Some properties of the Fibonacci sequence on an infinite alphabet ∗

The infinite Fibonacci sequence F, which is an extension of the classic Fibonacci sequence to the infinite alphabet N, is the fixed point of the morphism φ: (2i) 7→ (2i)(2i+ 1) and (2i+ 1) 7→ (2i+ 2) for all i ∈ N. In this paper, we study the growth order and digit sum of F, and give several decompositions of F using singular words.


Introduction
Let f = {f n } n 0 be the Fibonacci sequence that is the fixed point of the Fibonacci morphism σ defined by 0 → 01, 1 → 0. The Fibonacci sequence f occurs in the study of combinatorics on words, number theory, dynamical system, quasi-crystals, etc.; see [1,2,13,18] and references therein.It is well known that f is a characteristic Sturmian word of slope 1/φ 2 where φ = 1+ √ 5 2 is the golden ratio.In [9,10,11], several generalizations of the Fibonacci sequence were given, and they were proved to be characteristic Sturmian words of irrational slopes, whose simple continued fraction expansions are periodic sequences with period 1 and 2.
The previous extensions of f are all on the alphabet of 2 letters.There are also extensions on alphabets containing at least 3 letters.A famous one is the Tribonacci sequence, which is the fixed point of Rauzy substitution 0 → 01, 1 → 02, 2 → 0. Rauzy [8] studied the dynamical and geometrical aspects of the Tribonacci sequences; see also [18].For an extension to an alphabet of m letters, see [15].In this paper, we study the following extension of the Fibonacci sequence to an infinite alphabet.
The infinite Fibonacci sequence F = {F i } i 0 is the fixed point of φ starting by 0. The first several terms of F and f are F = 0 1 2 2 3 2 3 4 2 3 4 4 5 2 3 4 4 5 4 5 6 • • • f = 0 1 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 1 0 1 0 It is easy to see that F = f (mod 2), since φ is reduced to σ while modulo 2. In this sense, these two sequence are similar.Hence the infinite Fibonacci sequence F may inherit some combinatorial properties of the Fibonacci sequence f .For example, since '11' does not occur in f , two adjacent elements in F can not be both odd numbers.However, F also has some properties that f does not have.For example, one can find arbitrarily long palindromes in f ([27, Property 2]); while in F, there are no palindromes of length larger than 3 (see Proposition 27).In general, the sequence F (mod k) is a substitution sequence up to a coding (see Section 6).
In [27], Wen and Wen studied the Fibonacci sequence f by analyzing singular words of f .They proved that the adjacent singular words of the same order are positively separated, and gave a decomposition of f using singular words.Levé and Séébold [5] studied the singular factorization of the Fibonacci sequence and its conjugates.Melançon studied the Lyndon factorization of the Fibonacci sequence in [6,7].Tan and Wen [4] studied the singular words of Sturmian sequences and Tribonacci sequence.In [26], Huang and Wen studied the occurrence of arbitrary factors of f and gave a quite general decomposition of f .The study of singular words have applications in the study of combinatorics on words [3,16,22,23,24] and Padé approximation [14,21], etc.
There are many works devoted to the study of the substitution sequences and the automatic sequences over infinite alphabets.For the research of infinite-state automata see for example [12,20] and a survey [25].For the study of substitution sequences over an infinite alphabets, see for example [19].
In this paper, we first investigate the growth order of F n , and show that for all n 1, 0 F n c log n the electronic journal of combinatorics 24(2) (2017), #P2.52 where c is a constant (see Remark 6).Then we turn to study the digit sum problem of F. We give an exact form of the digit sum of φ n (0) (n 1) in Theorem 13, and give a formula for the digit sum of the first n digits of F in Theorem 15.Inspired by the work of Wen and Wen [27], we also study the decomposition (factorisation) of F. In fact, we give the following three decompositions.The first one is a decomposition using its (slightly modified) prefixes (see Theorem 16).The second one is a decomposition using singular words of letter 2 whose lengths are Fibonacci numbers (see Theorem 17).The third one is a decomposition using singular words of a fixed order (see Theorem 19).The main difference here is that the singular words (defined in Section 5) in F are defined to be classes of words, and while modulo 2, the singular words in F are only part of the singular words in f which is defined in [27].This paper is organized as follows.In Section 2, we state some basic notation and definitions.In Section 3 and Section 4, we study the growth order of the terms of the infinite Fibonacci sequence F and its digit sums.In Section 5, we give three decompositions of F. In the last section, several other properties of F are given.

