Partitions of the set of nonnegative integers with the same representation functions

Partitions of the set of nonnegative integers with the same representation functions Sándor Z. Kiss, Csaba Sándor arxiv:1606.0699v [math.nt] 10 Aug 016 Abstract For a set of nonnegative integers S let R S (n denote the number of unordered representations of the integer n as the sum of two different terms from S. In this paper we focus on partitions of the natural numbers into two sets affording identical representation functions. We solve a recent problem of Lev Chen. 000 AMS Mathematics subject classification number: 11B34. Key words phrases: additive number theory, representation functions, partitions of the set of natural numbers. 1 Introduction Let S be a set of nonnegative integers let R S (n denote the number of solutions of the equation s + s = n, where s, s S s < s. Let N denotes the set of nonnegative integers. The binary representation of an integer n is the representation of n in the number system with base. Let A be the set of those nonnegative integers which contains even number of 1 binary digits in its binary representation let B be the complement of A. The set A is called Thue-Morse sequence. The investigation of the partitions of the set of nonnegative integers with identical representation functions was a popular topic in the last few decades [1], [], [7], [8], [9]. By using the Thue - Morse sequence in 00 Dombi [5] constructed two sets of nonnegative integers with infinite symmetric difference such that the corresponding representation functions are identical. Namely, he proved the following theorem. Theorem 1. (Dombi, 00 The set of positive integers can be partitioned into two subsets C D such that R C (n = R D (n for every positive integer n. The complete description of the suitable partitions is the following. Institute of Mathematics, Budapest University of Technology Economics, H-159 B.O. Box, Hungary, kisspest@cs.elte.hu This author was supported by the OTKA Grant No. NK105645. This research was supported by the National Research, Development Innovation Office NKFIH Grant No. K11588. Institute of Mathematics, Budapest University of Technology Economics, H-159 B.O. Box, Hungary, csor@math.bme.hu. This author was supported by the OTKA Grant No. K109789. This paper was supported by the János Bolyai Research Scholarship of the Hungarian Academy of Sciences. 1

Theorem. Let C D be sets of nonnegative integers such that C D = N, C D = 0 C. Then R C (n = R D (n for every positive integer n if only if C = A D = B. As far as we know this theorem has never been formulated in the above form, but the nontrivial part was proved by Dombi, therefore this theorem is only a little extension of Dombi s result. We give an alternative proof of the previous theorem. A finite version of the above theorem is the following. Put A l = A [0, l 1] B l = B [0, l 1]. Theorem 3. Let C D be sets of nonnegative integers such that C D = [0,m] C D =, 0 C. Then R C (n = R D (n for every positive integer n if only if there exists an l natural number such that C = A l D = B l. If C = A [0,m] D = B [0,m] then by Theorem we have R C (n = R D (n for n m, therefore Theorem 3. implies the following corollary. Corollary 1. If C = A [0,m] D = B [0,m], where m is a positive integer not of the form l 1, then there exists a positive integer m < n < m such that R C (n R D (n. In Dombi s example the union of the set C D is the set of nonnegative integers, they are disjoint sets. Tang Yu [11] proved that if the union of the sets C D is the set of nonnegative integers the representation functions are identical from a certain point on, then at least one cannot have the intersection of the two sets is the non-negative integers divisible by 4 i.e., Theorem 4. (Tang Yu, 01 If C D = N C D = 4N, then R C (n R D (n for infinitely many n. Moreover, they conjectured that under the same assumptions the intersection cannot be the union of infinite arithmetic progressions. Conjecture 1. (Tang Yu, 01 Let m N R {0,1,...,m 1}. If C D = N C D = {r+km : k N,r R}, then R C (n = R D (n cannot hold for all sufficiently large n. Recently Tang extended Theorem 4. In particular, she proved the following theorem [10]: Theorem 5. (Tang, 016 Let k be an integer. If C D = N C D = kn, then R C (n R D (n for infinitely many n. Chen Lev [] disproved the above conjecture by constructing a family of partitions of the set of natural numbers such that all the corresponding representation functions are the same the intersection of the two sets is an infinite arithmetic progression properly contained in the set of natural numbers. Theorem 6. (Chen Lev, 016 Let l be a positive integer. There exist sets C D such that C D = N, C D = ( l 1+( l+1 1N R C (n = R D (n for every positive integer n. Their construction is based on the following lemma:

