2 Preliminary

We review the basic results on Salem numbers (cf. [Reference Bertin, Decomps-Guilloux, Grandet-Hugot, Pathiaux-Delefosse and Schreiber7, Reference Smyth28]). It is easy to show that a Salem number has even degree $2d \ge 4$ and its minimal polynomial is self-reciprocal. Thus $d_{\beta }(1)$ cannot be purely periodic for a Salem number $\beta $ , see [Reference Akiyama, Győry, Pethő and Sós1]. Let $P(x)\in \mathbb Z[x]$ be a monic irreducible self-reciprocal polynomial of even degree $2d$ . Putting $Q(x):=P(x)/x^{2d}$ , we have $Q(x)\in \mathbb R[x+x^{-1}]$ . We write $Q(x)=G(y)$ with $y=x+x^{-1}$ . Then $P(x)$ is a minimum polynomial of a Salem number if and only if $G(2)<0$ and $G(y)$ has $d-1$ distinct roots in $(-2,2)$ . Here, G is coined a trace polynomial of P in [Reference Boyd12]. The factorization

$$ \begin{align*} G(y)=(y-\gamma)(y+\alpha_1)\ldots (y+\alpha_{d-1})\end{align*} $$

with $\gamma>2$ and $\alpha _i\in (-2,2)$ corresponds to the factorization of $P(x)$ in $\mathbb R[x]$ :

(2) $$ \begin{align} P(x)=(x-\beta)\bigg(x-\frac 1{\beta}\bigg) \prod_{i=1}^{d-1} (x^2+\alpha_i x+1), \end{align} $$

where $\gamma =\beta +1/\beta $ and $x^2+\alpha _i x+1$ gives a root $\exp (\theta _i \sqrt {-1}) =\cos (\theta _i)\pm \sin (\theta _i)\sqrt {-1}$ of $P(x)$ with $\alpha _i=-2\cos (\theta _i)$ and $\theta _i\in (0,\pi )$ . It is well known that $1, \theta _1/\pi , \theta _2/\pi , \ldots , \theta _{d-1}/\pi $ are linearly independent over $\mathbb Q$ , that is, $\exp (\theta _i\sqrt {-1})\ (i=1,\ldots , d-1)$ are multiplicatively independent. This is shown by applying a conjugate map to the possible multiplicative relation among them, cf. [Reference Dubickas and Jankauskas13]. Note that this fact guarantees that $\beta ^m\ (m=1,2,\ldots )$ are Salem numbers of the same degree $2d$ . Applying this linear independence, we see that

(3) $$ \begin{align} \bigg(\frac{m \theta_1}{2\pi}, \ldots, \frac{m \theta_n}{2\pi}\bigg) \bmod{\mathbb Z^n}\end{align} $$

is uniformly distributed in $(\mathbb R/\mathbb Z)^n$ , that is, for any parallelepiped

$$ \begin{align*} I=[a_1,b_1]\times [a_2,b_2]\times \cdots \times [a_n,b_n], \end{align*} $$

we have

$$ \begin{align*} \lim_{M\to \infty} \frac 1M \sum_{m=1}^M \chi_I\bigg( \bigg(\frac{m \theta_1}{2\pi}, \ldots, \frac{m \theta_n}{2\pi}\bigg) \bmod{\mathbb Z^n}\bigg) = \prod_{i=1}^n (b_i-a_i), \end{align*} $$

where $\chi _I$ is the characteristic function of I. Indeed, this is shown by the higher dimensional Weyl criterion [Reference Kuipers and Niederreiter20]. Note that it is understood as unique ergodicity of the action

$$ \begin{align*} x\mapsto x + \bigg(\frac{\theta_1}{2\pi}, \ldots, \frac{\theta_n}{2\pi}\bigg) \end{align*} $$

on $(\mathbb R/\mathbb Z)^n$ . Since $(\mathbb R/\mathbb Z)^n$ is a compact metric group, minimality and unique ergodicity are equivalent [Reference Walters30], and minimality is a little easier to show.

3 Our strategy

Our idea is to find a nice region so that if complex conjugates of a Salem number $\beta>1$ fall into this region and $\beta $ is sufficiently large, then $d_{\beta }(1)$ is eventually periodic. We realize this idea in a general form which can be applied to the dominant real root of self-reciprocal polynomials, not only Salem numbers. The statement seems useless for a single $\beta $ (because it is easier to compute $d_{\beta }(1)$ directly) but we will find a nice application in the following sections.

Lemma 3. Let us fix constants $u,v$ with $0<u<v<1$ . If a monic polynomial $f(x)\in \mathbb Z[x]$ of degree $2n+2$ satisfies

$$ \begin{align*} f(\beta)=0,\quad f(0)=1,\quad \beta> \max\bigg\{ \frac 2u, \frac 1{1-v} \bigg\} \end{align*} $$

and

$$ \begin{align*} \frac{f(x)}{(x-\beta)(x-1/\beta)} = x^{2n}+1+\sum_{i=1}^{2n-1}g_i x^i \end{align*} $$

with $u<g_i<v$ for $i=1,\ldots ,2n-1$ , then $f(x)$ is self-reciprocal and $d_{\beta }(1)$ is $(1, 2n+1)$ -periodic.

Proof. Set $g_0=g_{2n}=1$ . Putting

$$ \begin{align*} f(x)=x^{2n+2}+1-\sum_{i=1}^{2n+1} c_i x^i, \end{align*} $$

we obtain

$$ \begin{align*} c_1=\beta+1/\beta-g_1,\quad c_{2n+1}=\beta+1/\beta-g_{2n-1}, \end{align*} $$

and

$$ \begin{align*} c_i=(\beta+1/\beta) g_{i-1} - g_{i-2}-g_{i} \end{align*} $$

for $i=2,\ldots , 2n$ . Since $c_i\in \mathbb Z$ , $g_1$ and $g_{2n-1}$ are uniquely determined by $\beta +1/\beta $ , and thus $g_1=g_{2n-1}$ and $c_1=c_{2n+1}$ . Moreover by induction, we see $c_{i}=c_{2n+2-i}$ for every $i=1,\ldots , 2n+1$ , that is, $f(x)$ is self-reciprocal. By assumption, we have

$$ \begin{align*} c_i\ge \beta u -2>0,\quad c_1-c_i\ge (\beta+1/\beta)(1-v) -1 >0 \end{align*} $$

for $i=2,\ldots ,2n$ . Therefore, we have

$$ \begin{align*} c_i\in \mathbb Z, \quad c_1> c_j> 0\ (j=2,\ldots, 2n). \end{align*} $$

One can write

$$ \begin{align*} \beta^{2n+2}+1-\sum_{i=1}^{2n+1} c_i \beta^i=0 \end{align*} $$

as a representation of zero in base $\beta $ :

$$ \begin{align*} (-1),c_{1},c_{2},\ldots, c_{n},c_{n+1},c_{n}, \ldots ,c_2, c_1, (-1). \end{align*} $$

Adding its $2n+1$ shifted form, we see

$$ \begin{align*} (-1),c_{1},c_{2},\ldots, c_{n+1}, \ldots , c_2, c_1-1, c_1-1, c_2, \ldots, c_{n+1}, \ldots ,c_2, c_1,(-1) \end{align*} $$

is another representation of zero. Iterating this shifted addition, we obtain an infinite expansion of $1$ :

$$ \begin{align*} c_1(c_2,c_3,\ldots, c_{n},c_{n+1},c_{n},\ldots, c_2,c_1-1,c_1-1)^{\infty}, \end{align*} $$

which satisfies the lexicographic condition in equation (1). (This implies $c_1<\lceil \beta \rceil -1$ as well.) Therefore, $d_{\beta }(1)$ is $(1,2n+1)$ - periodic.

