Partitions of Natural Numbers with the Intersection Not Empty

Abstract

Let ℕ be the set of nonnegative integers. For a given set 𝐴⊂ℕ the representation functions 𝑅₂(𝐴,𝑛), 𝑅₃(𝐴,𝑛) are defined as the number of solutions of the equation 𝑛=𝑎+𝑎^′,𝑎,𝑎^′∈𝐴 with condition 𝑎<𝑎^′, 𝑎≤𝑎^′, respectively. In this paper, we prove that if ℕ=𝐴∪𝐵 and 𝐴∩𝐵={8𝑘∶𝑘∈ℕ}, then 𝑅_𝑖(𝐴,𝑛)=𝑅_𝑖(𝐵,𝑛) cannot hold for all sufficiently large integers 𝑛 where 𝑖=2,3.

1. Introduction

Let ℕ be the set of nonnegative integers. For 𝐴⊂ℕ, let 𝑅₁(𝐴,𝑛), 𝑅₂(𝐴,𝑛), 𝑅₃(𝐴,𝑛) denote the number of solutions of 𝑎+𝑎^′=𝑛,𝑎,𝑎^′∈𝐴,𝑎+𝑎^′=𝑛,𝑎,𝑎^′∈𝐴,𝑎<𝑎^′,𝑎+𝑎^′=𝑛,𝑎,𝑎^′∈𝐴,𝑎<𝑎^′, (1) respectively. Sárközy asked ever whether there exist two sets 𝐴 and 𝐵 of nonnegative integers with infinite symmetric difference, that is: )⧵(𝑅|(𝐴∪𝐵𝐴∩𝐵)|=∞,_𝑖(𝐴,𝑛)=𝑅_𝑖(𝐵,𝑛),𝑛≥𝑛₀ (2) for 𝑖=1,2,3. As Dombi [1] has shown, the answer is negative for 𝑖=1 by the simple observation that 𝑅₁(𝐴,𝑛) is odd if and only if 𝑛=2𝑎 for some 𝑎∈𝐴. For 𝑖=2, Dombi presented a partition of the set of all positive integers into two disjoint subsets 𝐴 and 𝐵 such that 𝑅₂(𝐴,𝑛)=𝑅₂(𝐵,𝑛) for all sufficiently large 𝑛. For 𝑖=3, Chen and Wang [2] constructed sets 𝐴,𝐵 such that 𝑅₃(𝐴,𝑛)=𝑅₃(𝐵,𝑛) for all integers 𝑛≥3, where 𝐴 is the set of all those positive integers 𝑛 such that the number of zeroes in the binary representation of 𝑛−1 is even, and 𝐵 is the set of all those positive integers 𝑛 such that the number of zeroes in the binary representation of 𝑛−1 is odd. For the other related results, the reader is referred to see [3–7].

It is natural to ask the following: for 𝑖=2,3, are there subsets 𝐴,𝐵⊂ℕ with 𝑅_𝑖(𝐴,𝑛)=𝑅_𝑖(𝐵,𝑛) for all large enough integers 𝑛 such that ℕ=𝐴∪𝐵 and 𝐴∩𝐵≠∅? Recently, the authors of this paper [8] obtained some results in this direction.

In this paper, we obtain the following results.

Theorem 1. If ℕ=𝐴∪𝐵 and 𝐴∩𝐵={8𝑘∶𝑘∈ℕ}, then 𝑅₂(𝐴,𝑛)=𝑅₂(𝐵,𝑛) cannot hold for all sufficiently large integers 𝑛.

Theorem 2. If ℕ=𝐴∪𝐵 and 𝐴∩𝐵={8𝑘∶𝑘∈ℕ}, then 𝑅₃(𝐴,𝑛)=𝑅₃(𝐵,𝑛) cannot hold for all sufficiently large integers 𝑛.