Lemma 1. (Chen Lev, 016 If there exist sets C 0 D 0 such that C 0 D 0 = [0,m], C 0 D 0 = {r} R C0 (n = R D0 (n for every positive integer n then there exist sets C D such that C D = N, C D = r + (m+1n R C (n = R D (n for every positive integer n. When r = l 1 m = l+1 they proved the existence of two such sets C 0 D 0, which verifies the sufficiency of the following problem in [] (we use different notations then they applied. Problem 1. (Chen Lev, 016 Given R C (n = R D (n for every positive integer n, C D = [0,m], C D = {r} with integers r 0 m, must there exist an integer l 1 such that r = l 1, m = l+1, C = A l ( l 1 + B l D = B l ( l 1+A l? In this paper we solve this problem affirmatively. Theorem 7. Let C D be sets of nonnegative integers such that C D = [0,m] C D = {r}, 0 C. Then R C (n = R D (n for every n positive integer if only if there exists an l natural number such that C = A l ( l 1+B l D = B l ( l 1+A l. The previous theorem suggests that there are no other counterexample for Tang Yu s conjecture. Problem. (Chen Lev, 016 Given R C (n = R D (n for every n positive integer, C D = N, C D = r +mn with integers r 0 m, must there exist an integer l 1 such that r = l 1, m = l+1 1? Similar questions were investigated for unordered representation functions in [3], [4], [6]. Thoughout this paper the characteristic function of the set S is denoted by χ S (n, i.e., { 1, if n S χ S (n = 0, if n / S. For a nonnegative integer a a set of nonnegative integers S we define the sumset a+s by a+s = {a+b : b S}. Proof of Theorem. 3. First we prove that if there exists a natural number l such that C = A l D = B l, then R C (n = R D (n for every positive integer n. We prove by induction on l. For l = 1, A 1 = {0} B 1 = {1} thus R A1 (n = R B1 (n = 0. Assume the statement holds for any l we prove it to l + 1. By the definition of A B we have A l+1 = A l ( l +B l B l+1 = B l ( l +A l. Hence R Al+1 (n = R Al ( l +B l (n = {(a,a : a < a,a,a A l,a+a = n} + {(a,a : a A l,a l +B l,a+a = n} + {(a,a : a < a,a,a l +B l,a+a = n} 3

= R Al (n+ {(a,a : a A l,a B l,a+a = n l } +R Bl (n l+1. On the other h R Bl+1 (n = R Bl ( l +A l (n = {(a,a : a < a,a,a B l,a+a = n} + {(a,a : a B l,a l +A l,a+a = n} + {(a,a : a < a a,a l +A l,a+a = n} = R Bl (n+ {(a,a : a B l,a A l,a+a = n l } +R Al (n l+1, thus we get the result. Observe that if k l 1, then R Al (k = R A (k R Bl (k = R B (k. On the other h R Al (k = R Bl (k thus we have R A (k = R B (k for k l 1. This equality holds for every l, therefore we have R A (k = R B (k for every k. (1 To prove Theorem. 3. we need the following three claims. Claim 1. Let 0 < r 1 <... < r s m be integers. Then there exists at most one pair of sets (C,D such that C D = [0,m], 0 C, C D = {r 1,...,r s }, R C (k = R D (k for every k m. Proof of Claim 1. We prove by contradiction. Assume that there exist at least two pairs of different sets (C 1,D 1 (C,D which satisfies the conditions of Claim 1. Let v denote the smallest positive integer such that χ C1 (v χ C (v. It is clear that R C1 (v = R D1 (v R C (v = R D (v. We will prove that R D1 (v = R D (v but R C1 (v R C (v which is a contradiction. Obviously R D1 (v = {(d,d : d < d,d,d D 1,d+d = v}. As 0 / C D 0 C, we have d, d < v. We prove that D 1 [0,v 1] = D [0,v 1], which implies that R D1 (v = R D (v. Clearly we have C 1 [0,v 1] = C [0,v 1] [0,v 1] = (C 1 [0,v 1] (D 1 [0,v 1] [0,v 1] = (C [0,v 1] (D [0,v 1]. ( Let (C 1 [0,v 1] (D 1 [0,v 1] = {r 1,...,r t }. Thus we have D 1 [0,v 1] = [0,v 1]\(C 1 [0,v 1] {r 1,...,r t }. Similarly D [0,v 1] = ( [0,v 1]\(C [0,v 1] {r 1,...,r t }, which implies D 1 [0,v 1] = D [0,v 1]. On the other h as 0 C 1 C we have R C1 (v = {(c,c : c < c < v,c,c C 1,c+c = v} +χ C1 (v, R C (v = {(c,c : c < c < v,c,c C 1,c+c = v} +χ C (v, thus R C1 (v R C (v. Claim. Let (C,D be a pair of different sets, C D = [0,m], C D = {r 1,...,r s }, R C (n = R D (n for every n nonnegative integer if C = m C D = m D then C D = [0,m], C D = {m r s,...,m r 1 }, R C (n = R D (n for every positive integer n. Proof of Claim. Clearly, R C (k = {(c,c : c < c,c,c C,c+c = k} = {(c,c : c < c,m c,m c C,c+c = k} 4

