Rainbow arithmetic progressions

Transcription

1 Raibow arithmetic progressios Steve Butler 1 Craig Erickso 2 Leslie Hogbe 1,3 Kirste Hogeso 1 Lucas Kramer 4 Richard L. Kramer 1 Jephia Chi-Hug Li 1 Rya R. Marti 1,5 Derrick Stolee 1,6 Natha Warberg 7 Michael Youg 1,8,9 Jauary 4, 2016 Abstract I this paper, we ivestigate the ati-ramsey (more precisely, ati-va der Waerde) properties of arithmetic progressios. For positive itegers ad k, the expressio aw([], k) deotes the smallest umber of colors with which the itegers {1,..., } ca be colored ad still guaratee there is a raibow arithmetic progressio of legth k. We establish that aw([], 3) = Θ(log ) ad aw([], k) = 1 o(1) for k 4. For positive itegers ad k, the expressio aw(z, k) deotes the smallest umber of colors with which elemets of the cyclic group of order ca be colored ad still guaratee there is a raibow arithmetic progressio of legth k. I this settig, arithmetic progressios ca wrap aroud, ad aw(z, 3) behaves quite differetly from aw([], 3), depedig o the divisibility of. As show i [Jugić et al., Combi. Probab. Comput., 2003], aw(z 2 m, 3) = 3 for ay positive iteger m. We establish that aw(z, 3) ca be computed from kowledge of aw(z p, 3) for all of the prime factors p of. However, for k 4, the behavior is similar to the previous case, that is, aw(z, k) = 1 o(1). Keywords. arithmetic progressio; raibow colorig; ati-ramsey; Behred costructio. AMS subject classificatios. 05D10, 11B25, 11B30, 11B50, 11B75. 1 Itroductio Let G be a additive (abelia) group such as the itegers or the itegers modulo, ad let S be a fiite oempty subset of G. A k-term arithmetic progressio (k-ap) i S is a set of distict elemets of the form a, a + d, a + 2d,..., a + (k 1)d where d 1 ad k 2. A r-colorig of S is a fuctio c : S [r], where [r] := {1,..., r}. We say such a colorig is exact if c is surjective. Give a r-colorig c of S, the i th color class is C i := {x S : c(x) = i}. A arithmetic progressio is called raibow if the image of the progressio uder the r-colorig is ijective. Formally, give c : S [r] we say a k-term arithmetic progressio is raibow if {c(a+id) : i = 0, 1,..., k 1} has k distict values. The ati-va der Waerde umber aw(s, k) is the smallest r such that every exact r-colorig of S cotais a raibow k-term arithmetic progressio. Note that this tautologically defies aw(s, k) = S + 1 wheever 1 Departmet of Mathematics, Iowa State Uiversity, Ames, IA 50011, USA. {butler, lhogbe, kahoges, ricardo, chli, rymarti, dstolee, myoug}@iastate.edu 2 Departmet of Mathematics ad Computer Sciece, Grad View Uiversity, Des Moies, IA 50316, USA. (cerickso@gradview.edu) 3 America Istitute of Mathematics, 600 E. Brokaw Road, Sa Jose, CA 95112, USA (hogbe@aimath.org). 4 Mathematics Departmet, Bethel College, North Newto, KS 67117, USA. (lkramer@bethelks.edu) 5 Research supported i part by Natioal Security Agecy grat H Departmet of Computer Sciece, Iowa State Uiversity, Ames, IA 50011, USA. 7 Departmet of Mathematics ad Statistics, Uiversity of Wiscosi-La Crosse, La Crosse, WI 54601, USA. (warberg@uwlax.edu 8 Research supported i part by NSF DMS Correspodig author. 1

2 S < k, ad this defiitio retais the property that there is a colorig with aw(s, k) 1 colors that has o raibow k-ap. Sice aw(s, 2) = 2 for all S, we assume heceforth that k 3. Several importat results o the existece of raibow 3-APs implyig iformatio about aw([], 3) ad aw(z, 3) (i our otatio) have bee established by Jugić, et al. [8]. A prelimiary study of the ati-va der Waerde umber was doe by Uherka i [13]; it should be oted the otatio there is slightly differet, with AW (k, ) used to deote our aw([], k). Other results o balaced colorigs of the itegers with o raibow 3-AP have bee obtaied by Axeovich ad Fo-Der-Flaass [1] ad Axeovich ad Marti [2]. First, we cosider the set S = []. The value of aw([], 3) is logarithmic i : Theorem 1.1. For every iteger 9, log aw([], 3) log Moreover, aw([], 3) = log for {3, 4, 5, 6, 7} ad aw([8], 3) = 5. Theorem 1.1 is prove by Lemmas 2.3 ad 2.6 (for 9), ad Remark 2.1 gives exact values of aw([], 3) that justify the secod statemet. We cojecture that the lower boud is, essetially, correct: Cojecture 1.2. There exists a costat C such that aw([], 3) log 3 + C for all 3. The behavior of aw([], k) is, however, differet for k 4. Istead of logarithmic, it is almost liear: Theorem 1.3. For k 4, e O( log ) < aw([], k) e log log log ω(1). Theorem 1.3 is established by Lemma 2.8 ad Corollary Fially, we cosider arithmetic progressios i the cyclic group Z. Remark 1.4. For positive itegers ad k, aw(z, k) aw([], k), because every AP i [] correspods to a AP i Z. However, because progressios i Z may wrap aroud, there are additioal APs i Z, some of which may be raibow. Thus it is possible that every colorig of Z with aw([], k) 1 colors guaratees a raibow k-ap, so strict iequality is possible. As was show i [8, Theorem 3.5] (ad follows from Theorem 1.6 below), there are ifiitely may values of for which aw(z, 3) = 3, for example, whe is a power of two. Defiitio 1.5. Let 3 be a iteger. Defie f 2 () to be 0 if is odd ad 1 if is eve. Defie f 3 () to be the umber of odd prime factors p of that have aw(z p, 3) = 3 ad f 4 () to be the umber of odd prime factors p of that have aw(z p, 3) = 4, both couted accordig to multiplicity. Theorem 1.6. For every prime umber p, 3 aw(z p, 3) 4. For a iteger 2, the value of aw(z, 3) is determied by the values of aw(z p, 3) for the prime factors p: aw(z, 3) = 2 + f 2 () + f 3 () + 2f 4 (). For a iteger 2 havig every prime factor less tha 100, f 4 () is the umber of odd prime factors of i the set Q 4 := {17, 31, 41, 43, 73, 89, 97} ad f 3 () is the umber of odd prime factors of i Q 3, where Q 3 is the set of all odd primes less tha 100 ad ot i Q 4. Theorem 1.6 is established by Propositio 3.5, Corollary 3.15 ad Propositio For k 4, the bouds we obtai for aw(z, k) are the same as those for aw([], k): Theorem 1.7. For k 4, e O( log ) < aw(z, k) e log log log ω(1). 2

