Combinatorial structure and sumsets associated with Beatty sequences generated by powers of the golden ratio

1.   Introduction

Let $A$ and $B$ be nonempty subsets of $\mathbb Z$, $h\in \mathbb N$, and $x\in \mathbb Z$. The sumset $A+B$, the $h$-fold sumset $hA$, the translation $x+A$, and the dilation $x\ast A$ are defined by

The central problem in additive number theory is to determine whether a given subset $A$ of $\mathbb N$ (or $\mathbb N_0$) is an additive basis or an asymptotic additive basis of finite order, and if it is, then it is desirable to explicitly obtain positive integers $h$ and $N$ such that the $h$-fold sumset $hA$ contains all positive integers, or every positive integer larger than $N$. For additional details and references on sumsets and additive number theory, we refer the reader to the books by Freiman ^[1], Halberstam and Roth ^[2], Nathanson ^[3], Tao and Vu ^[4], and Vaughan ^[5].

Let $\alpha$ be the golden ratio, $B(\alpha)$ and $B(\alpha^2)$ the lower and upper Wythoff sequences, respectively, and in general, let $B(x)$ be the Beatty sequence (or Beatty set) generated by $x$. Previously, Pongsriiam and his coauthors ^[6] obtained various results on sumsets associated with $B(\alpha)$ and $B(\alpha^2)$ and showed that $2B(\alpha)$ and $3B(\alpha^2)$ are cofinite while $B(\alpha)$ and $2B(\alpha^2)$ are not. Using automata theory and a computer program called Walnut, Shallit ^[7,8] gave a different proof of this result and also calculated the Frobenius numbers for some classical automatic sequences. Dekking ^[9] introduced a different technique to compute the sumsets $2B(\alpha)$ and $3B(\alpha^2)$ by using combinatorics on words and the theory of two dimensional substitution; see also ^[10]. Napp Phunphayap, Pongsriiam, and Shallit ^[11] studied sumsets associated with $B(x)$ where $x$ is any irrational number in the interval $(1, 3)$. For more information on related research, we refer the reader to Fraenkel ^[12,13,14], Kawsumarng et al. ^[15], Kimberling ^[16,17], Pitman ^[18], Zhou ^[19], and in the online encyclopedia OEIS ^[20].

One of our motivations comes from the recent results on palindromes as an additive basis: Banks ^[21] showed that every positive integer can be written as the sum of at most 49 palindromes in base $10$; Cilleruelo, Luca, and Baxter ^[22] improved it by showing that if $b \geq 5$ is fixed, then every positive integer is the sum of at most three $b$-adic palindromes; Rajasekaran, Shallit, and Smith ^[23] completed the study by proving that the theorem of Cilleruelo, Luca, and Baxter ^[22] also holds when $b\in \{3, 4\}$, and if $b = 2$, then we need four summands to write every positive integer as a sum of $b$-adic palindromes. Comparing these complete results on palindromes ^[21,22,23] and those satisfactory but incomplete answers on the classical bases such as primes or powers of nonnegative integers in Goldbach's or Waring's problems, we are led to an idea of studying a new arithmetic or combinatorial sequence as an additive basis like they did for palindromes in ^[21,22,23].

Clearly, arithmetic progressions are bad for sumsets because the calculation is too easy and many of them cannot be a basis. For instance, the set of all positive integers that are congruent to $a$ modulo $m$ where $m$ and $\gcd(a, m)$ are larger than $1$ is not a basis. Being a generalization of arithmetic progressions, Beatty sequences $B(x)$ are nearly periodic but not really periodic when their generator $x$ is an irrational number. So it is interesting to replace arithmetic progressions by Beatty sequences and see what happen.

It seems that there are some connections between the sumsets of $B(x)$ and the members of a particular linear recurrence relation whose characteristic polynomial has $x$ as one of its root; see, for example, in Theorems 3.5, 3.16, 3.17, Remark 3.19, and open questions in the article by Kawsumarng et al. ^[15]. In particular, if $m\geq 3$ is an integer and $h = h(m)$ is the smallest positive integer such that $hB(\alpha^m)$ is cofinite, then it seems that there are infinitely many Fibonacci numbers that do not belong to $(h-1)B(\alpha^m)$. While many combinatorial properties of lower and upper Wythoff sequences have been extensively studied, there are only a few arithmetic results concerning sumsets associated with Beatty sequences, a generalization of Wythoff sequences. These motivate us to investigate more on this problem.

In this article, we continue the investigation on $B(\alpha^m)$ for $m\geq 3$. In particular, we provide some combinatorial structures of $B(\alpha^m)$ in Theorems 3.4 and 3.5 and use them to calculate, for each $m\geq 3$, a positive integer $h = h(m)$ such that $hB(\alpha^m)$ is cofinite. In addition, we explicitly give in Theorems 4.3 and 4.4 a positive integer $N = N(m)$, which is best possible when $m$ is small, such that $hB(\alpha^m)$ contains every integer that is larger than or equal to $N$. For example, we obtain that $12B(\alpha^5)$ contains every integer larger than or equal to $2684$ and that $2684$ is the smallest integer having this property. By some numerical evidence as shown in Remark 2, Corollary 1, and Question 1, we believe that our choices of the integers $h$ and $N$ are actually the smallest possible for all $m\in \mathbb N$. Nevertheless, the proof seems very long and difficult, and so we postpone this for future research.

We organize this article as follows. In Section 2, we recall some definitions and useful results for the reader's convenience. In Section 3, we give some combinatorial structures of $B(\alpha^m)$, and then in Section 4, we use them to obtain the desired results on sumsets.

2.   Preliminaries and lemmas

We first introduce the notation which will be used throughout this article as follows: $\alpha = (1+\sqrt 5)/2$ is the golden ratio, $\beta = (1-\sqrt 5)/2$, and if $x\in \mathbb R$, then $\lfloor x\rfloor$ is the largest integer less than or equal to $x$, $\{ x\} = x-\lfloor x\rfloor$, $\left\lceil x\right\rceil$ is the smallest integer larger than or equal to $x$, and with a little abuse of notation

where we consider $B(x)$ as a sequence $(\left\lfloor nx\right\rfloor)_{n\geq 1}$ when we show its combinatorial structure, and we treat $B(x)$ as a set when we give a result on sumsets. In addition, for $n\geq 0$, we write $F_n$ and $L_n$ to denote the $n$th Fibonacci and Lucas numbers, which are defined by the same recursive pattern $a_n = a_{n-1}+a_{n-2}$ for $n\geq 2$ but with different initial values $F_0 = 0$, $F_1 = 1$, $L_0 = 2$, and $L_1 = 1$. We can also extend them to negative indices by the formula $F_{-n} = (-1)^{n+1}F_n$ and $L_{-n} = (-1)^nL_n$. Furthermore, if $P$ is a mathematical statement, then the Iverson notation $[P]$ is defined by

We often use the following fact: $-1 < \beta < 0$, $(|\beta^n|)_{n\geq 1}$ is strictly decreasing, if $a_1 > a_2 > \cdots > a_r$ are even positive integers, then $0 < \beta^{a_1} < \beta^{a_2} < \cdots < \beta^{a_r}$, and if $b_1 > b_2 > \cdots > b_r$ are odd positive integers, then $0 > \beta^{b_1} > \beta^{b_2} > \cdots > \beta^{b_r}$. In addition, $\alpha$ and $\beta$ are roots of the equation $x^2-x-1 = 0$. So, for instance, $\alpha\beta = -1$, $\beta^2 = \beta+1$, and $\beta^2+\beta^4 = 4\beta+3$. Finally, we remark that we apply Lemmas 2.1 to 2.3 throughout this article sometimes without reference.

Lemma 2.1. For $n\in\mathbb{Z}$ and $x, y\in\mathbb{R}$, the following statements hold.