Logically we have to fix $u,v$ at first. However, this is often impractical. When we apply Lemma 3 to a Salem number $\beta $ of degree $2d$ , we take a self-reciprocal monic polynomial $R(x)\in \mathbb Z[x]$ and study $f(x)=R(x)P(x)$ of degree $2n+2$ , where $P(x)$ is the minimum polynomial of $\beta $ . Note that one can take $R(x)=1$ as well. We check if there exist u and v which satisfy our requirements. Since

$$ \begin{align*} \frac{f(x)}{(x-\beta)(x-1/\beta)}= \sum_{i=0}^{2n} g_i x^i= R(x) \prod_{i=1}^{d-1} (x^2+\alpha_i x+1) \end{align*} $$

with $\alpha _i\in (-2,2)$ , for a fixed $R(x)$ , the problem is reduced to the set of solutions $(\alpha _1,\ldots ,\alpha _{d-1})\in \mathbb R^{d-1}$ for the system of inequalities

$$ \begin{align*} 0<g_i<1 \ (i=1,2,\ldots, 2n-1),\quad -2<\alpha_j<2\ (j=1,\ldots,d-1), \end{align*} $$

over $d-1$ variables $\alpha _1,\ldots ,\alpha _{d-1}$ . A self-reciprocal monic polynomial $R(x)$ gives a certain choice of $u,v$ if and only if the set of solutions contains an inner point in the space $\mathbb R^{d-1}$ , and for every inner point, we can find $u,v$ . Solving this set of inequalities is not an easy task in general, but it is feasible for degree $6$ since all the inequalities are quadratic. In §5, we use this method to find good regions for $\beta $ . It is also useful to solve a slightly wider set of inequalities:

(4) $$ \begin{align} 0<g_i\le 1 \ (i=1,2,\ldots, 2n-1),\quad -2<\alpha_j<2\ (j=1,\ldots,d-1). \end{align} $$

The solution may contain the case $g_i=1$ . The proof of Lemma 3 works in the same way under a little more involved assumptions and we have the following lemma.

Lemma 4. Let us fix constants $u,v$ with $0<u<v<1$ . If a monic polynomial $f(x)\in \mathbb Z[x]$ of degree $2n+2$ satisfies

$$ \begin{align*} f(\beta)=0,\ f(0)=1,\quad \beta> \max\bigg\{ \frac 2u, \frac 1{1-v} \bigg\} \end{align*} $$

and

$$ \begin{align*} \frac{f(x)}{(x-\beta)(x-1/\beta)} = x^{2n}+1+\sum_{i=1}^{2n-1}g_i x^i \end{align*} $$

with

$$ \begin{align*} u<g_i<v \end{align*} $$

or

$$ \begin{align*} (g_i=1\ \text{and}\ g_1<g_{i-1}+g_{i+1}) \end{align*} $$

or

$$ \begin{align*} ( g_i=1\ \text{and}\ g_1=g_{i-1}+g_{i+1}\ \text{and} \ g_1-g_{i+1}>u) \end{align*} $$

hold for $i=1,\ldots ,2n-1$ , then $f(x)$ is self-reciprocal and $d_{\beta }(1)$ is $(1,2n+1)$ -periodic.

We shall see later that the effect of this small extension of Lemma 3 is pretty large both in theory and in practice. Indeed, Lemma 4 works very well with a discretized rotation algorithm, see the discussion in §6.

4 Proof of Theorem 1

We start with the following lemma.

Lemma 5. For a positive even integer $2n$ , all roots of the self-reciprocal polynomial

$$ \begin{align*} P(x)=x^{2n}+ d_{2n-1} x^{2n-1}+\cdots + d_1 x +1 \quad (d_{2n-i}=d_i) \end{align*} $$

with $d_i\in \mathbb R$ and $|d_i|<1/(2n-2)$ are on the unit circle.

Proof. Let $G_0(y):=(x^{2n}+1)/x^{n}$ and $G_1(y):=P(x)/x^{n}$ with $y=x+x^{-1}$ . Since

$$ \begin{align*}G_0(2\cos(\pi k /n))=2(-1)^k\end{align*} $$

for $k=0,1,2,\ldots , n$ , $G_0(y)$ has n real roots $\psi _i\ (i=1,2,\ldots , n)$ with

$$ \begin{align*} 2&>\psi_1>2\cos\bigg(\frac{\pi}n\bigg)>\psi_2>2\cos\bigg(\frac{2\pi}n\bigg)>\psi_3>\cdots\\ &\cdots> 2\cos\bigg(\frac{(n-2)\pi}n\bigg)>\psi_{n-1}> 2\cos\bigg(\frac{(n-1)\pi}n\bigg)>\psi_n>-2. \end{align*} $$

From $|d_i|<1/(2n-2)$ , we have $G_1(2\cos (\pi k /n))<0$ for odd k and $G_1(2\cos (\pi k /n))$ $>0$ for even k. By intermediate value theorem, $G_1(y)$ has n real roots $\psi ^{\prime }_i\ (i=1,\ldots ,n)$ in $(-2,2)$ satisfying the same inequality as $\psi _i$ . Therefore, we have

$$ \begin{align*}G_1(y)=\prod_{i=1}^{n} (y-\psi^{\prime}_i).\end{align*} $$

Coming back to $P(x)$ , we get the assertion.

For degree 4, we have nothing to do since Boyd [Reference Boyd11] showed that every Salem number of degree 4 is a $(1,p)$ -periodic Parry number for some $p\in \mathbb N$ . Consider a polynomial

$$ \begin{align*} h(x)=x^{2n}+ \frac 1{4(n-1)}\sum_{i=1}^{2n-1} x^i +1. \end{align*} $$

By Lemma 5, we have

$$ \begin{align*} h(x)=\prod_{i=1}^{n} (x-\exp(\eta_i \sqrt{-1}))(x-\exp(-\eta_i \sqrt{-1}))\end{align*} $$

with

(5) $$ \begin{align} 2&>2\cos(\eta_1)>2\cos\bigg(\frac{\pi}n\bigg)>2\cos(\eta_2)>2\cos\bigg(\frac{2\pi}n\bigg)>2\cos(\eta_3)>\cdots \nonumber\\ &\cdots> 2\cos\bigg(\frac{(n-2)\pi}n\bigg)>2\cos(\eta_{n-1})> 2\cos\bigg(\frac{(n-1)\pi}n\bigg)>2\cos(\eta_{n})>-2. \end{align} $$

This type of discussion is called ‘interlacing’ and efficiently used in the construction of Salem numbers having desired properties, see [Reference Akiyama and Kwon6, Reference McKee and Smyth22, Reference Smyth27] and its references.