3 Theorem 1.7 is established by Remark 1.4 ad Lemma The structure of the paper is as follows: Sectio 2 presets results pertaiig to aw([], k), with Theorem 1.1 proved i Sectio 2.1 ad Theorem 1.3 proved i Sectio 2.2. Results pertaiig to aw(z, k) appear i Sectio 3, with Theorem 1.6 proved i Sectio 3.1 ad Theorem 1.7 proved i Sectio 3.2. Sectio 4 describes the methods ad algorithms used to compute values of aw([], k) ad aw(z, k), while Sectio 5 cotais cojectures ad ope questios for future research. I the remaider of this sectio we establish a basic but ecessary observatio that aw(s, ) is mootoe i k. Observatio 1.8. Let G be a additive (abelia) group such as the itegers or the itegers modulo, let S be a fiite oempty subset of G, ad let k 3 be a iteger. The aw(s, k) aw(s, k + 1). Observatio 1.8 follows immediately from Propositio 1.9 below ad was oted oted by Uherka i [13] for the fuctio aw([], ). Propositio 1.9. Let G be a additive (abelia) group such as the itegers or the itegers modulo, let S be a fiite oempty subset of G, ad let k 3 be a iteger. If there is a exact r-colorig of S that has o raibow k-ap, the aw(s, k) r + 1. Proof. Let c be a exact r-colorig of S with color set {1,..., r} that has o raibow k-ap. We proceed by costructig a exact (r 1)-colorig of S with o raibow k-ap. For x S, defie { c(x) if c(x) {1,..., r 2}, ĉ(x) = r 1 if c(x) {r 1, r}. Note that ĉ is a exact (r 1)-colorig of S. Let K be a k-ap i S. Sice there is o raibow k-ap uder c there exists j, l K such that c(j) = c(l). It the follows that ĉ(j) = ĉ(l). Hece K is ot raibow uder the colorig ĉ. By the geerality of K, ĉ is a exact (r 1)-colorig of S that has o raibow k-ap. Repeatig this costructio we obtai a exact (r i)-colorig of S with o raibow k-ap for i {1, 2,..., r 1}. Therefore aw(s, k) r aw([], k) I this sectio we establish properties of aw([], k). Sectios 2.1 ad 2.2 establish our mai results for aw([], 3) ad aw([], k), k 4, respectively. Sectios 2.3 ad 2.4 cotai additioal results valid for all k ad specific to k = 3, respectively. I Table 1 we give our calculated values of aw([], k) for k 3. We have a larger list of kow values i the case of k = 3 that is icluded i Remark 2.1 below; i Table 1 we iclude oly the values aw([], 3) for which we have a value for aw([], 4) so that we may compare them. We also restrict, k 3, ad have stopped with k = 2 + 1, because aw([], k) = if ad oly if k (Propositio 2.16 below). The growth rates whe k = 3 ad whe k 4 appear to be differet based o data give i Table 1. The upper boud of log give i Propositio 2.6 for k = 3 ad the lower boud of 1 o(1) i Lemma 2.8 for k 4 cofirm that the growth rates are ideed radically differet. 2.1 Mai results for aw([], 3) Before we address Theorem 1.1, we show a summary of the computed data for this case i Remark 2.1 below. Remark 2.1. The exact values of aw([], 3) are kow from computer computatios (described i Sectio 4) for 58, ad are recorded here. 1. aw([], 3) = 2 for {1}. 2. aw([], 3) = 3 for {2, 3}. 3. aw([], 3) = 4 for {4,..., 7} {9}. 3

4 \ k Table 1: Values of aw([], k) for 3 k aw([], 3) = 5 for {8} {10,..., 21} {27}. 5. aw([], 3) = 6 for {22,..., 26} {28,..., 58}. Now we tur to the proof of Theorem 1.1, begiig with the lower boud. Propositio 2.2. Let be a positive iteger ad let s { 2, 1, 0, 1, 2}. The aw([3 s], 3) aw([], 3)+1 provided s. Proof. Let r = aw([], 3) ad s {0, 1, 2}. We costruct a exact r-colorig of [3 s] that does ot cotai a raibow 3-AP. By defiitio there exists a exact (r 1)-colorig, deoted c, of [] such that there is o raibow 3-AP i []. Color [3 s] i the followig maer: If i + s is divisible by 3, the ĉ(i) = c((i + s)/3), otherwise ĉ(i) = r. Cosider a 3-AP, K, i [3 s]. The either the three terms i K + s are all divisible by 3 or at least two of the terms i K + s are ot divisible by 3. If all terms i K + s are divisible by 3, the K is ot raibow uder ĉ, sice there is o raibow 3-AP uder c. If two terms of K + s are ot divisible by 3 the those two terms are both colored r ad K is ot raibow. Hece aw([3 s], 3) r + 1 for s {0, 1, 2}. For s { 2, 1}, use the same colorig as for s = 0. Usig Propositio 2.2, we establish the lower boud i Theorem 1.1. Lemma 2.3. Let be a positive iteger. The aw([], 3) log Proof. The proof is by iductio. The cases = 1, 2, 3 are true by ispectio. Suppose > 3 ad that aw([m], 3) log 3 m + 2 for all m satisfyig 1 m <. We show that aw([], 3) log First, we write = 3m s, where s {0, 1, 2} ad 2 m <. The by Propositio 2.2, aw([], 3) = aw([3m s], 3) aw([m], 3) + 1 log 3 m = log 3 (3m) + 2 log Example 2.4. Iductio ad the proof of Propositio 2.2 produce the followig exact (m + 1)-colorig of [3 m ] that does ot have a raibow 3-AP: For x [3 m ] with the prime factorizatio x = 2 e2 3 e3 5 e5 p ep, c(x) = m + 1 e 3. This attais the value i Lemma

5 To complete the proof of Theorem 1.1, we establish the upper boud. Lemma 2.5. Let c be a exact r-colorig of [] that does ot have a raibow 3-AP. For i [r], defie b i [] to be the least x such that the iduced colorig o [x] has exactly i colors. The for all i [r 1], b i+1 2b i. Furthermore, for ay 1 i j r, we have b j 2 j i b i. Proof. Observe that b 1 = 1. Suppose that b i+1 < 2b i for some i [r 1]. The {2b i b i+1, b i, b i+1 } is a raibow 3-AP. The last statemet follows by iductio, sice b i 2 1 b i b i+2 2 (j i) b j. Lemma 2.6. For 9, aw([], 3) log Proof. Suppose r = aw([], 3) 1, so there is a r-colorig with o raibow 3-AP. Lemma 2.5 implies that b r 2 r 1. Thus aw([], 3) log 2 + 2, which establishes the result for ot a power of 2. The case aw([2 m ], 3) m + 1 follows similarly by usig the fact that aw([2 m ], 3) = m + 1 for m = 4 ad m = 5 (see Remark 2.1); aw([16], 3) = 5 implies b 5 > 16 = 2 4 for ay raibow-free colorig with r 5. The for m > 5, Lemma 2.5 implies a r-colorig of 2 m has 2 m b r 2 r 5 b 5 > 2 r 1, so m r ad m + 1 aw(2 m, 3). This completes the proof of Theorem Mai results for aw([], k), k 4 I this sectio we specialize to the case k 4, focusig o lower ad upper bouds that give aw([], k) = 1 o(1). Lemma 2.8 gives the lower boud ad Corollary 2.14 gives the upper boud. Let sz(, k) deote the largest size of a set S [] such that S cotais o k-ap (similar otatio was itroduced i [5] i hoor of Szemerédi [12]). Determiig bouds o sz(, k) is a fudametal problem i the study of arithmetic progressios. Behred [3], Gowers [6], ad others [9, 10] have established various bouds o sz(, k). Propositio 2.7 provides a strog lik betwee sz(, k) ad our ati-va der Waerde umbers, allowig us to use kow results o sz(, k) to boud aw(, k). Propositio 2.7. For all > k 3, sz(, k/2 ) + 1 aw([], k) 1 sz(, k). Proof. If c is a exact r-colorig of [] that cotais o raibow k-ap, the selectig oe elemet of each color class creates a set S that cotais o k-ap; therefore aw([], k) 1 sz(, k). If S is a set i [] that cotais o k/2 -AP, the color [] by givig each elemet i S a distict color ad the elemets of [] \ S a ew color. If a k-ap {a 1, a 2,..., a k } is raibow i this colorig, the exactly oe such a i is i [] \ S. But this implies that the etries a j where j i (mod 2) form a AP i S with at least k/2 terms, a cotradictio Theorem 1.3: Proof of lower boud Lemma 2.6 ad Behred s results (stated i Theorem 2.10 ad Propositio 2.11 below) show that the upper boud i Propositio 2.7 is ot useful for k = 3. Observe that whe k {4, 5}, the lower boud i Propositio 2.7 is trivial but is i fact useful i the case of k 6. We provide a similar lower boud for k {4, 5} i Lemma 2.8 by carefully studyig Behred s origial costructio [3] of a relatively large set S [] that cotais o 3-AP, thus givig a lower boud o sz(, 3). Let {a 1, a 2, a 3, a 4 } be a 4-AP with a 1 a 2 a 3 a 4. A set A {a 1, a 2, a 3, a 4 } of size A = 3 is called a puctured 4-AP. If such a puctured 4-AP A is ot a 3-AP, the it is of the form A = {a 1, a 2, a 4 } or A = {a 1, a 3, a 4 }. We prove that Behred s costructio i fact cotais o puctured 4-AP (Propositio 2.9 below). This leads to Lemma 2.8 below. Lemma 2.8. There exists a absolute costat b > 0 such that for all, k 4, aw([], k) > e b log = 1 o(1). 5

