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On the nonvanishing of representation functions of some special sequences
Discrete Mathematics 338 (2015) 571–575
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
On the nonvanishing of representation functions of some special sequences Zhenhua Qu Department of Mathematics, Shanghai Key Laboratory of PMMP, East China Normal University, 500 Dongchuan Rd., Shanghai 200241, PR China
article
info
Article history: Received 17 February 2014 Received in revised form 15 November 2014 Accepted 18 November 2014
Keywords: Representation function Partition Sárközy problem
abstract For a given positive integer N, and any coloring function c : N → {0, 1} satisfying c (2k) = 1 − c (k), c (2k + 1) = c (k) for all k ≥ N, we show that for all n ≥ 20N, n has both a monochromatic representation and a multicolored representation, in other words, there exist x, y, u, v ∈ N, such that n = x + y = u +v , c (x) = c (y) and c (u) ̸= c (v). Similar results are obtained for another kind of coloring function c : N → {0, 1} satisfying c (2k) = c (k) and c (2k + 1) = 1 − c (k) for all k ≥ N. This answers a question of Y.-G. Chen on the values of representation functions. © 2014 Elsevier B.V. All rights reserved.
1. Introduction Let N be the set of nonnegative integers. For a set A ⊂ N and n ∈ N, we use R1 (A, n), R2 (A, n) and R3 (A, n) to denote the number of solutions of n = a + a′ with a, a′ ∈ A, n = a + a′ with a, a′ ∈ A, a < a′ , and n = a + a′ with a, a′ ∈ A, a ≤ a′ respectively. For i ∈ {1, 2, 3}, Sárközy asked whether there exist sets A, B ⊂ N with infinite symmetric difference such that Ri (A, n) = Ri (B, n) holds for all sufficiently large integers n. Dombi [4] observed that the answer is negative for i = 1 and is positive for i = 2. Chen and Wang [3] showed that N can be partitioned into two sets A and B such that R3 (A, n) = R3 (B, n) for all n ≥ 1. Later on, Lev [5], Sándor [6] and Tang [7] characterized, for each i ∈ {2, 3} and N ≥ 1, all sets A ⊂ N with the property that Ri (A, n) = Ri (N\A, n) holds for all n ≥ 2N − 1. More precisely, the following theorems are proved. Theorem A. Let N be a positive integer. Then R2 (A, n) = R2 (N\A, n) for all n ≥ 2N − 1 if and only if |A ∩ [0, 2N − 1]| = N, and 2k ∈ A ⇔ k ∈ A, 2k + 1 ∈ A ⇔ k ̸∈ A for all k ≥ N. Theorem B. Let N be a positive integer. Then R3 (A, n) = R3 (N\A, n) for all n ≥ 2N − 1 if and only if |A ∩ [0, 2N − 1]| = N, and 2k ∈ A ⇔ k ̸∈ A, 2k + 1 ∈ A ⇔ k ∈ A for all k ≥ N. Let D(n) denote the number of ones in the binary representation of n. Then A0 = {n ∈ N : 2 | D(n)} and A1 = {n ∈ N : 2 - D(n)} are the two sets satisfying Theorem A for N = 1. There have been some results on the asymptotic behavior of the representation functions of sets satisfying Theorem A or Theorem B. In particular, Chen and Tang [2] proved the following theorems. The reader is referred to [1] for more results. Theorem C. Let N be a positive integer and A ⊂ N satisfying R2 (A, n) = R2 (N\A, n) for all n ≥ 2N − 1. Then R2 (A, n) ≥ 1 for all n ≥ 12N 2 − 10N − 2, except for A = A0 or A = A1 . E-mail address: [email protected]. http://dx.doi.org/10.1016/j.disc.2014.11.011 0012-365X/© 2014 Elsevier B.V. All rights reserved.
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For A = A0 or A = A1 , we have R2 (A, 22n+1 − 1) = R2 (N\A, 22n+1 − 1) = 0 for all n ≥ 0. Theorem D. Let N be a positive integer and A ⊂ N satisfying R3 (A, n) = R3 (N\A, n) for all n ≥ 2N − 1. Then R3 (A, n) ≥ 1 for all n ≥ 12N 2 + 2N. It is asked whether the quadratic bounds in Theorems C and D can be improved. More precisely, Y.-G. Chen [1] raised the following problems. Problem 1. Does there exist an absolute constant c, such that for any positive integer N, A ⊂ N satisfying R2 (A, n) = R2 (N\A, n) for all n ≥ 2N − 1, and A ̸= A0 , A1 , we have R2 (A, n) ≥ 1 for all n ≥ cN? Problem 2. Does there exist an absolute constant c ′ , such that for any positive integer N and A ⊂ N satisfying R3 (A, n) = R3 (N\A, n) for all n ≥ 2N − 1, we have R3 (A, n) ≥ 1 for all n ≥ c ′ N? In this paper, we prove that both problems have an affirmative answer. In fact we consider slightly more general situation. For a positive integer N, consider coloring functions c : N → {0, 1} satisfying c (2k) = c (k) and c (2k + 1) = 1 − c (k) for all k ≥ N. Let DN denote the set of all such coloring functions. Similarly, consider coloring functions c : N → {0, 1} satisfying c (2k) = 1 − c (k) and c (2k + 1) = c (k) for all k ≥ N, and use TN to denote the set of all such coloring functions. For a coloring c : N → {0, 1} and n ∈ N, n is said to have a monochromatic representation (with respect to c) if there exist x, y ∈ N such that n = x + y and c (x) = c (y). Similarly, n is said to have a multicolored representation (with respect to c) if there exist u, v ∈ N such that n = u + v and c (u) ̸= c (v). Let c0 ∈ D1 be the coloring defined by c0 (0) = 0, c0 (1) = 1, c (2k) = c (k) and c (2k + 1) = 1 − c (k) for all k ≥ 1. The following results are proved. Theorem 1. Let N be a positive integer and c ∈ DN such that c ̸= c0 and c ̸= 1 − c0 . Then for every n ≥ 22N, n has a monochromatic representation n = x + y with x ̸= y, and n also has a multicolored representation. Corollary 1. Let N be a positive integer, A ⊂ N such that R2 (A, n) = R2 (N\A, n) for all n ≥ 2N − 1, and A ̸= A0 , A1 . Then R2 (A, n) = R2 (N\A, n) ≥ 1 for all n ≥ 22N. Theorem 2. Let N be a positive integer and c ∈ TN . Then for every n ≥ 20N, n has a monochromatic representation and a multicolored representation. Corollary 2. Let N be a positive integer and A ⊂ N such that R3 (A, n) = R3 (N\A, n) for all n ≥ 2N − 1. Then R3 (A, n) = R3 (N\A, n) ≥ 1 for all n ≥ 20N.
