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

Abstract

Let ${\left\{r\left(n\right)\right\}}_{n\ge 0}$ be the Rudin-Shapiro sequence, and let $\rho \left(n\right):=max\{{\sum }_{j=i}^{i+n-1}r\left(j\right)\mid i\ge 0\}+1$ be the abelian complexity function of the Rudin-Shapiro sequence. In this note, we show that the function $\rho \left(n\right)$ has many similarities with the classical summatory function $S_{\mathbf{r}}\left(n\right):={\sum }_{i=0}^{n}r\left(i\right)$. In particular, we prove that for every positive integer n, $\sqrt{3}\le \frac{\rho \left(n\right)}{\sqrt{n}}\le 3$. Moreover, the point set $\{\frac{\rho \left(n\right)}{\sqrt{n}}:n\ge 1\}$ is dense in $[\sqrt{3},3]$.

1. Introduction

In this note, we are concerned with the abelian complexity $\rho \left(n\right)=max\{{\sum }_{j=i}^{i+n-1}r\left(j\right)\mid i\ge 0\}+1$ of the Rudin-Shapiro sequence $\mathbf{r}$. The Rudin-Shapiro sequence $\mathbf{r}=r\left(0\right)r\left(1\right)\cdots \in {\{-1,1\}}^{\mathbb{N}}$ is given by the following recurrence relations:

$$r\left(0\right)=1,r\left(2n\right)=r\left(n\right),r(2n+1)={(-1)}^{n}r\left(n\right)(n\ge 0).$$

The Rudin-Shapiro sequence $\mathbf{r}$ is a typical 2-automatic sequence [1]. It has been proved in [2] that the sequence $\rho \left(n\right)$ satisfies $\rho \left(1\right)=2$, $\rho \left(2\right)=3$, $\rho \left(3\right)=4$ and for every $n\ge 1$,

$$\rho \left(4n\right)=2\rho \left(n\right)+1,\rho (4n+1)=2\rho \left(n\right),\rho (4n+2)=\rho \left(n\right)+\rho (n+1),\rho (4n+3)=2\rho (n+1).$$

Let $\mathbf{w}=w\left(0\right)w\left(1\right)w\left(2\right)\cdots$ be an infinite sequence with $w\left(i\right)\in \mathbb{Z}$ for every $i\ge 0$. There are many papers focusing on the summatory function $S_{\mathbf{w}}\left(n\right):={\sum }_{j=0}^{n}w\left(j\right)$. In [3,4,5], Brillhart and Morton studied the summatory function $S_{\mathbf{r}}\left(n\right):={\sum }_{i=0}^{n}r\left(i\right)$ of the Rudin-Shapiro sequence. The sequence $S_{\mathbf{r}}\left(n\right)$ satisfies $S_{\mathbf{r}}\left(0\right)=1$, $S_{\mathbf{r}}\left(1\right)=2$, $S_{\mathbf{r}}\left(2\right)=3$, $S_{\mathbf{r}}\left(3\right)=2$ and for every $n\ge 1$,

$$S_{\mathbf{r}}\left(4n\right)=2S_{\mathbf{r}}\left(n\right)+r\left(n\right),S_{\mathbf{r}}(4n+1)=S_{\mathbf{r}}(4n+3)=2S_{\mathbf{r}}\left(n\right),S_{\mathbf{r}}(4n+2)=2S_{\mathbf{r}}\left(n\right)+{(-1)}^{n}r\left(n\right).$$

In detail, Brillhart and Morton proved that for every $n\ge 1$,

$$\sqrt{3/5}\le \frac{S_{\mathbf{r}}\left(n\right)}{\sqrt{n}}\le \sqrt{6},$$

and $\{\frac{S_{\mathbf{r}}\left(n\right)}{\sqrt{n}}:n\ge 1\}$ is dense in $[\sqrt{3/5},\sqrt{6}].$ In [6], Lafrance, Rampersad and Yee introduced a Rudin-Shapiro-like sequence ${\left(l\left(n\right)\right)}_{n\ge 0}$ which satisfies $l\left(0\right)=1$ and for every $n\ge 0$,

$$l\left(4n\right)=l\left(n\right),l(4n+1)=l\left(2n\right),l(4n+2)=-l\left(2n\right)\mathrm{and}l(4n+3)=l\left(n\right).$$