6 The proof of Lemma 2.8 follows from Propositio 2.9, Theorem 2.10, Propositio 2.11 ad Propositio 2.12, which follow. Propositio 2.9. Suppose S [] does ot cotai ay puctured 4-APs. The aw([], k) > S + 1 for all k 4. Proof. Color each member of S a distict color, ad color each iteger i [] \ S with a ew color called zero. If there is a raibow 4-AP i this colorig, the at most oe of the elemets i this 4-AP is colored zero. Thus there must be a puctured 4-AP i the other colors, but S cotais o puctured 4-AP. There is a bijectio betwee vectors x = (x 1,..., x m ) Z m where x i {0, 1,..., 2d 2} for all i [m] ad elemets of {0, 1,..., (2d 1) m 1}, by viewig x as a (2d 1)-ary represetatio of a iteger: x = (x 1,..., x m ) a x = m x i (2d 1) i 1. Moreover, observe that if x, y Z m with x i, y i {0,..., d 1}, i = 1,..., m, are associated with a x, a y {0, 1,..., (2d 1) m 1} by this bijectio, the x + y has x i + y i {0,..., 2d 1} ad x + y is associated with a x+y = a x + a y {0, 1,..., (2d 1) m 1}. Recall that for a vector x R m, x 2 = m i=1 x2 i. Let m, l, d be positive itegers ad defie X l(m, d) to be the set of vectors x = (x 1,..., x m ) such that 1. x i {0,..., d 1} for all i {1,..., m}, ad 2. x 2 = l. The set S l (m, d) of itegers associated with the vectors i X l (m, d) via the map x a x forms a subset of itegers i {0, 1,..., (2d 1) m 1}. Behred [3] used the pigeohole priciple to prove the followig lemma; here we state the versio from [4]. Theorem [3, 4] There exist absolute costats b, b > 0 such that for all ad positive itegers m = m(), l = l(), ad d = d() such that S l (m, d) [] the followig iequality holds: i=1 b S l (m, d) e b log. 2 8 log2 (log ) 1/4 The importat property of S l (m, d) is that it avoids o-trivial arithmetic progressios. Behred s simple proof of this fact for completeess. Propositio [3] The set S l (m, d) cotais o 3-AP. We iclude Proof. Suppose {a x1, a x2, a x3 } is a 3-AP i S l (m, d). Let x 1, x 2, x 3 be the associated vectors i X l (m, d). Sice a x1 + a x3 = 2a x2, we also have that x 1 + x 3 = 2x 2. See Figure 1. However, by the triagle iequality, we have that 2 l = 2 x 2 = x 1 + x 3 x 1 + x 3 = 2 l, ad equality ca oly hold if 0, x 1, x 3 ad 2x 2 are colliear. However, sice x 1 = x 3, this would imply x 1 = x 3 ad thus a x1 = a x3, a cotradictio. Propositio The set S l (m, d) cotais o puctured 4-AP. Proof. Let {a x1, a x2, a x3, a x4 } be a 4-AP. Sice S l (m, d) cotais o 3-AP, it must be that oe of a x2 or a x3 is ot i S l (m, d). Assume by symmetry that a x2 / S l (m, d) ad a x1, a x3, a x4 S l (m, d). Let x 1, x 2, x 3, x 4 be the associated vectors where x 1, x 3, x 4 X l (m, d). Sice a x1 +a x3 = 2a x2, we have x 1 +x 3 = 2x 2. See Figure 1. However, as i the proof of Propositio 2.11, this implies that x 2 < l. Sice a x2 + a x4 = 2a x3, we have x 2 + x 4 = 2x 3. However, this implies that a cotradictio. 2 l = 2 x 3 = x 2 + x 4 x 2 + x 4 < 2 l, 6

7 x 3 2x 2 x 1 x 4 2x 3 x3 x 2 2x 2 x 1 l 2 l (a) Propositio l 2 l (b) Propositio Figure 1: Proofs of Propositios 2.11 ad Lemma 2.8 ow follows by combiig Propositios 2.9 ad It may be possible that the boud i b(log )1/4 Lemma 2.8 could be improved by usig the costructio of Elki [4, 7] that avoids 3-APs usig 2 8 log2 elemets for some costat b > 0. Sice this costructio avoids a 3-AP, we ca use Propositio 2.7 directly i order to obtai a colorig with o 6-AP, givig aw([], k) > b(log )1/4 2 8 log2 for all k 6. Further use of costructios of Raki [10] or Laba ad Lacey [9] of large sets that avoid k-aps could slightly improve the asymptotics of aw([], k), but these bouds are all of the form 1 o(1) Theorem 1.3: Proof of upper boud A theorem of Gowers, stated here as Theorem 2.13, provides a upper boud for aw([], k). However, must be very large compared to k for this upper boud to be sigificatly differet tha the aïve upper boud of itself. Theorem [6, Theorem 1.3] For every positive iteger k there is a costat b = b(k) > 0 such that every subset of [] of size at least (log 2 log 2 ) b cotais a k-ap. Moreover, b ca be take to be 2 2k+9. Corollary Let ad k be positive itegers. The there exists a costat b such that aw([], k) (log2 log 2 ) b. That is, for a fixed positive iteger k, the fuctio aw([], k) of is o( log log ). Proof. Cosider a exact t-colorig of [], where t := (log 2 log 2 ) b ad b = 2 2k+9. Sice the colorig is exact, there exists a set A [] of t differetly colored itegers. By Theorem 2.13, A cotais a k-ap. Therefore aw([], k) t. Note that the upper boud i Corollary 2.14 ca be expressed as e log log log ω(1). The combiig this upper boud o aw([], k) ad the lower boud from Lemma 2.8, we have that for k 4 This completes the proof of Theorem 1.3. e b log < aw([], k) e log log log ω(1). 2.3 Additioal results for aw([], k) valid for all k I this sectio we preset some additioal elemetary results for aw([], k). The ext propositio describes a relatioship betwee aw([], k) ad aw([ 1], k). Propositio Let ad k be positive itegers. The aw([], k) aw([ 1], k) + 1. Proof. Let r = aw([], k). Note that if < k our result follows from the defiitio. Suppose k. The there is some exact (r 1)-colorig of [] that has o raibow k-ap, ad without loss of geerality is colored r 1. Cosider this colorig restricted to [ 1]. The we have two cases: 7