Considering coefficients as a continuous function of roots, there exists a constant $\varepsilon>0$ that if

$$ \begin{align*} \psi_i \in [\eta_i-\varepsilon, \eta_i+\varepsilon] \end{align*} $$

for $i=1,2,\ldots , n$ , then we have

(6) $$ \begin{align} \prod_{i=1}^{n} (x-\exp(\psi_i \sqrt{-1}))(x-\exp(-\psi_i \sqrt{-1})) =x^{2n}+1+\sum_{i=1}^{2n-1} g_i x^i \end{align} $$

with ${1}/{6(n-1)}<g_i< {1}/{3(n-1)}$ . See Remark 6 for the choice of $\varepsilon $ .

Let $\beta $ be a Salem number of degree $2n+2$ with $n\ge 2$ and let $\theta _i{\kern-1pt}\in{\kern-1pt} (0,\pi )\, (i{\kern-1pt}={\kern-1pt}1,\ldots ,n)$ be the arguments of the conjugates of $\beta $ on the unit circle determined as in equation (2). Since $1, \theta _1/\pi ,\ldots , \theta _{n}/\pi $ are linearly independent over $\mathbb Q$ , by Kronecker’s approximation theorem, we find infinitely many positive integers m such that

(7) $$ \begin{align} \frac{m \theta_i}{2\pi} \pmod{\mathbb Z} \in [\eta_i-\varepsilon, \eta_i+\varepsilon] \end{align} $$

hold for $i=1,2,\ldots ,n$ . For an integer m with this property, the minimum polynomial of $\beta ^m$ has the form

$$ \begin{align*} (x-\beta^m)\bigg(x-\frac 1{\beta^m}\bigg)\bigg( x^{2n}+1+\sum_{i=1}^{2n-1} g^{(m)}_i x^i \bigg) \end{align*} $$

and $ {1}/{6(n-1)}<g^{(m)}_i <{1}/{3(n-1)}$ for $i=1,2,\ldots , 2n-1$ . By Lemma 3, we see that $d_{\beta ^m}(1)$ is $(1,2n+1)$ -periodic for sufficiently large m. Finally, we show that

$$ \begin{align*} \{m\in \mathbb N\ | \ d_{\beta^m}(1) \text{ is}\ (1,2n+1)\text{-periodic} \} \end{align*} $$

has positive lower natural density. Using the fact that equation (3) is uniformly distributed in $(\mathbb R/\mathbb Z)^n$ , we have

$$ \begin{align*} \liminf_{M\to \infty} \frac 1M \mathrm{Card}\{ m\in [1,M]\cap \mathbb N\ |\ d_{\beta^m}(1) \text{ is}\ (1,2n+1)\text{-periodic} \} \ge (2\varepsilon)^n \end{align*} $$

in view of equation (7), giving the lower bound of the natural density.

Remark 6. We give a lower bound of $c(d)$ from the above proof. First, we compute the asymptotic expansion of $\eta _i$ with respect to n, when $\eta _i$ is close to $\pi /2$ (cf. [Reference Akiyama3]). If n is odd, then we have

$$ \begin{align*} \eta_{\lfloor n/2 \rfloor}= \frac{\pi}2+ \frac {1}{8n^2}+\frac {9}{64n^3}+ O\bigg(\frac 1{n^4}\bigg), \end{align*} $$

which leads to $\tan (\eta _{\lfloor n/2 \rfloor })=-8 n^2+O(n)$ . Further, we have

$$ \begin{align*} \eta_{\lfloor n/2 \rfloor\pm 1}= \frac{\pi}2 \pm \frac{\pi}{n}+ \frac {1}{8n^2}+\frac {9\mp 8\pi}{64n^3}+ O\bigg(\frac 1{n^4}\bigg) \end{align*} $$

and

$$ \begin{align*} \tan(\eta_i)=O(n) \end{align*} $$

for $|i-\lfloor n/2\rfloor |>1$ in light of equation (5). If n is even, then we have

$$ \begin{align*} \eta_{n/2 \pm 1}= \frac{\pi}2 \pm \frac{\pi}{2n}+ \frac {1}{8n^2}+\frac {9\mp 4\pi}{64n^3}+ O\bigg(\frac 1{n^4}\bigg). \end{align*} $$

Thus, $\tan (\eta _i)=O(n)$ is valid for all i by equation (5).

Second, comparing the coefficients of $\prod _{i=1}^n (x^2+2\cos (\eta _i+\nu ) x+ 1)$ and those of $\prod _{i=1}^n (x^2+2\cos (\eta _i) x+ 1)$ , we see that the desired inequality of equation (6) holds if

(8) $$ \begin{align} \frac 23<\prod_{i=1}^n \bigg|\frac{\cos(\eta_i+\nu)}{\cos(\eta_i)}\bigg| < \frac 43 \end{align} $$

for $|\nu |<\varepsilon $ . Using an easy inequality

$$ \begin{align*} 1-|{\tan}(x)y|-\frac{y^2}2< \bigg|\frac{\cos(x+y)}{\cos(x)}\bigg|<1+|{\tan}(x) y| \end{align*} $$

for $\cos (x)\neq 0$ and the above estimates on $\eta _i$ , we see that there exists a positive constant $\kappa $ so that if $|\varepsilon |<\kappa /n^2$ , then equation (8) holds. Thus we obtain $c(d)\ge (\kappa /n^{2})^{n}$ with $d=2n+2$ . Making explicit the implied constants of Landau symbols, we see that $\kappa =1/32$ suffices. Thus we have $c(d)\ge (32(d/2)^2)^{-d/2}>(3d)^{-d}$ .

5 Proof of Theorem 2

For the rest of this paper, we deal with a Salem number $\beta $ of degree 6. For simplicity of presentation, we prove that $c(6)\ge 0.458$ . To show that $c(6)$ exceeds $0.5$ , we have to use more polynomials and the computation becomes much harder, see Appendix A.

To apply the discussion after Lemma 3, there exists an efficient practical way to find $R(x)$ which is similar to the shift radix system (cf. [Reference Akiyama, Borbély, Brunotte, Pethő and Thuswaldner4]). Pick a random $(\alpha _1,\alpha _2)\in (-2,2)^2$ and start with a coefficient vector $(1,c_1,c_2,c_1,1)$ of

$$ \begin{align*} (x^2+\alpha_1 x+1)(x^2+\alpha_2 x+1)=x^4+c_1x^3+c_2 x^2+c_1 x+1, \end{align*} $$

that is, $c_1=\alpha _1+\alpha _2$ , $c_2=\alpha _1 \alpha _2 +2$ . We wish to iterate the shifted addition as

$$ \begin{align*} (1,c_1,c_2,c_1,1) & \rightarrow (1,c_1,c_2,c_1,1,0)+k (0, 1,c_1,c_2,c_1,1)\\ &=(1, c_1+k, c_2+c_1 k, c_1+c_2 k, 1+c_1 k ,k), \end{align*} $$

with $k=-\lceil c_1-1\rceil $ , to find a longer coefficient vector where all entries except the first and the last ones fall in $(0,1]$ , as in Lemma 4. To make this idea into an algorithm, set $z_{-1}=(0,0,0,0)$ and for $z_n=(t_n(1),t_n(2),t_n(3),t_n(4))$ , we define

