p⌊nα⌋ + 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 ∆(⌊nϕ⌋) is equal to the Fibonacci word x2,1 = 21221212 . . . on the alphabet {2, 1}.

Generalized Beatty sequences

In the conjectures occur sequences V of the type V (n) = p⌊nα⌋ + 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 ∆(⌊nϕ⌋) is equal to the Fibonacci word x2,1 = 21221212 . . . on the alphabet {2, 1}. LEMMA (JPA & FMD, 2019) Let V (n) = p⌊nϕ⌋ + qn + r be a golden mean GBS. Then ∆V = x2p+q,p+q. Conversely, if xa,b is the Fibonacci word on the alphabet {a, b}, then any V with ∆V = xa,b is a GBS V = ((a − b)⌊nϕ⌋) + (2b − a)n + r) for some integer r.

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.

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.

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. LEMMA The morphism γ has the following properties i) |γn(A)| = Ln, for all n ≥ 2, where Ln is the nth Lucas number. ii) γn(A) = γn(C) and γn(A) = γn+1(B) for all n ≥ 2.

Lucas numbers

The Lucas numbers (Ln) = (2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, . . . ): L0 = 2, L1 = 1, Ln = Ln−1 + Ln−2 for n ≥ 2.

Lucas numbers

The Lucas numbers (Ln) = (2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, . . . ): L0 = 2, L1 = 1, Ln = Ln−1 + Ln−2 for n ≥ 2. The Lucas numbers have a particularly simple base phi representation. From the well-known formula L2n = ϕ2n + ϕ−2n, and the recursion L2n+1 = L2n + L2n−1, we have for all n ≥ 1 β(L2n) = 102n · 02n−11, β(L2n+1) = 1(01)n · (01)n.

Lucas numbers

The Lucas numbers (Ln) = (2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, . . . ): L0 = 2, L1 = 1, Ln = Ln−1 + Ln−2 for n ≥ 2. The Lucas numbers have a particularly simple base phi representation. From the well-known formula L2n = ϕ2n + ϕ−2n, and the recursion L2n+1 = L2n + L2n−1, we have for all n ≥ 1 β(L2n) = 102n · 02n−11, β(L2n+1) = 1(01)n · (01)n. Since β(L2n) consists of only 0’s between the exterior 1’s: LEMMA EVEN For all n ≥ 1 and k = 1, . . . , L2n−1 one has β(L2n + k) = β(L2n) + β(k) = 10 . . . 0 β(k) 0 . . . 01.

More Lucas numbers

As in Hart (’98), Hart & Sanchis (’99), Sanchis & Sanchis (’01): STRATEGY: partition the natural numbers in intervals [Ln + 1, Ln+1], and establish recursive relations for the β-expansions of the numbers in these intervals.

More Lucas numbers

As in Hart (’98), Hart & Sanchis (’99), Sanchis & Sanchis (’01): STRATEGY: partition the natural numbers in intervals [Ln + 1, Ln+1], and establish recursive relations for the β-expansions of the numbers in these intervals. However, an analogous formula as in previous LEMMA starting from an odd Lucas number does not exist. It seems that to obtain recursive relations the interval [L2n+1 + 1, L2n+2 − 1] has to be divided into three subintervals.

The odd Lucas number intervals

G.R. Sanchis and L.A. Sanchis, On the frequency of occurrence of αi in the α-expansions of the positive integers, Fibonacci Quart. 39 (2001), 123–173. LEMMA ODD For all n ≥ 2 and k = 1, . . . , L2n−2 − 1 β(L2n+1 + k) = 1000(10)−1β(L2n−1 + k)(01)−11001, β(L2n+1 + L2n−1 + k) = 10β(L2n−1 + k)(01)−10001. Moreover, for all n ≥ 2 and k = 0, . . . , L2n−3 β(L2n+1 + L2n−2 + k) = 10010(10)−1β(L2n−2 + k)(01)−1001001.

A proof of the conjectures

Conjecture 1 and Conjecture 2 are implied by: THEOREM d1d0 Let β(N) = (di(N)) be the base phi representation of a natural number N. Then: d0(N) = 1 ⇔ N = ⌊nϕ⌋ + 2n + 1 for some n, d1d0(N) = 10 ⇔ N = ⌊nϕ⌋ + 2n − 1 for some n, d1d0d−1(N) = 000 ⇔ N = ⌊nϕ⌋ + 2n for some n, d1d0d−1(N) = 001 ⇔ N = 3⌊nϕ⌋ + n + 1 for some n.