2. Proof of Theorem 1 We start with several lemmas that will be used in the proof of our main results. Observe first that if c ∈ DN or c ∈ TN , then c (k) ̸= c (k + 1) for all even numbers k ≥ 2N. Lemma 1. Let c ∈ DN (resp. c ∈ TN ) such that c (2k) ̸= c (2k + 1) for all k ∈ N. Then there exists a unique d ∈ Dm (resp. d ∈ Tm ) such that d(k) = c (2k) for all k ∈ N, where m = ⌈N /2⌉ denotes the smallest integer ≥ N /2. Furthermore, c = c0 ⇔ d = c0 , c = 1 − c0 ⇔ d = 1 − c0 .
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Proof. The uniqueness of d is clear. We prove the existence of d. For 0 ≤ k ≤ 2m − 1, let d(k) = c (2k). For k ≥ m, let d(2k) = d(k) and d(2k + 1) = 1 − d(k) (resp. d(2k) = 1 − d(k) and d(2k + 1) = d(k)), and then d ∈ Dm (resp. d ∈ Tm ). It remains to show that d(k) = c (2k) for all k ≥ 2m. Let k ≥ 2m ≥ N, and suppose that d(s) = c (2s) for all nonnegative integers s < k. If k is even, write k = 2t, m ≤ t < k, then d(k) = d(2t ) = d(t )
(resp. d(k) = d(2t ) = 1 − d(t )), and c (2k) = c (k) = c (2t ) = d(t )
(resp. c (2k) = 1 − c (k) = 1 − c (2t ) = 1 − d(t )), thus d(k) = c (2k). If k is odd, write k = 2t + 1, m ≤ t < k, then d(k) = d(2t + 1) = 1 − d(t )
(resp. d(k) = d(2t + 1) = d(t )), and c (2k) = c (k) = c (2t + 1) = 1 − c (2t ) = 1 − d(t )
(resp. c (2k) = 1 − c (k) = 1 − c (2t + 1) = c (2t ) = d(t )), thus d(k) = c (2k). This establishes the existence of d. For the second statement, since d is determined by c and vice versa, it suffices to show that c = c0 implies d = c0 , and c = 1 − c0 implies d = 1 − c0 . If c = c0 ∈ D1 , then d(0) = c0 (0) = 0, d(1) = c0 (2) = 1. Since d ∈ D1 , we have d = c0 . Similarly, if c = 1 − c0 , then d = 1 − c0 . Lemma 2. Let c1 ∈ D1 such that c1 (0) = c1 (1) = 0. Then for every n ≥ 4, n has a monochromatic representation n = x + y with x ̸= y. For every n ≥ 3, n has a multicolored representation. Proof. Let A1 = {n ∈ N : c1 (n) = 0} and B1 = N\A1 . By definition, A1 = {0, 1, 2, 4, 7, . . .} = {a1 < a2 < a3 < · · ·} and B1 = {3, 5, 6, . . .} = {b1 < b2 < b3 < · · ·}. Since |A1 ∩ {2k, 2k + 1}| = |B1 ∩ {2k, 2k + 1}| = 1 for all k ≥ 1, we have ai+1 − ai ≤ 3 and bi+1 − bi ≤ 3 for all i ≥ 1. Let n ≥ 4; then ai ≤ n < ai+1 for some i ≥ 4. Since n − ai ∈ {0, 1, 2}, n = ai +(n − ai ) is a monochromatic representation, and ai ≥ 4 > n − ai . Let m ≥ 3; then bi ≤ m < bi+1 for some i ≥ 1. Since m − bi ∈ {0, 1, 2}, m = bi +(n − bi ) is a multicolored representation. This completes the proof of the lemma. Proof of Theorem 1. First let l ∈ N, N = 2l , c ∈ DN such that c ̸= c0 and c ̸= 1 − c0 . We show that for every n ≥ 11N, n has a monochromatic representation n = x + y with x ̸= y, and n also has a multicolored representation. For l = 0, N = 1, D1 \{c0 , 1 − c0 } = {c1 , 1 − c1 } where c1 is defined as in Lemma 2, the result follows immediately from Lemma 2. Now consider l ≥ 1, N = 2l ≥ 2, and assume that the above statement is true for l − 1. We distinguish two cases. Case 1. n is even and n ≥ 11N. Suppose on the contrary that n does not have a monochromatic representation n = x + y with x ̸= y (resp. n does not have a multicolored representation). For c (4N + 1) = c (4N + 2), we consider n = (4N + 1) + (n − 4N − 1) = (4N + 2) + (n − 4N − 2). Since N ≥ 2, n − 4N − 2 ≥ 11N − 4N − 2 = 7N − 2 > 4N + 2. We must have c (n − 4N − 2) = c (n − 4N − 1) otherwise n has a required representation. However n − 4N − 2 is even and n − 4N − 2 ≥ 2N, we shall have c (n − 4N − 2) ̸= c (n − 4N − 1) by definition of c ∈ DN , which is a contradiction. For c (4N + 1) ̸= c (4N + 2), we have c (4N + 1) = c (4N + 3) since c (4N + 2) ̸= c (4N + 3). By definition, c (2N ) = 1 − c (4N + 1) = 1 − c (4N + 3) = c (2N + 1), again we obtain a contradiction.