(9) $$ \begin{align} z_{n+1}:=(t_n(2)+ c_1k_{n+1}, t_n(3)+ c_2k_{n+1}, t_n(4)+c_1 k_{n+1}, k_{n+1}) \end{align} $$

with $k_{n+1}=-\lceil t_n(1)-1 \rceil $ . Thus we see $k_0=1$ and $z_0=(c_1,c_2,c_1,1)$ . We stop this iteration when $0<t_n(i)\le 1$ for $i=1,2,3$ and $t_n(4)=1$ (that is, $k_{n}=1$ ) and obtain a candidate $R(x)=\sum _{i=0}^{n} k_i x^i$ . It is natural to set $k_{n+1}=k_{n+2}=k_{n+3}=k_{n+4}=0$ and we obtain

$$ \begin{align*}(x^4+c_1x^3+c_2x^2+c_1x+1)R(x)=\sum_{i=0}^{n+4} g_i x^i \end{align*} $$

with $g_0=g_{n+4}=1$ and $g_i=t_{i-1}(1)+k_i\in (0,1]\ (i=1,\ldots ,n+3)$ . If n became larger than a given threshold, then we restart with a different $(\alpha _1,\alpha _2)$ . Applying this random search, we find polynomials $R_i(x)\ (i=1,\ldots , 18)$ so that Lemma 3 gives relatively large regions $\mathcal {R}_i\ (i=1,\ldots ,18)$ defined by equation (4).

$$ \begin{align*} R_1&=1,\\ R_2&=1+x,\\ R_3&=1+x^2,\\ R_4&=1+2 x+x^2,\\ R_5&=1+x+x^2+x^3,\\ R_6&=1+2 x+2 x^2+x^3,\\ R_7&=1-x+x^2-x^3+x^4,\\ R_8&=1+2 x+2 x^2+2 x^3+2x^4+x^5,\\ R_9&=1+x-x^2-x^3+x^4+x^5,\\ R_{10}&=1+2 x^2+2 x^4+x^6,\\ R_{11}&=1+x+x^2+2 x^3+x^4+x^5+x^6,\\ R_{12}&=1+x+x^2+2 x^3+2 x^4+x^5+x^6+x^7,\\ R_{13}&=1+x+2 x^2+2 x^3+2 x^4+2 x^5+x^6+x^7,\\ R_{14}&=1+2 x+x^2-x^3-x^4+x^5+2 x^6+x^7, \\ R_{15}&=1+x^2-x^3+x^4-x^5+x^6+x^8,\\ R_{16}&=1-2x+2x^2-x^3+x^4-2x^5+3x^6-2x^7+x^8-x^9+2x^{10}-2x^{11}+x^{12},\\ R_{17}&=1+3x+4x^2+3x^3+x^4,\\ R_{18}&=1+3x+4x^2+2x^3-2x^4-4x^5-2x^6+2x^7+4x^8+3x^9+x^{10}. \end{align*} $$

For example,

$$ \begin{align*} (x^4+c_1x^3+c_2x^2+c_1x+1)R_8=x^9+(c_1+2)x^8+\cdots+(c_1+2)x+1 \end{align*} $$

gives the coefficient vector

$$ \begin{align*} &(1,c_1+2,2 c_1+c_2+2,3 c_1+2 c_2+2,4 c_1+2 c_2+3, 4 c_1+2 c_2+3,3 c_1+2 c_2+2,\\&\quad 2 c_1+c_2+2,c_1+2,1) \end{align*} $$

which gives rise to a system of linear inequalities

$$ \begin{align*} 0{\kern-1pt}<{\kern-1pt}c_1{\kern-1pt}+{\kern-1pt}2{\kern-1pt}\leq{\kern-1pt} 1,\quad 0{\kern-1pt}<{\kern-1pt}2 c_1{\kern-1pt}+{\kern-1pt}c_2+2{\kern-1pt}\leq{\kern-1pt} 1,\quad 0{\kern-1pt}<{\kern-1pt}3 c_1{\kern-1pt}+{\kern-1pt}2 c_2{\kern-1pt}+{\kern-1pt}2{\kern-1pt}\leq{\kern-1pt} 1,\quad 0{\kern-1pt}<{\kern-1pt}4 c_1{\kern-1pt}+{\kern-1pt}2 c_2{\kern-1pt}+{\kern-1pt}3{\kern-1pt}\leq{\kern-1pt} 1. \end{align*} $$

Solving this system, we obtain a triangular region:

$$ \begin{align*} \bullet&\ c_1>\frac{-2 c_2-3}{4} \quad\text{if }c_2\in \bigg(\frac 12,\frac 52\bigg);\\ \bullet&\ c_1 \leq -1 \qquad\quad\ \ \ \text{if }c_2\in \bigg(\frac{1}{2}, 1\bigg];\\ \bullet&\ c_1\leq \frac{-2 c_2-1}{3}\quad\text{if }c_2\in\bigg(1,\frac{5}{2}\bigg). \end{align*} $$

This collection of three sentences with $\bullet $ reads $c_2$ must be in at least one of the intervals, and we take the logical ‘and’ of the three. Replacing $(c_1,c_2)$ by $(\alpha _1+\alpha _2,\alpha _1\alpha _2+2)$ , we can confirm that $-2<\alpha _i<2$ for $i=1,2$ hold in this triangle. (For a general $R(x)$ , we compute the intersection with the region $-2<\alpha _i<2\ (i=1,2)$ . If this intersection is empty, then we have to restart with a different $(\alpha _1,\alpha _2)$ .) Thus we find the two curvilinear triangles $\mathcal {R}_{8}$ in Figure 1 bounded by segments and hyperbola. We also found polynomials $L_i\ (i=1,\ldots , 5)$ giving large regions $\mathcal {L}_i\ (i=1,\ldots , 5)$ where Lemma 3 does not apply and Lemma 4 is necessary.

$$ \begin{align*} L_1&=1-x^2+x^3+x^4-x^5+x^7,\\ L_2&=1-x+x^3-x^5+x^6, \\ L_3&=1-x^2+x^3+x^6-x^7+x^9,\\ L_4&=1-2x+2x^2-2x^4+3x^5-2x^6+2x^8-2x^9+x^{10},\\ L_5&=1-x+x^3-x^6+x^7+x^8-x^9+x^{12}-x^{14}+x^{15}. \end{align*} $$

For example, the coefficients of $x^3$ and $x^8$ in

$$ \begin{align*} &(x^4+c_1 x^3+c_2x^2+c_1 x+1)L_1\\ &\quad= 1+c_1 x-x^2+c_2 x^2+x^3+2 x^4+c_1 x^4-c_2 x^4-x^5+c_2 x^5-x^6+c_2 x^6\\ &\qquad + 2 x^7 +c_1 x^7-c_2 x^7+x^8-x^9+c_2 x^9+c_1 x^{10}+x^{11} \end{align*} $$

are equal to $1$ , so we have to use Lemma 4.