$\text{(i)}$ $\lfloor n+x\rfloor = n+\lfloor x\rfloor$ and $\left\lceil n+x\right\rceil = n+\left\lceil x\right\rceil$.

$\text{(ii)}$ $\{n+x\} = \{x\}$.

$\text{(iii)}$ $0\leq \{x\} < 1$ and if $x$ is not an integer, then $\{-x\} = 1-\{x\}$.

$\text{(iv)}$ $\lfloor x+y\rfloor = \begin{cases}\lfloor x\rfloor +\lfloor y\rfloor, & \mathit{\text{if}}~~ \{x\}+\{y\} < 1;\\ \lfloor x\rfloor +\lfloor y\rfloor+1, & \mathit{\text{if}}~~ \{x\}+\{y\}\geq 1.\end{cases}$

$\text{(v)}$ $\lfloor (n+1)x\rfloor - \lfloor nx\rfloor = \lfloor x\rfloor$ or $\lfloor x\rfloor+1$.

Proof. These are well known and can be proved easily. For more details, see for instance in [24,Chapter 3].

Lemma 2.2. The following statements hold for all nonnegative integers $m$ and $n$.

$\text{(i)}$ (Binet's formula) $F_n = \frac{\alpha^n-\beta^n}{\alpha-\beta}$ and $L_n = \alpha^n+\beta^n$.

$\text{(ii)}$ $\alpha^n = \alpha F_n+F_{n-1}$ and $\beta^{n} = \beta F_{n}+F_{n-1}$.

$\text{(iii)}$ $L_{n} = F_{n-1}+F_{n+1}$.

Proof. These are well known, not difficult to prove, and can be found on pages 78–80 in ^[25].

By Lemma 2.5 of ^[6], we know the formulas for $\left\lfloor F_n\alpha\right\rfloor$, $\left\lfloor F_n\alpha^2\right\rfloor$, $\{F_n\alpha\}$, and $\{F_n\alpha^2\}$. In this article, we extend those to the following lemma.

Lemma 2.3. Let $k$ and $m$ be positive integers. Then the following statements hold.

${\rm(i)}$ If $k\geq m$, then $\left\lfloor \beta^kF_m\right\rfloor = -[k\equiv 1\pmod 2]$.

${\rm(ii)}$ If $k < m$, then $\left\lfloor \beta^kF_m\right\rfloor = (-1)^kF_{m-k}-[m\equiv 1\pmod 2]$.

${\rm(iii)}$ If $k\geq m$, then $\left\lfloor F_k\alpha^m\right\rfloor = F_{k+m}-[k\equiv 0\pmod 2]$.

${\rm(iv)}$ If $k < m$, then $\left\lfloor F_k\alpha^m\right\rfloor = F_{k+m}+(-1)^{k+1}F_{m-k}-[m\equiv 0\pmod 2]$.

${\rm(v)}$ If $k\geq m$, then $\left\{F_k\alpha^m\right\} = \left\{-\beta^kF_m\right\} = [k\equiv 0\pmod 2]-\beta^kF_m$.

${\rm(vi)}$ If $k \geq m$, then $\left\{\beta^kF_m\right\} = [k\equiv 1\pmod 2]+\beta^kF_m$.

For ${\rm(vii)}$ and ${\rm(viii)}$, let $k < m$ and $g(m, k) = \beta^{m-k}((-1)^k-\beta^{2k})/\sqrt5$. Then

${\rm(vii)}$ $\left\{F_k\alpha^m\right\} = \left\{-\beta^kF_m\right\} = \{-g(m, k)\} = -g(m, k)+[m\equiv 0\pmod 2]$,

${\rm(viii)}$ $\left\{\beta^kF_m\right\} = \left\{g(m, k)\right\} = g(m, k)+[m\equiv 1\pmod 2]$.

Proof. By Binet's formula and the fact that $\alpha\beta = -1$, we see that

Case 1 $k\geq m$. Then $|\beta^kF_m|\leq |\beta^mF_m| < 1$. So if $k$ is even, then $0 < \beta^kF_m < 1$; if $k$ is odd, then $-1 < \beta^kF_m < 0$. Therefore $\left\lfloor \beta^kF_m\right\rfloor = -[k\equiv 1\pmod 2]$, $\left\lfloor -\beta^kF_m\right\rfloor = -[k\equiv 0\pmod 2]$, and (2.1) implies that

Therefore (i) and (iii) are proved.

Case 2 $k < m$. Let $\ell = m-k$ and $g(m, k) = \beta^{m-k}((-1)^k-\beta^{2k})/\sqrt5$. Then $m = k+\ell$ and

For convenience, let $A = g(m, k)$ throughout the remaining proof. Then

Therefore, if $k$ and $\ell$ are even, then $A = \beta^\ell(1-\beta^{2k})/\sqrt5\in (0, 1)$; if $k$ is even and $\ell$ is odd, then $A\in (-1, 0)$; if $k$ and $\ell$ are odd, then $A = -\beta^\ell(1+\beta^{2k})/\sqrt5\in (0, 1)$; if $k$ is odd and $\ell$ is even, then $A\in (-1, 0)$. These imply that

From (2.3) and (2.4), we obtain

Since $\ell = m-k$, we see that (2.5) implies (ii). In addition, we obtain (iv) from (2.1) and (2.6) as

Next, we use some of the above calculation to prove (v) to (viii). We obtain from (2.1) that $\{F_k\alpha^m\} = \{-\beta^kF_m\}$, and the analysis of $\beta^kF_m$ is already done. Suppose $k\geq m$. Then $|\beta^kF_m| < 1$. So if $k$ is even, then $\{\beta^kF_m\} = \beta^kF_m$ and $\{-\beta^kF_m\} = 1-\beta^kF_m$; if $k$ is odd, then $\{\beta^kF_m\} = 1+\beta^kF_m$ and $\{-\beta^kF_m\} = -\beta^kF_m$. These imply (v) and (vi). Next, let $k < m$. From (2.3), we know that $\{\beta^kF_m\} = \{A\}$, $\{-\beta^kF_m\} = \{-A\}$, and the analysis of $A$ is already done. We have

These imply (vii) and (viii). So the proof is complete.

Lemma 2.4. Let $m$ and $n$ be positive integers. Then the following statements hold.

$\text{(i)}$ $\left\lfloor \alpha^m\right\rfloor = L_m-[m\equiv 0\pmod 2]$.

$\text{(ii)}$ $\{\alpha^m\} = -\beta^m+[m\equiv 0\pmod 2]$.

$\text{(iii)}$ If $m$ is odd and $n < \alpha^m$, then $n\{\alpha^m\} = \{n\alpha^m\}$ and $\left\lfloor n\{\alpha^m\}\right\rfloor = 0$.

$\text{(iv)}$ If $m$ is even and $n < \alpha^m$, then $\{n\alpha^m\} = 1-n\beta^m$ and $\left\lfloor n\{\alpha^m\}\right\rfloor \leq \left\lfloor \alpha^m\right\rfloor - 1$.

$\text{(v)}$ If $m$ is odd, then $\left\lfloor \alpha^m\right\rfloor\{\alpha^m\} = 1-\beta^{2m}$, $\alpha^m\{\alpha^m\} = 1$, and $\left\lceil \alpha^m\right\rceil\{\alpha^m\} = 1-\beta^{2m}-\beta^m\in (1, 2)$.

$\text{(vi)}$ If $m$ is even and $n \in \left\{\left\lfloor \alpha^m\right\rfloor, \left\lceil \alpha^m\right\rceil\right\}$, then $\left\lfloor n\{\alpha^m\}\right\rfloor = \left\lfloor \alpha^m\right\rfloor-1$.

