Van der Waerden Numbers with Finite Gap Sets of Size Three

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1 Van der Waerden Numbers with Finite Gap Sets of Size Three Stephen Hardy Emily McLean Kurt Vinhage July 29, 2009 Abstract For a set D of positive integers, let w D (k; r) denote the least positive integer n (if it exists) such that every r-coloring of {1, 2, 3... n} contains a monochromatic k-term arithmetic progression of the form {x, x + d, x + 2d,... x + (k 1)d}, where d D. If no such n exists, we say that w D (k; r) =. If D = 2, the values of w D (k; r) are already established. We find that whenever D = 3, w D (3; 2) =, and consider cases for which w D (2; 3) =. We show that w D (2; 3) < whenever D is of the form {1, 2, 3k}, or {d 1, d 2, d 1 + d 2 } and 3 d 1, d 2 or d 1 + d 2. An explicit formula is provided for the first case, and an upper and lower bound for the other. 1 Introduction The Van der Waerden number w(k; r) is defined as the least positive integer n such that every r-coloring of {1, 2, 3,... n} (which will from this point be denoted by [1, n]) yields a monochromatic k-term arithmetic progression. While Van der Waerden Numbers have long been studied by mathematicians since they were first introduced, very little is actually known to date (for a summary, see [2]). In an effort to better understand these numbers, a variation on them, which limits the progressions allowed to those with a common difference belonging to a finite set of positive integers, was considered [1]: Definition 1.1. Let D be a set of positive integers. Then w D (k; r) is the least positive integer n (if it exists) such that every r-coloring of [1, n] yields a k-term arithmetic progression with common difference d D. If no such integer exists, we say that w D (k; r) = Definition 1.2. Let D be a set of positive integers, k 2 and S Z +. A coloring χ of S is called (D, k)-valid if it admits no monochromatic k-term arithmetic progression with common difference d D in S. If the values of D and k are understood, we say that the coloring is valid. Using this Definition 1.2, we note that w D (k; r) = if and only if there exists a valid coloring of the integers. In [1], the cases when D = 2 and 2 colors were developed and well-understood, but only partially understood for the cases in which D = 3. It was shown that w D (4; 2) = for every set D with three elements, and conjectured that w D (3; 2) =, which we will show. Furthermore, we show partial results for w D (2; 3) in an effort to extend previous results to more than two colors. 1

2 2 Useful Lemmas and Other Results It is important to note the following result established in [2] before the results of this paper are given: Lemma 2.1. Let D be a set of positive integers, and let k, t 1. Then, if w D (k; r) <, w td (k; r) = t[w D (k; r) 1] + 1. Otherwise, w td (k; r) =. Note that this allows us to limit our investigation of w D (k; r) to those D with gcd(d) = 1. We begin by showing that if w D (k; r) =, there exists a valid periodic coloring of the integers, an easy observation of the following Lemma: Lemma 2.2. Let D = {d 1, d 2,..., d n }, d 1 < d 2 <... < d n. If there is a (D, k)-valid r-coloring of the set I = [1, r 1+(k 1)dn + (k 1)d n + 1], then w D (k; r) =. Proof. Let χ be a valid r-coloring of I. Note that there are r 1+(k 1)dn + 1 distinct sets of 1 + (k 1)d n consecutive integers in I. Since there are r 1+(k 1)dn ways to r-color a set of 1 + (k 1)d n integers, by the pigeonhole principle there are