Figure 1

Period cells in $(\alpha _1,\alpha _2)$ coordinate.

We can check directly that those 23 sets are mutually disjoint. We will see later in §6 that Lemma 4 works fine for any period and this disjointness is natural. Indeed, if two period cells share an inner point, then their periods of the corresponding discretized rotations must coincide, since $d_{\beta ^m}(1)$ for a Salem number $\beta $ is uniquely determined by m. This implies that the period cells must be identical. See Figure 1 and the explicit computation of period cells below.

Choose a small $\varepsilon>0$ and consider the subset $\mathcal {R}_i(\varepsilon )$ of solutions of the system of inequalities

$$ \begin{align*} \varepsilon<g_i<1-\varepsilon \ (i=1,2,\ldots, 2n-1),\quad -2<\alpha_j<2\ (j=1,\ldots,d-1) \end{align*} $$

for a polynomial $R_i\ (i=1,\ldots , 18)$ . We also define $\mathcal {L}_i(\varepsilon )$ by the system of inequalities

$$ \begin{align*} \varepsilon<g_i<1-\varepsilon \ (i\in \{j \in [1,2n-1]\ |\ g_j\not = 1\}), \quad -2<\alpha_j<2\ (j=1,\ldots,d-1) \end{align*} $$

for a polynomial $L_i\ (i=1,\ldots ,5)$ . Since $g_i\in \mathbb Z+(\alpha _1+\alpha _2)\mathbb Z+\alpha _1\alpha _2\mathbb Z$ , $g_i$ is a constant if and only if $g_i=1$ . Therefore, unless $g_i=1$ , the boundary equalities $g_i=\varepsilon $ and $g_i=1-\varepsilon $ give one-parameter families of (linear or hyperbolic) curves, which continuously move along $\varepsilon $ . In light of Lemmas 3 and 4, there exist $m_0=m_0(\varepsilon )\in \mathbb N$ where if $m\ge m_0$ and $(-2\cos (m\theta _1/2\pi ),-2\cos (m\theta _2/2\pi ))$ falls into

$$ \begin{align*} S(\varepsilon):=\bigg(\bigcup_{i=1}^{18} \mathcal{R}_i(\varepsilon) \bigg) \cup \bigg(\bigcup_{i=1}^{5} \mathcal{L}_i(\varepsilon) \bigg), \end{align*} $$

then $d_{\beta ^m}(1)$ is $(1,p)$ -periodic with some $p\in \mathbb N$ . Since $(m \theta _1/2\pi ,m\theta _2/2\pi ) \pmod {\mathbb Z^2}$ is uniformly distributed and $S(\varepsilon )$ is Jordan measurable, the induced probability measure is computed by

$$ \begin{align*} \frac 1{\pi^2}\int\int_{S(\varepsilon)} \frac {d\alpha_1 d\alpha_2}{\sqrt{(4-\alpha_1^2)(4-\alpha_2^2)}}, \end{align*} $$

since the normalized Lebesgue measure ${1}/{2\pi } d\theta $ on the unit circle is projected 2 to 1 to the interval $[-2,2]$ by the map $\theta \mapsto -2\cos (\theta )=:\alpha $ and

$$ \begin{align*}\frac {2}{2\pi} d \theta = \frac {1}{\pi}d(\arccos(-\alpha/2)) = \frac {1}{\pi}d(\pi-\arccos(\alpha/2)) =\frac{1}{\pi} \frac{d \alpha}{\sqrt{4-\alpha^2}}. \end{align*} $$

Let

$$ \begin{align*}S := \bigg(\bigcup_{i=1}^{18} \mathcal{R}_i \bigg) \cup\bigg(\bigcup_{i=1}^{5} \mathcal{L}_i \bigg) \end{align*} $$

and $\mu $ be the two-dimensional Lebesgue measure. Since both $S(\varepsilon )$ and S have piecewise smooth boundaries, we have

$$ \begin{align*} \mu(S\setminus S(\varepsilon))\to 0 \end{align*} $$

as $\varepsilon \to 0$ . Thus the above measure converges to

$$ \begin{align*} \frac 1{\pi^2}\int\int_S \frac {d\alpha_1 d\alpha_2}{\sqrt{(4-\alpha_1^2)(4-\alpha_2^2)}}\approx 0.458895 \end{align*} $$

which proves our theorem. The explicit forms of the regions $\mathcal {R}_i\ (i=1,2,\ldots ,18)$ and $\mathcal {L}_i\ (i=1,2,\ldots ,5)$ are listed below, using the coordinate $(x,y)=(\alpha _1,\alpha _2)$ with $y<x$ . Each collection of sentences with $\bullet $ is read in the similar way as before.

$$\begin{align*} \begin{array}{rll}&\mathcal{R}_{1}{:}&\\ &\bullet\ y \leq -\displaystyle\frac{1}{x}&\hspace{-10pt} \mathrm{if}\ x\in\displaystyle\bigg[1,\frac{1+\sqrt{5}}{2}\bigg];\\[3pt] &\bullet\ y\leq 1-x &\hspace{-10pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{1+\sqrt{5}}{2},2\bigg] ;\\[3pt] &\bullet\ y>-x &\hspace{-10pt} \mathrm{if}\ x\in [1,\sqrt{2}] ;\\[3pt] &\bullet\ y>-\displaystyle\frac{2}{x} &\hspace{-10pt} \mathrm{if}\ x\in [\sqrt2,2].\\[12pt]&\mathcal{R}_{2}{:}&\\ &\bullet\ y\leq -1 & \mathrm{if}\ x\in[0,1];\\[3pt] &\bullet\ y\leq -x & \mathrm{if}\ x\in [1,\sqrt{2}] ;\\[3pt] &\bullet\ y>-1-x & \mathrm{if}\ x\in \displaystyle\bigg[0,\frac{\sqrt{5}-1}{2}\bigg] ;\\[3pt] &\bullet\ y>-\displaystyle\frac{2+x}{1+x} & \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{5}-1}{2},\sqrt{2}\bigg].\end{array}\end{align*} $$

