Anti-van der Waerden numbers of 3-term arithmetic progressions.

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1 Ant-van der Waerden numbers of 3-term arthmetc progressons. Zhanar Berkkyzy, Alex Schulte, and Mchael Young Aprl 24, 2016 Abstract The ant-van der Waerden number, denoted by aw([n], k), s the smallest r such that every exact r-colorng of [n] contans a ranbow k-term arthmetc progresson. Butler et. al. showed that log 3 n + 2 aw([n], 3) log 2 n + 1, and conjectured that there exsts a constant C such that aw([n], 3) log 3 n + C. In ths paper, we show ths conjecture s true by determnng aw([n], 3) for all n. We prove that for 7 3 m n 21 3 m 2, { m + 2, f n = 3 m aw([n], 3) = m + 3, otherwse. Keywords. arthmetc progresson; ranbow colorng; untary colorng; Behrend constructon. 1 Introducton Let n be a postve nteger and let G {[n], Z n }, where [n] = {1,..., n}. A k-term arthmetc progresson (k-ap) of G s a sequence n G of the form a, a + d, a + 2d,..., a + (k 1)d, where d 1. For the purposes of ths paper, an arthmetc progresson s referred to as a set of the form {a, a + d, a + 2d,..., a + (k 1)d}. An r-colorng of G s a functon c : G [r], and such a colorng s called exact f c s surjectve. Gven c : G [r], an arthmetc progresson s called ranbow (under c) f c(a + d) c(a + jd) for all 0 < j k 1. The ant-van der Waerden number, denoted by aw(g, k), s the smallest r such that every exact r-colorng of G contans a ranbow k-ap. If G contans no k-ap, then aw(g, k) = G + 1; ths s consstent wth the property that there s a colorng of G wth aw(g, k) 1 colors that has no ranbow k-ap. An r-colorng of G s untary f there s an element of G that s unquely colored. The smallest r such that every exact untary r-colorng of G contans a ranbow k-ap s denoted by aw u (G, k). Smlar to the ant-van der Waerden number, aw u (G, k) = G + 1 f G has no k-ap. Problems nvolvng countng and the exstence of ranbow arthmetc progressons have been well-studed. The man results of Axenovch and Fon-Der-Flaass [1] and Axenovch and Martn [2] deal wth the exstence of 3-APs n colorngs that have unformly szed color classes. Fox, Department of Mathematcs, Iowa State Unversty, Ames, IA 50011, USA (zhanarb@astate.edu) Department of Mathematcs, Iowa State Unversty, Ames, IA 50011, USA (aschulte@astate.edu) Department of Mathematcs, Iowa State Unversty, Ames, IA 50011, USA (myoung@astate.edu) 1

2 Jungć, Mahdan, Ne setrl, and Rado cć also studed ant-ramsey results of arthmetc progressons n [6]. In partcular, they showed that every 3-colorng of [n] for whch each color class has densty more than 1/6, contans a ranbow 3-AP. Fox et. al. also determned all values of n for whch aw(z n, 3) = 3. The specfc problem of determnng ant-van der Waerden numbers for [n] and Z n was studed by Butler et. al. n [4]. It s proved n [4] that for k 4, aw([n], k) = n 1 o(1) and aw(z n, k) = n 1 o(1). These results are obtaned usng results of Behrend [3] and Gowers [5] on the sze of a subset of [n] wth no k-ap. Butler et. al. also expand upon the results of [6] by determnng aw(z n, 3) for all values of n. These results were generalzed to all fnte abelan groups n [7]. Butler et. al. also provdes bounds for aw([n], 3), as well as many exact values (see Table 1). n \ k Table 1: Values of aw([n], k) for 3 k n+3 2. In ths paper, we determne the exact value of aw([n], 3), whch answers questons posed n [4] and confrms the followng conjecture: Conjecture 1. [4] There exsts a constant C such that aw([n], 3) log 3 n + C, for all n 3. Our man result, Theorem 2, also determnes aw u ([n], 3) whch shows the exstence of extremal colorngs of [n] that are untary. Theorem 2. For all ntegers n 2, { m + 2, f n = 3 m aw u ([n], 3) = aw([n], 3) = m + 3, f n 3 m and 7 3 m n 21 3 m 2. 2