They studied the properties of the summatory function $S_{\mathbf{l}}\left(N\right):={\sum }_{n=0}^{N}l\left(n\right)$. The sequence $S_{\mathbf{l}}\left(N\right)$ satisfies $S_{\mathbf{l}}\left(0\right)=1$ and for every $m\ge 0$,

$$S_{\mathbf{l}}\left(4m\right)=2S_{\mathbf{l}}\left(m\right)-l\left(m\right)=S_{\mathbf{l}}(4m+2),S_{\mathbf{l}}(4m+1)=2S_{\mathbf{l}}\left(m\right)-l\left(m\right)+l\left(2m\right),S_{\mathbf{l}}(4m+3)=2S_{\mathbf{l}}\left(m\right).$$

Moreover, Lafrance, Rampersad and Yee showed that

$$\underset{n\to +\infty }{lim\; sup}\frac{S_{\mathbf{l}}\left(n\right)}{\sqrt{n}}=\sqrt{2},\underset{n\to +\infty }{lim\; inf}\frac{S_{\mathbf{l}}\left(n\right)}{\sqrt{n}}=\frac{\sqrt{3}}{3}.$$

The sequences $S_{\mathbf{r}}\left(n\right)$ and $S_{\mathbf{l}}\left(n\right)$ 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 ${\left(s\left(n\right)\right)}_{n\ge 0}$ be a k-regular sequence over $\mathbb{Z}$. It was proved in [1] that there exists a constant c such that $s\left(n\right)=O\left(n^c\right)$. In general, it is a difficult task to compute the exact growth order of sequences satisfying certain recursive relations such as k-regular sequences.

In [7], Gawron and Ulas obtained the sequence $\left\{a\right(n)\mid n\in \mathbb{N}\}:=\{m\in \mathbb{N}\mid c(m)=1\}$ where ${\left(c\left(n\right)\right)}_{n\ge 0}$ is the sequence of coefficients of the compositional inverse of the generating function of the Thue-Morse sequence. The sequence ${\left(a\left(n\right)\right)}_{n\ge 0}$ satisfies that $a\left(0\right)=0,a\left(1\right)=1,a\left(2\right)=2,a\left(3\right)=7$ and for all $n\ge 1$, $a(4n+i)=a(4n-1)+i+1$ with $i\in [0,2]$, $a(8n+3)=a\left(8n\right)+7$ and $a(8n+7)=4a(4n+3)+3$. They proved that

$$\underset{n\to +\infty }{lim\; inf}\frac{a\left(n\right)}{n^2}=\frac{1}{6},\underset{n\to +\infty }{lim\; sup}\frac{a\left(n\right)}{n^2}=\frac{1}{2}$$

and $\{\frac{a\left(n\right)}{n^2}:n\ge 1\}$ is dense in $[\frac{1}{6},\frac{1}{2}]$. In [2], Chen, Wen, Wu and the first author studied the maximal digit sum sequence $M_{\mathbf{r}}\left(n\right):=max\{{\sum }_{j=i}^{i+n-1}r\left(j\right)\mid i\ge 0\}$ and proved that the abelian complexity $\rho \left(n\right)$ of the Rudin-Shapiro sequence satisfies $\rho \left(n\right)=M_{\mathbf{r}}\left(n\right)+1$ for every $n\ge 1$. It is remarkable that the authors in [2] just gave the recursive formulas for the sequence $M_{\mathbf{r}}\left(n\right)$ and proved the 2-regularity of the sequence $\rho \left(n\right)$. It is natural to ask whether the function $\rho \left(n\right)$ has similar properties as the summatory function $S_{\mathbf{r}}\left(n\right)$. In fact, it is of great interest to study the properties of sequences which satisfy certain recursive formulas.

This note focuses on the growth order of the abelian complexity ${\left(\rho \left(n\right)\right)}_{n\ge 1}$ of the Rudin-Shapiro sequence $\mathbf{r}.$ Firstly, by studying the maximal and minimal values of the function $\rho \left(n\right)$ in the interval $I_k:=[4^k,4^{k+1}-1]$ with $k\ge 0$, we got $\rho \left(n\right)=\Theta \left(\sqrt{n}\right)$. Then, we investigated two functions $s\left(k\right):=min\{n\mid \rho (n)=k\}$ and $\ell \left(k\right):=max\{n\mid \rho (n)=k\}$, and obtained the optimal lower and upper bound of the sequence ${\left(\frac{\rho \left(n\right)}{\sqrt{n}}\right)}_{n\ge 1}$. Finally, we showed that $\rho \left(n\right)$ is a quasi-linear function for 4. As a consequence, the set $\{\frac{\rho \left(n\right)}{\sqrt{n}}:n\ge 1\}$ is dense between its optimal lower bound and upper bound. In detail, we proved the following theorems.

