Research Article | Open Access

Bing Li, Chao Ma, "Finite and Infinite Arithmetic Progressions Related to Beta-Expansion", *Abstract and Applied Analysis*, vol. 2014, Article ID 678769, 6 pages, 2014. https://doi.org/10.1155/2014/678769

# Finite and Infinite Arithmetic Progressions Related to Beta-Expansion

**Academic Editor:**Douglas R. Anderson

#### Abstract

Let and be the -expansion of . Denote by the set of positions where the digit 1 appears in . We consider the sets of points such that contains arbitrarily long arithmetic progressions and includes infinite arithmetic progressions, respectively. Their sizes are investigated from the topological, metric, and dimensional viewpoints.

#### 1. Introduction

Let be a sequence of natural numbers. The lower asymptotic density of (the upper asymptotic density of , resp.) is defined as If , the common value is called the asymptotic density of , denoted by .

The concept of asymptotic density was extensively applied to number theory such as arithmetic progressions, the first of which is the following theorem due to Szemerédi [1].

Theorem 1 (see [1]). *If , then the sequence contains arbitrarily long arithmetic progressions.*

The issue to find what kind of sequence can contain arbitrarily long arithmetic progressions is very popular nowadays. For instance, Green and Tao [2] proved that the sequence of prime numbers contains arbitrarily long arithmetic progressions, which extends Theorem 1 since .

Assume that the sequence contains arbitrarily long arithmetic progressions; a further question is whether can contain an infinite arithmetic progression. In 1972, Wagstaff Jr. [3] found a sequence of natural numbers which contains arbitrarily long arithmetic progressions but does not have any infinite one. In fact, he even gave more information as the following.

Theorem 2 (see [3]). *For each with , there exists a sequence of natural numbers such that contains no infinite arithmetic progression and and .*

The topics of arithmetic progressions and binary expansion were firstly linked by Šalát and Tomanová [4]. They considered the sequence of the positions where the digit 1 appears in the binary expansion of real numbers in the unit interval. In this paper, we investigate the same questions for -expansion with and generalize the results in [4].

Let . The -transformation is defined as (mod 1) for any . Consider the greedy -expansion of with and for all , called -digits of . Here denotes the integer part of . It is clear that all the digits for any and .

Define that is, the set of natural numbers with . In other words, is the sequence of positions where the digit appears in the -expansion of .

Denote

Obviously, since an infinite arithmetical progression contains arbitrarily long ones. In Section 2, we will give some introduction to -expansion. In Sections 3 and 4, we will describe the sets , and their complements , from the viewpoints of topology, metric, and Hausdorff dimension.

#### 2. Preliminary for -Expansion

When , any sequence in except countably many elements is a -expansion of some point . It is not this case when is not an integer. Denote A 0-1 sequence is said to be admissible if and a 0-1 word is admissible if there exists one point such that . Denote by the set of all admissible words of length . Let be the closure of and then is a subshift of , where is the shift transformation of the symbolic space . Parry [5] gave a criterion to check if a sequence in is admissible where the -expansion of the number 1 (defined similarly with (2)) plays an essential role. If the -expansion of 1, denoted by , is finite, that is, ending with zeros, we call the modified periodic sequence the infinite -expansion of 1, where is the last nonzero element. If is infinite, namely, that there are infinitely many 1s in the sequence , then just let . We use the notation to represent the lexicographical order in the symbolic space .

Theorem 3 (see [5]). *Let . Then
*

For the case , ; that is, the coding space is the full shift space. When , is a proper subset of . If is finite, is a subshift of finite type (SFT) which can be obtained by forbidding finitely many words. The set of ’s with being SFT is dense in (see [5]). If , then and , which means that every -admissible sequence is -admissible.

*Definition 4. *Let . The set
is called a cylinder of order .

Any cylinder is a left-closed and right-open interval and its length We say that a cylinder is full if ; that is, any admissible sequence can be concatenated with the word . In other words, if . We remark that if is full. The following lemma, due to Li and Wu [6], will be used to prove Proposition 6.

Lemma 5 (see [6]). *Let . Then there exists such that the cylinder is full, where just depends on and ; means the word of consecutive zeros.*

When , the Lebesgue measure is a -invariant measure. If is not an integer, the Lebesgue measure is not -invariant; however Rényi [7] proved that there exists a unique invariant measure equivalent to . Parry [5] and Gel’fond [8] independently gave the density formula of and .