A proof of the conjectures

Conjecture 1 and Conjecture 2 are implied by: THEOREM d1d0 Let β(N) = (di(N)) be the base phi representation of a natural number N. Then: d0(N) = 1 ⇔ N = ⌊nϕ⌋ + 2n + 1 for some n, d1d0(N) = 10 ⇔ N = ⌊nϕ⌋ + 2n − 1 for some n, d1d0d−1(N) = 000 ⇔ N = ⌊nϕ⌋ + 2n for some n, d1d0d−1(N) = 001 ⇔ N = 3⌊nϕ⌋ + n + 1 for some n. Coding: T(N) = A iff d1d0(N) = 10, T(N) = B iff d1d0d−1(N) = 000, T(N) = C iff d0(N) = 1, T(N) = D iff d1d0d−1(N) = 001.

The 4 letter coding

N β(N) T(N) 1 1 C 2 10 · 01 A 3 100 · 01 B 4 101 · 01 C 5 1000 · 1001 D 6 1010 · 0001 A 7 10000 · 0001 B 8 10001 · 0001 C 9 10010 · 0101 A 10 10100 · 0101 B 11 10101 · 0101 C 12 100000 · 101001 D 13 100010 · 001001 A 14 100100 · 001001 B 15 100101 · 001001 C d1d0 = 10 A d1d0d−1 = 00 · 0 B d0 = 1 C d1d0d−1 = 00 · 1 D

The structure of the coding

THEOREM FIX The sequence (T(N))N≥2 is the unique fixed point of the morphism γ.

The structure of the coding

THEOREM FIX The sequence (T(N))N≥2 is the unique fixed point of the morphism γ. Theorem FIX is an immediate consequence of: THEOREM Let γ be the morphism given by A → AB, B → C, C → D, D → ABC. Then a) T(2) T(3) · · · T(Ln+1) = γn(A) for n ≥ 2, b) T(Ln+2) T(Ln+3) · · · T(Ln+1+1) = γn−1(A) for n ≥ 3.

The structure of the coding

THEOREM FIX The sequence (T(N))N≥2 is the unique fixed point of the morphism γ. Theorem FIX is an immediate consequence of: THEOREM Let γ be the morphism given by A → AB, B → C, C → D, D → ABC. Then a) T(2) T(3) · · · T(Ln+1) = γn(A) for n ≥ 2, b) T(Ln+2) T(Ln+3) · · · T(Ln+1+1) = γn−1(A) for n ≥ 3. This theorem is proved by induction, using LEMMA EVEN and LEMMA ODD.

Idea of the proof of THEOREM d1d0

Using Theorem FIX, we study the occurrences of C in T: digit d0(N) = 1 ⇔ T(N) = C. T := CABCABCD . . . is the fixed point of γ, prefixed by C.

Idea of the proof of THEOREM d1d0

Using Theorem FIX, we study the occurrences of C in T: digit d0(N) = 1 ⇔ T(N) = C. T := CABCABCD . . . is the fixed point of γ, prefixed by C.

Recall: γ(A) = AB, γ(B) = C, γ(C) = D, γ(D) = ABC.

The return words of C are CAB and CDAB.

Idea of the proof of THEOREM d1d0

Using Theorem FIX, we study the occurrences of C in T: digit d0(N) = 1 ⇔ T(N) = C. T := CABCABCD . . . is the fixed point of γ, prefixed by C.

Recall: γ(A) = AB, γ(B) = C, γ(C) = D, γ(D) = ABC.

The return words of C are CAB and CDAB. Code the return words by their lengths: 3 and 4.

Idea of the proof of THEOREM d1d0

Using Theorem FIX, we study the occurrences of C in T: digit d0(N) = 1 ⇔ T(N) = C. T := CABCABCD . . . is the fixed point of γ, prefixed by C.

Recall: γ(A) = AB, γ(B) = C, γ(C) = D, γ(D) = ABC.

The return words of C are CAB and CDAB. Code the return words by their lengths: 3 and 4. γ(CAB) = DABC, γ(CDAB) = DABCABC. Coded: 3 → 4, 4 → 43. (modulo conjugation by C) We see that the descendant morphism is the Fibonacci morphism

• n the alphabet {4, 3}, with fixed point x4,3.

Idea of the proof of THEOREM d1d0

Using Theorem FIX, we study the occurrences of C in T: digit d0(N) = 1 ⇔ T(N) = C. T := CABCABCD . . . is the fixed point of γ, prefixed by C.

Recall: γ(A) = AB, γ(B) = C, γ(C) = D, γ(D) = ABC.

The return words of C are CAB and CDAB. Code the return words by their lengths: 3 and 4. γ(CAB) = DABC, γ(CDAB) = DABCABC. Coded: 3 → 4, 4 → 43. (modulo conjugation by C) We see that the descendant morphism is the Fibonacci morphism

• n the alphabet {4, 3}, with fixed point x4,3.