Theorem 1.

For every integer $n\ge 1$, we have

$$\sqrt{3}\le \frac{\rho \left(n\right)}{\sqrt{n}}\le 3.$$

Theorem 2.

The set $\{\rho \left(n\right)/\sqrt{n}:n\ge 1\}$ is dense in $[\sqrt{3},3].$

The outline of this note is as follows. In Section 2, we compute the maximal and minimal values of the function $\rho \left(n\right)$ in the interval $I_k:=[4^k,4^{k+1}-1]$ for every $k\ge 0$. In Section 3, we give the proofs of Theorem 1 and Theorem 2.

2. Basic Properties of the Function $\mathbf{\rho }\left(\mathit{n}\right)$

In this section, we exhibit some basic properties of the abelian complexity function $\rho \left(n\right)$ of the Rudin-Shapiro sequence $\mathbf{r}$.

Following from ([2] Theorem 1 and Lemma 3), the abelian complexity function $\rho \left(n\right)$ of the Rudin-Shapiro sequence is given by the following formulas: $\rho \left(1\right)=2$, $\rho \left(2\right)=3$, $\rho \left(3\right)=4$ and for every integer $n\ge 1$,

$$\begin{array}{cccc}\hfill \rho \left(4n\right)& = 2\rho \left(n\right)+1,\hfill & \hfill \rho (4n+1)& = 2\rho \left(n\right),\hfill \\ \hfill \rho (4n+2)& = \rho \left(n\right)+\rho (n+1),\hfill & \hfill \rho (4n+3)& = 2\rho (n+1).\hfill \end{array}$$

(1)

Set $\rho \left(0\right):=1.$ For every integer $n\ge 0$, let $\Delta \rho \left(n\right):=\rho (n+1)-\rho \left(n\right)$. Then $\Delta \rho \left(0\right)=\Delta \rho \left(1\right)=\Delta \rho \left(2\right)=\Delta \rho \left(3\right)=1$, and for every integer $n\ge 1$,

$$\begin{array}{c}\hfill \Delta \rho \left(4n\right)=-1,\Delta \rho (4n+3)=1,\\ \hfill \Delta \rho (4n+1)=\Delta \rho (4n+2)=\Delta \rho \left(n\right).\end{array}$$

(2)

This implies that $\Delta \rho \left(n\right)\in \{-1,1\}$ for every integer $n\ge 0$. The first 16 terms of $\rho \left(n\right)$, starting with $n=1$, are listed in Table 1.

Table 1. The first 16 terms of the sequence $\rho \left(n\right).$

n	$\mathit{\rho }\left(\mathit{n}\right)$	n	$\mathit{\rho }\left(\mathit{n}\right)$
1	2	9	6
2	3	10	7
3	4	11	8
4	5	12	9
5	4	13	8
6	5	14	9
7	6	15	10
8	7	16	11

For simplicity of notation, for every integer $k\ge 0$, put $m_k:=\frac{4^{k+1}-1}{3},M_k:=4^{k+1}-1$ and $I_k:=[4^k,4^{k+1}-1].$ Then we have the following two lemmas which give the minimal and maximal values of the function $\rho \left(n\right)$ in the interval $I_k$ for every $k\ge 0$.

Lemma 1.

For every integer $k\ge 0$, the minimum value of $\rho \left(n\right)$ in $I_k=[4^k,4^{k+1}-1]$ is $2^{k+1}$. Moreover,

$$max\left\{n\in I_k:\rho \left(n\right)=2^{k+1}\right\}=m_k.$$

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 $\rho \left(n\right)$ 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 $m\in I_k$ and some $d\in \{0,1,2,3\}$. There are two cases to be considered.