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Case 2. n is odd and n ≥ 11N + 1. If c (2k) = c (2k + 1) for some nonnegative integer k ≤ N − 1, consider n = 2k + (n − 2k) = 2k + 1 + (n − 2k − 1). We must have c (n − 2k) = c (n − 2k − 1) otherwise n has a required representation. However n − 2k − 1 is even and n − 2k − 1 ≥ 2N, c (n − 2k − 1) ̸= c (n − 2k), a contradiction. If c (2k) ̸= c (2k + 1) for all k ≤ N − 1, then it holds for all k ≥ 0. By Lemma 1, there exists d ∈ D2l−1 such that d(k) = c (2k) for all k ≥ 0. Since c ̸= c0 and c ̸= 1 − c0 , we have d ̸= c0 and d ̸= 1 − c0 . Now 1 2
(n − 1) ≥
11 2
N = 11 · 2l−1 ,
it follows by the inductive hypothesis, there exists x, y ∈ N, such that 1 2
(n − 1) = x + y,
and d(x) ̸= d(y) (resp. d(x) = d(y)). Hence c (2x) ̸= c (2y) (resp. c (2x) = c (2y)), and c (2x) = c (2y + 1) (resp. c (2x) ̸= c (2y + 1)). We obtain a contradiction since n = 2x + (2y + 1) is a required representation. For any positive integer N, write 2l−1 ≤ N < 2l for some l ≥ 1. Let c ∈ DN such that c ̸= c0 and c ̸= 1 − c0 . Viewing c also as an element of D2l , we conclude that for every n ≥ 22N ≥ 11 · 2l , n has a monochromatic representation n = x + y with x ̸= y, and n also has a multicolored representation. This completes the proof of Theorem 1. Proof of Corollary 1. Define c : N → {0, 1} as follows: c (n) = 0 ⇔ n ∈ A. By Theorem A, c ∈ DN . Since A ̸= A0 and A ̸= A1 , we have c ̸= c0 and c ̸= 1 − c0 . By Theorem 1, every n ≥ 22N has a monochromatic representation n = x + y with x ̸= y. Equivalently, R2 (A, n) = R2 (N\A, n) ≥ 1 for all n ≥ 22N. 3. Proof of Theorem 2 Lemma 3. Let c2 ∈ T1 such that c2 (0) = 0 and c2 (1) = 1. Then, with respect to c2 , every n ≥ 10 has a monochromatic representation n = x + y with xy ̸= 0 and a multicolored representation n = u + v with uv ̸= 0. Let c3 ∈ T1 such that c3 (0) = c3 (1) = 0. Then, with respect to c3 , every n ≥ 10 has a monochromatic representation n = x + y with xy ̸= 0 and a multicolored representation n = u + v with uv ̸= 0. Proof. By definition, we set A2 = {n ∈ N : c2 (n) = 0} = {0, 2, 5, 6, 8, 11, 13, 14, 17, 18, . . .}, B2 = N\A2 = {1, 3, 4, 7, 9, 10, 12, 15, 16, 19, . . .}, A3 = {n ∈ N : c3 (n) = 0} = {0, 1, 3, 4, 7, . . .} = B2 ∪ {0}, B3 = N\A3 = A2 \{0}, therefore the two statements with respect to c2 and c3 are equivalent. It is verified straightforwardly that 10 = 5 + 5 = 4 + 6,
11 = 5 + 6 = 3 + 8,
12 = 6 + 6 = 4 + 8,
13 = 5 + 8 = 6 + 7,
14 = 7 + 7 = 5 + 9,
15 = 2 + 13 = 7 + 8,
16 = 8 + 8 = 6 + 10,
17 = 6 + 11 = 4 + 13,
18 = 9 + 9 = 1 + 17,
and 19 = 2 + 17 = 1 + 18. For 10 ≤ k ≤ 19, k has the required representations. By Lemma 1, there exists d ∈ T1 such that d(t ) = c2 (2t ) for all t ≥ 0. In fact, since d(0) = c2 (0) = 0, d(1) = c2 (2) = 0, we have d = c3 . In other words, c3 (t ) = c2 (2t ) for all t ≥ 0. Suppose k ≥ 20, and the statement is true for every integer t, 10 ≤ t < k. If k is even, write k = 2m, then 10 ≤ m < k. By assumption, there exist x, y, u, v ∈ N, such that xyuv ̸= 0, m=x+y=u+v and c3 (x) = c3 (y), c3 (u) ̸= c3 (v); hence 2m = 2x + 2y = 2u + 2v and c2 (2x) = c2 (2y), c2 (2u) ̸= c2 (2v). If k is odd, write k = 2m + 1, then 10 ≤ m < k. By assumption, there exist x, y, u, v ∈ N, such that xyuv ̸= 0, m=x+y=u+v