= {(m c,m c : c < c,m c,m c C,m (c+c = m k} = R C (m k. Similarly, R D (k = R D (m k, which implies R D (k = R D (m k = R C (m k = R C (k, as desired. Claim 3. If for some positive integer M, the integers M 1,M,M 4,M 8,...,M u, u = log M 1 are all contained in the set A, then M = u+1 1. Proof of Claim 3. Suppose that theintegers M 1,M,M 4,M 8,...,M u, u = log M 1 are all contained in the set A. If M is even then M is also an even M 1 = (M +1, therefore by the definition of A, χ A (M 1 χ A (M, thus we may assume that M is an odd positive integer. Assume that M is not of the form k 1. Obviously, χ A (M χ A (M 1. Let M = u i=0 b i i be the representation of M in the number system with base. Let x denote the largest index i such that b i = 0. Then x log M 1. Thus we have thus we have x 1 M = b i i + x+1 + i=0 x 1 M x = b i i + x + i=0 u+1 i=x+ u+1 i=x+ b i i, b i {0,1}, u+1 b i i = b i i, b i {0,1}, which implies that u i=0 b i = u i=0 b i. It follows from the definition of A that χ A (M = χ A (M x. On the other h χ A (M χ A (M 1 = χ A (M = χ A (M 4 =... = χ A (M x, which proves Claim 3. Claim 4. If for some positive integer M, the integers M 1,M,M 4,M 8,...,M u, u = log M 1 are all contained in the set B, then M = u+1 1. The proof of Claim 4. is completely the same as Claim 3. Theorem. is a consequence of (1 Claim 1. (for s = 0. In the next step we prove that if the sets C D satisfies C D = [0,m], C D = R C (n = R D (n for every positive integer n, then there exists an l positive integer such that C = A l D = B l. Then Claim 1. (1 imply that C = A [0,m] D = B [0,m]. Let C = m C D = m D. By Claim 1., Claim. (1 we have C = A [0,m] or D = B [0,m]. It follows that i=0 χ C ( 0 = χ C ( 1 = χ C ( =... = χ C ( u, where u = log M 1, which implies that χ C (m 1 = χ C (m = χ C (m 4 =... = χ C (m u. By Claim 3. Claim 4. we get m = u+1 1. The proof of Theorem 3. is completed. 5

3 Proof of Theorem 6. First, assume that there exists a positive integer l such that C = A l ( l 1+B l, D = B l ( l 1+A l. Obviously, C D = [0, l+1 ], C D = { l 1}, 0 C we will prove that for every positive integer n, R C (n = R D (n. It is easy to see that R C (n = {(c,c : c < c,c,c A l,c+c = n} + {(c,c : c A l,c l 1+B l,c+c = n} + {(c,c : c < c,c,c l 1+B l,c+c = n} = R Al (n+ {(c,c : c A l,c B l,c+c = n ( l 1} +R Bl (n ( l 1. Moreover, R D (n = {(d,d : d < d,d,d B l,d+d = n} + {(d,d : d B l,d l 1+A l,d+d = n} + {(d,d : d,d l 1+A l,d+d = n} = R Bl (n+ {(d,d : d A l,d B l,d+d = n ( l 1} +R Al (n ( l 1. It follows from Theorem 3. that R Al (m = R Bl (m for every positive integer m, thus we get the result. In the next step we prove that if C D = [0,m], C D = {r}, 0 C R C (n = R D (n, then there exists a positive integer l such that C = A l ( l 1 + B l, D = B l ( l 1 + A l m = l+1, r = l 1. By Claim. if C D = [0,m], C D = {r} R C (n = R D (n for every positive integer n, then for C = m C, D = m D wehavec D = [0,m], C D = {m r}byclaim. R C (n = R D (n, thus we may assume that r m/. Let p C (x = m χ C (ix i, p D (x = i=0 m χ D (ix i. ( i=0 As C D = [0,m], C D = {r} we have p D (x = 1 xm+1 1 x p C(x+x r. (3 Since R C (n = R D (n for every positive integer n, thus we have R C (nx n = n=0 An easy observation shows that 1 p C(x 1 p C(x = R C (nx n = n=0 It follows from (3 (4 that R D (nx n. n=0 R D (nx n = 1 p D(x 1 p D(x. (4 ( 1 x (p C (x p C (x m+1 ( 1 x = 1 x p C(x+x r m+ p 1 x C (x +x r. 6 n=0