two such sets which have the same coloring. This coloring can be considered as a string of integers between 1 and r. Let s be the repeated string of length 1 + (k 1)d n. Call the intervals which s appears in I 1 and I 2, so that, written as a string, χ(i 1 ) = χ(i 2 ) = s. The strings can appear in one of two ways: I 1 I 2 =, or I 1 I 2 Case 1. I 1 I 2 = Then χ is of the form xsysz for some (possibly empty) strings x, y and z. Consider the coloring sy (where a denotes a periodic coloring of the integers with period a ). Suppose, for a contradiciton, that there were a monochromatic k-term arithmetic progression in this coloring. If the first term of the k-term arithmetic progression is in s, the terms of the arithmetic progression must be contained in the string sys, since the difference of the highest and lowest term of the progression is no more than (k 1)d n < s. Then there is a corresponding k-term arithmetic progression in χ, which contradicts our assumption that χ is valid. Likewise if the first term of the arithmetic progression is in the string y the entire arithmetic progression must be contained in the string ys in χ, again arriving at a contradiction. So the periodic coloring sy is valid, and w D (k; r) =. Case 2. I 1 I 2 Letting u be the overlapping string, with u = I 1 I 2 = m, we must have the first m integers in the string s colored the same as the last m. Then, written as a string, χ(i 1 I 2 ) = ututu. Thus we have s = utu and χ is of the form xututuy for some strings x and y. Consider the coloring ut. Suppose, for contradiction, that there is a k-term arithmetic progression in this coloring. If the first term is in the string u then since the difference between the highest and lowest terms is no more than (k 1)d n = s = utu, there would be a corresponding k-arithmetic progression in the string ututu in χ. Likewise 2

3 if the first term of the arithmetic progression is in the string t then there would be a corresponding arithmetic progression in the string tutu in χ. So in both cases, we have a valid periodic coloring, and w D (k; r) =. Corollary 2.1. If w D (k; r) =, there is a valid periodic coloring of Z +. Proof. If there is a valid coloring of the integers, it is also valid on I. The result follows directly from the construction of the colorings in Lemma 2.2. The following propositions allow us to eliminate certain cases for k, r and D: Proposition 2.1. If D is any subset of the integers and no element of D is divisible by n for 1 < n r then w D (2; r) =. Proof. The n-coloring χ: N {1, 2,..., n} given by defining χ(x) to be the unique integer y such that y x mod n and y [1, n] will avoid any 2-term arithmetic progressions with gap in D, since no element d D has d 0 mod n, so the color of x + d always differs from that of x. This shows w D (2; n) = immediately implying w D (2; r) =. Proposition 2.2. For any gap set D = {d 1, d 2,..., d n }, w D (2; n + 1) =. Proof. We can define a valid (n + 1)-coloring χ recursively, because for all integers x, χ(x) only needs to be different from at most n other colors {χ(x d 1 ), χ(x d 2 ),..., χ(x d n )} to avoid 2-term arithmetic progressions with gap in D. Proposition 2.3. Fix a pair of integers 1 a < n. Let D Z + have the property that if d D, then d a mod n or d a mod n. Then if the largest power of 2 dividing n does not divide a, w D (2; 2) =. Otherwise, we have w D (3; 2) = w D (2; 3) =. Proof. Define the 2-coloring χ 0 on the interval [1, n] by defining it first on the group Z n. Note