$\text{(vii)}$ If $m$ is even, then $\left\lceil \alpha^m\right\rceil\beta^m = 1+\beta^{2m}$ and $\left\lfloor \alpha^m\right\rfloor\beta^m = 1-\beta^m+\beta^{2m}$.

Proof. For (i), we obtain by Lemmas 2.1 and 2.2 that

Substituting $L_m = \alpha^m+\beta^m = \left\lfloor \alpha^m\right\rfloor+\{\alpha^m\}+\beta^m$ in (i) leads to (ii). For (iii), suppose $m$ is odd and $n < \alpha^m$. Then $\{n\alpha^m\} = \{n\left\lfloor \alpha^m\right\rfloor+n\{\alpha^m\}\} = \{n\{\alpha^m\}\}$. In addition, $0 < n\{\alpha^m\} = -n\beta^m < -\alpha^m\beta^m = 1$, and so $\{n\{\alpha^m\}\} = n\{\alpha^m\}$ and $\left\lfloor n\{\alpha^m\}\right\rfloor = 0$. This proves (iii). For (iv), suppose $m$ is even and $n < \alpha^m$. Then similar to (iii), we have

Since $0 < n\beta^m < \alpha^m\beta^m = 1$, we obtain $\{n\alpha^m\} = 1-n\beta^m$. In addition, $n\{\alpha^m\} < \alpha^m(1-\beta^m) = \alpha^m-1$, which implies $\left\lfloor n\{\alpha^m\}\right\rfloor \leq \left\lfloor \alpha^m\right\rfloor-1$. For (v), suppose $m$ is odd. Then $\left\lfloor \alpha^m\right\rfloor\{\alpha^m\} = L_m\{\alpha^m\} = (\alpha^m+\beta^m)(-\beta^m) = 1-\beta^{2m}$, $\alpha^m\{\alpha^m\} = -\alpha^m\beta^m = 1$, and

In addition, we have $0 > \beta^m+\beta^{2m} > \beta^m \geq \beta > -1$. Therefore $1-\beta^{2m}-\beta^m$ lies in the interval $(1, 2)$, as required. For (vi), suppose $m$ is even and $n = \left\lfloor \alpha^m\right\rfloor$ or $\left\lceil \alpha^m\right\rceil$. If $n = \left\lfloor \alpha^m\right\rfloor$, then

which implies $\left\lfloor n\{\alpha^m\}\right\rfloor = \left\lfloor \alpha^m\right\rfloor-1$, by (iv). Next, suppose $n = \left\lceil \alpha^m\right\rceil$. Then by (i) and (ii), we obtain $n = L_m = \alpha^m+\beta^m$, $\{\alpha^m\} = 1-\beta^m$, and so $n\{\alpha^m\} = L_m-1-\beta^{2m}$. Therefore $\left\lfloor n\{\alpha^m\}\right\rfloor = L_m-2 = \left\lfloor \alpha^m\right\rfloor-1$. This proves (vi). For (vii), we have $(\left\lfloor \alpha^m\right\rfloor+1)\beta^m = L_m\beta^m = (\alpha^m+\beta^m)\beta^m = 1+\beta^{2m}$, which implies the first part of (vii). Subtracting both sides of the above equation by $\beta^m$, we obtain the second part. This completes the proof.

3.   Combinatorial structure of $B(\alpha^m)$

Let $A = (a_n)_{n\geq 1}$ be a sequence of real numbers. We say that $C$ is a segment of $A$ if $C$ is a finite sequence of consecutive terms of $A$, that is, $C = (a_{k}, a_{k+1}, \ldots, a_{k+m})$ for some $k, m \in \mathbb N$. In this case, the length of $C$ is $m+1$, and if $a_k = a_{k+1} = \cdots = a_{k+m}$, then we call $C$ a constant segment. We often refer to the following segments:

where $a_i = \left\lfloor \alpha^m\right\rfloor$ and $b_i = \left\lceil \alpha^m\right\rceil$ for all $i$. In addition, we define ${\rm{Diff}}(A)$ to be the sequence of the difference between consecutive elements of $A$, that is,

In particular, ${\rm{Diff}}(B(x)) = \left(\left\lfloor (n+1)x\right\rfloor-\left\lfloor nx\right\rfloor\right)_{n\geq 1}$. Our purpose in this section is to give a structure of ${\rm{Diff}}(B(\alpha^m))$ in terms of its segments. We begin with the following lemma.

Lemma 3.1. Let $m\geq 3$ and $n\geq 1$ be integers. Then the following statements hold.

$\text{(i)}$ If $m$ is odd and $\left\lfloor (n+1)\alpha^m\right\rfloor-\left\lfloor n\alpha^m\right\rfloor = \left\lceil \alpha^m\right\rceil$, then

$\text{(ii)}$ If $m$ is even and $\left\lfloor (n+1)\alpha^m\right\rfloor-\left\lfloor n\alpha^m\right\rfloor = \left\lfloor \alpha^m\right\rfloor$, then

Proof. For convenience, let $\ell = \left\lfloor \alpha^m\right\rfloor$ and $u = \left\lceil \alpha^m\right\rceil$. By Lemma 2.1, the difference between consecutive terms of $B(\alpha^m)$ is either $\ell$ or $u$. For (i), suppose $m$ is odd and $\left\lfloor (n+1)\alpha^m\right\rfloor-\left\lfloor n\alpha^m\right\rfloor = u$ but (3.5) does not hold. Then

On the other hand, we obtain by Lemma 2.1 that

Writing $\alpha^m = \left(\ell-2\right)+2+\{\alpha^m\}$ and applying Lemmas 2.1 and 2.4, we see that the right-hand side of (3.7) is

which contradicts (3.6). Hence (i) holds. Similarly, suppose (ii) does not hold. Since $m$ is even, we obtain by Lemma 2.4 that $\left\lfloor \ell\{\alpha^m\}\right\rfloor = \ell-1$. Similar to (3.6) and (3.7), we obtain

which implies $\left\lfloor \ell\{\alpha^m\}\right\rfloor \leq \ell-2$, a contradiction. Hence the proof is complete.

Lemma 3.2. Let $m\geq 3$ be an integer. Then the following statements hold.

$\text{(i)}$ If $m$ is odd, then the list of the first $\left\lfloor \alpha^m\right\rfloor$ elements of ${\rm{Diff}}(B(\alpha^m))$ is the segment $S$ given in $(3.1)$.

$\text{(ii)}$ If $m$ is even, then the list of the first $\left\lfloor \alpha^m\right\rfloor$ elements of ${\rm{Diff}}(B(\alpha^m))$ is the segment $T_0$ given in $(3.4)$.

Proof. For convenience, let $\ell = \left\lfloor \alpha^m\right\rfloor$. We first consider the case that $m$ is odd. To prove (i), it is enough to show that $\left\lfloor (n+1)\alpha^m\right\rfloor-\left\lfloor n\alpha^m\right\rfloor = \left\lfloor \alpha^m\right\rfloor$ for each $n = 1, 2, \ldots, \ell-1$ and that $\left\lfloor (\ell+1)\alpha^m\right\rfloor-\left\lfloor \ell\alpha^m\right\rfloor = \left\lceil \alpha^m\right\rceil$. So suppose that $1\leq n\leq \ell-1$. Then $n+1 < \alpha^m$ and we obtain by Lemma 2.4 that

By Lemma 2.1, we obtain $\left\lfloor (n+1)\alpha^m\right\rfloor - \left\lfloor n\alpha^m\right\rfloor = \left\lfloor \alpha^m\right\rfloor$. Next, suppose $n = \ell$. Again, by Lemma 2.4, we have $\{n\alpha^m\} = n\{\alpha^m\}$ and so

Thus $\left\lfloor (n+1)\alpha^m\right\rfloor-\left\lfloor n\alpha^m\right\rfloor = \left\lceil \alpha^m\right\rceil$. This proves (i). For (ii), assume that $m$ is even. Similar to (i), if $1\leq n\leq \ell-1$, then we obtain by Lemma 2.4 that

and thus $\left\lfloor (n+1)\alpha^m\right\rfloor-\left\lfloor n\alpha^m\right\rfloor = \left\lceil \alpha^m\right\rceil$. Next, if $n = \ell$, then

implying $\left\lfloor (n+1)\alpha^m\right\rfloor - \left\lfloor n\alpha^m\right\rfloor = \left\lfloor \alpha^m\right\rfloor$, as desired. This completes the proof.