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and c3 (x) ̸= c3 (y), c3 (u) = c3 (v); hence 2m = 2x + (2y + 1) = 2u + (2v + 1) and c2 (2x) = c2 (2y + 1), c2 (2u) ̸= c2 (2v + 1). In either case, k has the required representations. This completes the proof of the lemma. We are now ready to prove Theorem 2. Proof of Theorem 2. Let l ∈ N, N = 2l , and c ∈ TN . We first show that for every n ≥ 10N, n has a monochromatic representation and a multicolored representation. If l = 0, N = 1, then T1 = {c2 , 1 − c2 , c3 , 1 − c3 } where c2 , c3 are defined as in Lemma 3, and the result follows by Lemma 3. Now suppose l ≥ 1, N = 2l ≥ 2, c ∈ TN , and the result is true for l − 1. We distinguish two cases. Case 1. n is even and n ≥ 10N. Suppose on the contrary that n does not have a monochromatic representation (resp. multicolored representation). For c (4N + 1) = c (4N + 2), we consider n = (4N + 1) + (n − 4N − 1) = (4N + 2) + (n − 4N − 2). We must have c (n − 4N − 2) = c (n − 4N − 1) otherwise n has a required representation. However n − 4N − 2 is even and n − 4N − 2 ≥ 2N, we shall have c (n − 4N − 2) ̸= c (n − 4N − 1) by definition of c ∈ TN , which is a contradiction. For c (4N + 1) ̸= c (4N + 2), we have c (4N + 1) = c (4N + 3) since c (4N + 2) ̸= c (4N + 3). By definition, c (2N ) = c (4N + 1) = c (4N + 3) = c (2N + 1), again we obtain a contradiction. Case 2. n is odd and n ≥ 10N + 1. If c (2k) = c (2k + 1) for some nonnegative integer k ≤ N − 1, consider n = 2k + (n − 2k) = 2k + 1 + (n − 2k − 1). We must have c (n − 2k) = c (n − 2k − 1); otherwise n has a required representation. However n − 2k − 1 is even and n − 2k − 1 ≥ 2N, c (n − 2k − 1) ̸= c (n − 2k), a contradiction. If c (2k) ̸= c (2k + 1) for all k ≤ N − 1, then it holds for all k ≥ 0. By Lemma 1, there exists d ∈ T2l−1 such that d(k) = c (2k) for all k ≥ 0. Since 1 2
(n − 1) ≥ 5N = 10 · 2l−1 ,
it follows by the inductive hypothesis, there exist x, y ∈ N, such that 1 2
(n − 1) = x + y,
and d(x) ̸= d(y) (resp. d(x) = d(y)). Hence c (2x) ̸= c (2y) (resp. c (2x) = c (2y)), and c (2x) = c (2y + 1) (resp. c (2x) ̸= c (2y + 1)). We obtain a contradiction since n = 2x + (2y + 1) is a required representation. For any positive integer N, write 2l−1 ≤ N < 2l for some l ≥ 1. Let c ∈ TN . Viewing c also as an element of T2l , we conclude that for every n ≥ 20N ≥ 10 · 2l , n has a monochromatic representation and a multicolored representation. This completes the proof of Theorem 2. Proof of Corollary 2. Define c : N → {0, 1} as follows: c (n) = 0 ⇔ n ∈ A. By Theorem B, c ∈ TN . By Theorem 2, every n ≥ 20N has a monochromatic representation n = x + y. Equivalently, R3 (A, n) = R3 (N\A, n) ≥ 1 for all n ≥ 20N. Acknowledgments This work was supported by the National Natural Science Foundation of China, Grant No. 11101152. The author would like to thank the referees for their careful review and helpful comments. References [1] [2] [3] [4] [5] [6] [7]
Y.-G. Chen, On the values of representation functions, Sci. China Math. 54 (2011) 1317–1331. Y.-G. Chen, M. Tang, Partitions of natural numbers with the same representation functions, J. Number Theory 129 (2009) 2689–2695. Y.-G. Chen, B. Wang, On additive perperties of two special sequences, Acta Arith. 110 (2003) 299–303. G. Dombi, Additive properties of certain sets, Acta Arith. 103 (2002) 137–146. V.F. Lev, Reconstructing integer sets from their representation functions, Electron. J. Combin. 11 (2004) R78. C. Sándor, Partitions of natural numbers and their representation functions, Integers 4 (2004) A18. M. Tang, Partitions of the set of natural numbers and their representation functions, Discrete Math. 308 (2008) 2614–2616.
