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A Construction of Arithmetic Progression-Free A CONSTRUCTION OF ARITHMETIC PROGRESSION-FREE SEQUENCES by BRIAN L. MILLER, B.S. A THESIS IN MATHEMATICS Submitted to the Graduate Faculty of Texas Tech University in Partial Fulfillment of the Requirements for the Degree of MASTER OF SCIENCE Approved December, 2004 ACKNOWLEDGMENTS I first and foremost want to thank Dr. Chris Monico for his patience and dedication to this work. I also want to thank my undergraduate professors, Dr. Hagler, Dr. Waugh, and Dr. Hodge for their time, effort, and challenges to become a better mathematician. I also want to thank my family, especially my mom and dad, for their support and encouragement. And to all my friends, Robert, Tanya, and family, the Paz family, and Jake, thanks for remaining behind me during all of this. 51 xony noGjiaroAapnxb CTacio CnnnKC. H Taic^e xony no6.JiaroAapHTb Mnxaiuia Xape;i,H sa ero nocBHni,eHHe H no^n^epxcKy MoeMy o6pa30BaHHio BO Bcex OTHomenHax. 11 CONTENTS ACKNOWLEDGMENTS ii ABSTRACT iv I INTRODUCTION 1 II ATTACK OF THE PROBLEM BACKGROUND 2 2.1 Introduction 2 2.2 Known Results 2 2.3 Span and Its Impact 3 2.3.1 Definitions and Motivations 3 2.3.2 Extension of Values 4 2.3.3 Partitions and Compositions 5 III DESCRIPTION OF THE CONSTRUCTION 13 3.1 Introduction 13 3.2 The Formal Description of the Construction 13 3.3 Justification of the Construction 15 IV CONSEQUENCES OF THE CONSTRUCTION 17 V ANALYSIS OF THE CONSTRUCTION 22 APPENDIX 32 ill ABSTRACT We describe a particular greedy construction of an arithmetic progression-free sequence from a finite composition. We also give an analysis on the properties of the resulting sequence. IV CHAPTER I INTRODUCTION One of the many open problems that remains in number theory addresses the question of existence of arbitrarily long finite arithmetic progressions of primes. Erdos asked a more general question: If A is an infinite set of positive integers such that the series Y^^eA ^~^ diverges, then must A contain arbitrarily long finite arithmetic progressions? If the answer to his question is yes, then this would definitely imply the existence of arbitrarily long finite arithmetic progressions of primes since /Jp~^ p is known tp diverge. An arithmetic progression is commonly defined a.s s-\-nd for fixed s, d and consec­ utive n; that is, the set S = {s,s-\-d,s + 2d,s + 3d,...,s + {n — l)d} is an arithmetic progression. We use the term progression, without the qualifier arithmetic, to mean a finite sequence. In the case that S is infinite, we will say arithmetic progression sequence. An arithmetic progression free sequence is then rightly defined as an infi­ nite set of integers such that no choice of any k > 2 elements falls into the previously described formulation. Note that k must be greater than two; otherwise such a se­ quence would never exist. Furthermore, a A;-term A.P. (A.P. will stand for arithmetic progression from this point forward), is a finite sequence in which k terms follow the prescription s,s-\-d,s-\-2d,... ,s +{k— l)d. Observe that a sequence with no 3-term A.P. will not contain a 4-term A.P. and so forth. The main concern of this paper is approaching Erdos' conjecture by the contra- positive: If A does not contain arbitrarily long finite arithmetic progressions, does y^ a~^ necessarily converge? More specifically, we mainly concentrate on sets that aeA do not contain any A.P. Hence it suffices to concentrate on sequences without a 3- term A.P. We develop a construction to produce A.P.-free sequences; we then analyze and oflter examples of the construction. Before we divulge the construction, we first traverse through the ideas and formulations that lead to it. CHAPTER II ATTACK OF THE PROBLEM - BACKGROUND 2.1 Introduction The study of arithmetic progressions, and a lack thereof, has been considered by several well known authors, including Erdos and Turan [4], Guy [11], Odlyzko [15], Newman [14], Grosswald [8] and [9], Gerver [6], and many others. Before proceeding with our examination of the special case of Erdos' conjecture, we summarize some of the known related results from the literature. 2.2 Known Results Gerver [6] has shown that for each integer A: > 3, there exists a set Sk of positive integers containing no A.P. of k terms, such that ^ n~^ > (1 - ^)k log A; for large neSk k and e e R+-^. The S'fc's are generated recursively by the greedy algorithm in the cases where k is prime. That is, n e ^jt if and only if {m e ^fc : m < n - 1} U {n} contains no arithmetic progression of k terms. Gerver and Ramsey [7] give a heuristic formula, supported by computational evi­ dence, that describes the asymptotic density of Sk in the case where k is composite. However, with a couple of assumptions, was shown to imply that the greedy algorithm would not maximize TJ?