Lemma 3.3. Let $m\geq 3$ be an integer. Then the following statements hold.

$\text{(i)}$ If $m$ is odd and ${\rm{Diff}}(B(\alpha^m))$ contains the constant segment $(\left\lfloor \alpha^m\right\rfloor$, $\left\lfloor \alpha^m\right\rfloor$, $\ldots$, $\left\lfloor \alpha^m\right\rfloor)$ of length $k$, then $k\leq \left\lfloor \alpha^m\right\rfloor$.

$\text{(ii)}$ If $m$ is even and ${\rm{Diff}}(B(\alpha^m))$ contains the constant segment $(\left\lceil \alpha^m\right\rceil$, $\left\lceil \alpha^m\right\rceil$, $\ldots$, $\left\lceil \alpha^m\right\rceil)$ of length $k$, then $k\leq \left\lfloor \alpha^m\right\rfloor$.

Remark 1. By Theorem 3.4 to be proved later, we see that the inequality $k\leq \left\lfloor \alpha^m\right\rfloor$ in Lemma 3.3 is sharp in the sense that there exists such a constant segment of length $\left\lfloor \alpha^m\right\rfloor$ in ${\rm{Diff}}(B(\alpha^m))$.

Proof of Lemma 3.3. For (i), suppose for a contradiction that $m$ is odd but the sequence ${\rm{Diff}}(B(\alpha^m))$ contains a constant segment $(\left\lfloor \alpha^m\right\rfloor, \left\lfloor \alpha^m\right\rfloor, \ldots, \left\lfloor \alpha^m\right\rfloor)$ of length $k$ with $k > \left\lfloor \alpha^m\right\rfloor$. By considering a shorter segment (if necessary), we can choose $k = \left\lceil \alpha^m\right\rceil$. This implies that $B(\alpha^m)$ contains a finite arithmetic progression

where $a = \left\lfloor n\alpha^m\right\rfloor$ for some $n\in \mathbb N$, $d = \left\lfloor \alpha^m\right\rfloor$, $a+d = \left\lfloor (n+1)\alpha^m\right\rfloor$, $\ldots$, $a+kd = \left\lfloor (n+k)\alpha^m\right\rfloor$. Therefore $kd = a+kd-a$, which is equal to

where the last equality is obtained by using Lemma 2.4. This is a contradiction. So (i) is proved. The proof of (ii) is similar, so we omit some details. If (ii) is not true, we would obtain an arithmetic progression $a, a+d, \ldots, a+kd$, where $k = \left\lceil \alpha^m\right\rceil$, $a = \left\lfloor n\alpha^m\right\rfloor$ for some $n\in \mathbb N$, $d = \left\lceil \alpha^m\right\rceil$, and $a+kd = \left\lfloor (n+k)\alpha^m\right\rfloor$, and therefore

which is a contradiction. So (ii) is verified and the proof is complete.

Before proceeding further, we need to define a concept that is similar to a concatenation of words as in combinatorics. Suppose $A = (a_{k+1}, a_{k+2}, \ldots, a_{k+m})$ and $B = (b_{\ell+1}, b_{\ell+2}, \ldots, b_{\ell+r})$ are finite sequences. Then we define the concatenation of $A$ and $B$, and the $n$ copies of $A$ by

For example, $(1, 2, 3)^{(2)} = (1, 2, 3, 1, 2, 3)$, and $((-1)^n)_{1\leq n\leq 10} = (-1, 1)^{(5)}$. We now give a structure of ${\rm{Diff}}(B(\alpha^m))$ in terms the segments given in (3.1) to (3.4). We first deal with the case that $m$ is odd.

Theorem 3.4. Let $m\geq 3$ be an odd integer. Then the first $\left\lfloor \alpha^m\right\rfloor^2+\left\lceil \alpha^m\right\rceil$ elements of ${\rm{Diff}}(B(\alpha^m))$ can be written as

where $S$ and $S_0$ are the segments given in $(3.1)$ and $(3.2)$, and there are exactly $\left\lfloor \alpha^m\right\rfloor$ of $S$ appearing before $S_0$ in $(3.10)$.

Proof. By Lemma 3.2, the first $\left\lfloor \alpha^m\right\rfloor$ elements of ${\rm{Diff}}(B(\alpha^m))$ is indeed the segment $S$. To prove (3.10), it is enough to show that if we write ${\rm{Diff}}(B(\alpha^m))$ as

then $S$ follows $S^{(b)}$ if $1\leq b < \left\lfloor \alpha^m\right\rfloor$, and $S_0$ follows $S^{(b)}$ if $b = \left\lfloor \alpha^m\right\rfloor$, where $S^{(b)}$ is the $b$ copies of $S$ as defined in (3.9) and written in (3.11). So suppose that (3.11) holds with $b < \left\lfloor \alpha^m\right\rfloor$. Since the last element of $S$ is $\left\lceil \alpha^m\right\rceil$, we obtain by Lemma 3.1 that $S$ is followed by the constant segment $(\ell, \ell, \ldots, \ell)$ of length $\ell-1$, where $\ell = \left\lfloor \alpha^m\right\rfloor$. Therefore (3.11) implies that

where $x = \left\lfloor \alpha^m\right\rfloor$ or $\left\lceil \alpha^m\right\rceil$, and we need to show that $x = \left\lceil \alpha^m\right\rceil$ so that the segment $(\ell, \ell, \ldots, \ell, x)$ in (3.12) is indeed $S$.

Before proceeding further, let us explain the idea to be used in the proof. By the definition of ${\rm{Diff}}(B(\alpha^m))$, the segment such as $S$ implies that there exists a sequence

where $a = \left\lfloor n\alpha^m\right\rfloor$ for some $n\in \mathbb N$, $d = \left\lfloor \alpha^m\right\rfloor = k$, $a+jd = \left\lfloor (n+j)\alpha^m\right\rfloor$ for $j = 1, 2, \ldots, k-1$, and $a+kd+1 = \left\lfloor (n+k)\alpha^m\right\rfloor$. If $S$ appears as the first segment, then we can choose $n = 1$; but the above argument can be used whenever $S$ appears. Although $d = k$, we think of $d$ as the difference and $k$ as the number of terms.

Repeatedly applying the above argument to (3.12), we see that there exists a sequence

where $a = \left\lfloor \alpha^m\right\rfloor = d = k = \ell$, $a+d = \left\lfloor 2\alpha^m\right\rfloor$, $a+2d = \left\lfloor 3\alpha^m\right\rfloor$, $\ldots$, $a+bkd+b = \left\lfloor (bk+1)\alpha^m\right\rfloor$, $\ldots$, $a+bkd+b+(\ell-1)\ell+x = \left\lfloor (bk+\ell+1)\alpha^m\right\rfloor$. Subtracting the last element in (3.13) by the first element in (3.13) and substituting $d = k$ and $\ell = k$, we see that

Writing $\alpha^m = k+\{\alpha^m\}$ and letting $z = \left\lfloor (bk+k+1)\{\alpha^m\}\right\rfloor$, we see that the right-hand side of (3.14) is $(bk+k+1)k+z-k$, which implies that $b-k+x = z$. To calculate $z$, we recall from Lemma 2.4 that $\{\alpha^m\} = -\beta^m$ and $k\{\alpha^m\} = 1-\beta^{2m}$. Therefore $z = \left\lfloor b+1-(\beta^m+\beta^{2m}+b\beta^{2m})\right\rfloor$. Since $b\leq \left\lfloor \alpha^m\right\rfloor-1 < \alpha^m-1$, we see that

Therefore $z\geq b+1$. Since $b-k+x = z$, we obtain $x\geq k+1 = \left\lceil \alpha^m\right\rceil$. Since $x = \left\lfloor \alpha^m\right\rfloor$ or $\left\lceil \alpha^m\right\rceil$, we conclude that $x = \left\lceil \alpha^m\right\rceil$, as required.