Preliminary
Let A = {a 0 , a 1 , • • • , a n , • • • } be an (infinite) alphabet.A k denotes the set of all words of length k on A, and A * denotes the set of all finite words on A. The length of a finite word w ∈ A * is denoted by |w|.For any V, W ∈ A * , we write V ≺ W when the finite word V is a factor of the word W , that is, when there exist words U, U ∈ A * , such that W = U V U .We say that V is a prefix (resp.suffix ) of a word W , and we write V W (resp. V W ) if there exists a word . This makes sense in A * , since the reduced word associated with W V −1 belongs to A * .This abuse of language will be very useful in what follows.
For a sequence u = (u n ) n 0 and an integer k 0, we denote u ⊕ k := (u n + k) n 0 .
3 Growth order of the infinite Fibonacci sequence In this part, we shall study the growth order of the terms of the infinite Fibonacci sequence.
For this purpose, we first give some basic rules of the digits in F, which reveal the growth order of the sequence F.
Theorem 1.For any k 1, we have for any i, j 0.
the electronic journal of combinatorics 24(2) (2017), #P2.52 Proof.When k = 1, the result follows from the definition of φ.Now suppose the result holds for all k n, we shall prove it for n + 1. Suppose j is even.For any i 0, When j is odd, we have for any i 0, This completes the proof.
Letting j = 0 in Theorem 1, we have the following result.
Corollary 2. For all k 1 and i 0, Recall that for any n 0, the Fibonacci number L n counts the number of 1's in σ n (0) (or sums the digits of σ n (0)), and L n+2 represents the length of σ n (0).It is worth to remark that for n 1, where L 0 = 0 and L 1 = 1.Proposition 4. For any n 2, i 0, Moreover, n + i is the largest digit in φ n (i).
Proof.We first prove this result for i = 0. Note that φ n (0) is a prefix of F, and according to Lemma 3, |φ n (0)| = L n+2 .So F L n+2 −1 and F L n+2 −2 are the last and the penultimate digits of φ n (0).Now we will show by induction that for all n 2, The largest digit of φ n (0) is n; It is easy to check that (1) holds for n = 2. Assume that (1) holds for all m n.We shall check it for m = n + 1.Note that the electronic journal of combinatorics 24(2) (2017), #P2.52 and by the induction hypothesis the largest digits of φ n (0) and φ n−1 (0) ⊕ 2 are n and n + 1 respectively.So the largest digit of φ n+1 (0 , we have which implies Therefore, (1) holds.Suppose α and β are the last two digits of φ n (i).When i = 2k + 1, we have where the last equality follows from Theorem 1.So The above result implies that the digits in F can be arbitrary large.However, there are also infinitely many small digits, say '2' and '3', in F. Proposition 5.For any n 4, F Ln = 2 and F Ln+1 = 3.

Proof. By the definition of F, for any
Remark 6 (The growth order of {F n } n 0 ).Using Proposition 5, we have On the other hand, Proposition 4 gives 2 .This implies the limit of F n /n does not exist.However, we have lim inf n→∞ F n log n = 0 and lim sup n→∞ The first one is trivial.For the limsup we know from Proposition 4 that the electronic journal of combinatorics 24(2) (2017), #P2.52 Hence, we have 0 F n c log n, for any n 1, where c is a constant.
Next, we will show that the block of length L n+1 in F from position Corollary 8.For any n 4, 0 i L n − 1, we have Moreover, for n = 1 and n = 2, Proof.For 0 i L n−1 − 1, we have known By Proposition 7, we have By Proposition 7, we have and Then Thus, F i+Ln − F i = 0.
It is natural to ask the location of 'n − 1' in φ n (0).For this purpose, we need the following decomposition of φ n (0).Lemma 9.For any n 0, let k = n−1
Remark 11.In the same process, one can also locate 'n − 2' in φ n (0) and then 'n − 3' and so on.However, the formulae would not be nice to read.