3 In secton 2, we provde lemmas that are useful n provng Theorem 2 and secton 3 contans the proof of Theorem 2. 2 Lemmas In [4, Theorem 1.6] t s shown that 3 aw(z p, 3) 4 for every prme number p and that f aw(z p, 3) = 4 then p 17. Furthermore, t s shown that the value of aw(z n, 3) s determned by the values of aw(z p, 3) for the prme factors p of n. We have ncluded ths theorem below wth some notaton change. Theorem 3. [4] Let n be a postve nteger wth prme decomposton n = 2 e 0 p e 1 1 pe 2 2 pes s for e 0, = 0,..., s, where prmes are ordered so that aw(z p, 3) = 3 for 1 l and aw(z p, 3) = 4 for l + 1 s. Then 2 + l e j + s 2e j, f n s odd j=1 j=l+1 aw(z n, 3) = 3 + l e j + s 2e j, f n s even. j=1 j=l+1 We use Theorem 3 to prove the followng lemma. Lemma 4. Let n 3, then aw(z n, 3) log 3 n + 2 wth equalty f and only f n = 3 j or 2 3 j for j 1. Proof. Suppose n = 2 e 0 p e 1 1 pe pes s wth e 0 for = 0,..., s, where prmes p 1, p 2,..., p s are ordered so that aw(z p, 3) = 3 for 1 l and aw(z p, 3) = 4 for l + 1 s. We consder two cases dependng on party of n. Case 1. Suppose n s odd, that s e 0 = 0. Then aw(z n, 3) = 2 + l e j + s 2e j by j=1 j=l+1 Theorem 3. Snce aw(z p, 3) = 3 for odd prmes p 13, we have p 17 for l+1, and clearly p 3 for l, therefore 3 aw(zn,3) = 3 2+ l e j + j=1 s j=l+1 2e j = 9 3 e1 3 el 9 e l+1 9 es 9 p e 1 1 pes s = 9n. Note that the equalty holds f and only f n s a power of 3, that s e j = 0 for 2 j s. Therefore, aw(z n, 3) log 3 n + 2 for odd n, wth equalty f and only f n = p e 1 1. Case 2. Suppose n s even, that s e 0 1. Then aw(z n, 3) = 3 + l e j + j=1 s j=l+1 2e j by Theorem 3. If n = 2 e0 3 j for j 1, then by drect computaton aw(z n, 3) = 3+j 2+ log 3 n, wth equalty f and only f e 0 = 1. So suppose there s such that p 3, and let h = n. If l then p 5, and so 3 3 e odd, by the prevous case < 2 e 0 p e 2 e 0 p e for all e 0 1 and e 1. Therefore, snce h s 3 aw(zn,3) = 3 3 e 3 aw(z h,3) 3 3 e 9h < 2 e 0 p e 9h = 9n. If l + 1 then p 17, and so 3 9 e prevous case < 2 e 0 p e 3 for all e 0 1 and e 1. Then by the

4 3 aw(zn,3) = 3 9 e 3 aw(z h,3) 3 9 e 9h < 2 e 0 p e 9h = 9n. A set of consecutve nteger I n [n] s called an nterval and l(i) s the number of ntegers n I. Gven a colorng c of some fnte nonempty subset S of [n], a color class of a color under c n S s denoted c (S) := {x S : c(x) = }. A colorng c of [n] s specal f n = 7q + 1 for some postve nteger q, c(1) and c(n) are both unquely colored, and there are two colors α and β such that c α ([n]) = {q + 1, 2q + 1, 4q + 1} and c β ([n]) = {3q + 1, 5q + 1, 6q + 1}. Lemma 5. Let N be an nteger and c be an exact r-colorng of [N] wth no ranbow 3-AP, where 1 and N are colored unquely. Then ether the colorng c s specal or {c(x) : x (mod 3) and x [N]} r 1 for = 1 or = N. Proof. Observe that N s even, otherwse {1, (N + 1)/2, N} s a ranbow 3-AP. We partton the nterval [N] nto four subntervals I 1 = {1,..., N/4 }, I 2 = { N/4 + 