8 1. This is a exact (r 1)-colorig of [ 1]. 2. The oly iteger i [] with the color r 1 is, so this is a exact (r 2)-colorig of [ 1]. Note that sice [] had o raibow k-ap i both of our cases we still do ot have a raibow k-ap. So by Propositio 1.9 we have aw([ 1], k) r 1 = aw([], k) 1 ad the result follows. I the ext propositio we characterize the values of k for which aw([], k) =. Propositio Let ad k be positive itegers with k. k The aw([], k) = if ad oly if Proof. Suppose k We show that aw([], k) > 1. Color 2 ad with the same color ad all the remaiig itegers with uique colors. This is a exact ( 1)-colorig. Sice k 2 + 1, the itegers i ay k-ap must be cosecutive itegers, ad the values 2 ad must be cotaied i ay k-ap. Hece o k-ap is raibow. For the coverse, suppose aw([], k) =. Color [] with 1 colors such that there is o raibow k-ap. Therefore exactly oe color class has size two ad the rest have size oe. Deote the color class of size two as C = { 1, 2 }, 1 < 2. The every k-ap cotais both 1 ad 2, or else we would have a raibow k-ap. Suppose that k 2. The {1, 2,..., k} ad { k + 1, k + 2,..., } are k-aps. Note that {1, 2,..., k} {1, 2,..., 2 } ad { k + 1, k + 2,..., } { 2 + 1,..., }. The 1, 2 {1, 2,..., 2 } { 2 + 1,..., }. This itersectio is empty or cotais oe elemet depedig o whether is eve or odd. I both cases, this cotradicts the fact that 1 2 ad 1, 2 {1, 2,..., 2 } { 2 + 1,..., }. Therefore k The followig upper boud was proved by Uherka [13]; we iclude the brief proof for completeess. Propositio [13] Let, k, 1, ad 2 be positive itegers such that k 1 2 ad =. The aw([], k) aw([ 1 ], k) + aw([ 2 ], k) 1. Proof. Let r = aw([ 1 ], k) + aw([ 2 ], k) 1, ad cosider a arbitrary exact r-colorig c of []. Let r 1 = c([ 1 ]) ad r 2 = c({ 1 + 1,..., }). Sice =, r r 1 + r 2. This implies that r 1 aw([ 1 ], k) or r 2 aw([ 2 ], k). Clearly r 1 aw([ 1 ], k) implies c has a raibow k-ap. By traslatig c({ 1 + 1,..., }) to a colorig o [ 2 ], we also see that c has a raibow k-ap if r 2 aw([ 2 ], k). Thus aw([], k) r = aw([ 1 ], k) + aw([ 2 ], k) Additioal results for aw([], 3) I this sectio we establish additioal bouds o aw([], 3) i Propositios 2.18 ad 2.19, ad use Propositio 2.19 together with Remark 2.1, Propositio 2.2, ad Lemma 2.3 to compute (at least) 93 additioal exact values for aw([], 3). Propositio For 2, there exists m 2 such that aw([], 3) aw ([m], 3) + 1. Proof. We may assume that 3, sice the case = 2 follows by ispectio. Let r = aw([], 3). The there exists a (r 1)-colorig, amely c, of [] that has o raibow 3-AP. Let t be the legth of a shortest cosecutive iteger sequece i [] that cotais all r 1 colors, say the iterval is {s + 1, s + 2,..., s + t} for some s. Defie ĉ to be a (r 1)-colorig of [t] = {1, 2,..., t} so that ĉ(j) := c(s + j) for 1 j t. Notice that ĉ(1) ad ĉ(t) caot be the same color ad each must be the oly elemet of its color class, or else we could fid a smaller t. Let ĉ(1) = a ad defie b i to be the smallest elemet of [t] such that [b i ] has i+1 colors for 1 i r 2. Note that if b i is odd, i.e., b i = 2x + 1, the {1, x + 1, 2x + 1} is a raibow 3-AP. So the set of eve umbers of [t] are colored with exactly r 2 colors with o raibow 3-AP. Defie m = t 2 2 ad cosider the colorig c of [m] iduced by the colorig ĉ of the eve itegers i [t]. The colorig c uses at least r 2 colors ad has with o raibow 3-AP, so aw([], 3) 1 = (r 2) + 1 aw ([m], 3). 8

9 Propositio Let m,, ad l be positive itegers. If m < < 2 l (m+1), the aw([], 3) aw([m], 3)+ l. Proof. Suppose ot. The there exists m, l 1 ad with m < < 2 l (m + 1) such that there is a colorig c o [] usig exactly r = aw([m], 3) + l colors that does ot have a raibow 3-AP. For i [r], let b i [] be the least x such that the iduced colorig o [x] has exactly i colors. Sice r l = aw([m], 3), we must have b r l m + 1, sice otherwise the iduced colorig o [m] cotais at least aw([m], 3) colors, which is impossible. Thus by Lemma 2.5, b r 2 l b r l 2 l (m + 1), which cotradicts our assumptio o. Corollary aw([], 3) = 7 for Proof. Sice aw([m], 3) = 6 for 22 m 26, ad = 64 ad = 80, we see that aw([], 3) 7, by Propositio 2.2. Sice 27 < < 112 = 4 28, we have aw([], 3) aw([27], 3) + 2 = = 7 by Propositio 2.19 ad Remark 2.1. Corollary aw([], 3) = 7 for Proof. Sice 27 < < 112 = 4 28, we have aw([], 3) aw([27], 3) + 2 = = 7 by Propositio 2.19 ad Remark 2.1. Also, sice 3 4 = 81 < 111 < 243 = 3 5, we have 4 < log 3 5, so that aw([], 3) log = = 7 by Lemma 2.3. Corollary aw([], 3) = 8 for Proof. Sice aw([m], 3) = 7 for 64 m 80, ad = 190 ad = 242, we see that aw([], 3) 8 for , by Propositio 2.2. For 58 < < 236 = 2 2 (58 + 1), we see that aw([], 3) aw([58], 3) + 2 = = 8, by Propositio 2.19 ad Remark 2.1. Fially we combie the upper ad lower bouds. Propositio If 3 u < < 2 3 u +2, the u+3 aw([], 3) aw([3 u ], 3)+1. If 2 3 u +1 < < 4 3 u +4, the u + 3 aw([], 3) aw([3 u ], 3) + 2. Proof. The lower bouds follow immediately from Lemma 2.3, ad the first upper boud follows immediately from Propositio For the secod, apply Propositio 2.19 with m = 2 3 u + 1 ad l = 1 to obtai aw([], 3) aw([m], 3) + 1 ad sice 3 u < m < 2 3 u + 2, aw([m], 3) aw([3 u ], 3) aw(z, k) I this sectio we establish properties of aw(z, k). Sectios 3.1 ad 3.2 establish our mai results for aw(z, 3) ad aw(z, k), k 4, respectively. Sectio 3.3 cotais additioal results. Please ote that for x Z, we will also use x to deote the equivalece class {x + i : i Z} i Z. Because arithmetic progressios may wrap aroud i the group Z, we call attetio to the fact that we cosider oly k-aps that iclude k distict members of Z. Naturally, oe of our first questios about aw(z, k) cocers its relatioship with aw([], k). Theorem 3.3(a) below ad Lemma 2.3 show that aw(z, k) eed ot be asymptotic to aw([], k) for k = 3 ad = 2 m. However, we do have the simple boud aw(z, k) aw([], k) (already stated i Remark 1.4). 9