$$ \begin{align*}\begin{array}{rll}&\mathcal{R}_{3}{:}&\\ &\bullet\ y\leq \displaystyle\frac{-2}{x} &\hspace{-36pt} \mathrm{if}\ x\in\displaystyle\bigg[\sqrt{2},\frac{1+\sqrt{33}}{4}\bigg];\\[3pt]&\bullet\ y\leq \displaystyle\frac12-x &\hspace{-36pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{1+\sqrt{33}}{4},2\bigg] ;\\[3pt] &\bullet\ y>-x &\hspace{-36pt} \mathrm{if}\ x\in [\sqrt{2},\sqrt{3}] ;\\[3pt] &\bullet\ y>\displaystyle\frac{-3}{x} &\hspace{-36pt} \mathrm{if}\ x\in [\sqrt{3},2].\\[13pt]&\mathcal{R}_{4}{:}&\\ &\bullet\ y\leq -\displaystyle\frac{3+2x}{2+2x} &\hspace{-23pt} \mathrm{if}\ x\in\displaystyle\bigg[0,\frac{\sqrt{3}-1}{2}\bigg];\\[3pt] &\bullet\ y\leq -1-x &\hspace{-23pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{3}-1}{2},\frac{\sqrt{5}-1}{2}\bigg] ;\\[3pt] &\bullet\ y>-\displaystyle\frac{3+2x}{2+x} &\hspace{-23pt} \mathrm{if}\ x\in \displaystyle\bigg[0,\frac{\sqrt{5}-1}{2}\bigg].\\[13pt]&\mathcal{R}_{5}{:}&\\ &\bullet\ y\leq -\displaystyle\frac{2+x}{1+x} &\hspace{-24pt} \mathrm{if}\ x\in\displaystyle\bigg[\frac{\sqrt{5}-1}{2},\sqrt{2}\bigg];\\[3pt] &\bullet\ y\leq -x &\hspace{-25pt} \mathrm{if}\ x\in [\sqrt{2},\sqrt{3}] ;\\[3pt] &\bullet\ y>-\displaystyle\frac{3+2x}{2+x} &\hspace{-25pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{5}-1}{2},\sqrt{3}\bigg].\\[13pt]&\mathcal{R}_{6}{:}&\\ &\bullet\ y\leq -\displaystyle\frac{3+2x}{2+x} &\hspace{-21pt}\mathrm{if}\ x\in\displaystyle \bigg[-1,\frac{\sqrt{5}-1}{2}\bigg];\\[3pt] &\bullet\ y>-2-x &\hspace{-21pt} \mathrm{if}\ x\in \displaystyle\bigg [-1,\frac{\sqrt{2}-2}{2}\bigg] ;\\[3pt] &\bullet\ y>-\displaystyle\frac{5+3x}{3+2x} &\hspace{-21pt} \mathrm{if}\ x\in\displaystyle \bigg[\frac{\sqrt{2}-2}{2},\displaystyle\frac{\sqrt{5}-1}{2}\bigg].\\[12pt]&\mathcal{R}_{7}{:}&\\ &\bullet\ 1-x<y\leq \displaystyle\frac{2-x}{2+x} & \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{5}+1}{2},2\bigg).\\[13pt]&\mathcal{R}_{8}{:}&\\ &\bullet\ y\leq -\displaystyle\frac{5+3x}{3+2x} &\hspace{-23pt} \mathrm{if}\ x\in\displaystyle \bigg[\frac{\sqrt{2}-2}{2},\displaystyle\frac{\sqrt{5}-1}{2}\bigg];\\[3pt]&\bullet\ y\leq -1-x &\hspace{-23pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{5}-1}{2},\displaystyle\frac{\sqrt{7}-1}{2}\bigg] ;\\[3pt] &\bullet\ y>-\displaystyle\frac{7+4x}{4+2x} &\hspace{-23pt} \mathrm{if}\ x\in\displaystyle \bigg[\frac{\sqrt{2}-2}{2},\displaystyle\frac{\sqrt{7}-1}{2}\bigg].\end{array}\end{align*} $$

$$ \begin{align*} \begin{array}{rll}&\kern-4.7pc\mathcal{R}_{9}{:}&\\&\kern-4.7pc\bullet\ -1<y\leq -x &\mathrm{if}\ x\in (0,1].\end{array}\end{align*} $$

$$ \begin{align*} \begin{array}{rll} &\mathcal{R}_{10}{:}&\\ &\bullet\ y\leq \displaystyle\frac{-3}{x} &\hspace{-46pt} \mathrm{if}\ x\in\displaystyle\bigg[\sqrt{3},\frac{1+\sqrt{193}}{8}\bigg];\\[4pt] &\bullet\ y\leq \displaystyle\frac14-x &\hspace{-46pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{1+\sqrt{193}}{8},2\bigg] ;\\[4pt] &\bullet\ y>-x &\hspace{-46pt}\mathrm{if}\ x\in [\sqrt{3},\sqrt{\tfrac72}] ;\\[4pt] &\bullet\ y>\displaystyle\frac{-7}{2x} &\hspace{-46pt} \mathrm{if}\ x\in [\sqrt{\tfrac72},2].\\[13pt] &\mathcal{R}_{11}{:}&\\ &\bullet\ y\leq -\displaystyle\frac{5+2x}{2+2x} &\hspace{-34pt} \mathrm{if}\ x\in\displaystyle\bigg[\frac{\sqrt{111}-3}{8},\frac{\sqrt{33}-1}{4}\bigg];\\[4pt] &\bullet\ y\leq -\displaystyle\frac{3+2x}{2+x} &\hspace{-34pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{33}-1}{4},\sqrt{3}\bigg];\\[4pt] &\bullet\ y>-\displaystyle\frac{4+3x}{3+x} &\hspace{-34pt} \mathrm{if}\ x\in\displaystyle\bigg[\frac{\sqrt{111}-3}{8},\frac{\sqrt{41}-1}{4}\bigg];\\[4pt] &\bullet\ y>-\displaystyle\frac{3+x}{1+x} &\hspace{-34pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{41}-1}{4},\sqrt{3}\bigg].\\[15pt] &\mathcal{R}_{12}{:}&\\ &\bullet\ y\leq -\displaystyle\frac{4+3x}{3+x} &\hspace{-34pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{5}-1}{2},\frac{\sqrt{19}-1}{3}\bigg];\\[4pt] &\bullet\ y>-1-x &\hspace{-34pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{5}-1}{2},\frac{\sqrt{7}-1}{2}\bigg] ;\\[4pt] &\bullet\ y>-\displaystyle\frac{6+3x}{3+2x} &\hspace{-34pt} \mathrm{if}\ x\in\displaystyle\bigg[\frac{\sqrt{7}-1}{2},\frac{\sqrt{19}-1}{3}\bigg].\\[15pt] &\mathcal{R}_{13}{:}&\\ &\bullet\ y\leq -\displaystyle\frac{3+x}{1+x}&\hspace{-35pt} \mathrm{if}\ x\in\displaystyle\bigg [\frac{\sqrt{41}-1}{4},\sqrt{3}\bigg];\\[4pt] &\bullet\ y\leq -x &\hspace{-35pt} \mathrm{if}\ x\in [\sqrt{3},\sqrt{\tfrac72}] ;\\[4pt] &\bullet\ y>-\displaystyle\frac{7+4x}{4+2x} &\hspace{-35pt} \mathrm{if}\ x\in\displaystyle\bigg [\frac{\sqrt{41}-1}{4},\sqrt{\tfrac72}\bigg].\\[15pt] &\mathcal{R}_{14}{:}&\\ &\bullet\ -\displaystyle\frac{3+2x}{2+x}<y\leq -1 & \mathrm{if}\ x\in [-1,0]. \\[12pt] &\mathcal{R}_{15}{:}&\\ &\bullet\ \displaystyle\frac12-x<y\leq \displaystyle\frac{3-x}{1-x} &\hspace{-10pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{41}+1}{4},2\bigg).\\[12pt] &\mathcal{R}_{16}{:}&\\ &\bullet\ \displaystyle\frac{6-3x}{3-x}<y\leq \frac{7-4x}{4-3x} & \mathrm{if}\ x\in \displaystyle\bigg[\frac{11+\sqrt{61}}{10},2\bigg). \end{array} \end{align*} $$