The function maps to its -expansion. Conversely there is a projection from the subshift to defined as for any . In particular, . We have the following commutative diagram; that is, :

#### 3. Topological Properties of and

Let be the sum of first digits of the -expansion of . Define

Proposition 6. *The set is of the first category.*

*Proof. *The set can be written as
For any , the set is the union of some cylinders of order satisfying . Note that the cylinder is a left-closed and right-open interval; denote by its closure, which is obtained by adding the right endpoint of . Let
Choose large enough such that there exists with . Such existence is because the sequence is admissible where with and . Notice that
Since the set is closed, is a closed set; it suffices to show that the interior of is empty. Suppose it is not empty; then there exists one interval inside , which implies that there exists some cylinder lying in this interval and then . Thus there exists only depending on and such that the cylinder is full by Lemma 5. Therefore the sequence is admissible; put
Thus which implies contradicting
So we complete the proof.

*Remark 7. *By the similar argument with the proof of Proposition 6, we can prove that the set
is of the first category for any . When , the set may be empty. For instance, ; the set can be characterized as that
So for any , which indicates if .

Theorem 8. *The set is residual in .*

*Proof. *Firstly we claim that . Indeed, if , by Theorem 1, we know that the sequence contains arbitrarily long arithmetical progressions; that is, .

Secondly, combining the facts that
and the set is of the first category (Proposition 6), we know that the set is of the first category. So the set is residual.

Theorem 9. *The set is of the first category in .*

*Proof. *Let and
Then . Note that , where
It suffices to show that, for any , the set is nowhere dense. That is, for any interval , there exists an interval such that . Indeed, for given , we can find a cylinder such that when is large enough. Put . Obviously . Since, for any , the word of consecutive zeros appears in the -expansion of , we have . Therefore is nowhere dense.

The combination of Theorems 8 and 9 implies the following.

Corollary 10. *The set is residual in .*

#### 4. Metric and Dimensional Properties of and

Theorem 11. * and .*

*Proof. *Note that , where is the indicator function. By Birkhoff’s ergodic theorem, we have, for -almost every (a.e.) ,
Note that is equivalent to the Lebesgue measure ; then, for —a.e. , the asymptotic density of
that is, . By Theorem 1, we know
Thus .

Let be natural numbers. Define
Then , where . To prove that , it suffices to show that for any . In fact, let
and then
The set consists of finitely many cylinders with . Now we estimate the length of by the following two cases.

If the word is admissible, then the cylinder is full by Lemma 2.9(i) in [6] and consequently . Since , , and , we have
That is, .

If the word is not admissible, then , which is not contained in .

Therefore,

When , note that and consists of the cylinders with . A similar argument with the estimate of implies
for any . Therefore,
Repeating this argument, we obtain
which implies together with (25). Then we complete the proof.

The following is a consequence of Theorem 11.

Corollary 12. *Consider .*

Lemma 13. *Let such that is a SFT. Then .*

*Proof. *Since is a SFT, there exists such that it is enough to forbid finitely many words of length . Note that for any . Now we construct a Cantor subset of with large Hausdorff dimension when is large.

Let with and define
where means for simplicity. Then the set can be regarded as a Cantor-like set as the following: the interval of level is and the set of level consists of the cylinders , for any , where is any admissible word for . Arrange such cylinders from left to right and denote them by , where is the number of admissible words of length ; that is, . Then

Let be any cylinder of level and let be its father cylinder of level . Note that both and are full cylinders; we know that their lengths are and , respectively. So the ratio between and is . Therefore is a self-similar set. Obviously satisfies strong separation condition (SSC). By the Hausdorff dimension formula for the self-similar set with SSC (see Falconer [9], Page 130), we know that is the unique solution of the equation
That is, . Note that ; then
We obtain by letting .

Theorem 14. *Consider*(i)*;*(ii)*. *

*Proof. * Since , it suffices to show that the Hausdorff dimension of the right-hand side set is zero. In fact, note that ; by the main theorem in [10], we have
where is the measure-theoretical entropy of with respect to . Since the -invariant measure with is just the Dirac measure at the origin 0, the measure-theoretical entropy , which implies . Therefore .

Let be a sequence of numbers satisfying which are SFTs and when . Note that ; write . Define the function as for any .

We claim that . Indeed, on the one hand, for any , we know by the definition of the function . Note that ; then , which leads to . On the other hand, for any , take . Clearly and , which implies . So .

By Theorem in [11], the function is -Hölder which implies
Since , we have
which concludes the result by letting and then , together with (34) in Lemma 13.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors thank the three anonymous reviewers for many helpful comments and suggestions that led to significant improvement for the paper. In addition, Bing Li was supported by NSFC (nos. 11201155 and 11371148) and Fundamental Research Funds for the Central Universities SCUT (2013ZZ0085). Chao Ma was supported by the Science and Technology Development Fund of Macau (no. 069/2011/A).

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#### Copyright

Copyright © 2014 Bing Li and Chao Ma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.