Digit sum
This section is devoted to study the digit sum of the Fibonacci sequence and the generalized Fibonacci sequences.For any integer sequence c = c 0 c 1 • • • , the digit sum of first n-terms is denoted by Let (n) F := a r a r−1 • • • a 0 be the unique Fibonacci representation of a non-negative integer n (see [13,Section 3.8]).That is where a i = 0 or 1, a r = 0 and a i a i+1 = 0 for 0 i r.
For any n 0 with Fibonacci representation (n In fact, consider the word w := σ i 0 (0 , where the indexes {i k } t k=0 satisfy i 0 r and for all k, a i k = 0 and i k+1 < i k .Moreover, Since w σ(w), w is the prefix of f .Therefore (4) holds.

Digit sum of F
For the digit sum of the infinite Fibonacci sequence F, we need a number, say F (i, n), which sums the digits of φ n (i).In the case of i = 0, F n := F (0, n).It is easy to see that for all i 1, Lemma 12.For any n 2, In the matrix form, the above equation is the electronic journal of combinatorics 24(2) (2017), #P2.52 Proof.Since φ n (0 . Then, by (5), 2 .For any n 1, Proof.Denote the 4 × 4 matrix in ( 6) by A and Then by Lemma 12, we have Thus, to evaluate F n , we only need to compute A n .This can be done by using the Jordan form of where for all n 1.The inverse of P is .
the electronic journal of combinatorics 24(2) (2017), #P2.52 Since F 1 = (1, 0, 1, 0) t and F n+1 = A n F 1 = P J n P −1 F 1 , we can give the closed form of L n and F n : Remark 14.The integer sequence { Fn } n 0 also satisfies the recurrence relation with the initial values 0, 1, 3, 8. Using this fact, one can show that its generating function is rational.In fact, Theorem 15 (Digit sum of F).Let n ∈ N and (n) F = a r a r−1 • • • a 0 .Collect all the index i such that a i = 0 and arrange them in descending order.Those indexes are denoted by Furthermore, So we only need to show that w F.

Singular words decomposition
In this section we present three decompositions (factorisations) of F. These are given in Theorems 16, 17 and 19.We will state the theorems first and introduce some notation, then give the three proofs.
, where we define φ −2 (2) to be 0 and φ −1 (2) to be 1.To state the second decomposition theorem, we introduce the singular words.For all n 1 and i 0, the word n is called a singular word of order n (of letter 2i), where α and β are two letters satisfying αβ φ n (i).In addition, we define In the above two decompositions, the lengths of the components are unbounded.In the following, we will decompose F into singular words of a fixed order and gaps between them; and the gaps have only two different lengths.
Before stating the next result, we shall say some words on the positions of p-th occurrence of letter 0 and 1 in Fibonacci sequence f .Those positions turn out to be very useful in locating singular words in F. Let Λ 0 (resp.Λ 1 ) be the set of positions of letter 0 (resp.1) in f .Namely, the electronic journal of combinatorics 24(2) (2017), #P2.52 where λ 0 (i) (resp.λ 1 (i)) is the position of i-th 0 (resp.1) in f .In fact, Λ 0 and Λ 1 are the sequences A022342 and A003622 in OEIS respectively (see [17]).Moreover, the closed form of the sequence A022342 is given in [17], i.e., Note that the indices of elements in f starts from 0. Thus λ 0 (1) = 0. Let m f := {m f (i)} i 0 where for all i 0, Of course f λ 0 (i) = 0 for all i, but writing the definition in this way will be useful in what follows.
Proof.The first equality is the definition.For the second one, we only need to show that for any i 1, . By the definition of λ 0 (i) and λ 1 (i), we have In Fibonacci sequence f , the i-th 1 is generated by iterating the i-th 0, and φ(F Thus, by the definition of φ, F λ Fix a k 1.For any p 1, let S k,p be the p-th occurrence of singular words of order k, and let G k,p be the gap word between two consecutive singular words S k,p and S k,p+1 .In addition, G k,0 is defined to be the prefix of F before S k,1 .Then, we have the following decomposition theorem: and For example, when k = 2, we have the following initial values