Lemma JPA&FMD then gives that this word, written as a sequence equals (⌊nϕ⌋ + 2n + 1).

Idea of the proof of THEOREM d1d0 , part II

Recall: γ(A) = AB, γ(B) = C, γ(C) = D, γ(D) = ABC.

Positions of D?

Idea of the proof of THEOREM d1d0 , part II

Recall: γ(A) = AB, γ(B) = C, γ(C) = D, γ(D) = ABC.

Positions of D? Return words of D are: DABC and DABCABC Code the return words by their lengths: 4 and 7. γ(DABC) = ABCABCD, γ(DABCABC) = ABCABCDABCD. Coded: 7 → 4, 7 → 74. (modulo conjugation by D) Descendant morphism is Fibonacci on {7, 4}.

Idea of the proof of THEOREM d1d0 , part II

Recall: γ(A) = AB, γ(B) = C, γ(C) = D, γ(D) = ABC.

Positions of D? Return words of D are: DABC and DABCABC Code the return words by their lengths: 4 and 7. γ(DABC) = ABCABCD, γ(DABCABC) = ABCABCDABCD. Coded: 7 → 4, 7 → 74. (modulo conjugation by D) Descendant morphism is Fibonacci on {7, 4}. Lemma JPA&FMD then gives that this word, written as a sequence, equals (3⌊nϕ⌋ + n + 1)n≥1.

A general result

Recall that we partitioned the natural numbers in Lucas intervals Λ2n := [L2n, L2n+1] and Λ2n+1 := [L2n+1 + 1, L2n+2 − 1].

A general result

Recall that we partitioned the natural numbers in Lucas intervals Λ2n := [L2n, L2n+1] and Λ2n+1 := [L2n+1 + 1, L2n+2 − 1]. The basic idea behind this partition is that if β(N) = dLdL−1 . . . d1d0 · d−1d−2 . . . dR+1dR, then the index L = L(N) and the index R = R(N) satisfy L = 2n+ 1, R = −2n ⇔ N ∈ Λ2n; L = 2n+ 2 = −R ⇔ N ∈ Λ2

n+1.

A general result

Recall that we partitioned the natural numbers in Lucas intervals Λ2n := [L2n, L2n+1] and Λ2n+1 := [L2n+1 + 1, L2n+2 − 1]. The basic idea behind this partition is that if β(N) = dLdL−1 . . . d1d0 · d−1d−2 . . . dR+1dR, then the index L = L(N) and the index R = R(N) satisfy L = 2n+ 1, R = −2n ⇔ N ∈ Λ2n; L = 2n+ 2 = −R ⇔ N ∈ Λ2

n+1.

THEOREM Let β(N) = (di(N)) be the base phi representation

• f a natural number N, and let k ≥ 2. Then dk(N) = 1 if and only

if N is a member of one of the generalized Beatty sequences (⌊nϕ⌋Lk + nLk−1 + r), where r = r1, r1+ 1, . . . , r1+ |Λk|− 1, with r1 = −Lk−1 if k is even, and r1 = −Lk−1+1 if k is odd.

Example: k=2

N β(N) 1 1 2 10 · 01 3 100 · 01 4 101 · 01 5 1000 · 1001 6 1010 · 0001 7 10000 · 0001 8 10001 · 0001 9 10010 · 0101 10 10100 · 0101 11 10101 · 0101 12 100000 · 101001 13 100010 · 001001 14 100100 · 001001 15 100101 · 001001 Λ2 = [L2, L3] = [3, 4], and L1 = 1. The numbers N with d2(N) = 1

• ccur as N = 3⌊nϕ⌋ + n − 1
• r N = 3⌊nϕ⌋ + n.

(⌊nϕ⌋) = 1, 3, 4, 6, 8, 9, 11, ...

Digits with negative indices

A result similar to the general theorem will hold for digits dN(k) with k negative, but the situation is somewhat more complex.

Digits with negative indices

A result similar to the general theorem will hold for digits dN(k) with k negative, but the situation is somewhat more complex. One has, for example, digit d−2(N) = 1 if and only if N = 4⌊nϕ⌋ + 3n + r for r = 2, 3 or 4 and some n = 0, 1, 2, . . . .

Digits with negative indices

A result similar to the general theorem will hold for digits dN(k) with k negative, but the situation is somewhat more complex. One has, for example, digit d−2(N) = 1 if and only if N = 4⌊nϕ⌋ + 3n + r for r = 2, 3 or 4 and some n = 0, 1, 2, . . . . Wanted: simple proof!

Thanks!