Next, suppose that (3.11) holds with $b = \left\lfloor \alpha^m\right\rfloor$. Similar to the previous case, we obtain by Lemma 3.1 that $S$ is followed by the constant segment $(\ell, \ell, \ldots, \ell)$ of length $\ell-1$, and therefore (3.11) implies

where $\ell = \left\lfloor \alpha^m\right\rfloor$, the number of $S$ in (3.15) is $b = \left\lfloor \alpha^m\right\rfloor$, and $x, y$ are $\left\lfloor \alpha^m\right\rfloor$ or $\left\lceil \alpha^m\right\rceil$. By Lemma 3.1, we know that $x, y$ cannot be both $\left\lceil \alpha^m\right\rceil$. By Lemma 3.3, $x, y$ cannot be both $\left\lfloor \alpha^m\right\rfloor$. Therefore

If $x = \left\lfloor \alpha^m\right\rfloor$ and $y = \left\lceil \alpha^m\right\rceil$, then the segment $(\ell, \ell, \ldots, \ell, x, y)$ in (3.15) is $S_0$, and we are done. By (3.16), it is enough to show that $x\neq \left\lceil \alpha^m\right\rceil$. Applying the same argument to (3.15) instead of (3.12), we can still obtain the sequence as in (3.13). Then (3.14) holds and the calculation after (3.14) still works, and therefore $b-k+x = z$. Since we now have $b = k$, we obtain $x = z$. The first part of the calculation of $z$ in the previous case still works too. Therefore

Since $b = \left\lfloor \alpha^m\right\rfloor$, we obtain by Lemma 2.4 that

and therefore $\beta^m+\beta^{2m}+b\beta^{2m} = \beta^{2m}+\beta^{3m}\in (0, 1)$. So (3.17) implies that $x = b\neq \left\lceil \alpha^m\right\rceil$, as required. This completes the proof.

Next, we give an analogue of Theorem 3.4 when $m$ is even.

Theorem 3.5. Let $m\geq 4$ be an even integer. Then the first $\left\lceil \alpha^m\right\rceil^2-1$ elements of ${\rm{Diff}}(B(\alpha^m))$ can be written as

where $T$ and $T_0$ are the segments given in $(3.3)$ and $(3.4)$, and there are exactly $\left\lfloor \alpha^m\right\rfloor-1$ of $T$ appearing after the first $T_0$ in $(3.18)$.

Proof. Since the proof of this theorem uses the same idea as that of Theorem 3.4, we sometimes skip some details. For convenience, let $\ell = \left\lfloor \alpha^m\right\rfloor$ and $u = \left\lceil \alpha^m\right\rceil$. By Lemma 3.2, the first $\ell$ elements of ${\rm{Diff}}(B(\alpha^m))$ is the segment $T_0$. By Lemma 3.1, we know that $\left\lceil \alpha^m\right\rceil$ follows the segment $T_0$, so in particular, $\left\lceil \alpha^m\right\rceil$ follows the second $T_0$ appearing in (3.18). Therefore, to prove (3.18), it is enough to show that $T$ follows the first $T_0$ and if we write ${\rm{Diff}}(B(\alpha^m))$ as

then $T$ follows $T^{(b)}$ if $1\leq b < \ell-1$, and $T_0$ follows $T^{(b)}$ if $b = \ell-1$, where $T^{(b)}$ is the $b$ copies of $T$ as defined in (3.9) and written in (3.19).

Step 1. By Lemma 3.1, we know that $T_0$ is followed by the constant segment $(u, u, \ldots, u)$ of length $\ell-1$. So we can write

where $x, y\in \{\ell, u\}$. To show that $x = u$ and $y = \ell$, we apply the same argument as in the proof of Theorem 3.4 to obtain from (3.20) the sequence

where $a = \left\lfloor \alpha^m\right\rfloor$, $d = \left\lceil \alpha^m\right\rceil$, $k = \left\lfloor \alpha^m\right\rfloor$, $a+d = \left\lfloor 2\alpha^m\right\rfloor$, $\ldots$, $a+(k-1)d = \left\lfloor k\alpha^m\right\rfloor$, $a+kd-1 = \left\lfloor (k+1)\alpha^m\right\rfloor$, $\ldots$, $a+kd-1+(\ell-1)u+x = \left\lfloor (k+\ell+1)\alpha^m\right\rfloor$. Subtracting the last element in (3.21) by the first element in (3.21), we see that

To calculate the right-hand side of (3.22), we first apply Lemma 2.4 to obtain

Thus the right-hand side of (3.22) is equal to

Substituting $k = \ell$, $d = \ell+1$, $u = \ell+1$ in (3.22) and applying (3.23) and Lemma 2.4, we obtain

which implies $x = \ell+1 = u$, as required. By Lemma 3.3, $y$ cannot be $u$. So $y = \ell$ and the segment $(u, u, \ldots, u, x, y)$ in (3.20) is $T$, as desired.

Step 2. Next, suppose that (3.19) holds with $b < \ell-1$. Again, we know that $T$ is followed by the constant segment $(u, u, \ldots, u)$ of length $\ell-1$. Therefore (3.19) implies that

where $x, y \in \{\ell, u\}$ and we only need to show that $x = u$ so that the segment $(u, u, \ldots, u, x, y)$ in (3.24) is indeed $T$. For convenience, let

Then we obtain from (3.24) the sequence

where $a = \left\lfloor \alpha^m\right\rfloor$, $d = u$, $k = \ell$, and $s+(\ell-1)u+x = \left\lfloor (bk+k+b+\ell+1)\alpha^m\right\rfloor$. Subtracting the last element in (3.25) by the first element in (3.25) and substituting $d = u = k+1$ and $\ell = k$, we obtain

Writing $\alpha^m = k+\{\alpha^m\}$ and letting $z = \left\lfloor(bk+2k+b+1)\{\alpha^m\}\right\rfloor$, we see that the right-hand side of (3.26) is $bk^2+2k^2+bk+z$, and so (3.26) reduces to $bk+k-2+x = z$. By Lemma 2.4, we see that

and therefore

Observe that we have not used the assumption $1\leq b < \ell-1$ in any calculation above. So far, we only use $b$ for the number of $T$ appearing in (3.19). This observation will be used in Step 3 too. We now consider the last term in (3.27). We have the equivalence

It is easy to see that $1+\beta^m-(b+2)\beta^{2m} < 2$. Therefore (3.27) and (3.28) imply

Since $bk+k-2+x = z$, it follows that $x = k+1 = u$, as desired.

Step 3. This step is similar to Step 2. Suppose (3.19) holds with $b = \ell-1$. As already observed in Step 2, we did not use the assumption $b < \ell-1$ before the end of Step 2. Therefore (3.24) to (3.28) still hold in this case. But we now have $b = \ell-1$, so (3.28) implies that $1+\beta^m-(b+2)\beta^{2m} < 1$, and it is easy to verify by applying Lemma 2.4 that $1+\beta^m-(b+2)\beta^{2m} > 0$. Therefore (3.27) implies $z = (b+2)(L_m-2)+b = bk+2k-2$. Since $bk+k-2+x = z$, we obtain $x = k = \left\lfloor \alpha^m\right\rfloor$, as desired. This completes the proof.