that a, as an element of Z n generates the subgroup a. Color the subgroup recursively, with χ 0 (a) = 0, and the other elements as: χ 0 (ia + a) = 1 χ 0 (ia) For 1 i < a, where a denotes the order of the element a in the group Z n. Note that every element of Z n belongs to some coset of a. That is, every element can be written as x = ia + j, where 0 j < Z n : a. Then for every x Z n, define χ 0 (x) = χ 0 (ia + j) = χ 0 (ia). Because every element of Z n has exactly one such representation, this is a welldefined coloring. Suppose that the largest power of 2 dividing n does not divide a. Note that a = a = Z n Z n : a = n gcd(n, a) So 2 a, and since χ 0 (2i a) = 1 for every i, we know that χ 0 (n) = χ 0 ( a a) = 1 χ 0 (a). 3

4 Now define the coloring χ of Z + by extending the coloring of Z n as follows. Denote for each x Z +, the element ˆx Z n such that x ˆx mod n. Then let χ(x) = χ 0 (ˆx). Let d D, so that d ±a mod n. Then let x, x+d be an arbitrary 2-term progression. Note that χ(x+d) = χ 0 (ˆx±a) and χ(x) = χ 0 (ˆx). Let ˆx = ia+j. Then χ(x±d) = χ 0 ((i±1)a) and χ(x) = χ 0 (ia). The way the coloring is defined, these two will always differ, so we know that x and x + d must differ in color, and the coloring admits no monochromatic 2-term arithmetic progressions with common difference in D. That is, w D (2; 2) =. If the largest power of 2 dividing n does divide a, then we do not have 2 a, and χ 0 (0) = χ 0 (a). If x 1, x 2, x 3 is a three-term arithmetic progression, then we know that χ(x 2 ) = χ 0 ( ˆx 1 ± a) and χ(x 3 ) = χ(x 2 ± d) = χ(x 1 ± 2d) = χ 0 ( ˆx 1 ± 2a). Since, with the exception of 0 and a (and any corresponding elements of the coset of a ), χ 0 ((i + 1)a) χ 0 (ia), we know each color must appear once, so this is a (D, 3)-valid coloring, and w D (3; 2) =. To show that w D (2; 3) =, define a 3-coloring χ of the integers as: χ (x) = { χ(x) if x 0 mod n 2 if x 0 mod n Note that if x, x + d is a 2-term arithmetic progression with d D, and neither x nor x+d is divisible by n, they are not monochromatic by the construction of χ. If one of them is divisible by n, then they must be different colors, since d 0 mod n. So χ does not admit a monochromatic 2-term arithmetic progression, and w D (2; 3) =. 3 D = 3 with 2-term Arithmetic Progressions and 3 Colors Note that Proposition 2.1 shows that if D = 3 and no element of D is divisible by 2 or no element is divisible by 3, then w D (2; 3) =. In addition, Proposition 2.3 shows that if every element of D is congruent some number a or its opposite mod n, where a n, then w D (2; 3) =. We now show further results for specific cases: Proposition 3.1. For every k Z +, if D = {1, 2, 3k}, then w D (2; 3) = 3k + 1 Proof. We first show that w D (2; 3) 3k + 1. Let χ be the 3-coloring of [1, 3k] given by defining χ(x) to be the unique element y such that y x mod 3 and y [1, 3]. Then χ yields no two-term arithmetic progressions with gaps 1, 2 or 3k (since x + 1 and x + 2 are always different mod 3 and there are no two elements with gap 3k). Thus w D (2; 3) 3k+1. We now show the reverse inequality. Suppose for the sake of contradiction that χ is a valid 3-coloring of [1, 3k + 1]. We claim that for all n [0, k], χ(3n + 1) = χ(1). This can be shown by induction on n. It is trivially true when n = 0. Suppose χ(3n + 1) = χ(1), with 1 3n + 1 3k + 1. Since χ is valid, χ(3n + 1), χ(3n + 2) and χ(3n + 3) must all be distinct colors. Since χ(3n + 4) χ(3n + 2) and χ(3n + 4) χ(3n + 3), we must have χ(3n + 4) = χ(3n + 1) = χ(1). Thus χ(1) = χ(3n + 1) for all n [0, k], and most 4