• When $4^k\le m\le 4^{k+1}-2$, (1) yields that for every $d\in \{0,1,2,3\}$

  $$\rho \left(n\right)=\rho (4m+d)\ge 2min\{\rho \left(m\right):m\in I_k\}=2^{k+2}.$$
  The last equality is true under the inductive assumption.
• When $m=4^{k+1}-1$, it follows from (2) that $\Delta \left(m\right)=1,$ which implies

  $$\rho (m+1)=\rho (4^{k+1}-1)+1=\rho \left(m\right)+1.$$
  Hence, for every $d\in \{0,1,2,3\},$ using (1) again, we have

  $$\rho \left(n\right)=\rho (4m+d)\ge 2\rho \left(m\right)>2min\{\rho \left(m\right):m\in I_k\}=2^{k+2}.$$

  (3)

At the same time, using the fact $m_{k+1}=4m_k+1,$ it is easy to check that

$$\rho \left(m_{k+1}\right)=\rho (4m_k+1)=2\rho \left(m_k\right)=2^{k+2}.$$

Now it suffices to show that

$$m_{k+1}=\max \left\{n\in I_{k+1}:\rho \left(n\right)=2^{k+2}\right\}.$$

Following from the inductive assumption, for every $m\in I_k$ satisfying $m>m_k$, we have $\rho \left(m\right)\ge \rho \left(m_k\right)+1$. By (1), we can get

$$\begin{array}{cc}\hfill \rho (m_{k+1}+1)& = \rho (4m_k+2) = \rho \left(m_k\right)+\rho (m_k+1)\ge 2^{k+2}+1,\hfill \\ \hfill \rho (m_{k+1}+2)& = \rho (4m_k+3) = 2\rho (m_k+1)\ge 2^{k+2}+2.\hfill \end{array}$$

Now we only need to consider the case $n=4m+d\ge 4m_k+4$ with $d\in \{0,1,2,3\}.$ In fact, for every $m_k+1\le m\le 4^{k+1}-2$ and $d\in \{0,1,2,3\},$ it follows from (1) that

$$\rho (4m+d)\ge 2min\{\rho \left(m\right):m_k+1\le m\le 4^{k+1}-1\}\ge 2^{k+2}+2.$$

By (3), the case $n=4m+d$ for $m=4^{k+1}-1$ holds, which completes the proof. □

Lemma 2.

Let k be a non-negative integer. The maximum value of $\rho \left(n\right)$ in $I_k=[4^k,4^{k+1}-1]$ is $3·2^{k+1}-2$ and this value occurs only at the point $n=M_k=4^{k+1}-1$.

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 $m\in I_k$ and $d\in \{0,1,2,3\}$. Similarly with the proof of Lemma 1, we divide it into two cases.

• When $4^k\le m<4^{k+1}-1$. By (1) and the inductive assumption, we have

  $$\rho \left(n\right)=\rho (4m+d)\le 2max\{\rho \left(m\right):m\in I_k\}+1<3·2^{k+2}-2.$$

  (4)
• When $m=M_k=4^{k+1}-1$. Following from (1) and (2), we have

  $$\begin{array}{cc}\hfill \rho \left(4M_k\right)& = 2\rho \left(M_k\right)+1=3·2^{k+2}-3,\hfill \\ \hfill \rho (4M_k+1)& = 2\rho \left(M_k\right)=3·2^{k+2}-4,\hfill \\ \hfill \rho (4M_k+2)& = \rho \left(M_k\right)+\rho (M_k+1)=3·2^{k+2}-3,\hfill \\ \hfill \rho (4M_k+3)& = 2\rho (M_k+1)=3·2^{k+2}-2.\hfill \end{array}$$

  (5)
  This implies that $\rho \left(n\right)\le \rho (4M_k+3)=\rho (4^{k+2}-1)=3·2^{k+2}-2.$

Following from (4) and (5), we can obtain that $M_{k+1}=4M_k+3$ is the unique point in $I_{k+1}$ which attains the maximal value of $\rho$ in the interval $I_{k+1}$. This completes the proof. □

Remark 1.

From Lemma 1, we have that for every $n\ge 1,$

$$\rho \left(n\right)\ge 2.$$

Remark 2.

If $n\in I_k=[4^k,4^{k+1}-1]$, Lemma 1 gives us

$$\frac{\rho \left(n\right)}{\sqrt{n}}>\frac{2^{k+1}}{\sqrt{4^{k+1}}}=1,$$

while Lemma 2 implies that

$$\frac{\rho \left(n\right)}{\sqrt{n}}<\frac{3·2^{k+1}-2}{\sqrt{4^k}}<6.$$

Thus, for every integer $n\ge 1$, $1\le \frac{\rho \left(n\right)}{\sqrt{n}}\le 6$, and so $\rho \left(n\right)$ is roughly a constant times $\sqrt{n}$. However, these bounds are not optimal. Note that $\rho \left(n\right)\ge 2$ for every $n\ge 1.$ It is easy to verify that

$$\underset{k\to \infty }{lim}\frac{\rho \left(M_k\right)}{\sqrt{M_k}}=\underset{k\to \infty }{lim}\frac{3·2^{k+1}-2}{\sqrt{4^{k+1}-1}}=3$$

and

$$\underset{k\to \infty }{lim}\frac{\rho \left(m_k\right)}{\sqrt{m_k}}=\underset{k\to \infty }{lim}\frac{2^{k+1}}{\sqrt{\frac{4^{k+1}-1}{3}}}=\sqrt{3}.$$