4.   Sumsets associated with $B(\alpha^m)$

In this section, we give various results on sumsets associated with $B(\alpha^m)$. We begin with a simple but useful result for general Beatty sequences as follows.

Theorem 4.1. If $x > 1$ is an irrational number, $h\in \mathbb N$, and the $h$-fold sumset $hB(x)$ contains $\left\lceil x\right\rceil$ consecutive integers, then $(h+1)B(x)$ is cofinite. More precisely, if $hB(x)$ contains consecutive integers $m, m+1, \ldots, m+\left\lfloor x\right\rfloor$, then $(h+1)B(x)$ contains every integer that is larger than or equal to $m+\left\lfloor x\right\rfloor$.

Proof. Suppose $hB(x)$ contains consecutive integers $m, m+1, \ldots, m+k$, where $k = \left\lfloor x\right\rfloor$. For each $i = 0, 1, 2, \ldots, k$, let $A_i = (m+i)+B(x)$. Then $\bigcup_{i = 0}^k A_i$ is a subset of $(h+1)B(x)$. So it is enough to show that

If $y \in \bigcup_{i = 0}^k A_i$, then $y \in A_i$ for some $i$, and so $y = m+i+\left\lfloor nx\right\rfloor$ for some $n\in \mathbb N$, and therefore $y\geq m+\left\lfloor x\right\rfloor$. Conversely, suppose $y\in \mathbb N$ and $y\geq m+k$. Since

we see that $m+\left\lfloor nx\right\rfloor \leq y < m+\left\lfloor (n+1)x\right\rfloor$ for some $n\in \mathbb N$. Recall that $\left\lfloor (n+1)x\right\rfloor-\left\lfloor nx\right\rfloor \leq \left\lceil x\right\rceil$ for every $n\in \mathbb N$. Therefore $y = m+\left\lfloor nx\right\rfloor+j$ for some $j = 0, 1, 2, \ldots, \left\lceil x\right\rceil-1$. Thus $y\in A_j\subseteq \bigcup_{i = 0}^k A_i$. This completes the proof.

In the proof of the next theorem, we write $[a, b]$ to denote the interval of integers, that is,

For example, if $A$ is a set of an arithmetic progression $a, a+d, a+2d, \ldots, a+kd$, then $A = a+d\ast[0, k]$ and the $h$-fold sumset $hA = ha+d\ast[0, hk]$. We would like to show that $\left\lceil \alpha^m\right\rceil B(\alpha^m)$ is always cofinite for all $m\geq 1$. If $m\leq 2$, then we already have a proof in ^[6]. If $m\geq 3$ and $m$ is odd, then we have a short proof as follows.

Theorem 4.2. Let $m\geq 3$ be an odd integer and $h = \left\lfloor \alpha^m\right\rfloor$. Then $hB(\alpha^m)$ contains $\left\lceil \alpha^m\right\rceil$ consecutive integers. In addition, $(h+1)B(\alpha^m)$ is cofinite, or more precisely, $(h+1)B(\alpha^m)$ contains every integer that is larger than or equal to $h^3+h\left\lfloor (h^2+1)\alpha^m\right\rfloor = h^4+h^3+2h^2$.

Proof. Let $k = d = \left\lfloor \alpha^m\right\rfloor$. We consider $d$ as the first $k$ terms in the segment $S_0$ and $k+1$ the number of terms in $S_0$. By Theorem 3.4, ${\rm{Diff}}(B(\alpha^m))$ contains the segment $S_0$. This implies that $B(\alpha^m)$ contains the segment $a, a+d, a+2d, \ldots, a+kd, a+(k+1)d+1$, where $a = \left\lfloor n\alpha^m\right\rfloor$ for some $n\in \mathbb N$. Let $A_1 = a+d\ast[0, k]$, $b = a+(k+1)d+1$, and $A = A_1\cup\{b\}$. Then $hA = \bigcup_{\ell = 0}^{h-1}(\ell b+(h-\ell)A_1)\cup \{hb\}$. Let $0\leq \ell < h$ be integers. Then

Since $\ell\leq h-1$ and $h = k$, we have $hk-\ell(k+1)\geq hk-(h-1)(k+1) = 1$. Therefore $\ell(k+1)\leq hk-1\leq hk+\ell$. By (4.2), $\ell b+(h-\ell)A_1$ contains the integer $x_\ell$ where $x_\ell = \ell+ha+d(hk-1)$, and therefore $x_\ell\in hA$. Since $0\leq \ell < h$ is arbitrary, we obtain the consecutive integers $x_0, x_1, x_2, \ldots, x_{h-1}$ in $hA$. Furthermore, $hA_1$ contains the integer $x_h = :ha+dhk$ which is equal to $1+x_{h-1}$. Therefore $x_0, x_1, x_2, \ldots, x_{h-1}, x_h$ are consecutive integers in $hA$. Since $A\subseteq B(\alpha^m)$, we see that $hB(\alpha^m)$ contains $\left\lceil \alpha^m\right\rceil$ consecutive integers $x_0, x_1, \ldots, x_{h}$. Therefore we obtain by Theorem 4.1 that $(h+1)B(\alpha^m)$ is cofinite and contains every integer that is larger than or equal to $x_h$.

So it only remains to write $x_h$ in the desired form. First, since $x_h = ha+dhk$ and $d = k = h$, we obtain $x_h = ha+h^3$. Secondly, we see from Theorem 3.4 that there are $h^2$ elements in ${\rm{Diff}}(B(\alpha^m))$ appearing before the segment $S_0$. Therefore $a = \left\lfloor n\alpha^m\right\rfloor = \left\lfloor (h^2+1)\alpha^m\right\rfloor$. Hence $x_h = h^3+h\left\lfloor (h^2+1)\alpha^m\right\rfloor$.

Alternatively, we can write $a = \left\lfloor (h^2+1)\alpha^m\right\rfloor$ in terms of $h$ only. Since the first $k^2$ elements of ${\rm{Diff}}(B(\alpha^m))$ are $S^{(k)}$, it follows that the $(k^2+1)$-th element of $B(\alpha^m)$ is $\left\lfloor \alpha^m\right\rfloor+(kd+1)k = k^3+2k = h^3+2h$. Therefore $x_h = ha+h^3 = h(h^3+2h)+h^3 = h^4+h^3+2h^2$. This completes the proof.

Theorem 4.3. Let $m\geq 4$ be an even integer and $h = \left\lfloor \alpha^m\right\rfloor$. Then $hB(\alpha^m)$ contains $\left\lceil \alpha^m\right\rceil$ consecutive integers and $(h+1)B(\alpha^m)$ contains every integer that is larger than or equal to

Proof. Let $a = \left\lfloor \alpha^m\right\rfloor = k$ and $d = \left\lceil \alpha^m\right\rceil$. We consider $a$ as the first term in $B(\alpha^m)$, $k$ the number of terms in the segment $T_0$, and $d$ the first $k$ elements of the segment $T$ and also the first $k-1$ elements of the segment $T_0$. By Theorem 3.5, we obtain the first $\left\lceil \alpha^m\right\rceil^2-1$ elements of ${\rm{Diff}}(B(\alpha^m))$, and so we know the first $\left\lceil \alpha^m\right\rceil^2$ terms of $B(\alpha^m)$. We write these $\left\lceil \alpha^m\right\rceil^2 = (k+1)^2$ terms of $B(\alpha^m)$ as entries of a matrix $\begin{bmatrix} a_{ij}\end{bmatrix}$ where $0\leq i, j\leq k$, $a_{0, 0} = \left\lfloor \alpha^m\right\rfloor$ is the first term, $a_{0, 1} = \left\lfloor 2\alpha^m\right\rfloor$ is the second term, $\ldots$, and $a_{k, k} = \left\lfloor (k+1)^2\alpha^m\right\rfloor$ is the $(k+1)^2$-th term of $B(\alpha^m)$. For clarity, we write $a_{i, j}$ instead of $a_{ij}$. So, for instance, we have