$$ \begin{align*} \begin{array}{rll} &\mathcal{R}_{17}{:}&\\ &\bullet\ y\leq -\displaystyle\frac{8+5x}{5+3x} &\hspace{-17pt} \mathrm{if}\ x\in\displaystyle\bigg[-1,\frac{\sqrt{2}-3}{3}\bigg];\\[3pt] &\bullet\ y\leq -2-x &\hspace{-17pt} \mathrm{if}\ x\in \displaystyle\bigg[\frac{\sqrt{2}-3}{3},\frac{\sqrt{2}-2}{2}\bigg];\\[3pt] &\bullet\ y>-\displaystyle\frac{6+3x}{3+x} &\hspace{-17pt} \mathrm{if}\ x\in\displaystyle \bigg[-1,\frac{\sqrt{13}-7}{6}\bigg];\\[3pt] &\bullet\ y>-\displaystyle\frac{5+3x}{3+2x} &\hspace{-17pt} \mathrm{if}\ x\in\displaystyle\bigg [\frac{\sqrt{13}-7}{6},\frac{\sqrt{2}-2}{2}\bigg].\\[10pt] &\mathcal{R}_{18}{:}&\\ &\bullet\ y\leq -2-x &\hspace{-17pt} \mathrm{if}\ x\in \displaystyle\bigg[-1,\frac{-1}{2}\bigg];\\[3pt] &\bullet\ y>-\displaystyle\frac{8+5x}{5+3x} &\hspace{-17pt} \mathrm{if}\ x\in \displaystyle\bigg[-1,\frac{\sqrt{21}-11}{10}\bigg];\\[3pt] &\bullet\ y>-\displaystyle\frac{7+5x}{5+4x} &\hspace{-17pt} \mathrm{if}\ x\in\displaystyle\bigg[\frac{\sqrt{21}-11}{10},\frac{-1}{2}\bigg].\\[10pt] &\mathcal{L}_1{:}&\\ &\bullet\ -x<y\le 1 &\hspace{-17pt} \mathrm{if}\ x\in (0,1].\\[10pt] &\mathcal{L}_2{:}&\\ &\bullet\ 0<y\le 2-x &\hspace{-17pt} \mathrm{if}\ x\in (1,2).\\[10pt] &\mathcal{L}_3{:}&\\ &\bullet\ -\displaystyle\frac{1}{x}<y\leq 1-x & \mathrm{if}\ x\in\displaystyle\bigg(1,\frac{1+\sqrt{5}}{2}\bigg].\\[10pt] &\mathcal{L}_4{:}&\\ &\bullet\ y\leq 1 &\hspace{-27pt} \mathrm{if}\ x\in [1,2];\\[3pt] &\bullet\ y>2-x &\hspace{-27pt} \mathrm{if}\ x\in \displaystyle\bigg[1,\frac{3+\sqrt{2}}{3}\bigg];\\[3pt] &\bullet\ y>\displaystyle\frac{6-4x}{4-3x} &\hspace{-27pt} \mathrm{if}\ x\in\displaystyle\bigg[\frac{3+\sqrt{2}}{3},2\bigg].\\[10pt] &\mathcal{L}_5{:}&\\ &\bullet\ y\leq 0 &\hspace{-27pt} \mathrm{if}\ x\in [1,2];\\[3pt] &\bullet\ y>1-x &\hspace{-27pt} \mathrm{if}\ x\in \displaystyle\bigg[1,\frac{1+\sqrt{5}}{2}\bigg];\\[3pt] &\bullet\ y>\displaystyle\frac{2-x}{1-x} &\hspace{-27pt} \mathrm{if}\ x\in\displaystyle\bigg[\frac{1+\sqrt{5}}{2},2\bigg]. \end{array} \end{align*} $$

Remark 7. By examining the beta expansion of 23 899 Salem numbers of degree 6 and trace at most 19, there are 18 250 (approximately $76\%$ ) Salem numbers that satisfy

$$ \begin{align*} - 2 < \alpha_1 < 0 < \alpha_2 < 2. \end{align*} $$

They are Parry numbers with relatively small orbit size $(\max (m,p)<1000)$ .

A heavy computational effort may be required to substantially improve Theorem 2 (or Proposition B.2 in Appendix B. For example, for $ -r<\alpha _1<0<\alpha _2<r<2<\gamma $ with $r>1$ quite close to 1, the orbit size starts taking many different values. For instance, the following polynomials:

$x^6-7x^5-3x^4-11x^3-3x^2-7x+1$ , $x^6-9x^5-x^4-11x^3-x^2-9x+1$ and $x^6-8x^5+10x^4-15x^3+10x^2-8x+1$ satisfy respectively: $\alpha _1\approx -1.08$ , $\alpha _1\approx -1.05$ and $\alpha _2\approx 1.1$ , but $(m,p)$ equals to $(6,23)$ , $(6,35)$ and $(1,119)$ , respectively.

6 Four-dimensional discretized rotation

Substituting the variables of the algorithm (9) in §5 by

$$ \begin{align*} \begin{pmatrix} t_n(1)\cr t_n(2)\cr t_n(3) \cr t_n(4)\end{pmatrix} = \begin{pmatrix} 1& c_1 & c_2 & c_1 \cr 0& 1 & c_1 & c_2\\ 0 & 0& 1 & c_1\\ 0 & 0 & 0 & 1\end{pmatrix} \begin{pmatrix} k_{n-3}\cr k_{n-2}\cr k_{n-1} \cr k_n\end{pmatrix}, \end{align*} $$

we obtain an integer sequence $(k_n)_{n\ge -4}$ which satisfies

(10) $$ \begin{align} 0<k_{n+4}+c_1k_{n+3}+c_2 k_{n+2}+c_1k_{n+1}+k_n\le 1, \end{align} $$

where $c_1=\alpha _1+\alpha _2$ and $c_2=\alpha _1\alpha _2+2$ . Here $\alpha _i\in (-2,2)$ are arbitrary chosen constants. The bijective map T on $\mathbb Z^4$ :

$$ \begin{align*} (k_{n},k_{n+1},k_{n+2},k_{n+3}) \mapsto (k_{n+1},k_{n+2},k_{n+3}, -\lceil c_1 k_{n+3}+c_2 k_{n+2}+c_1 k_{n+1}+k_n -1\rceil ) \end{align*} $$

is conjugate to equation (9). By bijectivity, the orbit is purely periodic if and only if it is eventually periodic. Moreover, the periodicity is equivalent to the boundedness of the orbit. We are interested in the recurrence of the orbit of $(k_{-4},k_{-3},k_{-2},k_{-1}) =(0,0,0,0)$ by T. This map approximates a linear map $\Phi $ defined by

$$ \begin{align*} \begin{pmatrix} x_1\\ x_2\\ x_3 \cr x_4\end{pmatrix} \mapsto \begin{pmatrix} 0& 1 & 0 & 0 \cr 0& 0 & 1 & 0\\ 0 & 0& 0 & 1\\ -1 & -c_1 & -c_2 & -c_1\end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3 \cr x_4\end{pmatrix} \end{align*} $$

for $(x_1,x_2,x_3,x_4)^T\in \mathbb R^4$ . The map $\Phi $ has four eigenvalues $\exp (\pm \theta _i \sqrt {-1})$ with $\alpha _i=-2\cos (\theta _i)$ for $i=1,2$ . Therefore, the map $T:\mathbb Z^4\rightarrow \mathbb Z^4$ is understood as a discretized version of rotation. A simpler case: the discretized rotation in $\mathbb Z^2$ is extensively studied in the literature. It is defined similarly by a recurrence