1,..., N/2}, I 3 = {N/2 + 1,..., 3N/4 }, and I 4 = { 3N/4 + 1,..., N}. Notce that every color other than c(1) and c(n) must be used n the subnterval I 2. To see ths, assume s the mssng color n I 2 dstnct from c(1) and c(n). Let x be the largest nteger n c (I 1 ). Snce N s even, we have 2x 1 2 N/4 1 N/2, and so 2x 1 I 2 and c(2x 1). Therefore the 3-AP {1, x, 2x 1} s a ranbow. If there s no such nteger x n I 1, then the ntegers colored wth must be n the second half of the nterval [N], so we choose the smallest such nteger y n c (I 3 I 4 ). Then {2y N, y, N} s a ranbow 3-AP snce c(2y N), because 2y N I 1 I 2. Smlarly, every color other than c(1) and c(n) must be used n the subnterval I 3. Throughout the proof we mostly drop (mod 3) and just say congruent even though we mean congruent modulo 3. We consder the followng three cases. Case 1: N 0 (mod 3). Assume {c(x) : x (mod 3) and x [N]} < r 1 for both = 1 and = N. So there are two colors, say red and blue, such that no nteger n [N] colored wth red s congruent to 1, and no nteger n [N] colored wth blue s congruent to 0. We further partton the nterval I 2 nto subntervals I 2() and I 2() so that l(i 2() ) l(i 2() ) l(i 2() ) + 1, and partton the nterval I 3 nto subntervals I 3() and I 3() so that l(i 3() ) l(i 3() ) l(i 3() ) + 1. Then we have the followng observatons: () x 0 for all x c red (I 3 I 4 ) and y 1 for all y c blue (I 1 I 2 ). If there s an nteger r n I 3 I 4 colored wth red and congruent to 2, then 2r N 1, and so c(2r N) s not red by our assumpton. Therefore the 3-AP {2r N, r, N} s ranbow. Smlarly, f there s an nteger b n I 1 I 2 colored wth blue and congruent to 2, then 2b 1 0, and so c(2b 1) s not blue, formng a ranbow 3-AP {1, b, 2b 1}. () x 2 for all x c red (I 2 ) and y 2 for all y c blue (I 3 ). If there s an nteger r n c red (I 2 ) congruent to 0, then 2r 1 2 and 2r 1 I 3 I 4 snce 2r 1 N/ Therefore, 2r 1 s not colored wth red by the prevous observaton, and so the 3-AP {1, r, 2r 1} s a ranbow. Smlarly, f there s an nteger b n c blue (I 3 ) congruent to 1, then usng N we obtan the ranbow 3-AP {2b N, b, N}, because 2b N 2 and 2b N N/2. () c red (I 3() ) = c blue (I 2() ) =. If there s an nteger r n I 3() colored wth red, then 2r N 0, by observaton (). Furthermore, 2r N N/2 and 2r N 2(N/2 + l(i 3() ) + 1) N (2l(I 3() ) + 1) + 1 N/ So 4

5 2r N I 2 and hence t s not colored wth red by observaton (). Therefore, {2r N, r, N} s a ranbow 3-AP. Smlarly, f there s an nteger b n I 2() colored wth blue, then 2b 1 1 and N/ b 1 3N/4. So 2b 1 I 3 and hence t s not colored wth blue by observaton (). Therefore, {1, b, 2b 1} s a ranbow 3-AP. (v) c red (I 2() ) = c blue (I 3() ) =. Suppose there s an nteger r n I 2() colored wth red. Snce the colorng of I 2 contans both red and blue and there s no nteger n I 2() colored wth blue, by (), there must be an nteger b n I 2() colored wth blue. By () and (), b 1 and r 2. Wlog, suppose b > r. Then 2r b 0 and 2r b I 2 snce l(i 2() ) l(i 2() ) + 1. So 2r b s not colored red or blue and hence the 3-AP {2r b, r, b} s ranbow. Therefore, there s no nteger n I 2() that s colored wth red. Smlarly, there s no nteger n I 3() that s colored wth blue. Recall that every color other than c(1) and c(n) s used n both ntervals I 2 and I 3. Therefore, sets c red (I 2() ), c blue (I 2() ), c red (I 3() ), and c blue (I 