5 importantly, χ(1) = χ(3k + 1), a contradiction. Thus any 3-coloring of [1, 3k + 1] yields a 2-term arithmetic progression with gap in D, and w D (2; 3) 3k + 1. We now provide both the lower and upper bound for the cases when D = {a, b, a + b}. Theorem 3.1. If a < b Z + and D = {a, b, a + b} then w D (2; 3) a + b + 1. Proof. For each x [1, a + b], define φ: [1, a + b] {1, 2, 3} by 1 if x b and x i (mod 2a), where 0 i < a φ(x) = 2 if x b and x i (mod 2a), where a i < 2a 3 if x > b. The 3-coloring φ clearly avoids any 2-term arithmetic progressions with gaps in D, so w D (2; 3) a + b + 1. Theorem 3.2. If D = {a, b, a + b}, gcd (a, b) = 1 and at least one of a, b, a + b is divisible by 3, then w D (2; 3) 2(a + b) Proof. Let D = {a, b, a + b}, gcd (a, b) = 1, and suppose 3 divides at least one of a, b, and a + b. Further suppose, without loss of generality, that a < b. Note that 3 cannot divide b a (since otherwise a b (mod 3), and we either have a b a + b 0 (mod 3), contradicting our assumption that gcd (a, b, a + b) = 1, or a b 1 (mod 3) and a + b 2 (mod 3), then none of a, b, or a + b are divisible by 3, or a b 2 (mod 3) and a + b 1 (mod 3), arriving at the same contradiction.) Also note gcd (a, b a) = gcd (a, b) = 1. Thus a generates the additive group Z b a. Suppose for sake of contradiction that χ is a valid 3-coloring of [1, 2a + 2b]. We claim that χ(x) = χ(x + b a) for all x [1 + a, b + 2a]. Assume, again for contradiction that χ(x) χ(x+b a) for an arbitrary x. Then χ(x), χ(x+b a), and χ(x a) are all distinct colors. Then χ(x + b) cannot be colored without causing a 2-term arithmetic progression with gap in D. Since x + b 2a + 2b, this contradicts our assumption that χ is a valid 3-coloring of [1, 2a + 2b]. This implies that if x y (mod b a) and x, y [1 + a, 2b + a], then χ(x) = χ(y). Let [x] denote the set {y : y [1 + a, 2b + a] and y x (mod b a)}. Hence, χ([x]) is a well-defined function. Note that [1 + a, b] contains exactly one representative of each equivalence class modulo b a. For all integers t, let ˆt satisfy ˆt t (mod b a) and ˆt [1 + a, b]. Thus ˆt + a [1+2a, a+b] [1+a, a+2b] and χ(ˆt) χ(ˆt+a). We can then conclude that χ([t+a]) χ([t]). Likewise, ˆt + a + b [1 + 2a + b, a + 2b] [1 + a, a + 2b], so χ([t]) χ([t + a + b]). By the same reasoning, χ([t + a]) χ([t + a + b]). Note that since t + a + b t + 2a (mod b a), this also means χ([t+a]) χ([t+2a]). Thus χ([t]), χ([t+a]), and χ([t+2a]) are all distinct colors. 5

6 Let x Z +. Letting t = x yields χ([x]), χ([x + a]), and χ([x + 2a]) are also all distinct colors. Likewise, letting t = x + a yields χ([x + a]), χ([x + 2a]), and χ([x + 3a]) are also all distinct colors, so it follows that χ([x]) = χ([x + 3a]) for all integers x. Recall that b a is the order of the group element a in Z b a and is not a multiple of three, so gcd (3, b a) = 1. Then 3a also generates the group Z b a, so every equivalence class modb a must be the same color, and all of [1 + a, b + 2a] must be monochromatic. But then {1 + a, 1 + 2a} is a two term monochromatic arithmetic progression with gap a. Thus there is no valid 3-coloring of [1, 2a + 2b] and w D (2; 3) 2a + 2b. Note that by Lemma 2.1, this result can be generalized to all sets of the form {a, b, a+b} with at least two different powers of 3 dividing a, b and a + b. 4 D = 3 with 3-term Arithmetic Progressions and 2 Colors The Conjecture of this section is the following: Conjecture 4.1. For every D Z + with D = 3, w D (3; 2) = The authors of this report believe to have found colorings for an arbitrary D, broken down into 8 distinct, exhaustive cases. One of the cases is presented in this report. Before the proof of the case is presented, we introduce variables to aid in the description of the cases. Let D = {a, b, c}, a < b < c and a, b, c Z +. Define the set S 1 as S 1 = {kb: kb < 2c, k Z + }. Then let β = max(s 1 ), so that β is the highest multiple of b before 2c. Furthermore, let S 2 = {b ka: b ka > 0, k Z + }. Then let d = min(s 2 ), so that d is the difference between the highest multiple of a before b and b. Finally, define α in the following manner: α = { 2c β if 2c β + a 2c β a if 2c >β + a Lemma 4.1. If b < 2a and 2c < a d (mod 2b) (thus α < a d and β is an even multiple of b), then the following coloring will always prevent a 3-term a.p. of gap in D: f(x) = 1 if x a (1) f(x) f(x a) if a < x b (2) f(x) f(x b) if b < x β b + a d = β 2d (3) f(x) = 1 if β 2d < x 2c (4) f(x) f(x 2c) if x > 2c (5) Proof. We begin with a remark. Let x be an arbitrary positive integer such that x β 2d. Note that if x [1, a] (mod ) or if x [b + a + 1, 2b] (mod 2b), then f(x) = 1. Otherwise, f(x) = 0. Let x 1, x 2, x 3 be a 3-term arithmetic progression where x 1 < x 2 < x 3. Also note that, since the coloring is periodic mod 4c, and the 2c integers following the first 2c integers are colored in the opposite way, it suffices to show that no monochromatic 6

7 arithmetic progression with gap a or b exists with x 1 [1, 2c] (One cannot exist with gap c, since if it did χ(x 3 ) = χ(x 1 + 2c) χ(x 1 )). Case 1. x 3 x 2 = x 2 x 1 = a Subcase 1.1. x 1 β b d. Recall that β is a multiple of b. Since d is the difference between the highest multiple of a before b and b, we know that x 1 a (mod b). If x 1 > b a (mod b), then x 2 = x 1 + a falls into the interval [1, a] mod b, at least one of the x 1, x 2 must be less than or equal to a b d (mod b). Let x i be the element that falls into this interval. Then f(x i+1 ) = f(x i + a) f(x i ), by defenition of the coloring up to β b d. Subcase 1.2. x 1 [β b d + 1, β b] Let x 1 [β b d + 1, β b], then x 2 [β b d + a, β b + a + 1] = [β 2d + 1, β d]. This implies that x 1 [a, b] (mod 2b) and x 2 [b + a d, b + a] (mod 2b). By the opening remark, f(x 1 ) = 0 and f(x 2 ) = 1. Thus x 1, x 2, and x 3 cannot be monochromatic when x 1 [β b d + 1, β b]. Subcase 1.3. x 1 [β b + 1, β b + a d] = [β b + 1, β 2d] Let x 1 [β b + 1, β b + a d] = [β b + 1, β 2d], then x 1 [b + 1, b + a d] (mod 2b) and therefore f(x 1 ) = 0, by the opening remark. This forces x 2 [β b + a + 1, β 2d + a] = [β b + a + 1, β 2d + a]. We know from (4) that f(x 2 ) = 1 when x 2 [β d + 1, 2c]. Thus x 1 and x 2 are both 0 if x 2 [β b, 2c], or rather when x 2 [2c + 1, β 2d + a] [2c + 1, 2c + a] (note that β 2d + a < 2c + a). Consider x 3 when x 3 [2c + a + 1, 2c + 2a] = [2c + a + 1, 2c + b + a d]. Note that 2c < x 3 < 4c, and f(x 3 ) f(x 3 2c). Since b + a d < b + a, f(x 3 2c) = 0, by the opening remark, so f(x 3 ) = 1. Thus x 1, x 2, and x 3 cannot be monochromatic when x 1 [1, β 2d]. Subcase 1.4. x 1 [β b + a d + 1, 2c] = [β 2d + 1, 2c] Let x 1 [β b + a d + 1, 2c] = [β 2d + 1, 2c], then f(x 1 ) = 1 by (4). Thus x 2 [β 2d + a + 1, 2c + a]. We know that x 1 and x 2 cannot be monochromatic if x 2 > 2c, since then x 2 [2c + 1, 2c + a], and f(x 2 ) f(x 2 2c) = 1. Note that if 2d + α a, then β 2d + a < 2c and x 2 [β 2d + a + 1, 2c]. Then f(x 2 ) = 1. Yet when x 3 [β 2d + 2a + 1, 2c + a], f(x 3 ) = 0 since β 2d + 2a = β + 2(a d) > β + α = 2c (recall that α < a d). Thus x 1, x 2, and x 3 cannot be monochromatic when x 1 [1, 2c]. Case 2. x 3 x 2 = x 2 x 1 = b Subcase 2.1. x 1 β b 2d It is obvious by (3) that x 1 and x 2 will not be monochromatic. Thus x 1, x 2, and x 3 cannot be monochromatic when x 1 [1, β b 2d]. Subcase 2.2. x 1 [β b 2d + 1, β b d] 7