10 Table 2: Computed values of aw(z, 3) for = 3,..., 99 (the row label gives the rage of ad the colum headig is the oes digit withi this rage). 3.1 Mai results for aw(z, 3) Whe we tur to the special case k = 3, may values of aw(z, 3) ca be computed, ad ew pheomea appear. Our mai results i this case are described by Theorem 1.6, which we establish i this sectio. Curretly available computatioal data is give i Table 2; the row label displays the rage of for which the values of aw(z, 3) are reported i that row, ad the colum headig is the oes digit withi this rage. This data led to the discovery of several results established i this sectio ad is used to establish the secod statemet i Theorem 1.6 that cocers itegers havig all prime factors less tha oe hudred. May odd primes p have aw(z p, 3) = 3 (see Table 2 above). However, there are several examples of odd primes p for which aw(z p, 3) = 4. I Example 3.1 below we exhibit a explicit exact colorig that establishes aw(z 17, 3) 4. Example 3.1. Colorig the elemets of Z 17 i order as is a exact 3-colorig that does ot cotai a raibow 3-AP. Computatios establish that equality holds ad so aw(z 17, 3) = 4. Defiitio 3.2. Whe dealig with a colorig c of Z st, the i th residue class modulo s is R i := {j Z st : j i (mod s)} ad the i th residue palette modulo s is P i := {c(l) : l R i }. For a positive iteger t, we call the elemets of the two residue classes, R 0 ad R 1, modulo 2 i Z 2t the eve umbers ad the odd umbers, respectively Cosequeces of results i Jugić et al. I this sectio we state two importat results of Jugić et al. [8] ad derive implicatios. These are used i the proof of Theorem 1.6. The ext result is a equivalet form of Theorem 3.5 i that paper. Theorem 3.3. [8, Theorem 3.5] Let be a positive iteger. The aw(z, 3) = 3 if ad oly if oe of the followig coditios is satisfied: a) is a power of 2, b) is prime ad 2 is a geerator of the multiplicative group Z, c) is prime, 1 2 is odd, ad the order of 2 i Z is 1 2. Theorem 3.4. [8, Theorem 3.2] Let be a odd positive iteger ad q be the smallest prime factor of. The every 3-colorig of Z i which every color class has at least q + 1 elemets cotais a raibow 3-AP. A colorig c of Z is a extremal colorig if c is a exact (aw(z, 3) 1)-colorig of Z with o raibow 3-AP. A colorig c of Z is a sigleto colorig if some color is used exactly oce. 10

11 Propositio 3.5. Let p be a prime positive iteger. The 3 aw(z p, 3) 4, ad aw(z p, 3) = 4 implies every extremal colorig of Z p is a sigleto colorig. Proof. First we suppose aw(z p, 3) 5 ad let c be a extremal colorig with r = aw(z p, 3) 1 4 colors. That is, c does ot have a raibow 3-AP. Hece, there is at least oe color class with more tha oe elemet. We ca defie a 3-colorig ĉ by partitioig the color classes of c ito three sets ad defiig the color classes of ĉ to be the uios of the color classes i the sets. Clearly ĉ is a 3-colorig of Z p that does ot have a raibow 3-AP. By Theorem 3.4, there exists a color class of ĉ with < p p + 1 = 2 elemets. This meas that all but oe of the color classes has a sigle elemet (ad r = 4). Without loss of geerality, let the sigleto colors be i positios 0, x, ad y, with 0 < x < y < p (whe viewed as itegers rather tha elemets of Z p ). I order to avoid the raibow 3-AP i c cosistig of 0, x 2, x, x must odd, ad similarly y must be odd as well. But the x, y x 2, y is a a raibow 3-AP i c, cotradictig aw(z p, 3) 5. Next, suppose that aw(z p, 3) = 4 ad let c be a extremal colorig of r = 3 colors. Sice c has o raibow 3-AP, Theorem 3.4 gives that there is a color class with oe elemet. That is, c must be a sigleto colorig. Sice aw(z p, 3) = 3 implies Z p has the sigleto extremal colorig c(0) = 1 ad c(i) = 2 for every i 0 (mod p), the ext corollary is immediate. Corollary 3.6. Every prime p has a sigleto extremal colorig of Z p. Sice aw(z 2 m, 3) = 3, there are ifiitely may values of for which aw(z, 3) = 3. As stated i Theorem 1.6, aw(z, 3) ca be be made arbitrarily large ad computed from the values of aw(z p, 3) for the prime factors p of. For primes p, aw(z p, 3) > 3 seems rare from the data i Table 2. However, it follows from Theorem 3.3 that there are ifiitely may primes p such that aw(z p, 3) = 4: Corollary 3.7. If p is a prime ad p 1 (mod 8), the aw(z p, 3) = 4. There are ifiitely may such primes. Proof. Sice p 1 (mod 8), p 1 2 is eve. Also, 2 must be a square i Z p, which implies it is ot a geerator of Z p. So by Theorem 3.3, aw(z p, 3) 3. The by Propositio 3.5, aw(z p, 3) = 4. By Dirichlet s Theorem there are ifiitely may primes p 1 (mod 8) Proof of Theorem 1.6 I this sectio we preset a series of results that lead to equivalet lower ad upper bouds o aw(z, 3) i terms of the prime factorizatio of, establishig Theorem 1.6. The ext result gives our mai recursive upper boud for aw(z, 3). Propositio 3.8. Suppose s is odd, ad either t is odd or t = 2 m. The aw(z st, 3) aw(z t, 3) + aw(z s, 3) 2. Propositio 3.8 is established by Propositios 3.10 (t odd) ad 3.13 (t = 2 m ) below, after the proofs of ecessary prelimiaries. Propositio 3.9. Let s be a odd positive iteger. Suppose c is a colorig of Z st that does ot have a raibow 3-AP. Let R 0, R 1,..., R s 1 be the residue classes modulo s i Z st with associated residue palettes P i. Let m be a idex such that P m P i for all i. The P i \ P m 1 for all i. Proof. For arbitrary oegative itegers h ad j, we show that P h+j \P h 2 implies P h = P h+2j. Assume P h+j \ P h 2. Suppose first that P h+2j \ P h is ot empty ad z P h+2j \ P h. Sice P h+j \ P h 2, we ca pick some y P h+j \ P h other tha z. Let l y, l z Z st with l y R h+j, l z R h+2j ad c(l y ) = y, c(l z ) = z. Defie l x :=2l y l z R h, so x:= c(l x ) is a color i P h. By the choice of y, y z; z x sice z P h+2j \ P h ad x P h ; x y sice y P h+j \ P h ad x P h. Thus l x, l y, l z is a raibow 3-AP, a cotradictio. Therefore we coclude that P h+2j P h. With this coditio, we cosider the case P h \ P h+2j is ot empty. 11