3() ) are nonempty. Usng above observatons we next show that n fact these ntegers colored wth blue and red n each subnterval are unque. Let B = {b 1,..., b 2 } be the shortest nterval n I 2() whch contans all ntegers colored wth blue and let R = {r 1,..., r 2 } be the shortest nterval n I 3() whch contans all ntegers colored wth red. Choose the largest nteger x n c red (I 2() ) and consder two 3-APs {x, b 1, 2b 1 x} and {x, b 2, 2b 2 x}. Snce x s congruent to 2 and both b 1 and b 2 are congruent to 1, we have that both 2b 1 x and 2b 2 x are congruent to 0 and are contaned n I 3, otherwse the 3-APs are ranbow. Snce all ntegers colored wth blue n I 3 are congruent to 2 by (), we have that 2b 1 x and 2b 2 x are both colored wth red and so contaned n R. Therefore, 2l(B) 1 l(r). Now usng the smallest nteger n c blue (I 3() ), we smlarly have that 2l(R) 1 l(b). Snce l(b) 1 and l(r) 1, we have that l(r) = l(b) = 1,.e. there are unque ntegers b n c blue (I 2() ) and r n c red (I 3() ). Now for any nteger r from c red (I 2() ) the nteger 2 r 1 must be colored wth red, otherwse the 3-AP {1, r, 2 r 1} s ranbow. Snce 2 r 1 I 3, t must be equal to the unque red colored nteger r of I 3. Therefore, there s exactly one such r n I 2(),.e. c red (I 2() ) = { r}. Smlarly, usng N there s a unque nteger b n I 3() colored wth blue. Snce {1, r, r}, { r, b, r}, {b, r, b}, and {b, b, N} are all 3-APs, N = 7(l({b,..., r}) 1) + 1 = 7(r b) + 1. Observe that f r s even, the nteger ( r + N)/2 n 3-AP { r, ( r + N)/2, N} must be red and congruent to 1 snce r 2 by (), contradctng our assumpton. So r s odd, and hence the nteger r = ( r + 1)/2 n I 1 must be colored wth red. Notce that there cannot be another nteger x larger than r n c red (I 1 ), otherwse 2x 1 wll be another nteger colored wth red n I 2 dstnct from r. Now, snce l({r,..., r}) = l({b,..., r}) we have that {r, r, N} s a 3-AP, and so r must be even. Suppose there are ntegers smaller than r n c red (I 1 ), and let z be the largest of them. Then 2z 1 s also n c red (I 1 ) and must be equal to or larger than r n I 1. However, that s mpossble because r s even and there s no nteger n c red (I 1 ) larger than r. So r s a unque nteger n I 1 colored wth red. Smlarly, there s a unque nteger b n I 4 colored wth blue. Therefore the 8-AP can be formed usng ntegers 1, r, r, b, r, b, b, N snce l({1,..., r }) = l({r,..., r}) = l({ r,..., b}) = l({b,..., r}) = l({r,..., b}) = l({ b,..., b }) = l({b,..., N}). In order for ths colorng to be specal, t remans to show that c blue (I 1 ) = c red (I 4 ) =. If c blue (I 1 ), then choose the largest nteger y n t and consder the 3-AP {1, y, 2y 1}. Snce 2y 1 must be n c blue (I 2 ) and the only nteger n ths set s b, we have 2y 1 = b. However, we know that b s even because b = 2 b N, a contradcton. Smlarly, f c red (I 4 ) choose the 5

6 smallest nteger x n t and consder the 3-AP {2x N, x, N}. Snce 2x N must be n c red (I 3 ) and the only nteger n ths set s r, we have 2x N = r. However, we know that r s odd because r = 2 r 1, a contradcton. Ths mples that c red ([N]) = {r, r, r} and c blue ([N]) = {b, b, b }, so the colorng s specal. Case 2: N 2 (mod 3). Ths case s analogous to Case 1. Case 3: N 1 (mod 3). Assume {c(x) : x (mod 3) and x [N]} < r 1.e. there are two colors, say red and blue, such that no nteger n [N] colored wth red or blue s