In other words, 3 and $\sqrt{3}$ are two accumulation points of the set $\{\frac{\rho \left(n\right)}{\sqrt{n}}:n\ge 1\}$. In the following section, we will prove that 3 and $\sqrt{3}$ are the optimal upper and lower bound for the sequence ${\left(\frac{\rho \left(n\right)}{\sqrt{n}}\right)}_{n\ge 1}$ respectively.

3. 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 $\rho \left(n\right)$ has a fixed value k. When $\rho \left(n\right)=k$, for a fixed k, the ratio $\rho \left(n\right)/\sqrt{n}$ 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 $k\in \mathbb{N}$ with $k\ge 1$, let us focus on the smallest and largest values of n such that $\rho \left(n\right)=k$. For this purpose, we introduced two auxiliary functions $s\left(k\right)$ and $\ell \left(k\right)$.

Definition 1.

Given an integer $k\ge 1$, let $s\left(k\right)$ and $\ell \left(k\right)$ be the smallest and largest values of n such that $\rho \left(n\right)=k$ respectively, i.e.,

$$\begin{array}{c}\hfill s\left(k\right):=min\{n:\rho (n)=k\},\\ \hfill \ell \left(k\right):=max\{n:\rho (n)=k\}.\end{array}$$

Following from (1), Table 1 and $\rho \left(0\right)=1$, the initial 8 terms of the sequences $s\left(k\right)$ and $\ell \left(k\right)$ are given in Table 2.

Table 2. The initial values for the sequences $s\left(k\right)$ and $\ell \left(k\right)$.

k	1	2	3	4	5	6	7	8
$s\left(k\right)$	0	1	2	3	4	7	8	11
$\ell \left(k\right)$	0	1	2	5	6	9	10	21

For the sequences $s\left(k\right)$ and $\ell \left(k\right)$, we have the following results.

Lemma 3.

The sequence $\ell \left(k\right)$ satisfies $\ell \left(1\right)=0$ and for every integer $k\ge 1$,

$$\ell \left(2k\right)=4\ell \left(k\right)+1,\ell (2k+1)=4\ell \left(k\right)+2.$$

Proof. 

The assertion holds for $k=1$ by Table 2. Fix some integer $k\ge 2,$ assume $\ell \left(k\right)=n.$ Then $\rho \left(n\right)=k.$ By the definition of $\ell \left(n\right)$ and (2), for every integer $m>n$, $\rho \left(m\right)\ge \rho \left(n\right)+1=k+1$ and

$$\rho (n+1)=\rho \left(n\right)+1=k+1.$$

Therefore, for every $m>n$ and $d\in \{0,1,2,3\}$,

$$\rho (4m+d)\ge 2min\left\{\rho \right(m):m>n\}=2k+2.$$

At the same time, when $m=n$, it is obvious that

$$\begin{array}{cc}\hfill \rho \left(4n\right)& = 2\rho \left(n\right)+1=2k+1,\hfill \\ \hfill \rho (4n+1)& = 2\rho \left(n\right)=2k,\hfill \\ \hfill \rho (4n+2)& = \rho \left(n\right)+\rho (n+1)=2k+1,\hfill \\ \hfill \rho (4n+3)& = 2\rho (n+1)=2k+2.\hfill \end{array}$$

This implies that $\ell \left(2k\right)=4n+1=4\ell \left(k\right)+1\mathrm{and}\ell (2k+1)=4n+2=4\ell \left(k\right)+2.$ □

Lemma 4.

The sequence $s\left(k\right)$ satisfies $s\left(1\right)=0$, $s\left(2\right)=1$, $s\left(3\right)=2$ and for every $k\ge 2,$

$$s\left(2k\right)=4s\left(k\right)-1,s(2k+1)=4s\left(k\right).$$

Proof. 