We write the entries in column $C_0$, $C_1$, $C_2$ and $C_{k-2}$, $C_{k-1}$, $C_k$ separately as follows:

There are two patterns that are helpful in the following calculation. First, in each row $R_i$ for $i = 0, 1, 2, \ldots, k-1$, we have $a_{i, j}+d = a_{i, j+1}$ for $0\leq j\leq k-2$, and $a_{i, k-1}+d-1 = a_{i, k}$, and in row $R_k$, we have $a_{k, j}+d = a_{k, j+1}$ for $0\leq j\leq k-3$, $a_{k, k-2}+d-1 = a_{k, k-1}$, and $a_{k, k-1}+d = a_{k, k}$. Secondly, in each column $C_j$ for $0\leq j\leq k$ and $j\neq k-1$, we have

and in column $C_{k-1}$, we have

These two patterns are used throughout the remaining proof without further reference.

Let $A = \{a_{i, j}\mid 0\leq i, j\leq k\}$ be the set of the first $(k+1)^2$ elements of $B(\alpha^m)$. We construct consecutive integers $x_1, x_2, \ldots, x_{k+1}$ that are in the $h$-fold sumset $hA$ as follows. Let $x_1 = (h-1)a_{k, k}+a_{0, 0}$, which is clearly an element of $hA$. Next, let $x_2 = x_1+1$. To write $x_2$ as the sum of $h$ elements of $A$, we observe that for $0\leq j\leq k-3$,

Therefore $x_2 = (h-2)a_{k, k}+a_{k, k}+a_{0, 0}+1 = (h-2)a_{k, k}+a_{k, k-2}+a_{0, 2}$. In general, for $j = 1, 2, 3, \ldots, \left\lfloor \frac{k+1}{2}\right\rfloor$, let

Then, for each $j = 1, 2, \ldots, \left\lfloor \frac{k+1}{2}\right\rfloor$, we see that $x_j$ is a sum of $h-j+j-1+1 = h$ elements of $A$, and so $x_j\in hA$. In addition, for $2\leq j\leq \left\lfloor \frac{k+1}{2}\right\rfloor$, we have

Therefore $x_1, x_2, x_3, \ldots, x_{\left\lfloor \frac{k+1}{2}\right\rfloor}$ are consecutive integers in $hA$. Next, we divide the construction into two cases according to the parity of $k$.

Case 1. $k$ is even. So we already have consecutive integers $x_1, x_2, \ldots, x_{\frac k2}$, and we need to construct $x_{\frac k2+1}, x_{\frac k2+2}, \ldots, x_k, x_{k+1}$. Let $x_{\frac k2+1} = x_{\frac k2}+1$. Observe that

By (4.3), we have $x_{\frac k2} = \frac k2a_{k, k}+\left(\frac k2-1\right)a_{k, k-2}+a_{0, k-2}$. This and (4.4) imply that

which is a sum of $\frac k2-1+\frac k2-1+1+1 = k = h$ elements of $A$, and so it is an element of $hA$. In general, for $1\leq j\leq \frac k2$, let

Clearly, $x_{\frac k2+j}$ is a sum of $\frac k2-j+\frac k2-j+2j-1+1 = k = h$ elements of $A$. In addition, for $2\leq j\leq \frac k2$, we have

This shows that $x_1, x_2, \ldots, x_{\frac k2}, x_{\frac k2+1}, x_{\frac k2+2}, \ldots, x_k$ are consecutive integers contained in $hA$. It remains to construct $x_{k+1}$. Let $x_{k+1} = x_k+1$. By (4.5), we know $x_k$, and so

which is a sum of $k-1+1 = k = h$ elements of $A$. Hence, we obtain consecutive integers $x_1, x_2, x_3, \ldots, x_{k+1}$ in $hA$, as desired.

Case 2. $k$ is odd. So we already have $\frac{k+1}{2}$ consecutive integers $x_1, x_2, x_3, \ldots, x_{\frac{k+1}{2}}$ in $hA$. Therefore we need to construct $x_{\frac{k+1}{2}+1}, x_{\frac{k+1}{2}+2}, \ldots, x_{k+1}$. Let $x_{\frac{k+1}{2}+1} = x_{\frac{k+1}{2}}+1$. By (4.3), we know $x_{\frac{k+1}{2}}$, and so we know that

For $0\leq j\leq k-3$, we have

From (4.6) and (4.7), we obtain

which is a sum of $\frac{k-1}{2}-1+\frac{k-1}{2}-1+2+1 = k = h$ elements of $A$. In general, for each $j = 1, 2, 3\ldots, \frac{k-1}{2}$, let

Clearly, $x_{\frac{k+1}{2}+j}$ is a sum of $\frac{k-1}{2}-j+\frac{k-1}{2}-j+2j+1 = k = h$ elements of $A$ for every $j = 1, 2, 3, \ldots, \frac{k-1}{2}$. In addition, for $2\leq j\leq \frac{k-1}{2}$, we obtain from (4.8) and (4.7) that

This shows that $x_1, x_2, \ldots, x_k$ are consecutive integers in $hA$. So it remains to construct $x_{k+1}$. Let $x_{k+1} = x_k+1$. By (4.8), we know $x_k$, and so we obtain

which is a sum of $k-1+1 = k = h$ elements of $A$. Hence, we obtain consecutive integers $x_1, x_2, x_3, \ldots, x_{k+1}$, as desired.

In both cases, we obtain $\left\lceil \alpha^m\right\rceil$ consecutive integers in $hA$. Since $A\subseteq B(\alpha^m)$, we see that $hB(\alpha^m)$ contains $\left\lceil \alpha^m\right\rceil$ consecutive integers. By Theorem 4.1, $(h+1)B(\alpha^m)$ is cofinite and contains every integer that is larger than or equal to

Recall that $a_{k, k}$ is the $(k+1)^2$-th element of $B(\alpha^m)$ and $a_{0, 0}$ is the first element of $B(\alpha^m)$, and $k = h$. Therefore $a_{k, k} = \left\lfloor (h+1)^2\alpha^m\right\rfloor$, $a_{0, 0} = \left\lfloor \alpha^m\right\rfloor = h$, and $x_{k+1} = (h-1)\left\lfloor (h+1)^2\alpha^m\right\rfloor+2h$. Alternatively, we know from the list of $a_{i, j}$ that $a_{0, 0} = a = h$, $a_{k, k} = a+(k^2+2k)d-k-1 = (h^2+2h)(h+1)-1$, and therefore $x_{k+1} = h^4+2h^3-h^2-h+1$. This completes the proof.

We can use the argument from the proof of Theorem 4.3 to get a sharper version of Theorem 4.2 as follows.

Theorem 4.4. Let $m\geq 3$ be an odd integer and $h = \left\lfloor \alpha^m\right\rfloor$. Then $(h+1)B(\alpha^m)$ contains every integer that is larger than or equal to $2h^3+2h$.

Proof. Since the proof of this theorem is similar to that of Theorem 4.3, we skip some details. Let $a = \left\lfloor \alpha^m\right\rfloor = k = d$ and consider $a$ as the first term in $B(\alpha^m)$, $k$ the number of terms in the segment $S$, and $d$ the first $k$ terms of $S_0$ (and the first $k-1$ terms of $S$). By Theorem 3.4, we can write the first $\left\lfloor \alpha^m\right\rfloor^2+\left\lceil \alpha^m\right\rceil+1$ elements of $B(\alpha^m)$ as entries of the matrix $\begin{bmatrix} a_{ij}\end{bmatrix}$ as follows.