congruent to 1. Recall that every color other than c(1) and c(n) appears n I 2 and I 3. Frst, notce that all ntegers colored wth red or blue n I 2 must be congruent modulo 3. Otherwse, choosng a red colored nteger and a blue colored nteger, we obtan a 3-AP whose thrd term s colored wth red or blue and s congruent to 1 contradctng our assumpton. Smlarly, ths s also the case for I 3. So suppose all ntegers n c red (I 2 ) c blue (I 2 ) and c red (I 3 ) c blue (I 3 ) are congruent modulo 3 to ntegers p 1 and q 1, respectvely. Pck the largest ntegers from c red (I 2 ) and c blue (I 2 ) and form a 3-AP whose thrd term s n I 3. Then the thrd term s colored wth red or blue and s congruent to p. Therefore, p q 1. We further partton the nterval I 2 nto subntervals I 2() and I 2(), so that l(i 2() ) l(i 2() ) l(i 2() ) + 1. If there exsts x c red (I 2() ) c blue (I 2() ), the nteger 2x 1 must be colored wth c(x) and contaned n I 3, so 2x 1 p whle x p 1, a contradcton. So c red (I 2() ) c blue (I 2() ) =. However, then the smallest ntegers of c red (I 2() ) and c blue (I 2() ) form a 3-AP whose frst term s contaned n I 2() and s colored wth red or blue, a contradcton. Ths completes the proof of the lemma. 3 Proof of Theorem 2 Gven a postve nteger n, defne the functon f as follows: { m + 2, f n = 3 m f(n) = m + 3, f n 3 m and 7 3 m n 21 3 m 2. In ths secton, we prove Theorem 2 by showng that aw([n], 3) = f(n) for all n. Frst, we show that f(n) aw u ([n], 3) by nductvely constructng a untary colorng of [n] wth f(n) 1 colors and no ranbow 3-AP. The result s true for n = 1, 2, 3, by nspecton. Suppose n > 3 and that the result holds for all postve ntegers less than n. Let n = 3h s, where s {0, 1, 2} and 2 h < n. Let r = aw u ([h], 3). So there s an exact untary (r 1)-colorng c of [h] wth no ranbow 3-AP. Let red be a color not used n c. Defne the colorng c 1 of [n] such that f x 1 (mod 3), then c 1 (x) = c((x + 2)/3), otherwse color x wth red. When s 0, defne the colorng c 2 of [n] as follows: f x 0 (mod 3) then color x wth red; f x 0 (mod 3) then c 2 (x) = c(x/3 + 1) when c(h) s the only unque color n c and c 2 (x) = c(x/3) otherwse. Notce that c 2 s a untary aw u ([h 1], 3)-colorng when s 0 and c 1 s a untary r-colorng of [n]. Now consder a 3-AP {a, b, 2b a} n [n]. If a b 1, then a and b are colored wth red, and so the 3-AP s not a ranbow. If a b 1, then 2b a 1, so ths set corresponds to a 3-AP n [h] wth colorng c, and hence the 3-AP s not ranbow. If a b, then 2b a s not congruent to a or b, so two of the terms of the 3-AP are colored wth red, and hence the 3-AP s not ranbow under c 1. Smlarly, ths 3-AP s not ranbow under c 2. Therefore, c 1 and c 2 are untary colorngs of [n] wth no ranbow 6

7 3-AP. Also note that aw u ([n], 3) aw u ([h], 3) + 1 under c 1 and aw u ([n], 3) aw u ([h 1], 3) + 1 under c 2. We proceed wth three cases determned by n 3. Case 1. Frst suppose 7 3 m n 3 m 3 or 3 m n 21 3 m 2. By the nducton hypothess and usng the colorng c 1, aw u ([n], 3) aw u ([h], 3) + 1 f(h) + 1 = f(n). Case 2. Suppose n = 3 m t where t {1, 2}. Notce that h = 3 m 1, so by nducton and usng colorng c 2, aw u ([n], 3) aw u ([h 1], 3) + 1 f(h 1) + 1 = f(3 m 1 1) + 1 = (m + 2) + 1 = f(n). The upper bound, aw([n], 3) f(n), s also proved by nducton on n. For small n, the result follows from Table 1. Assume the statement