The initial values $s\left(1\right)$, $s\left(2\right)$ and $s\left(3\right)$ can be easily verified by Table 2. For an integer $k\ge 2$, suppose $s\left(k\right)=n$ for some integer n. Then $\rho \left(n\right)=k$. By the definition of $s\left(k\right)$ and (2), we have that $\rho \left(m\right)<\rho \left(n\right)$ whenever $m<n$ and

$$\rho (n-1)=\rho \left(n\right)-1=k-1.$$

Therefore, following from (1), for every $0\le m<n-1$ and $d\in \{0,1,2,3\},$ we have

$$\rho (4m+d)\le 2max\left\{\rho \right(m):0\le m\le n-1\}+1=2k-1.$$

Note that the value of $4m+d$ ranges from 0 to $4n-5$. At the same time, it follows from (1) that

$$\begin{array}{cc}\hfill \rho (4n-4)& = 2\rho (n-1)+1=2k-1,\hfill \\ \hfill \rho (4n-3)& = 2\rho (n-1)=2k-2,\hfill \\ \hfill \rho (4n-2)& = \rho (n-1)+\rho \left(n\right)=2k-1,\hfill \\ \hfill \rho (4n-1)& = 2\rho \left(n\right)=2k.\hfill \end{array}$$

This implies that $s\left(2k\right)=4n-1=4s\left(k\right)-1.$ The other formula $s(2k+1)=4s\left(k\right)$ follows from the fact that $\rho \left(4n\right)=2\rho \left(n\right)+1=2k+1$ and $\rho \left(i\right)\le 2k$ for every $0\le i\le 4n-1$. This ends the proof. □

The following two propositions show the upper bound for the sequence $\ell \left(k\right)$ and the lower bound for the sequence $s\left(k\right)$. For the sake of simplicity, for every integer $k\ge 2$, let

$$k=\sum _{j=0}^m{k}_j{2}^j:={[k_m{k}_{m-1}\cdots k_0]}_2$$

be the binary expansion of k with $m\ge 1$ and $k_m=1.$ For every $x\ge 0$, let $\lfloor x\rfloor$ be the greatest integer which is no more than x.

Proposition 1.

For every integer $k\ge 2$, we have

$$\ell \left(k\right)\le \frac{k^2}{3}.$$

Proof. 

For every $k\ge 2$, let the binary expansion of k be ${[k_m{k}_{m-1}\cdots k_0]}_2.$ Following from Lemma 3, we have

$$\ell \left(k\right)=\ell \left({[k_m{k}_{m-1}\cdots k_0]}_2\right)=4\ell \left({[k_m{k}_{m-1}\cdots k_1]}_2\right)+k_0+1.$$

(6)

Now we apply (6) by replacing k with ${[k_m{k}_{m-1}\cdots k_1]}_2$, which yields

$$\ell \left({[k_m{k}_{m-1}\cdots k_1]}_2\right)=4\ell \left({[k_m{k}_{m-1}\cdots k_2]}_2\right)+k_1+1.$$

Repeating this progress m times, by the fact that $k_j=\lfloor \frac{k}{2^j}\rfloor -2\lfloor \frac{k}{2^{j+1}}\rfloor$ for every $0\le j\le m-1$, we can obtain that

$$\begin{array}{cc}\hfill \ell \left(k\right)& = \ell \left({[k_m{k}_{m-1}\cdots k_0]}_2\right)\hfill \\ \hfill & = 4\ell \left({[k_m{k}_{m-1}\cdots k_1]}_2\right)+k_0+1\hfill \\ \hfill & = 4^2\ell \left({[k_m{k}_{m-1}\cdots k_2]}_2\right)+4(k_1+1)+k_0+1\hfill \\ \hfill & = \cdots \cdots \hfill \\ \hfill & = 4^m\ell \left(k_m\right)+\sum _{j=0}^{m-1}{4}^j(k_j+1)\hfill \\ \hfill & = 4^m\ell \left(1\right)+\sum _{j=0}^{m-1}{4}^j(\lfloor \frac{k}{2^j}\rfloor -2\lfloor \frac{k}{2^{j+1}}\rfloor +1)\hfill \\ \hfill & = \frac{4^m-1}{3}+\sum _{j=0}^{m-1}{4}^j(\lfloor \frac{k}{2^j}\rfloor -2\lfloor \frac{k}{2^{j+1}}\rfloor )\hfill \\ \hfill & \le \frac{4^m}{3}+k-2\lfloor \frac{k}{2}\rfloor +4\lfloor \frac{k}{2}\rfloor -8\lfloor \frac{k}{2^2}\rfloor +\cdots +4^{m-1}\lfloor \frac{k}{2^{m-1}}\rfloor -2·4^{m-1}\lfloor \frac{k}{2^m}\rfloor \hfill \\ \hfill & = \frac{4^m}{3}-2·4^{m-1}+k+2\sum _{j=1}^{m-1}{4}^{j-1}\lfloor \frac{k}{2^j}\rfloor \hfill \\ \hfill & \le \frac{4^m}{3}-2·4^{m-1}+k+2\sum _{j=1}^{m-1}{4}^{j-1}\frac{k}{2^j}\hfill \\ \hfill & = 2^{m-1}k-\frac{2}{3}·4^{m-1}.\hfill \end{array}$$