Remark 3. For 𝑟=1,2,3,4,5,6,7, similarly, we can prove that if ℕ=𝐴∪𝐵 and 𝐴∩𝐵={8𝑘+𝑟∶𝑘∈ℕ}, then 𝑅_𝑖(𝐴,𝑛)=𝑅_𝑖(𝐵,𝑛) cannot hold for all sufficiently large integers 𝑛, where 𝑖=2,3.

2. Proof of Theorem 1

Proof. Suppose that there exist integer 𝑛₀ and sets 𝐴,𝐵⊂ℕ with 𝐴∪𝐵=ℕ and 𝐴∩𝐵={8𝑘∶𝑘∈ℕ} such that 𝑅₂(𝐴,𝑛)=𝑅₂(𝐵,𝑛) for all 𝑛≥𝑛₀. Without loss of generality, we may assume that 𝑛₀=16𝑁−1, 𝑁 is a positive integer; then there exists a polynomial 𝑝(𝑥) of degree at most 16𝑁−2 such that ^∞_𝑛=0𝑅₂(𝐴,𝑛)−𝑅₂(𝐵,𝑛)𝑥^𝑛=𝑝(𝑥). (3) Let 𝑓(𝑥)=_𝑎∈𝐴𝑥^𝑎,𝑔(𝑥)=_𝑏∈𝐵𝑥^𝑏,𝜀_𝑖=1,𝑖∈𝐴,0,𝑖∉𝐴. (4) Then we have 𝑓(𝑥)=_𝑎∈𝐴𝑥^𝑎=^∞_𝑖=0𝜀_𝑖𝑥^𝑖,_∞𝑛=0𝑅₂(𝐴,𝑛)𝑥^𝑛=12𝑓²(𝑥)−𝑓𝑥². (5) Moreover, 𝑔(𝑥)=_𝑏∈𝐵𝑥^𝑏=_{()∪()𝑏∈ℕ⧵𝐴𝐴∩𝐵}𝑥^𝑏=^∞_𝑛=0𝑥^𝑛−𝑓(𝑥)+^∞_𝑘=0𝑥^8𝑘=1+11−𝑥1−𝑥⁸−𝑓_∞𝑛=0(𝑥),𝑅₂(𝐵,𝑛)𝑥^𝑛=12𝑔²(𝑥)−𝑔𝑥²=121+11−𝑥1−𝑥⁸−𝑓(𝑥)²−11−𝑥²+11−𝑥¹⁶−𝑓𝑥²=𝑥(1−𝑥)1−𝑥²+𝑥⁸1−𝑥⁸1−𝑥¹⁶+1(1−𝑥)1−𝑥⁸−𝑓(𝑥)−𝑓(𝑥)1−𝑥1−𝑥⁸+12𝑓²(𝑥)+12𝑓𝑥². (6) Therefore, ^∞_𝑛=0𝑅₂(𝐴,𝑛)−𝑅₂(𝐵,𝑛)𝑥^𝑛=𝑓(𝑥)+𝑓(𝑥)1−𝑥1−𝑥⁸−𝑓𝑥²−𝑥(1−𝑥)1−𝑥²−𝑥⁸1−𝑥⁸1−𝑥¹⁶−1(1−𝑥)1−𝑥⁸. (7)
Write 𝑝(𝑥)1−𝑥⁸=^16𝑁+6_𝑖=0𝛼_𝑖𝑥^𝑖. (8) By (3) and (7), we have ⁷_𝑖=0𝑥^𝑖𝑓(𝑥)=+11−𝑥⁸𝑓𝑥²+1+𝑥+𝑥³+𝑥⁵+𝑥⁷+𝑥1−𝑥₈1−𝑥¹⁶+𝑝(𝑥)1−𝑥⁸. (9) Writing both sides of (9) in power series, then we have ^∞_𝑖=02𝜀_𝑖𝑥^𝑖+^∞_𝑖=0𝜀_𝑖𝑥^𝑖+1+⋯+^∞_𝑖=0𝜀_𝑖𝑥^𝑖+7=^∞_𝑖=0𝜀_𝑖𝑥^2𝑖−^∞_𝑖=0𝜀_𝑖𝑥^2𝑖+8+^∞_𝑖=0𝑥^𝑖+^∞_𝑖=0𝑥^𝑖+1+^∞_𝑖=0𝑥^𝑖+3+^∞_𝑖=0𝑥^𝑖+5+^∞_𝑖=0𝑥^𝑖+7+^∞_𝑖=0𝑥^16𝑖+8+^16𝑁+6_𝑖=0𝛼_𝑖𝑥^𝑖. (10)