An easy calculation shows that p C (x = 1 xm+ 1 x ( +p C (x 1 xm+1 1 x m+1 x 1 x r 1 xm+1 1 x 1 x +xr p C (x. (5 We will prove that r must be odd. If r were even r k r m is also even then it is easy to see from ( the coefficient of x k in (5 we have ( k χ C = 1+ χ C (i (k +1 +χ C (k r, (6 i k If k +1 r m, then from the coefficient of x k+1 in (5 we have 0 = χ C (i (k + +χ C (k +1 r. (7 i k+1 By (6 - (7 dividing by we get that ( k χ C = 1 χ C (k +1+χ C (k r χ C (k +1 r. In view of Claim 1., k+1 r < r, k r is even, C [0,r 1] = A [0,r 1] by the definitionofaweget χ C (k+1 r+χ C (k r = 1thuswe haveχ C (k r χ C (k+1 r = ±1. Ifχ C (k r χ C (k+1 r = 1thenwegetthatχ C (k r = 1, χ C (k+1 r = 0, which (k yields χ C (k+1 = 1 χ C = 1. On the other h if χ C (k r χ C (k+1 r = 1 then we get that χ C (k r = 0, χ C (k + 1 r = 1, which yields χ C (k + 1 = 0 (k (k χ C = 0. This gives that χ C = χ C (k r. Putting k = r i+1, where i+1 log r we obtain that χ C (r i+1 = χ C (r i. It follows that χ C (r 1 = χ C (r = χ C (r 4 =... = χ C (r t. for t = log r 1. Moreover, Claim 3. Claim 4. imply that r = l 1, which contradicts the assumption that r is even. It follows that r must be odd. By using the same argument as before we prove that r = l 1. If r k < r m k is even then from the coefficient of x k in (5 we have ( k χ C = 1+ χ C (i (k +1 +χ C (k r. i k In this case k 1 is odd, k 1 r, therefore from the coefficient of x k 1 in (5 we have 0 = χ C (i k +χ C (k 1 r. i k 1 As before, subtracting the above equalities dividing by we get that ( k χ C = χ C (k+χ C (k r χ C (k 1 r. If r is odd, then k 1 r is even, we know from Claim 1. that C [0,r 1] = A [0,r 1] by definition of A we get χ C (k 1 r+χ C (k r = 1 thus we have 7

χ C (k r χ C (k 1 r = ±1. If χ C (k r χ C (k 1 r = 1 then we get that (k χ C (k r = 1, χ C (k 1 r = 0, which yields χ C (k = 0 χ C = 1. If χ C (k r χ C (k 1 r = 1 then we get that χ C (k r = 0, χ C (k 1 r = 1, (k which yields χ C (k = 1 χ C = 0. This gives that χ C (k r = χ C when r k < r k is even. Put k = r i+1, where i + 1 log r implies that χ C (r i+1 = χ C (r i. It follows that χ C (r 1 = χ C (r = χ C (r 4 =... = χ C (r log r 1, which yields by Claim 3. Claim 4. that r = u 1. If k = r, then from the coefficient of x k in (5 we have 0 = χ C (i (r +1 +. (8 i r On the other h if k = r 1, then from the coefficient of x k 1 in (5 we have Since r C, it follows from (8 - (9 that ( (r 1 we get that 0 = χ C = χ A ( r 1 χ C = 1+ χ C (i r. (9 i r 1 ( r 1 χ C = +χ C (r = 0, r 1, thus r 1 = u 1 1, where u 1 is odd, so that r = l 1. As r m/, Claim 1. the first part of Theorem 3. imply that C [0,r] = A l ( l 1+B l D [0,r] = B l ( l 1+A l. We will show that m < 3 l. We prove by contradiction. Assume that m > 3 l. We will prove that C [0,3 l 3] = A l ( l 1+B l ( l+1 1+(B l [0, l ] D [0,3 l 3] = B l ( l 1+A l ( l+1 1+(A l [0, l ]. Define the sets E F by E = A l ( l 1+B l ( l+1 1+(B l [0, l ], F = B l ( l 1+A l ( l+1 1+(A l [0, l ]. In this case R E (m = R F (m for every m l+1. If l+1 < n 3 l 3, then we have R E (n = {(c,c : c A l,c l 1+B l,c+c = n} + {(c,c : c A l,c l+1 1+(B l [0, l ],c+c = n} + 8