Proof of the first two decomposition theorems
Proof of Theorem 16.We only need to show for all n 0. When k = 0, φ 0 (0) = 0 = φ −2 (2).Assume that it is true for all k < n.Since the formula (9) holds for k = n.So, by induction, the result follows.
Now, we will prove Theorem 17.
Proof of Theorem 17.Firstly, we show φ n (0 n−2 for all n 1.When n = 1, S where the last equality follows from the definition of singular words and Proposition 4. Since for all n 0, φ n (0)n −1 is a prefix of F, and |φ n (0)n −1 | → ∞ when n → ∞, we have

Proof of Theorem 19
In [27], Wen and Wen define the k-th singular word of f to be s k = ασ k (0)β −1 , where αβ σ k (0), and they proved the following result.
Moreover, s k occurs only once in σ k (10) and We have a similar characterization of the singular words of order k in F.
the electronic journal of combinatorics 24(2) (2017), #P2.52 Lemma 21.For any i 1, k 1 and Proof.From the definition of singular words, we know that where αβ φ k (2i).By Proposition 4, we have α where u i ≡ v i ∈ {0, 1} (mod 2) for i = 0, 1.By Lemma 20, we have v 1 v 2 = 10 and Therefore, the first and the last digit of S (2i) k are the last digit of φ k (u 1 ) and the penultimate digit of φ k (u 2 ) respectively.Combining this fact and Proposition 4, we have Remark 22.By Lemma 21, for any k, i 1, Remark 22 shows that there is a (unique) singular word of order k in the (k + 2)-th iteration of any even number (2i).The following two lemmas will show that this is the only place that singular words of order k can occur.where 0 j i.
Lemma 24.Let u 1 u 2 be a factor of length two in F. We have the following: (1) If u 1 ≡ u 2 mod 2, then singular words of order k occur in φ k+2 (u 1 u 2 ) only once; (2) If u 1 ≡ u 2 mod 2, then singular words of order k occur in φ k+2 (u 1 u 2 ) only twice.
Proof.By (11), we know singular words of order k occur in φ k+2 (u 1 u 2 ), where u 1 u 2 is a factor of length two in F and at least one of u 1 and u 2 is an even number.
the electronic journal of combinatorics 24(2) (2017), #P2.52 In the following, we will discuss the properties of the gap sequence {G k,p } p 1 for any k 1.
Proof.Combining (11) and Lemma 24, we know that for any p 1, the singular word S k,p ≺ φ k+2 (F λ 0 (p) ).Then, by Lemma 21, For any p 1, F λ 0 (p) and F λ 0 (p+1) either occur in F consecutively or separate by only one odd number.Then for some t ∈ Λ 1 .In the first case, by (12), In the second case, by (12), So for any p 1, By (8), which is the Fibonacci sequence over {1, 2}.This implies that {|G k,p |} p 1 is the Fibonacci sequence over {2L k+3 , L k+3 }.Now we will give the explicit expression of the gap words.We denote gap words of length 2L k+3 and L k+3 by G L k, * and G S k, * where the superscript 'L' and 'S' stand for long and short.Then Combining ( 13) and ( 14), we have the electronic journal of combinatorics 24(2) (2017), #P2.52 That is Proof of Theorem 19.The proof is composed by the following three steps.Fix a k 1.
Step 1.We will show that S k,p = S k ⊕ m f (p − 1), for any p 1.In fact, combining (11) and Lemma 24, we know that for any p 1, the singular word S k,p ≺ φ k+2 (F λ 0 (p) ).So by (12) and the definition of {m f (p)} p 1 , we have S k,p = S k,1 ⊕ m f (p − 1), where S k,1 = S (2) k .
Step 2. We will show that So there are no palindrome of length 5 in F. (v) If there are palindromes of length larger than or equal to 6, then we can find palindromes of length 4 or 5.However, according to (iii) and (vi), there are no palindromes of length 4 and 5, so there are no palindromes of length larger than or equal to 6 in F.
In the last part, we give some observations of the sequence F (mod k) where k 2. When k is an even number, the sequence F ( mod k) is a pure substitution sequence which is generated by the substitution φ (mod k), i.e., When k is an odd number, the sequence F (mod k) is a substitution sequence under a coding where the substitution φ (mod 2k), i.e., and the coding is τ : i (i + k) → i for every 0 i < k.