(11) $$ \begin{align} 0\le a_{n+2}+\unicode{x3bb} a_{n+1}+a_n\ <1,\quad a_n\in \mathbb Z, \end{align} $$

with a fixed $\unicode{x3bb} \in (-2,2)$ . A notorious conjecture states that any sequence produced by this recursion is periodic for any initial vector $(a_0,a_1)\in \mathbb Z^2$ . The validity is known only for $11$ values of $\unicode{x3bb} $ , see [Reference Akiyama, Brunotte, Pethő and Steiner5, Reference Kouptsov, Lowenstein and Vivaldi19, Reference Lowenstein, Hatjispyros and Vivaldi21].

Note that if $\alpha _1=\alpha _2$ , then there is an unbounded real sequence $ (b_n)_{n\in \mathbb N}$ which satisfies

$$ \begin{align*} b_{n+4}+c_1b_{n+3}+c_2 b_{n+2}+c_1b_{n+1}+b_n=0 \end{align*} $$

because of the shape of general terms of this recurrence. In particular, if $\alpha _1=\alpha _2\in \{-1,0,1\}$ , then $\{T^n(0,0,0,0)|\ n\in \mathbb N\}$ is unbounded. Thus we can not expect the boundedness of the orbits of T when $\alpha _1=\alpha _2$ . (However, there are points with $\alpha _1=\alpha _2$ where the T-orbits of $(0,0,0,0)$ are periodic, e.g., $(\alpha ,\alpha )$ with $\alpha \in (1-\sqrt {3},-2/3]\cup [-1/4,-2/9]$ .) Excluding these cases, a natural generalization of the above conjecture for equation (11) would be the following conjecture.

This is because $c_2>2|c_1|-2$ and $c_2-2<{c_1^2}/4<4$ is equivalent to $\alpha _i\in (-2,2)$ for $i=1,2$ and $\alpha _1\neq \alpha _2$ .

We are pessimistic about its validity, due to the existence of very large orbits. However, even if Conjecture 8 may not hold, it could be true for almost all cases.

Let us restrict ourselves to the orbit of the origin. Since periodic orbits are often dominant in zero entropy systems, unbounded orbits may not give a contribution of positive measure in §5 and period cells would exhaust the total square $(-2,2)^2$ in measure. We propose a weaker conjecture.

Here our measure is equivalent to the two-dimensional Lebesgue measure. See Figure 2 for period cells. Black dots are the points where the orbit of the origin might be unbounded.

Figure 2

Period cells occupy more than half in measure. The orbit of the origin does not form a period until 50 000 iterations at the black dots.

Note that once we find a period of T starting from the origin, Lemmas 3 and 4 give us a (possibly degenerated) period cell. This fact is clear when $g_i$ does not visit $1$ and we can apply Lemma 3 with

$$ \begin{align*} u=\tfrac 12 \min_{i} g_i, \quad v=\tfrac 12(1+\max_i g_i). \end{align*} $$

For Lemma 4, it looks like we have additional constraints. We shall show that this is not the case. Taking the period $p\in \mathbb N$ with $T^p((0,0,0,0))=(0,0,0,0)=(k_{-4},k_{-3},k_{-2},k_{-1})$ , we have

$$ \begin{align*} g_n=t_{n-1}(1)+k_n=k_{n-4}+c_1 k_{n-3} + c_2 k_{n-2} + c_1k_{n-1}+ k_{n} \end{align*} $$

for $n=0,\ldots ,p$ and

$$ \begin{align*} k_0=1,\quad k_1=-\lceil c_1-1 \rceil. \end{align*} $$

If $g_i=1$ occurs as a polynomial of $\mathbb Z[\alpha _1,\alpha _2]$ with $1\le i \le p-1$ , then $k_{i-2}=k_{i-3}$ $+\,k_{i-1}=0$ and $k_{i-4}+k_i=1$ . From $g_{i\pm 1}\in (0,1]$ , we have

$$ \begin{align*} 0<g_{i-1}=k_{i-5}+c_1 k_{i-4} + (c_2-1) k_{i-3}\le 1 \end{align*} $$

and

$$ \begin{align*} 0<g_{i+1}=(1-c_2)k_{i-3} + c_1(1-k_{i-4})+ k_{i+1}\le 1. \end{align*} $$

These imply

$$ \begin{align*} k_{i-5}=-\lceil c_1 k_{i-4} + (c_2-1) k_{i-3}-1 \rceil \end{align*} $$

and

$$ \begin{align*} k_{i+1}=-\lceil (1-c_2)k_{i-3} + c_1(1-k_{i-4}) -1 \rceil. \end{align*} $$

We see

$$ \begin{align*} k_1\le k_{i-5}+k_{i+1} \end{align*} $$

from an easy fact

$$ \begin{align*} \lceil x \rceil + \lceil y \rceil - \lceil x+y+1 \rceil \in \{-1,0\} \end{align*} $$

for any $x,y\in \mathbb R$ . Therefore, when $g_i=1$ , we have

$$ \begin{align*} g_{i-1}+g_{i+1}=k_{i-5} + c_1+ k_{i+1} \ge c_1+k_1=g_1. \end{align*} $$

Since $g_1=g_{i+1}+g_{i-1}$ implies $g_1>g_{i+1}$ , we can apply Lemma 4 in any case with

$$ \begin{align*} u=\tfrac 12 \min\bigg(\min_i g_i, \min_{\substack{g_i=1\\ g_1=g_{i-1}+g_{i+1}}} (g_1-g_{i+1})\bigg), \quad v=\tfrac 12\Big(1+\max_{g_i\neq 1} g_i\Big). \end{align*} $$

Therefore, we can apply Lemma 4 for every period starting from the origin. Since we expect such period cells to cover $(-2,2)^2$ in measure, as in Conjecture 9, for any $\varepsilon>0$ , we will find a finite union of period cells whose measure is not less than $1-\varepsilon $ . Following the same proof as Theorem 2, we arrive at a plausible

For general Salem numbers $\beta $ , we numerically observe many $(m,p)$ -periodic $d_{\beta }(1)$ with $m>1$ . Interestingly, we find no role of $(m,p)$ -periods with $m>1$ in the above discussion. Every period of T gives rise to $(1,p)$ -periodic $d_{\beta ^m}(1)$ and other orbits of T are aperiodic if they exist. Of course this does not cause any contradiction, since we are studying sufficiently large $\beta $ with respect to the location of the conjugates.