A Note on the Abelian Complexity of the Rudin-Shapiro Sequence

: Let { r ( n ) } n ≥ 0 be the Rudin-Shapiro sequence, and let ρ ( n ) : = max { ∑ i + n − 1 j = i 0 } + 1 be the abelian complexity function of the Rudin-Shapiro sequence. In this note, we show that the function ρ ( n ) has many similarities with the classical summatory function S r ( n ) : = ∑ ni = 0 r ( i ) . In particular, we prove that for every positive integer n , √ 3 ≤ ρ ( n ) √ n ≤ 3. Moreover, the point set { ρ ( n ) √ n : n ≥ 1 } is dense in [ √ 3,3

They studied the properties of the summatory function S l (N) := ∑ N n=0 l(n). The sequence S l (N) satisfies S l (0) = 1 and for every m ≥ 0, Moreover, Lafrance, Rampersad and Yee showed that The sequences S r (n) and S l (n) are both 2-regular sequences (in the sense of Allouche and Shallit [1]). For the definition and properties of k-regular sequences, one can refer to [1]. Let (s(n)) n≥0 be a k-regular sequence over Z. It was proved in [1] that there exists a constant c such that s(n) = O(n c ). In general, it is a difficult task to compute the exact growth order of sequences satisfying certain recursive relations such as k-regular sequences.
This note focuses on the growth order of the abelian complexity (ρ(n)) n≥1 of the Rudin-Shapiro sequence r. Firstly, by studying the maximal and minimal values of the function ρ(n) in the interval I k := [4 k , 4 k+1 − 1] with k ≥ 0, we got ρ(n) = Θ( √ n). Then, we investigated two functions s(k) := min{n | ρ(n) = k} and (k) := max{n | ρ(n) = k}, and obtained the optimal lower and upper bound of the sequence ( ρ(n) √ n ) n≥1 . Finally, we showed that ρ(n) is a quasi-linear function for 4. As a consequence, the set { ρ(n) √ n : n ≥ 1} is dense between its optimal lower bound and upper bound. In detail, we proved the following theorems. Theorem 1. For every integer n ≥ 1, we have The outline of this note is as follows. In Section 2, we compute the maximal and minimal values of the function ρ(n) in the interval I k := [4 k , 4 k+1 − 1] for every k ≥ 0. In Section 3, we give the proofs of Theorem 1 and Theorem 2.

Basic Properties of the Function ρ(n)
In this section, we exhibit some basic properties of the abelian complexity function ρ(n) of the Rudin-Shapiro sequence r.
Proof. We will prove this by induction on the variable k. For k = 0, it follows from Table 1 that this assertion is true. Assume the assertion holds for the interval I k .
We first show that 2 k+2 is the lower bound for ρ(n) in I k+1 . If n lies in I k+1 = [4 k+1 , 4 k+2 − 1], then we can write n = 4m + d for some m ∈ I k and some d ∈ {0, 1, 2, 3}. There are two cases to be considered.

1.
When The last equality is true under the inductive assumption.
Proof. We will prove this by induction on the variable k. For k = 0, this assertion holds following from Table 1. Assume the assertion is true for the interval I k . When n lies in I k+1 , let n = 4m + d for some m ∈ I k and d ∈ {0, 1, 2, 3}. Similarly with the proof of Lemma 1, we divide it into two cases.
However, these bounds are not optimal. Note that ρ(n) ≥ 2 for every n ≥ 1. It is easy to verify that In other words, 3 and √ 3 are two accumulation points of the set { ρ(n) √ n : n ≥ 1}. In the following section, we will prove that 3 and √ 3 are the optimal upper and lower bound for the sequence ( ρ(n) √ n ) n≥1 respectively.

Proofs of Theorems 1 and 2
Following that M k and m k both go to infinity with k tending to infinity (by Lemmas 1 and 2), we can see that there are only finite number of places n such that ρ(n) has a fixed value k. When ρ(n) = k, for a fixed k, the ratio ρ(n)/ √ n will be the smallest when n is largest while it will be largest if n is smallest. This leads us to the following idea: for a fixed k ∈ N with k ≥ 1, let us focus on the smallest and largest values of n such that ρ(n) = k. For this purpose, we introduced two auxiliary functions s(k) and (k). Following from (1), Table 1 and ρ(0) = 1, the initial 8 terms of the sequences s(k) and (k) are given in Table 2. For the sequences s(k) and (k), we have the following results.
The following two propositions show the upper bound for the sequence (k) and the lower bound for the sequence s(k). For the sake of simplicity, for every integer k ≥ 2, let be the binary expansion of k with m ≥ 1 and k m = 1. For every x ≥ 0, let x be the greatest integer which is no more than x.
Proof. For every k ≥ 2, let the binary expansion of k be [k m k m−1 · · · k 0 ] 2 . Following from Lemma 3, we have Now we apply (6) by replacing k with [k m k m−1 · · · k 1 ] 2 , which yields Repeating this progress m times, by the fact that k j = k 2 j − 2 k 2 j+1 for every 0 ≤ j ≤ m − 1, we can obtain that It suffices to show that Note that 2 m ≤ k < 2 m+1 . This implies that Consider the function f k (x) = kx − 2 3 x 2 . It is not hard to check that f k (x) is strictly increasing on the interval ( k 4 , k 2 ] with fixed k ≥ 1. Hence which is the desired result.
By Remark 2 and Theorem 1, we get the following corollary.
Corollary 1. The numbers √ 3 and 3 are the optimal lower and upper bounds for the set { ρ(n) √ n : n ≥ 1} respectively.
To prove Theorem 2, we need an auxiliary notation which was firstly introduced in [8]. Let b ≥ 2 be an integer and f : N → Z be a discrete function (or an integer sequence). For every x ≥ 0, let then we call f a quasi-linear function for b.
For the quasi-linear functions, we have the following lemma. For more details about this, see [8].
Therefore ρ(n) is a quasi-linear function for b = 4. It follows from Remark 2 that √ 3 and 3 are two accumulation points of the set {ρ(n)/ √ n : n ≥ 1}. Hence we obtained the desired result by Lemma 5. Institutional Review Board Statement: Not applicable.
Informed Consent Statement: Not applicable.

Data Availability Statement:
The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest:
The authors declare no conflict of interest.