8 Let x 1 [β b 2d + 1, β b d], so that x 2 [β b 2d + b + 1, β b d + b] = [β 2d, β d] and x 3 [β 2d + b + 1, β d + b] = [β + a d, β + a]. This implies that x 1 [a d + 1, a] (mod 2b), so f(x 1 ) = 1. Since α < a d, β + a d > β + α = 2c. Thus x 3 [2c+1, 2c+a] and therefore f(x 3 ) = 0. Thus x 1, x 2, and x 3 cannot be monochromatic when x 1 [β b 2d + 1, β b d]. Subcase 2.3. x 1 [β b d + 1, β b + α] = [β b d + 1, 2c b] Let x 1 [β b d+1, β b+α] = [β b d+1, 2c b], then x 1 [a+1, b+α] (mod 2b) and therefore f(x 1 ) = 0 since alpha < a. Since this forces x 2 [β b d + b + 1, β b + α + b] = [β d + 1, β + α] = [β d + 1, 2c], we know from (4) that f(x 2 ) = 1. Thus x 1, x 2, and x 3 cannot be monochromatic when x 1 [1, β b + α]. Subcase 2.4. x 1 [2c b + 1, 2c] Note that x 2 [2c + 1, 2c + b], so by defenition of the coloring, f(x 2 2c) f(x 2 ) and f(x 3 2c) f(x 3 ). Then since, these colors must be different since they differ by b and are less than β 2d (since β is an even multiple of b and β 2b), f(x 2 ) f(x 3 ). 5 Concluding Remarks and Further Questions The following was introduced in [2]: Definition 5.1. Let the function m(n) denote the least number K such that if D Z + and D = n, w D (K; 2) = The Conjecture of Section 4 shows that m(3) 3. It was shown (also in [2]) that m(3) 3, so we now conjecture that m(3) = 3. More information about this function would be nice to know, specifically a nice upper and lower bound. The values of m(1) and m(2) were given to be exactly 2 and 3. We conjecture the following as an upper bound: Conjecture 5.1. For every n 1, m(n) n + 1 Furthermore, to extend the notion to multiple colors, we define the following: Definition 5.2. Let m r (n) denote the least number K such that if D Z + and D = n, w D (K; r) = Note then that m(n) = m 2 (n). The results of Section 3 show that m 3 (3) 3, and since every 2-coloring is a 3-coloring as well,the results of Section 4 would imply that m 3 (3) 3. So we conjecture that m 3 (3) = 3. In general, we have that if r 1 < r 2, m r1 (n) m r2 (n). Note that Proposition 2.2 shows that m 3 (1) = m 3 (2) = 2. While the lower bound provided for w D (2; 3) when D = {a, b, a + b} is sharp (take, for example, the case when D = {1, 2, 3}), the upper bound does not appear to be. We conjecture that the upper bound provided could be lowered to w D (3; 2) 2(a + b) 1. It cannot be lowered past this value, as when D = {3, 5, 8}, w D (2; 3) = 15. 8

9 Conjecture 5.2. If a, b Z + and D = {a, b, a + b} with 3 dividing at least one of the a, b or a + b, w D (2; 3) 2(a + b) 1 Another conjecture, supported by computer output for every 3-element subset of [1, 25] is the following: Conjecture 5.3. If D = 3 and D is not of the form {a, b, a + b} with at least one element divisible by 3 or of the form {1, 2, 3k}, then w D (2; 3) = References [1] B. Landman. On avoiding arithmetic progressions whose common differences belong to a given small set. The Journal of Combinatorial Mathematics and Combinatorial Computing, 30: , [2] B. Landman and A. Robertson. Ramsey Theory on the Integers, volume 24 of Student Mathematical Library. American Mathematical Society,