^"^ over all S with no k-texva A.P. They then show that for neS all e > 0, |5fc| < n is greater than (1 — e)y/2n for sufficiently large n. Odlyzko and Stanely [15] construct sequences of positive integers, S{k), using the greedy algorithm. Their sequence is constructed so that QQ = 0, ai — k and each a^^i such that a„ < a„+i creates an A.P.-free sequence QQI «!, • • • < a„, a^+i. The greedy algorithm does not, however, produce particularly dense A.P.-free sequences. Dense sequences are of interest because it is easy to construct A.P.-free sequences whose sum of reciprocals converge. For a quick example, consider any geometric series. However, is it possible to construct A.P.-free sequences whose sum of reciprocals diverge? If we can answer no, then consequently the answer to our question is fulfilled. Hence we try to construct A.P.-free sequences in which the terms are "close" together as possible. This leads us into the next idea. 2.3 Span and Its Impact 2.3.1 Definitions and Motivations If we define the span of a set 5, as in Guy [11], to be span(S')= max |si - S2I, or ai,S2eS sp(5), this gives us two tools. We can first compare the density of progressions. For example, given the sets 5 = {1,5,10,17} and T = (1,2,4,5} we see that sp(5) = 17 1 = 16 while sp(T) = 5 1 = 4. Therefore, we say that T is denser than S. Secondly, we may also "shift" sequences and progressions. Consider the sets S = {1,2,4,5} and T = {6, 7,9,10}. Notice that sp(5) = sp(r) and |5| = \T\. Thus to us, the sets are equivalent. We cay say this because the properties for which we are concerned are invariant under translation. That is, we can define an equivalence relation on the set of finite subsets of N by S^T if and only if sp(5) = sp(T) and |5| = |T|. It can be easily verified that this is indeed an equivalence relation. Although shifting a sequence has no bearing on its convergence property, we shift sets such as T down to sequences that start with one merely for convenience. Also, since we were concen­ trating on finding dense sequences that contain no A;-term A.P., we started looking at sets of n integers containing no 3-term A.P. that yielded the smallest span, defined as sp(3, n). The chase to answer the question "given n, what is sp(3, n)?" yielded two out­ comes. First, we were able to extend the values for sp(3, n) in Guy [11]. Second, in the search to further extend the values given, the idea of partitioning a number to construct the progressions came into play. 2.3.2 Extension of Values In a paper written by Erdos and Tiiran [4], they provide an upper bound on the possible value of sp(3, n) given n. They define r{N) to be the maximum number of elements in a finite sequence of integers ai < a2 < • • • < N containing no 3-term A.P. exists. They give the upper bound as r(2N) < N if N >8. Translating this into our language, we get sp(3, N) < . They also give the bound r{N) < ( - +e) iV for e > 0 and A^ > A'o(e). Notice that these bounds agree with the values given in Table ILL In spite of the bounds given by Erdos and Tiian, we wanted to find an explicit function of n to describe sp(3, n). Because if we could, then that function could be used to evaluate the convergence of the sum of reciprocals of A.P.-free sequences. This idea inspired us to write a computer program to compute sp(3, n) for small values of n. The program found sp(3, n) by recursion. The first element was chosen to be one. Then the next five possible values were computed. A sequence was built from each of these choices. From these choices, the next five choices were created. This process repeats until a sequence with the number of desired elements with the shortest span is reached. Note that there are a finite number of choices; one can easily construct the greedy algorithm sequence and use it as the comparison. That is, consider building a finite sequence with five elements. The greedy algorithm gives a span of 9. So, when building a finite sequence that already has 3 elements and has a span of 8, you can quit building that sequence since adding any more elements into the sequence will only make the span larger.
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