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
On the nonvanishing of representation functions of some special sequences Zhenhua Qu Department of Mathematics, Shanghai Key Laboratory of PMMP, East China Normal University, 500 Dongchuan Rd., Shanghai 200241, PR China
article
info
Article history: Received 17 February 2014 Received in revised form 15 November 2014 Accepted 18 November 2014
Keywords: Representation function Partition Sárközy problem
abstract For a given positive integer N, and any coloring function c : N → {0, 1} satisfying c (2k) = 1 − c (k), c (2k + 1) = c (k) for all k ≥ N, we show that for all n ≥ 20N, n has both a monochromatic representation and a multicolored representation, in other words, there exist x, y, u, v ∈ N, such that n = x + y = u +v , c (x) = c (y) and c (u) ̸= c (v). Similar results are obtained for another kind of coloring function c : N → {0, 1} satisfying c (2k) = c (k) and c (2k + 1) = 1 − c (k) for all k ≥ N. This answers a question of Y.-G. Chen on the values of representation functions. © 2014 Elsevier B.V. All rights reserved.
1. Introduction Let N be the set of nonnegative integers. For a set A ⊂ N and n ∈ N, we use R1 (A, n), R2 (A, n) and R3 (A, n) to denote the number of solutions of n = a + a′ with a, a′ ∈ A, n = a + a′ with a, a′ ∈ A, a < a′ , and n = a + a′ with a, a′ ∈ A, a ≤ a′ respectively. For i ∈ {1, 2, 3}, Sárközy asked whether there exist sets A, B ⊂ N with infinite symmetric difference such that Ri (A, n) = Ri (B, n) holds for all sufficiently large integers n. Dombi [4] observed that the answer is negative for i = 1 and is positive for i = 2. Chen and Wang [3] showed that N can be partitioned into two sets A and B such that R3 (A, n) = R3 (B, n) for all n ≥ 1. Later on, Lev [5], Sándor [6] and Tang [7] characterized, for each i ∈ {2, 3} and N ≥ 1, all sets A ⊂ N with the property that Ri (A, n) = Ri (N\A, n) holds for all n ≥ 2N − 1. More precisely, the following theorems are proved. Theorem A. Let N be a positive integer. Then R2 (A, n) = R2 (N\A, n) for all n ≥ 2N − 1 if and only if |A ∩ [0, 2N − 1]| = N, and 2k ∈ A ⇔ k ∈ A, 2k + 1 ∈ A ⇔ k ̸∈ A for all k ≥ N. Theorem B. Let N be a positive integer. Then R3 (A, n) = R3 (N\A, n) for all n ≥ 2N − 1 if and only if |A ∩ [0, 2N − 1]| = N, and 2k ∈ A ⇔ k ̸∈ A, 2k + 1 ∈ A ⇔ k ∈ A for all k ≥ N. Let D(n) denote the number of ones in the binary representation of n. Then A0 = {n ∈ N : 2 | D(n)} and A1 = {n ∈ N : 2 - D(n)} are the two sets satisfying Theorem A for N = 1. There have been some results on the asymptotic behavior of the representation functions of sets satisfying Theorem A or Theorem B. In particular, Chen and Tang [2] proved the following theorems. The reader is referred to [1] for more results. Theorem C. Let N be a positive integer and A ⊂ N satisfying R2 (A, n) = R2 (N\A, n) for all n ≥ 2N − 1. Then R2 (A, n) ≥ 1 for all n ≥ 12N 2 − 10N − 2, except for A = A0 or A = A1 . E-mail address: [email protected]. http://dx.doi.org/10.1016/j.disc.2014.11.011 0012-365X/© 2014 Elsevier B.V. All rights reserved.
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For A = A0 or A = A1 , we have R2 (A, 22n+1 − 1) = R2 (N\A, 22n+1 − 1) = 0 for all n ≥ 0. Theorem D. Let N be a positive integer and A ⊂ N satisfying R3 (A, n) = R3 (N\A, n) for all n ≥ 2N − 1. Then R3 (A, n) ≥ 1 for all n ≥ 12N 2 + 2N. It is asked whether the quadratic bounds in Theorems C and D can be improved. More precisely, Y.-G. Chen [1] raised the following problems. Problem 1. Does there exist an absolute constant c, such that for any positive integer N, A ⊂ N satisfying R2 (A, n) = R2 (N\A, n) for all n ≥ 2N − 1, and A ̸= A0 , A1 , we have R2 (A, n) ≥ 1 for all n ≥ cN? Problem 2. Does there exist an absolute constant c ′ , such that for any positive integer N and A ⊂ N satisfying R3 (A, n) = R3 (N\A, n) for all n ≥ 2N − 1, we have R3 (A, n) ≥ 1 for all n ≥ c ′ N? In this paper, we prove that both problems have an affirmative answer. In fact we consider slightly more general situation. For a positive integer N, consider coloring functions c : N → {0, 1} satisfying c (2k) = c (k) and c (2k + 1) = 1 − c (k) for all k ≥ N. Let DN denote the set of all such coloring functions. Similarly, consider coloring functions c : N → {0, 1} satisfying c (2k) = 1 − c (k) and c (2k + 1) = c (k) for all k ≥ N, and use TN to denote the set of all such coloring functions. For a coloring c : N → {0, 1} and n ∈ N, n is said to have a monochromatic representation (with respect to c) if there exist x, y ∈ N such that n = x + y and c (x) = c (y). Similarly, n is said to have a multicolored representation (with respect to c) if there exist u, v ∈ N such that n = u + v and c (u) ̸= c (v). Let c0 ∈ D1 be the coloring defined by c0 (0) = 0, c0 (1) = 1, c (2k) = c (k) and c (2k + 1) = 1 − c (k) for all k ≥ 1. The following results are proved. Theorem 1. Let N be a positive integer and c ∈ DN such that c ̸= c0 and c ̸= 1 − c0 . Then for every n ≥ 22N, n has a monochromatic representation n = x + y with x ̸= y, and n also has a multicolored representation. Corollary 1. Let N be a positive integer, A ⊂ N such that R2 (A, n) = R2 (N\A, n) for all n ≥ 2N − 1, and A ̸= A0 , A1 . Then R2 (A, n) = R2 (N\A, n) ≥ 1 for all n ≥ 22N. Theorem 2. Let N be a positive integer and c ∈ TN . Then for every n ≥ 20N, n has a monochromatic representation and a multicolored representation. Corollary 2. Let N be a positive integer and A ⊂ N such that R3 (A, n) = R3 (N\A, n) for all n ≥ 2N − 1. Then R3 (A, n) = R3 (N\A, n) ≥ 1 for all n ≥ 20N.