It suffices to show that

$$2^{m-1}k-\frac{2}{3}·4^{m-1}\le \frac{k^2}{3}.$$

Note that $2^m\le k<2^{m+1}.$ This implies that

$$\frac{k}{4}<2^{m-1}\le \frac{k}{2}.$$

Consider the function $f_k\left(x\right)=kx-\frac{2}{3}{x}^2.$ It is not hard to check that $f_k\left(x\right)$ is strictly increasing on the interval $(\frac{k}{4},\frac{k}{2}]$ with fixed $k\ge 1.$ Hence

$$kx-\frac{2}{3}{x}^2\le f_k\left(\frac{k}{2}\right)=\frac{1}{3}{k}^2,$$

which is the desired result. □

Proposition 2.

For every integer $k\ge 2$, we have

$$s\left(k\right)\ge \frac{k^2}{9}.$$

Proof. 

The assertion holds when $k=2$ and $k=3$ since $s\left(2\right)=1>4/9$ and $s\left(3\right)=2>1.$ For every integer $k\ge 4$, let the binary expansion of k be ${[k_m{k}_{m-1}\cdots k_0]}_2$ with $m\ge 2$ and $k_m=1$. Following from Lemma 4, we have

$$s\left(k\right)=s\left({[k_m{k}_{m-1}\cdots k_0]}_2\right)=4s\left({[k_m{k}_{m-1}\cdots k_1]}_2\right)+k_0-1.$$

Arguing analogously as in the proof of Proposition 1, we have

$$\begin{array}{cc}\hfill s\left(k\right)& = s\left({[k_m{k}_{m-1}\cdots k_0]}_2\right)\hfill \\ \hfill & = 4^{m-1}s\left({\left[k_m{k}_{m-1}\right]}_2\right)+\sum _{j=0}^{m-2}{4}^j(k_j-1)\hfill \\ \hfill & = 4^{m-1}s(2+k_{m-1})+\sum _{j=0}^{m-2}{4}^j(\lfloor \frac{k}{2^j}\rfloor -2\lfloor \frac{k}{2^{j+1}}\rfloor -1)\hfill \\ \hfill & = 4^{m-1}s(2+k_{m-1})-\frac{4^{m-1}-1}{3}+\sum _{j=0}^{m-2}{4}^j(\lfloor \frac{k}{2^j}\rfloor -2\lfloor \frac{k}{2^{j+1}}\rfloor )\hfill \\ \hfill & = 4^{m-1}s(2+k_{m-1})-\frac{4^{m-1}-1}{3}-2·4^{m-2}\lfloor \frac{k}{2^{m-1}}\rfloor +k+2\sum _{j=1}^{m-2}{4}^{j-1}\lfloor \frac{k}{2^j}\rfloor \hfill \\ \hfill & \ge 4^{m-1}s(2+k_{m-1})-\frac{4^{m-1}}{3}-2·4^{m-2}\lfloor \frac{k}{2^{m-1}}\rfloor +k+2\sum _{j=1}^{m-2}{4}^{j-1}(\frac{k}{2^j}-1)\hfill \\ \hfill & \ge 4^{m-1}s(2+k_{m-1})-2·4^{m-2}(3+k_{m-1})+2^{m-2}k\hfill \\ \hfill & = \left\{\begin{array}{cc}{2}^{m-2}k-2·4^{m-2}\hfill & \mathrm{if}{k}_{m-1}=0,\hfill \\ 2^{m-2}k\hfill & \mathrm{if}{k}_{m-1}=1.\hfill \end{array}\right.\hfill \end{array}$$

The last equality uses the fact that $s\left(2\right)=1$ and $s\left(3\right)=2.$