For instance, in row $R_0$, we have

and in row $R_{k+1}$, we have

and for $j = 2, 3, \ldots, k-1$, the number $a_{k+1, j}$ is unspecified and is not used in the proof. The patterns that are helpful in the following calculation are as follows.

First, in each row $R_i$ for $i = 0, 1, \ldots, k$, we have $a_{i, j}+d = a_{i, j+1}$ for $0\leq j\leq k-2$ and $a_{i, k-1}+d+1 = a_{i+1, 0}$ if $i\neq k$, and $a_{k, k-1}+d = a_{k+1, 0}$. In row $R_{k+1}$, we have $a_{k+1, 0}+d+1 = a_{k+1, 1}$. Secondly, in each column $C_j$ for $j = 0, 1, 2, \ldots, k-1$, we have $a_{i, j}+kd+1 = a_{i+1, j}$ for $i = 0, 1, 2, \ldots, k-1$, and $a_{k, 0}+kd = a_{k+1, 0}$, and $a_{k, 1}+kd+1 = a_{k+1, 1}$.

Let $A = \{a_{i, j}\mid 0\leq i\leq k\; \text{and}\; 0\leq j\leq k-1\}\cup\{a_{k+1, 0}, a_{k+1, 1}\}$ be the first $k^2+k+2$ elements of $B(\alpha^m)$ given above. We construct consecutive integers $x_0, x_1, \ldots, x_k$ as follows. Let

and for $j = 1, 2, \ldots, k-2$, let

It is easy to see that $x_j$ is a sum of $h$ elements of $A$ for every $j = 0, 1, 2, \ldots, k-2$. In addition, $x_1-x_0$ is equal to

Furthermore, for $j = 2, 3, \ldots, k-2$, we have

Therefore $x_0, x_1, x_2, \ldots, x_{k-2}$ are consecutive integers in $hA$. Next, let $x_{k-1} = 1+x_{k-2}$, which is equal to

and therefore $x_{k-1}\in hA$. Next, let $x_k = 1+x_{k-1}$, which is equal to

and thus $x_k\in hA$. Hence $x_0, x_1, x_2, \ldots, x_k$ are consecutive integers in $hA$. By Theorem 4.1, $(h+1)B(\alpha^m)$ contains every integer that is larger than or equal to $x_k$. We have

This completes the proof.

Remark 2. The integer $x = x(m, h) = h^4+2h^3-h^2-h+1$ in Theorem 4.3, and the integer $x = x(m, h) = 2h^3+2h$ in Theorem 4.4 are best possible when $m\leq 5$. For example, if $m = 3$, then $x(m, h)$ in Theorem 4.4 is equal to $2h^3+2h = 136$, and so $5B(\alpha^3)$ contains every integer that is larger than or equal to $136$. Then we can either straightforwardly check or use a computer to verify that $135\notin 5B(\alpha^m)$, and thus $136$ is best possible.

More generally, for a cofinite proper subset $A$ of $\mathbb N$, let $G(A) = \mathbb N\setminus A$, $g(A) = |G(A)|$, $f(A) = \max G(A)$, and $c(A) = f(A)+1$. Although this may be slightly different from a more restrictive definition in algebraic geometry or numerical semigroup theory, we may call $G(A)$ the set of gaps of $A$, $g(A)$ the genus of $A$, $f(A)$ the Frobenius number of $A$, and $c(A)$ the conductor of $A$. Then Theorems 4.3 and 4.4 actually give some information on the genus, the Frobenius number, and the conductor of $\left\lceil \alpha^m\right\rceil B(\alpha^m)$. We record these as a corollary.

Corollary 1. Let $m$ be a positive integer and $h(m) = \left\lfloor \alpha^m\right\rfloor$. Let $g(m)$, $f(m)$, $c(m)$ be the genus, the Frobenius number, and the conductor of the $(h+1)$-fold sumset $(h+1)B(\alpha^m)$, respectively, as defined in Remark 2. Then the following statements hold:

Furthermore, if $m\geq 6$ and $m$ is even, then $f(m) \leq h^4+2h^3-h^2-h$; if $m\geq 6$ and $m$ is odd, then $f(m) \leq 2h^3+2h-1$. In particular, the values of $f(m)$ and $c(m)$ obtained from (or implied by) Theorems 4.3 and 4.4 are best possible for $3\leq m\leq 5$. In fact, simply substituting $h = h(1) = \left\lfloor \alpha\right\rfloor = 1$ in Theorem 4.4 and $h = h(2) = \left\lfloor \alpha^2\right\rfloor = 2$ in Theorem 4.3, and comparing them with the results in ^[6], we see that they are also best possible for $m = 1, 2$.

Proof. When $m = 1$ or $2$, we obtain this corollary from Theorems 3.1 and 3.8 in ^[6]. For $3\leq m\leq 5$, we apply Theorems 4.3 and 4.4 to obtain an explicit constant $N = N(m)$ depending only on $m$ such that the $\left\lceil \alpha^m\right\rceil$-fold sumset $\left\lceil \alpha^m\right\rceil B(\alpha^m)$ contains every integer that is larger than or equal to $N$. Then we can either straightforwardly check or use a computer to verify that $N-1$ is not in $\left\lceil \alpha^m\right\rceil B(\alpha^m)$. So $N$ is the smallest integer such that $[N, \infty)\cap\mathbb Z$ is contained in $\left\lceil \alpha^m\right\rceil B(\alpha^m)$. This completes the proof.

We conclude this paper with the following conjecture.

Conjecture 1. Let $h = \left\lfloor \alpha^m\right\rfloor$ for each $m\geq 3$. Then the integers $h^4+2h^3-h^2-h+1$ and $2h^3+2h$ given in Theorems 4.3 and 4.4 are best possible for all $m\geq 3$. In other words, the Frobenius number $f(m)$ of $\left\lceil \alpha^m\right\rceil B(\alpha^m)$ satisfies

Conjecture 2. For each $m\in \mathbb N$, the number $\left\lceil \alpha^m\right\rceil$ is the smallest positive integer such that $\left\lceil \alpha^m\right\rceil B(\alpha^m)$ is cofinite.

Question 1. Numerical evidence suggests (but does not prove) that there are infinitely many Fibonacci numbers that are not an element of $\left\lfloor \alpha^m\right\rfloor B(\alpha^m)$. Is this true in a more general situation? If $\alpha > 0 > \beta$, $|\alpha| > \max\{1, |\beta|\}$, and $\alpha$, $\beta$ are roots of the characteristic polynomial $x^2-ax-b$ of the Lucas sequence of the first kind $(U_n(a, b))_{n\geq 1}$, is there an interesting connection between the sumsets of $B(\alpha^m)$ and $U_n$? Here $U_n$ is defined by the recurrence relation $U_0 = 0$, $U_1 = 1$, and $U_n = aU_{n-1}+bU_{n-2}$ for $n\geq 2$, where $a$, $b$ are fixed integers, $(a, b) = 1$, and $a^2+4b > 0$.

Acknowledgements

This project is funded by the National Research Council of Thailand (NRCT), grant number NRCT5-RSA63021-02. It will also be funded by the Tosio Kato Fellowship of the Mathematical Society of Japan in the near future. Napp Phunphayap has helped me with computer programming which led me to a sharp result as noted in Remark 2. I thank them all for the support. I am also grateful to the referees for reading my manuscript very carefully and for giving me kind comments and useful suggestions which improve the quality of this paper.

Conflict of interest

The authors declare there is no conflicts of interest.