2. Proof of Theorem 1 We start with several lemmas that will be used in the proof of our main results. Observe first that if c ∈ DN or c ∈ TN , then c (k) ̸= c (k + 1) for all even numbers k ≥ 2N. Lemma 1. Let c ∈ DN (resp. c ∈ TN ) such that c (2k) ̸= c (2k + 1) for all k ∈ N. Then there exists a unique d ∈ Dm (resp. d ∈ Tm ) such that d(k) = c (2k) for all k ∈ N, where m = ⌈N /2⌉ denotes the smallest integer ≥ N /2. Furthermore, c = c0 ⇔ d = c0 , c = 1 − c0 ⇔ d = 1 − c0 .
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Proof. The uniqueness of d is clear. We prove the existence of d. For 0 ≤ k ≤ 2m − 1, let d(k) = c (2k). For k ≥ m, let d(2k) = d(k) and d(2k + 1) = 1 − d(k) (resp. d(2k) = 1 − d(k) and d(2k + 1) = d(k)), and then d ∈ Dm (resp. d ∈ Tm ). It remains to show that d(k) = c (2k) for all k ≥ 2m. Let k ≥ 2m ≥ N, and suppose that d(s) = c (2s) for all nonnegative integers s < k. If k is even, write k = 2t, m ≤ t < k, then d(k) = d(2t ) = d(t )
(resp. d(k) = d(2t ) = 1 − d(t )), and c (2k) = c (k) = c (2t ) = d(t )
(resp. c (2k) = 1 − c (k) = 1 − c (2t ) = 1 − d(t )), thus d(k) = c (2k). If k is odd, write k = 2t + 1, m ≤ t < k, then d(k) = d(2t + 1) = 1 − d(t )
(resp. d(k) = d(2t + 1) = d(t )), and c (2k) = c (k) = c (2t + 1) = 1 − c (2t ) = 1 − d(t )
(resp. c (2k) = 1 − c (k) = 1 − c (2t + 1) = c (2t ) = d(t )), thus d(k) = c (2k). This establishes the existence of d. For the second statement, since d is determined by c and vice versa, it suffices to show that c = c0 implies d = c0 , and c = 1 − c0 implies d = 1 − c0 . If c = c0 ∈ D1 , then d(0) = c0 (0) = 0, d(1) = c0 (2) = 1. Since d ∈ D1 , we have d = c0 . Similarly, if c = 1 − c0 , then d = 1 − c0 . Lemma 2. Let c1 ∈ D1 such that c1 (0) = c1 (1) = 0. Then for every n ≥ 4, n has a monochromatic representation n = x + y with x ̸= y. For every n ≥ 3, n has a multicolored representation. Proof. Let A1 = {n ∈ N : c1 (n) = 0} and B1 = N\A1 . By definition, A1 = {0, 1, 2, 4, 7, . . .} = {a1 < a2 < a3 < · · ·} and B1 = {3, 5, 6, . . .} = {b1 < b2 < b3 < · · ·}. Since |A1 ∩ {2k, 2k + 1}| = |B1 ∩ {2k, 2k + 1}| = 1 for all k ≥ 1, we have ai+1 − ai ≤ 3 and bi+1 − bi ≤ 3 for all i ≥ 1. Let n ≥ 4; then ai ≤ n < ai+1 for some i ≥ 4. Since n − ai ∈ {0, 1, 2}, n = ai +(n − ai ) is a monochromatic representation, and ai ≥ 4 > n − ai . Let m ≥ 3; then bi ≤ m < bi+1 for some i ≥ 1. Since m − bi ∈ {0, 1, 2}, m = bi +(n − bi ) is a multicolored representation. This completes the proof of the lemma. Proof of Theorem 1. First let l ∈ N, N = 2l , c ∈ DN such that c ̸= c0 and c ̸= 1 − c0 . We show that for every n ≥ 11N, n has a monochromatic representation n = x + y with x ̸= y, and n also has a multicolored representation. For l = 0, N = 1, D1 \{c0 , 1 − c0 } = {c1 , 1 − c1 } where c1 is defined as in Lemma 2, the result follows immediately from Lemma 2. Now consider l ≥ 1, N = 2l ≥ 2, and assume that the above statement is true for l − 1. We distinguish two cases. Case 1. n is even and n ≥ 11N. Suppose on the contrary that n does not have a monochromatic representation n = x + y with x ̸= y (resp. n does not have a multicolored representation). For c (4N + 1) = c (4N + 2), we consider n = (4N + 1) + (n − 4N − 1) = (4N + 2) + (n − 4N − 2). Since N ≥ 2, n − 4N − 2 ≥ 11N − 4N − 2 = 7N − 2 > 4N + 2. We must have c (n − 4N − 2) = c (n − 4N − 1) otherwise n has a required representation. However n − 4N − 2 is even and n − 4N − 2 ≥ 2N, we shall have c (n − 4N − 2) ̸= c (n − 4N − 1) by definition of c ∈ DN , which is a contradiction. For c (4N + 1) ̸= c (4N + 2), we have c (4N + 1) = c (4N + 3) since c (4N + 2) ̸= c (4N + 3). By definition, c (2N ) = 1 − c (4N + 1) = 1 − c (4N + 3) = c (2N + 1), again we obtain a contradiction.