12 Let x P h \ P h+2j. Similarly, it is possible to pick y P h+j \ P h. Let l x, l y Z st with l x R h, l y R h+j, ad c(l x ) = x, c(l y ) = y. Thus l z :=2l y l x R h+2j ad so z := c(l z ) is a color i P h+2j. Agai, x y by the choice of y; x z sice x P h \ P h+2j ad z P h+2j ; y z sice y P h+j \ P h ad z P h+2j P h. Sice we agai have a cotradictio, P h = P h+2j. Next we show that P h+j \P h 2 implies P h \P h+j 1. Suppose P h+j \P h 2 ad P h \P h+j 2, ad the show this leads to a cotradictio. By the result just established, P h = P h+2j. Sice P h+2j \ P h+j = P h \ P h+j 2, P h+j = P h+3j. Therefore P h = P h+qj wheever q is eve ad P h+j = P h+qj wheever q is odd. Sice s is odd, the order d of j i Z s is also a odd umber. That meas P h = P h+dj = P h+j, which is a cotradictio. Fially, sice P m is chose to be maximum, P m \ P j 2 wheever P j \ P m 2, which is impossible. Hece P j \ P m 1. Propositio Suppose s ad t are both odd. The aw(z st, 3) aw(z s, 3) + aw(z t, 3) 2. Proof. Let c be a colorig of Z st that does ot have a raibow 3-AP. Cosider the residue classes ad residue palettes modulo s ad without loss of geerality assume P 0 P i for all i. We claim that s 1 P i (aw(z s, 3) 1) + (aw(z t, 3) 1) 1. (1) i=0 The proof is by cotradictio. Assume that (1) is false, i.e., assume s 1 P i (aw(z s, 3) 1) + (aw(z t, 3) 1) (2) i=0 ad defie a colorig ĉ of Z s = {0, 1,..., s 1} i the followig way: Let α be a color ot i s 1 i=1 (P i \ P 0 ) ad defie { α if P ĉ(i) = i P 0, the elemet of P i \ P 0 if P i P 0. Note that Propositio 3.9 implies that the required elemet i P i \P 0 is uique, so this colorig is well-defied. Sice c does ot have a raibow 3-AP, we kow P 0 aw(z t, 3) 1 so s 1 s 1 (P i \ P 0 ) P i (aw(z t, 3) 1) (aw(z s, 3) 1)+(aw(Z t, 3) 1) (aw(z t, 3) 1) = aw(z s, 3) 1. i=1 i=0 Note that every color that is ot i P 0, together with α, is used i ĉ, so ĉ uses at least aw(z s, 3) colors. Thus a raibow 3-AP exists i ĉ. We show that a raibow 3-AP i ĉ implies a raibow 3-AP i c, providig a cotradictio ad establishig that (1) is true. Let x, y, z be a raibow 3-AP i Z s usig colorig ĉ, with y = x + d (mod s) ad z = x + 2d (mod s). Sice x, y, z is raibow, ĉ(u) ĉ(v) for all distict u, v {x, y, z}, ad so at most oe u {x, y, z} has ĉ(u) = α. Note that by defiitio ĉ(u) P u or ĉ(u) = α for u {x, y, z}. Case 1: ĉ(z) α ad ĉ(y) α. The we ca fid g 2 ad g 3 such that c(g 2 s+y) = ĉ(y) ad c(g 3 s+z) = ĉ(z). Defie d := (g 3 s + z) (g 2 s + y). The (g 3 s + z) d = (g 2 s + y) y (mod s) (g 3 s + z) 2d = 2g 2 s + 2y g 3 s z 2y z 2(x + d) (x + 2d) x (mod s). With l:= (g 3 s+z) 2d, cosider the 3-AP {l, (g 3 s+z) d, (g 3 s+z)}. We show that this 3-AP is raibow: Note that ĉ(y) / P 0 ad ĉ(z) / P 0. If ĉ(x) = α, the P x P 0, so l R x implies c(l) ĉ(y) = c(g 2 s + y) ad c(l) ĉ(z) = c(g 3 s + z). If ĉ(x) α, the ĉ(x) is the uique elemet of P x \ P 0 ad ĉ(x) ĉ(y), ĉ(z), 12

13 so l R x implies c(l) ĉ(y) ad c(l) ĉ(z). Thus c has a raibow 3-AP, cotradictig our assumptio (2). The case where both ĉ(x) α ad ĉ(y) α is symmetric to Case 1. So oly Case 2 remais. Case 2: ĉ(y) = α. The ĉ(x) P x \P 0 ad ĉ(z) P z \P 0, so we ca fid g 1 ad g 3 such that c(g 1 s+x) = ĉ(x) ad c(g 3 s + z) = ĉ(z), ad defie e := (g 3 s + z) (g 1 s + x). Sice st is odd, 2 is ivertible i Z st ad there exists d such that 2d e (mod st), ad hece 2d e (mod s). Also, e z x 2d (mod s). Thus 2d 2d (mod s) ad so d d (mod s) sice s is odd. The (g 1 s + x) + 2d (g 1 s + x) + ((g 3 s + z) (g 1 s + x)) = g 3 s + z z (mod s) (g 1 s + x) + d x + d y (mod s). With l := (g 1 s + x) + d, the 3-AP {(g 1 s + x), l, (g 1 s + x) + 2d } is raibow, because l R y ad P y P 0, so c(l) ĉ(x) = c(g 1 s + x) ad c(l) ĉ(z) = c(g 3 s + z). I all cases, c has a raibow 3-AP, cotradictig our assumptio (2). Next we prove two techical propositios used i the proof of Propositio 3.13, Propositios 3.11 ad Propositio Let m ad s be positive itegers with s odd. Suppose c is a colorig of Z 2 m s usig at least r := aw(z s, 3) + 1 colors that does ot have a raibow 3-AP. Let R 0, R 1,..., R s 1 be the residue classes modulo s i Z 2 m s with associated residue palettes P i. The 1 P i 2 for i = 0,..., s 1, ad all palettes P i of size two share a commo color. Proof. Sice P i is oempty, 1 P i. Observe that the colorig c of R i iduces a colorig o Z 2 m that uses oly the colors i P i ad caot cotai a raibow 3-AP. Thus P i 2 by Theorem 3.3, establishig the first statemet. By Propositio 3.9, each pair of residue palettes of size two must itersect. Suppose the palettes of size two do ot all itersect i a commo color. The there are exactly three colors α, β, γ that are used by all the palettes of size two, ad there are exactly three distict palettes of size two, each cosistig of two of these three colors. We show this cofiguratio leads to a cotradictio. Create a colorig ĉ of Z s by the followig method: c(i) if P i = 1, ĉ(i) = β if P i = {α, β}, the uique elemet of P i \ {γ} if P i = 2 ad γ P i. Observe that ĉ uses r colors if there exists i such that P i = {γ} ad r 1 = aw(z s, 3) colors otherwise, so i either case ĉ must have a raibow 3-AP. Suppose that {x, y, z} is a raibow 3-AP for the colorig ĉ of Z s. Sice ĉ(x), ĉ(y), ad ĉ(z) are distict colors, at least oe of the palettes P x, P y, P z cotais oly oe color. Cosider the sizes of P x, P y, ad P z. Case 1: P z = 1. Observe that ĉ(i) is always a elemet i P i by our defiitio of ĉ(i). Pick 1 R x ad 2 R y such that c( 1 ) = ĉ(x) ad c( 2 ) = ĉ(y). Thus 3 := is a elemet i R z ad so c( 3 ) = ĉ(z). Sice ĉ(x), ĉ(y), ĉ(z) are all distict, { 1, 2, 3 } is a raibow 3-AP. The case P x = 1 is symmetric. Case 2: P x = P z = 2 ad P y = 1. Sice ĉ(x) ĉ(z), it must be that {ĉ(x), ĉ(z)} = {α, β}. Without loss of geerality, we assume that ĉ(x) = β ad ĉ(z) = α. By the defiitio of ĉ, P z = {α, γ}. The P x is oe of {α, β} or {β, γ}. If ĉ(y) P x P z, the ay 3-AP { 1, 2, 3 } where 1 R x ad c( 1 ) = β, 2 R y, ad 3 R z is a raibow 3-AP i the origial colorig. Thus, ĉ(y) P x P z {α, β, γ}, but ĉ(y) / {α, β} = {ĉ(x), ĉ(z)}, so ĉ(y) = γ. Note that this implies ĉ uses all r colors. Sice this is the fial case, ad all previous cases led to cotradictios, every raibow 3-AP i Z s give by the colorig ĉ must be of the { form {x, y, z} where {ĉ(x), ĉ(z)} = {α, β} ad ĉ(y) = γ. Create a ew colorig c of Z s where c c(i) if ĉ(i) γ, (i) = β if ĉ(i) = γ. 13