• If $k_{m-1}=0,$ then we have $2^m\le k<3·2^{m-1},$ and it follows that

  $$\frac{k}{6}<2^{m-2}\le \frac{k}{4}.$$
  Put $g_k\left(x\right):=xk-2x^2.$ It is obvious that $g_k\left(x\right)$ is strictly increasing on the interval $(\frac{k}{6},\frac{k}{4}]$ with fixed $k.$ Hence

  $$2^{m-2}k-2·4^{m-2}\ge g_k\left(\frac{k}{6}\right)=\frac{k^2}{9}.$$
• If $k_{m-1}=1,$ then we have $3·2^{m-1}\le k<2^{m+1},$ and it follows that

  $$\frac{k}{8}<2^{m-2}\le \frac{k}{6},$$

  which implies

  $$2^{m-2}k\ge \frac{k^2}{8}\ge \frac{k^2}{9}.$$

This ends the proof. □

Now, combining Proposition 1 and Proposition 2, we can prove Theorem 1.

Proof of Theorem 1.

For every integer $n\ge 1$, assume $\rho \left(n\right)=k$ for some integer k. It follows from Remark 1 that $k\ge 2$. By Proposition 1 and the definition of $\ell \left(k\right)$, we have

$$\frac{\rho \left(n\right)}{\sqrt{n}}\ge \frac{k}{\sqrt{\ell \left(k\right)}}\ge \sqrt{3}.$$

Following from Proposition 2 and the definition of $s\left(k\right)$, we have

$$\frac{\rho \left(n\right)}{\sqrt{n}}\le \frac{k}{\sqrt{s\left(k\right)}}\le 3.$$

This completes the proof. □

By Remark 2 and Theorem 1, we get the following corollary.

Corollary 1.

The numbers $\sqrt{3}$ and 3 are the optimal lower and upper bounds for the set $\{\frac{\rho \left(n\right)}{\sqrt{n}}:n\ge 1\}$ respectively.

To prove Theorem 2, we need an auxiliary notation which was firstly introduced in [8]. Let $b\ge 2$ be an integer and $f:\mathbb{N}\to \mathbb{Z}$ be a discrete function (or an integer sequence). For every $x\ge 0$, let

$${\delta }_f\left(x\right):=\underset{n\to +\infty }{lim\; sup}\frac{\left|f\right(n\left)\right|}{n^x}.$$

Write $\Delta f\left(n\right):=f(n+1)-f\left(n\right)$. Set

$$\alpha =\alpha \left(f\right):=inf\{x\ge 0\mid {\delta }_f\left(x\right)=0\}\mathrm{and}\beta =\beta \left(f\right):=inf\{x\ge 0\mid {\delta }_{\Delta f}\left(x\right)=0\}.$$

Definition 2.

(Quasi-linear function) Let $f:\mathbb{N}\to \mathbb{Z}$. If $\alpha \left(f\right)>\beta \left(f\right)$ and there exists a constant $C_1>0$ and an integer $b\ge 2$ such that for all positive integers n and $0\le i\le b-1$,

$$|f(bn+i)-b^{\alpha }f\left(n\right)|\le C_1{n}^{\beta },$$

(7)

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].

Lemma 5

([8]). Given an integer function $f\left(n\right)$, set $\alpha :=\alpha \left(f\right)$. Suppose $a_1$ and $a_2$ are two accumulation points of $\{f\left(n\right)/n^{\alpha }:n\ge 1\}$. If f is a quasi-linear function, then $\{f\left(n\right)/n^{\alpha }:n\ge 1\}$ is dense in $[a_1,a_2]$.

Proof of Theorem 2.

Following from the fact that $\Delta \rho \left(n\right)\in \{-1,1\}$ for every $n\ge 0$, we have $\beta \left(\rho \right)=0$. By Theorem 1, $\alpha \left(\rho \right)=\frac{1}{2}$. Moreover, for every $0\le i\le 3$ and $n\ge 1$, (1) yields

$$\left|\rho \right(4n+i)-2\rho (n\left)\right|\le 2\left|\rho \right(n+1)-\rho (n\left)\right|=2.$$

Therefore $\rho \left(n\right)$ is a quasi-linear function for $b=4$.

It follows from Remark 2 that $\sqrt{3}$ and 3 are two accumulation points of the set $\{\rho \left(n\right)/\sqrt{n}:n\ge 1\}.$ Hence we obtained the desired result by Lemma 5. □