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Case 2. n is odd and n ≥ 11N + 1. If c (2k) = c (2k + 1) for some nonnegative integer k ≤ N − 1, consider n = 2k + (n − 2k) = 2k + 1 + (n − 2k − 1). We must have c (n − 2k) = c (n − 2k − 1) otherwise n has a required representation. However n − 2k − 1 is even and n − 2k − 1 ≥ 2N, c (n − 2k − 1) ̸= c (n − 2k), a contradiction. If c (2k) ̸= c (2k + 1) for all k ≤ N − 1, then it holds for all k ≥ 0. By Lemma 1, there exists d ∈ D2l−1 such that d(k) = c (2k) for all k ≥ 0. Since c ̸= c0 and c ̸= 1 − c0 , we have d ̸= c0 and d ̸= 1 − c0 . Now 1 2
(n − 1) ≥
11 2
N = 11 · 2l−1 ,
it follows by the inductive hypothesis, there exists x, y ∈ N, such that 1 2
(n − 1) = x + y,
and d(x) ̸= d(y) (resp. d(x) = d(y)). Hence c (2x) ̸= c (2y) (resp. c (2x) = c (2y)), and c (2x) = c (2y + 1) (resp. c (2x) ̸= c (2y + 1)). We obtain a contradiction since n = 2x + (2y + 1) is a required representation. For any positive integer N, write 2l−1 ≤ N < 2l for some l ≥ 1. Let c ∈ DN such that c ̸= c0 and c ̸= 1 − c0 . Viewing c also as an element of D2l , we conclude that for every n ≥ 22N ≥ 11 · 2l , n has a monochromatic representation n = x + y with x ̸= y, and n also has a multicolored representation. This completes the proof of Theorem 1. Proof of Corollary 1. Define c : N → {0, 1} as follows: c (n) = 0 ⇔ n ∈ A. By Theorem A, c ∈ DN . Since A ̸= A0 and A ̸= A1 , we have c ̸= c0 and c ̸= 1 − c0 . By Theorem 1, every n ≥ 22N has a monochromatic representation n = x + y with x ̸= y. Equivalently, R2 (A, n) = R2 (N\A, n) ≥ 1 for all n ≥ 22N. 3. Proof of Theorem 2 Lemma 3. Let c2 ∈ T1 such that c2 (0) = 0 and c2 (1) = 1. Then, with respect to c2 , every n ≥ 10 has a monochromatic representation n = x + y with xy ̸= 0 and a multicolored representation n = u + v with uv ̸= 0. Let c3 ∈ T1 such that c3 (0) = c3 (1) = 0. Then, with respect to c3 , every n ≥ 10 has a monochromatic representation n = x + y with xy ̸= 0 and a multicolored representation n = u + v with uv ̸= 0. Proof. By definition, we set A2 = {n ∈ N : c2 (n) = 0} = {0, 2, 5, 6, 8, 11, 13, 14, 17, 18, . . .}, B2 = N\A2 = {1, 3, 4, 7, 9, 10, 12, 15, 16, 19, . . .}, A3 = {n ∈ N : c3 (n) = 0} = {0, 1, 3, 4, 7, . . .} = B2 ∪ {0}, B3 = N\A3 = A2 \{0}, therefore the two statements with respect to c2 and c3 are equivalent. It is verified straightforwardly that 10 = 5 + 5 = 4 + 6,
11 = 5 + 6 = 3 + 8,
12 = 6 + 6 = 4 + 8,
13 = 5 + 8 = 6 + 7,
14 = 7 + 7 = 5 + 9,
15 = 2 + 13 = 7 + 8,
16 = 8 + 8 = 6 + 10,
17 = 6 + 11 = 4 + 13,
18 = 9 + 9 = 1 + 17,
and 19 = 2 + 17 = 1 + 18. For 10 ≤ k ≤ 19, k has the required representations. By Lemma 1, there exists d ∈ T1 such that d(t ) = c2 (2t ) for all t ≥ 0. In fact, since d(0) = c2 (0) = 0, d(1) = c2 (2) = 0, we have d = c3 . In other words, c3 (t ) = c2 (2t ) for all t ≥ 0. Suppose k ≥ 20, and the statement is true for every integer t, 10 ≤ t < k. If k is even, write k = 2m, then 10 ≤ m < k. By assumption, there exist x, y, u, v ∈ N, such that xyuv ̸= 0, m=x+y=u+v and c3 (x) = c3 (y), c3 (u) ̸= c3 (v); hence 2m = 2x + 2y = 2u + 2v and c2 (2x) = c2 (2y), c2 (2u) ̸= c2 (2v). If k is odd, write k = 2m + 1, then 10 ≤ m < k. By assumption, there exist x, y, u, v ∈ N, such that xyuv ̸= 0, m=x+y=u+v