14 Now, every 3-AP that was previously o-raibow i ĉ remais o-raibow i c ad the raibow 3-APs (which ecessarily used the colors α, β, ad γ) are o loger raibow. Thus, this colorig c does ot have a raibow 3-AP, but c uses r 1 = aw(z s, 3) colors, a cotradictio. The above cases show that havig o commo color amog the palettes of size two leads to a cotradictio. Therefore, all of the residue palettes of size two share a commo color. Propositio Suppose c is a colorig of Z 2t (t 1) that does ot have a raibow 3-AP. Let A ad B deote the residue palettes modulo 2 i Z 2t associated with the eve ad odd umbers, respectively. The A \ B 1 ad B \ A 1. Proof. It suffices to show that A \ B 1 for every such colorig c because if B \ A 2, the the colorig defied by the rotatio c (x) := c(x + 1) has the roles of A ad B reversed. Suppose ot, so there exist two colors α, γ that appear oly i A. Let 1 = 2m 1 ad 3 = 2m 3 be eve elemets such that c( 1 ) = α ad c( 3 ) = γ. We ca select m 1 ad m 3 such that 0 m 1 < m 3 < t. Performig arithmetic i the itegers, we ca choose m 3 m 1 to be miimum with respect to the fact that the set of colors {c(2m 1 ), c(2m 3 )} is {α, γ}. Let 2 = m 1 + m 3 ad observe that { 1, 2, 3 } is a 3-AP ad hece is ot raibow. Therefore, 2 must have the color α or γ ad thus is eve. However, this implies that 2 = 2m 2 ad m 1 < m 2 < m 3, while oe of the sets of colors {c(2m 1 ), c(2m 2 )} or {c(2m 2 ), c(2m 3 )} is {α, γ}, so oe of the pairs (m 1, m 2 ), (m 2, m 3 ) violates our extremal choice. Propositio Let m ad s be positive itegers with s odd. The aw(z 2m s, 3) aw(z s, 3) + 1. Proof. The result is immediate for s = 1 because Theorem 3.3 gives that aw(z 2 m, 3) = 3 ad because aw(z s, 3) = s + 1 for s < 3, so assume s 3. We proceed by iductio o m. Suppose c is a exact r-colorig of Z 2m s with r = aw(z s, 3) + 1 that does ot have a raibow 3-AP. Let A ad B deote the residue palettes of the eve ad odd umbers, respectively. By Propositio 3.12, A \ B 1 ad B \ A 1, so B r 1 ad A r 1. The base case m = 1 is the immediate, because the colorig of the eve umbers of Z 2s iduces a colorig of Z s, so a raibow 3-AP ecessarily exists, producig a cotradictio. Now cosider m > 1. As usual R i, i = 0,..., s 1, are the residue classes modulo s of Z 2 m s ad P i, i = 0,..., s 1, are the residue palettes. Recall that by Propositio 3.11, 1 P i 2 for all i. For 0 i s 1, let A i = P i A be the colors appearig o the eve umbers i R i, ad let B i = P i B be the colors appearig o the odd umbers i R i. Thus, P i = A i B i, A = s 1 i=0 A i, ad B = s 1 i=0 B i. We claim that A = B = r 1. To see this, observe that the eve elemets iduce a colorig of Z 2 m 1 s, so if A = r, the a raibow 3-AP ecessarily exists, sice r aw(z 2 m 1 s, 3) by the iductio hypothesis. Thus A r 1, ad so A = r 1. The proof that B = r 1 is aalogous. Sice A = B = r 1, there exist colors α, β such that A \ B = {α} ad B \ A = {β}. Assume α A u ad β B v. Let j = v u, hece β B u+j = B v. Sice there is o raibow 3-AP, u + 2j must have a color i palette A, α A u+2j, which the implies β B u+3j = B v+2j. Iteratig this process gives that α A u+2lj ad β B v+2lj for all l 0. Sice s is odd, we have that for all q 0, A u+qj is of the form A u+2lj for some l ad similarly, every B u+qj = B v+(q 1)j is of the form B v+2lj for some l. Therefore, P u+qj = {α, β} for all q 0. By Propositio 3.11, there is a commo color for palettes of size two, ad thus oe of α or β is this commo color. Without loss of geerality, assume that α is the commo color for all palettes of size 2. This implies that B i = 1 for all 0 i s 1. Hece, defiig ĉ(i) to be the uique color i B i defies a exact (r 1)-colorig of Z s that avoids raibow 3-APs. However, r 1 = aw(z s, 3), a cotradictio. Propositio 3.8 is ow established from Propositio 3.10 ad Propositio We ow tur our attetio to establishig the lower boud. Propositio Suppose s is odd ad Z s has a sigleto extremal colorig. The for t 2, aw(z st, 3) aw(z t, 3) + aw(z s, 3) 2. 14

15 Proof. Let c s be a sigleto extremal colorig of Z s. Note that we ca shift c s so that c s (0) is the color that is used exactly oce. Choose a colorig c t of Z t usig aw(z t, 3) 1 colors ot used by c s that does ot have a raibow 3-AP. Let R 0, R 1,..., R s 1 be the residue classes modulo s i Z st. Defie a colorig ĉ of Z st as follows: For i = 1,..., s 1 ad l R i, ĉ(l) := c s (i), ad for 0 j t 1, ĉ(js) := c t (j). Notice that we ow have a exact aw(z s, 3) 2 + aw(z t, 3) 1 colorig of Z st because we have removed color c s (0). Clearly, if a 3-AP is withi some residue class it is ot raibow. Because s is odd, d 0 (mod s) implies 2d 0 (mod s) ad 2d d (mod s), so a 3-AP that is ot etirely withi oe residue class has elemets i three differet residue classes. But a raibow 3-AP with elemets i three differet residue classes would imply a raibow 3-AP i c s, which does ot exist. So we have foud a colorig of Z st usig aw(z t, 3) + aw(z s, 3) 3 colors that does ot have a raibow 3-AP. Thus aw(z st, 3) aw(z t, 3) + aw(z s, 3) 2. Corollary For a iteger 2, aw(z, 3) = 2 + f 2 () + f 3 () + 2f 4 (). Proof. By Propositio 3.5, every odd prime factor p has 3 aw(z p, 3) 4. Apply Propositio 3.8, removig oe odd prime s at a time ad observig that for aw(z s, 3) = 3, aw(z s, 3) 2 adds oe to the total, whereas for aw(z s, 3) = 4, aw(z s, 3) 2 adds two to the total. Thus aw(z, 3) 2 + f 2 () + f 3 () + 2f 4 (). For the reverse iequality, suppose p is a odd prime. The every extremal colorig of Z p is a sigleto colorig by Propositio 3.5. So we ca apply Propositio 3.14 to remove oe odd prime at a time to show that aw(z, 3) = 2 + f 2 () + f 3 () + 2f 4 (). Remark The costructive proof of Propositio 3.14 gives a sigleto extremal colorig of Z from the sigleto extremal colorigs of the prime factors of. Sice Z 2 m has the sigleto extremal colorig c(0) = 1 ad c(i) = 2 for every i 0 (mod 2 m ), every positive iteger has a sigleto extremal colorig. Propositio For all primes p < 100, aw(z p, 3) = 3 if p / Q 4 aw(z p, 3) = 4 if p Q 4. := {17, 31, 41, 43, 73, 89, 97} ad Proof. The statemet that for ay prime p < 100, aw(z p, 3) = 3 if p / Q 4 ad aw(z, 3) = 4 if p Q 4 has bee verified computatioally (see Table 2). The ext example illustrates the use of Corollary 3.15 to compute aw(z, 3) i the case that every prime factor of is less tha 100. Example Let = 14, 582, 937, 583, 067, 568. Sice = , aw(z, 3) = = Mai results for aw(z, k), k 4 I this sectio, we specialize to the case where k 4 ad prove Theorem 1.7. Corollary 3.19 below, which follows from Corollary 2.14 ad Remark 1.4, gives us e log log log ω(1) as a upper boud for aw(z, k). ( ) Corollary For every fixed positive iteger k, aw(z, k) = o. log log Our lower boud for aw(z, k) whe > 12 is preseted i Lemma Lemma There exists a absolute costat b > 0 such that for all c > 3, c 4 ad k 4, ( ) aw(z, k) > e b log(/c) = e b log(/c) log c = 1 o(1). c Lemma 3.20 is prove usig the Behred costructio from Sectio 2.2 ad usig Propositio 3.21 below. The Behred costructio i the itegers {1,..., m} has o puctured 4-AP ad size me b log m for some absolute costat b. 15