{(c,c : c < c,c,c l 1+B l,c+c = n} = {(c,c : c A l,c B l,c+c = n ( l 1} + {(c,c : c A l,c B l,c+c = n ( l+1 1} + R Bl (n ( l 1 R F (n = {(d,d : d B l,d l 1+A l,d+d = n} + {(d,d : d B l,d l+1 1+(A l [0, l ],d+d = n} {(d,d : d < d,d,d l 1+A l,d+d = n} = {(d,d : d B l,d A l,d+d = n ( l 1} + {(d,d : d B l,d A l,d+d = n ( l+1 1} +R Al (n ( l 1, which imply R E (n = R F (n for every positive integer n by Theorem 3., therefore by Claim 1., this is the only possible starting of C D. We will prove that 3 l C. We prove by contradiction. Assume that 3 l D, that is C [0,3 l ] = A l ( l 1+B l ( l+1 1+B l D [0,3 l ] = B l ( l 1+A l ( l+1 1+A l. We have a solution 3 l = ( l+1 1+( l 1 in set D, thus we have R D (3 l = 1+ {(d,d : d B l,d l+1 1+A l,d+d = 3 l } On the other h + {(d,d : d < d,d,d l 1+A l,d+d = 3 l } = 1+ {(d,d : d B l,d A l,d+d = l 1} +R Al ( l. R C (3 l = {(c,c : c A l,c l+1 1+B l,c+c = 3 l } + {(c,c : c < c,c,c l 1+B l,c+c = 3 l } = {(c,c : c A l,c B l,c+c = l 1} +R Bl ( l, therefore by Theorem 3 we have R D (3 l > R C (3 l, which is a contradiction. We may assume that 3 l C. Using the fact 1 C we have R C (3 l 1 = {(c,c : c A l,c l+1 1+B l,c+c = 3 l 1} + {(c,c : c < c,c,c l 1+B l,c+c = 3 l 1} +χ C (3 l 1 = {(c,c : c A l,c B l,c+c = l } +R Bl ( l +1+χ C (3 l 1. On the other h using 1 B l, l 1 A l 3 l C we get R D (3 l 1 = {(d,d : d B l,d l+1 1+A l [0, l ],d+d = 3 l 1} 9

+ {(d,d : d < d,d,d l 1+A l,d+d = 3 l 1} = {(d,d : d B l,d A l,d+d = l } +R Al ( l +1 1, therefore by Theorem 3 we have R C (3 l 1 > R D (3 l 1, which is a contradiction, that is we have m < 3 l. It follows that C = A l ( l 1+B l ( l+1 1+(B l [0,m ( l+1 1] D = B l ( l 1+A l ( l+1 1+(A l [0,m ( l+1 1]. Wewillprovethatm = l+1. Assumethatm > l+1. Ifm ( l+1 1 k 1, then by Corollary 1., there exists an m ( l+1 1 < u < (m ( l+1 1 such that R A [0,m ( l+1 1](u R B [0,m ( l+1 1](u. Since m+ l+1 1 < u+( l+1 1 < m we obtain R C (( l+1 1+u = R l+1 1+B l [0,m ( l+1 1](( l+1 1+u = R Bl [0,m ( l+1 1](u. Similarly R D (( l+1 1+u = R Al [0,m ( l+1 1](u, which is a contradiction. Thus we may assume that m ( l+1 1 = k 1, where k < l. Hence C = A l ( l 1+B l ( l+1 1+B k D = B l ( l 1+A l ( l+1 1+A k. If k = 0, then C,D [0, l+1 1] l+1, l+1 1 D, therefore R C ( l+ 3 = 0 R D ( l+ 3 = 1, a contradiction. Thus we may assume that k > 0. Then if C = l+1 + k C D = l+1 + k D itfollowsthatc D = [0, l+1 + k ], C D = { l + k 1}byClaim. R C (n = R D (nforeverypositiveintegern. Thus we have C [0, l + k ] = A [0, l + k ] or C [0, l + k ] = B [0, l + k ]. We prove that C [0, l + k ] = A [ l + k ]. Assume that C [0, l + k ] = B [ l + k ]. Then C [0, k 1] = B k C [ k, k+1 1] = k +A k, which is a contradiction because C [ k, k+1 1] = k +B k. We know C [0, k 1] = A k, then C [ k, k+1 1] = k +B k, C [ k+1,3 k ] = k+1 +B k [0, k ]. On the other h for C = l+1 (A l ( l 1+B l we have C [0, l ] = B [0, l ], therefore C [0, k 1] = B k C [ k, k+1 ] = k +A k [0, k ], which is a contradiction because C = A k ( k +C. The proof of Theorem 6. is completed. References [1] Y. G. Chen. On the values of representation functions, Sci. China Math., 54(011, 1317-1331. 10

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