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and c3 (x) ̸= c3 (y), c3 (u) = c3 (v); hence 2m = 2x + (2y + 1) = 2u + (2v + 1) and c2 (2x) = c2 (2y + 1), c2 (2u) ̸= c2 (2v + 1). In either case, k has the required representations. This completes the proof of the lemma. We are now ready to prove Theorem 2. Proof of Theorem 2. Let l ∈ N, N = 2l , and c ∈ TN . We first show that for every n ≥ 10N, n has a monochromatic representation and a multicolored representation. If l = 0, N = 1, then T1 = {c2 , 1 − c2 , c3 , 1 − c3 } where c2 , c3 are defined as in Lemma 3, and the result follows by Lemma 3. Now suppose l ≥ 1, N = 2l ≥ 2, c ∈ TN , and the result is true for l − 1. We distinguish two cases. Case 1. n is even and n ≥ 10N. Suppose on the contrary that n does not have a monochromatic representation (resp. multicolored representation). For c (4N + 1) = c (4N + 2), we consider n = (4N + 1) + (n − 4N − 1) = (4N + 2) + (n − 4N − 2). We must have c (n − 4N − 2) = c (n − 4N − 1) otherwise n has a required representation. However n − 4N − 2 is even and n − 4N − 2 ≥ 2N, we shall have c (n − 4N − 2) ̸= c (n − 4N − 1) by definition of c ∈ TN , which is a contradiction. For c (4N + 1) ̸= c (4N + 2), we have c (4N + 1) = c (4N + 3) since c (4N + 2) ̸= c (4N + 3). By definition, c (2N ) = c (4N + 1) = c (4N + 3) = c (2N + 1), again we obtain a contradiction. Case 2. n is odd and n ≥ 10N + 1. If c (2k) = c (2k + 1) for some nonnegative integer k ≤ N − 1, consider n = 2k + (n − 2k) = 2k + 1 + (n − 2k − 1). We must have c (n − 2k) = c (n − 2k − 1); otherwise n has a required representation. However n − 2k − 1 is even and n − 2k − 1 ≥ 2N, c (n − 2k − 1) ̸= c (n − 2k), a contradiction. If c (2k) ̸= c (2k + 1) for all k ≤ N − 1, then it holds for all k ≥ 0. By Lemma 1, there exists d ∈ T2l−1 such that d(k) = c (2k) for all k ≥ 0. Since 1 2
(n − 1) ≥ 5N = 10 · 2l−1 ,
it follows by the inductive hypothesis, there exist x, y ∈ N, such that 1 2
(n − 1) = x + y,
and d(x) ̸= d(y) (resp. d(x) = d(y)). Hence c (2x) ̸= c (2y) (resp. c (2x) = c (2y)), and c (2x) = c (2y + 1) (resp. c (2x) ̸= c (2y + 1)). We obtain a contradiction since n = 2x + (2y + 1) is a required representation. For any positive integer N, write 2l−1 ≤ N < 2l for some l ≥ 1. Let c ∈ TN . Viewing c also as an element of T2l , we conclude that for every n ≥ 20N ≥ 10 · 2l , n has a monochromatic representation and a multicolored representation. This completes the proof of Theorem 2. Proof of Corollary 2. Define c : N → {0, 1} as follows: c (n) = 0 ⇔ n ∈ A. By Theorem B, c ∈ TN . By Theorem 2, every n ≥ 20N has a monochromatic representation n = x + y. Equivalently, R3 (A, n) = R3 (N\A, n) ≥ 1 for all n ≥ 20N. Acknowledgments This work was supported by the National Natural Science Foundation of China, Grant No. 11101152. The author would like to thank the referees for their careful review and helpful comments. References [1] [2] [3] [4] [5] [6] [7]
Y.-G. Chen, On the values of representation functions, Sci. China Math. 54 (2011) 1317–1331. Y.-G. Chen, M. Tang, Partitions of natural numbers with the same representation functions, J. Number Theory 129 (2009) 2689–2695. Y.-G. Chen, B. Wang, On additive perperties of two special sequences, Acta Arith. 110 (2003) 299–303. G. Dombi, Additive properties of certain sets, Acta Arith. 103 (2002) 137–146. V.F. Lev, Reconstructing integer sets from their representation functions, Electron. J. Combin. 11 (2004) R78. C. Sándor, Partitions of natural numbers and their representation functions, Integers 4 (2004) A18. M. Tang, Partitions of the set of natural numbers and their representation functions, Discrete Math. 308 (2008) 2614–2616.