16 Propositio Let c > 3 be a real umber, ad let [ ] c deote the first c cosecutive residues i Z. Suppose S [ ] c does ot cotai ay puctured 4-APs. The aw(z, k) > S + 1 for all k 4. Proof. Color each member of S a distict color, ad color each member of Z \ S with a ew color called zero. Each i Z with c < i < will be colored zero. If K = {a 1, a 2, a 3, a 4 } is a raibow 4-AP i Z, the at most oe elemet of K is ot i S. Without a loss of geerality, assume K is ordered as a 1, a 2, a 3, a 4 ad a 3, a 4 S. The there exists d Z such that d a 4 a 3 (mod ) ad d c. Suppose a 2 S. Because d c < 2, we must have that a 2, a 3, a 4 is a 3-AP i [ ] c. This cotradicts the fact that S cotais o puctured 4-APs, so we must have a 2 S ad a 1 S. However, sice 2 d 2 c < (c 1) c, we must have that a 1, a 3, a 4 is a puctured 4-AP i [ ] c. This is a cotradictio, so a 1 S. This meas that K could ot have bee raibow, so we have a ( S + 1)-colorig of Z with o raibow 4-APs. We use the boud for the Behred costructio i Lemma 2.8 to obtai the bouds for aw(z, k), k 4: This completes the proof of Theorem Additioal results for aw(z, k) e b log(/c) log c < aw(z, k) e log log log ω(1). I this sectio, we preset computed data for aw(z, k), k 4, establish the value of aw(z, k) for k =, 1, ad 2, ad preset some examples that show some additioal results fail to exted from [] to Z. Table 3 below lists the computed values of aw(z, k) for k = 4,..., i the row labeled. \ k Next we examie aw(z, k) for k close to. Table 3: Computed values of aw(z, k) for k 4. Propositio For positive ad k we have aw(z, k) = if ad oly if k =. Proof. If k = the result is obvious. Now suppose that k < ad cosider a exact ( 1)-colorig of Z. The there are two umbers with the same color ad all other umbers are colored distictly. Suppose x ad y are the the two umbers with the same color. The {x + 1,..., x + k} is a k-ap that does ot cotai x, ad so is raibow. Therefore aw(z, k) 1. Corollary For positive, aw(z, 1) = 1. A patter ca be observed i the values of aw(z, 2), ad this is established i Propositio

17 Propositio For positive 5, if is prime the aw(z, 2) = 2; otherwise aw(z, 2) = 1. Proof. We trivially have a lower boud of 2 for aw(z, 2). First we assume is prime. We claim that for ay two distict elemets x ad y there is a ( 2)-AP that misses x ad y. To see this, simply form the -AP with a = x ad d = (y x), this will cover all of Z ad ow removig the first two terms leaves us with a ( 2)-AP that does ot cotai x or y. So suppose we have a exact ( 2)-colorig. The either there is oe color that occurs three times or two colors that each occur twice, ad i either case all other colors occur exactly oce. I either case we ca choose two umbers to avoid ad the the remaiig 2 umbers are raibow, but as just oted above the remaiig 2 umbers are a arithmetic progressio. Therefore every ( 2)-colorig cotais a raibow progressio. Whe is ot a prime, let p be the smallest prime divisor of ad cosider the ( 2)-colorig formed by colorig 0, p ad 2p moochromatically, with the remaiig umbers all give distict colors. This is a ( 2)-colorig (sice 2p < by assumptio that 5). We claim this colorig has o raibow ( 2)-AP (alog with the upper boud of 1, this claim establishes the result). Suppose that K = {a, a + d,..., a + ( 3)d} is a raibow ( 2)-AP, so all the elemets of K are distict ad K ecessarily misses two of 0, p, 2p. Sice Z caot have a proper subgroup of order 2, extedig K to a -AP ecessarily produces all elemets of Z ad thus {a + ( 2)d, a + ( 1)d} {0, p, 2p}. But the we have that p divides d = ((a + ( 1)d) (a + ( 2)d)), showig that this arithmetic progressio ca have at most p < 2 terms, which is a cotradictio. Propositio 3.22 shows that the if directio of Theorem 2.16 (k implies aw([], k) = ) does ot exted to Z. Example 3.25 below shows that the extesio of Propositio 2.15 to Z, which would assert that aw(z, k) aw(z 1, k) + 1, is ot true i geeral. There are couterexamples i both the cases k = 3 ad k 4. Example By Corollary 3.15, aw(z 30, 3) = 5, ad aw(z 29, 3) = 3 (see Table 2 i Sectio 3.1). Furthermore, aw(z 8, 4) = 6 ad aw(z 7, 4) = 4 (see Table 3). Example 3.26 below shows that Theorem 2.17, which bouds the ati-va der Waerde umber of a sum i terms of the ati-va der Waerde umbers of the summads, does ot exted to Z. Example Accordig to our computed data (see Table 3), aw(z 12, 4) = 8 > = aw(z 5, 4) + aw(z 7, 4) 1. There are also examples for k = 3, such as aw(z 54, 3) = 6 > = aw(z 47, 3) + aw(z 7, 3) 1. 4 Computatio May of the results we have proved i this paper were first cojectured from examiatio of data. I this sectio, we briefly discuss a efficiet algorithm to fid a exact r-colorig of [] or Z that avoids a raibow k-ap, if such a colorig exists. For the sake of brevity, we will focus o the case of colorig [] sice this case has a few extra properties that the Z case does ot. Specifically, we have [m] [] for all m while Z cotais a copy of Z m if ad oly if m divides. Fix k,, ad r ad assume that all values of aw([m], k) have bee computed for k m <. Let c : [] [r] { } be a fuctio called a partial r-colorig, where every positio i has color c(i) [r] or c(i) = ad i is ucolored. By startig with all positios ucolored, we recursively attempt to exted a partial r-colorig c where the positios i [i] are colored to a exact r-colorig c that avoids raibow k-aps. We brach at each recursive call for all possible choices of color for c(i + 1) such that o k-ap withi [i + 1] is colored with k distict colors. To guaratee that o chose color creates a raibow k-ap, we maitai a list of sets D(j) [r] that cotai all of the possible colors for the positio j. Specifically, assigig c(j) to be ay color i [r] \ D(j) will immediately create a raibow k-ap. Wheever a color is assiged to a positio i, we cosider a k-ap, X, whose secod-to-last elemet is i. If the set c(x) = {c(i ) : i X max X} cotais k 1 distict colors, we say that X is a almost-raibow k-ap ad the color for max X must be 17