s true for all value less than n, and let 7 3 m n 21 3 m 2 for some m. Let aw([n]) = r + 1, so there s an exact r-colorng ĉ of [n] wth no ranbow 3-AP. We need to show that r f(n) 1. Let [n 1, n 2,..., n N ] be the shortest nterval n [n] contanng all r colors under ĉ. Defne c to be an r-colorng of [N] so that c(j) = ĉ(n j ) for j {1,..., N}. By mnmalty of N the colors of 1 and N are unque. If [N] has at least r 1 colors congruent to 1 or N, then [n] has at least r 1 colors congruent to n 1 or n N, respectvely, so r aw( n/3 ) and by nducton r f( n/3 ) f(n) 1. So suppose that s not the case, then by Lemma 5 we have that the colorng c s specal. Let N = 7q+1 for some q 1, and let the 8-AP n ths specal colorng be {1, r 1, r 2, b 1, r 3, b 2, b 3, N}, where r 1, r 2, r 3 are the only ntegers colored red, b 1, b 2, b 3 are the only ntegers colored blue and q = r 1 1. If n 9q, then the 8-AP can be extended to a 9-AP n n by addng the 9th element to ether the begnnng or the endng. Wlog, suppose {1, r 1, r 2, b 1, r 3, b 2, b 3, N, 2N b 3 } correspond to a 9-AP n [n]. Snce the colorng has no ranbow 3-AP, the color of 2N b 3 s blue or c(n), so we have a 4-colorng of ths 9-AP. However, aw([9], 3) = 4 and hence there s a ranbow 3-AP n ths 9-AP whch s n turn a ranbow 3-AP n [n]. Therefore, n 9q 1. By unqueness of red colored nteger r 1 n nterval {1,..., r 2 1}, the colors of ntegers n nterval {r 1 + 1,..., r 2 1} s the same as the reversed colors of ntegers n {2,..., r 1 1},.e. c(r 1 +) = c(r 1 ) for = 1,..., q 1. Smlarly, colorng of ntegers n nterval {r 2 +1,..., b 1 1} s the reversed of the colorng of ntegers n nterval {r 1 + 1,..., r 2 1}, and so on. Ths gves a ranbow 3-AP-free (r 2)-colorng of Z 2q. Therefore, r 2 aw(z 2q, 3) 1. If q = 3 for some, then n can not be a power of 3 because n Suppose n = 3 m, then 2q s not twce a power of 3 and clearly 2q s not a power of 3. Therefore, by Lemma 4 we have r aw(z 2q, 3)+1 log 3 (2q) +2 log 3 (2n/7) +2 = log 3 (2 3 m /7) +2 = m+1 f(n) 1. Suppose now that n 3 m. If q = 3 for some then m 2. Otherwse, f m 1 then q 3 m 1 1/7n whch contradcts the fact that q < 1/7n. Therefore, 2q 2 3 m 2 = 18 3 m 4 and so by nducton and Lemma 4, r aw(z 2q, 3) + 1 = aw([2q], 3) + 1 m + 2 f(n) 1. If q s not a power of 3, then agan usng Lemma 4, r aw(z 2q, 3) + 1 aw([2q], 3). Notce that 6 3 m 3 + 2/7 2n/ m 3, and so aw([2q], 3) m + 2 by nducton. Therefore, r m + 2 f(n) 1. 7

8 References [1] M. Axenovch, D. Fon-Der-Flaass, On ranbow arthmetc progressons. Electronc Journal of Combnatorcs 11 (2004), no. 1, Research Paper 1, 7pp. [2] M. Axenovch and R.R. Martn, Sub-Ramsey numbers for arthmetc progressons. Graphs Comb. 22 (2006), no. 1, [3] F.A. Behrend, On sets of ntegers whch contan no three terms n arthmetcal progresson. Proc. Nat. Acad. Sc. USA 32 (1946), [4] S. Butler, C. Erckson, L. Hogben, K. Hogenson, L. Kramer, R. Kramer, J. Ln, R. Martn, D. Stolee, N. Warnberg, and M. Young, Ranbow arthmetc progressons. arxv preprnt arxv: (2014). [5] W.T. Gowers, A new proof of Szemeréd s theorem. Geom. Funct. Anal. 11 (2001), no. 3, [6] V. Jungć, J. Lcht (Fox), M. Mahdan, J. Ne setrl, and R. Rado cć, Ranbow arthmetc progressons and ant-ramsey results. Comb. Probab. Comput. 12 (2003), no 5-6, [7] M. Young, Ranbow arthmetc progressons n fnte abelan groups. Manuscrpt avalable at arxv: [math.co]. 8