Comparing the coefficients of 𝑥^16𝑁+𝑡,𝑡=7,9,11,13,15,17,19,21,23,25 on the both sides of (10), we have 2𝜀_16𝑁+𝑡+𝜀_{16𝑁+𝑡−1}+⋯+𝜀_{16𝑁+𝑡−7}=5,𝑡=7,9,…,25. (11) Then we have 2𝜀_16𝑁+7+𝜀_16𝑁+6+⋯+𝜀_16𝑁=5,2𝜀_16𝑁+9=𝜀_16𝑁+7+𝜀_16𝑁+1,2𝜀_16𝑁+11+𝜀_16𝑁+10=𝜀_16𝑁+9+𝜀_16𝑁+3+𝜀_16𝑁+2,2𝜀_16𝑁+13+𝜀_16𝑁+12=𝜀_16𝑁+11+𝜀_16𝑁+5+𝜀_16𝑁+4,2𝜀_16𝑁+15+𝜀_16𝑁+14=𝜀_16𝑁+13+𝜀_16𝑁+7+𝜀_16𝑁+6,2𝜀_16𝑁+17=𝜀_16𝑁+15+𝜀_16𝑁+9,2𝜀_16𝑁+19+𝜀_16𝑁+18=𝜀_16𝑁+17+𝜀_16𝑁+11+𝜀_16𝑁+10,2𝜀_16𝑁+21+𝜀_16𝑁+20=𝜀_16𝑁+19+𝜀_16𝑁+13+𝜀_16𝑁+12,2𝜀_16𝑁+23+𝜀_16𝑁+22=𝜀_16𝑁+21+𝜀_16𝑁+15+𝜀_16𝑁+14,2𝜀_16𝑁+25=𝜀_16𝑁+23+𝜀_16𝑁+17. (12)(13)(14)(15)(16)(17)(18)(19)(20)(21) Since 𝐴∩𝐵={8𝑘∶𝑘∈ℕ}, we have 𝜀_16𝑁+𝑚=1,𝑚=0,8,16,24. Then, by (12), (𝜀_16𝑁+1,𝜀_16𝑁+2,𝜀_16𝑁+3,𝜀_16𝑁+4,𝜀_16𝑁+5,𝜀_16𝑁+6,𝜀_16𝑁+7) has the following 30 cases: (),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),(),().0,0,0,0,1,1,10,0,0,1,0,1,10,0,0,1,1,0,10,0,1,0,0,1,10,0,1,0,1,0,10,0,1,1,0,0,10,0,1,1,1,1,00,1,0,0,0,1,10,1,0,0,1,0,10,1,0,1,0,0,10,1,0,1,1,1,00,1,1,0,0,0,10,1,1,0,1,1,00,1,1,1,0,1,00,1,1,1,1,0,01,0,0,0,0,1,11,0,0,0,1,0,11,0,0,1,0,0,11,0,0,1,1,1,01,0,1,0,0,0,11,0,1,0,1,1,01,0,1,1,0,1,01,0,1,1,1,0,01,1,0,0,0,0,11,1,0,0,1,1,01,1,0,1,0,1,01,1,0,1,1,0,01,1,1,0,0,1,01,1,1,0,1,0,01,1,1,1,0,0,0 (22) By (13), we have 𝜀_16𝑁+9=𝜀_16𝑁+7=𝜀_16𝑁+1=0, (23) or 𝜀_16𝑁+9=𝜀_16𝑁+7=𝜀_16𝑁+1=1. (24) Thus (𝜀_16𝑁+2,𝜀_16𝑁+3,𝜀_16𝑁+4,𝜀_16𝑁+5,𝜀_16𝑁+6,𝜀_16𝑁+7,𝜀_16𝑁+9) has the following 10 cases: (),(),(),(),(),(),(),(),(),().0,1,1,1,1,0,01,0,1,1,1,0,01,1,0,1,1,0,01,1,1,0,1,0,01,1,1,1,0,0,00,0,0,0,1,1,10,0,0,1,0,1,10,0,1,0,0,1,10,1,0,0,0,1,11,0,0,0,0,1,1 (25) By (14) and (25), we have (𝜀_16𝑁+4,𝜀_16𝑁+5,𝜀_16𝑁+6,𝜀_16𝑁+7,𝜀_16𝑁+9,𝜀_16𝑁+10,𝜀_16𝑁+11) which has the following 8 cases: (),(),(),(),(),(),(),().1,1,1,0,0,1,00,1,1,0,0,0,11,0,1,0,0,0,11,1,0,0,0,0,10,0,1,1,1,1,00,1,0,1,1,1,01,0,0,1,1,1,00,0,0,1,1,0,1 (26) By (15) and (26), we have (𝜀_16𝑁+6,𝜀_16𝑁+7,𝜀_16𝑁+9,𝜀_16𝑁+10,𝜀_16𝑁+11,𝜀_16𝑁+12,𝜀_16𝑁+13) which has the following 6 cases: (),(),(),(),(),().1,0,0,1,0,0,11,0,0,0,1,0,10,0,0,0,1,1,11,1,1,1,0,0,00,1,1,1,0,1,00,1,1,0,1,1,0 (27) By (16) and (27), we have (𝜀_16𝑁+9,𝜀_16𝑁+10,𝜀_16𝑁+11,𝜀_16𝑁+12,𝜀_16𝑁+13,𝜀_16𝑁+14,𝜀_16𝑁+15) which has the following 6 cases: (),(),(),(),(),().0,1,0,0,1,0,10,0,1,0,1,0,10,0,1,1,1,1,01,1,0,0,0,0,11,1,0,1,0,1,01,0,1,1,0,1,0 (28) By (17) and (28), we have (𝜀_16𝑁+10,𝜀_16𝑁+11,𝜀_16𝑁+12,𝜀_16𝑁+13,𝜀_16𝑁+14,𝜀_16𝑁+15,𝜀_16𝑁+17) which has the following 2 cases: (0,1,1,1,1,0,0),(1,0,0,0,0,1,1). (29) By (18) and (29), we have (𝜀_16𝑁+12,𝜀_16𝑁+13,𝜀_16𝑁+14,𝜀_16𝑁+15,𝜀_16𝑁+17,𝜀_16𝑁+18,𝜀_16𝑁+19) which has the following 2 cases: (1,1,1,0,0,1,0),(0,0,0,1,1,0,1). (30) By (19) and (30), we have (𝜀_16𝑁+14,𝜀_16𝑁+15,𝜀_16𝑁+17,𝜀_16𝑁+18,𝜀_16𝑁+19,𝜀_16𝑁+20,𝜀_16𝑁+21) which has the following 2 cases: (1,0,0,1,0,0,1),(0,1,1,0,1,1,0). (31) By (20) and (31), we have (𝜀_16𝑁+17,𝜀_16𝑁+18,𝜀_16𝑁+19,𝜀_16𝑁+20,𝜀_16𝑁+21,𝜀_16𝑁+22,𝜀_16𝑁+23) which has the following 2 cases: (0,1,0,0,1,0,1),(1,0,1,1,0,1,0). (32) By (21), we have 𝜀_16𝑁+25=𝜀_16𝑁+23=𝜀_16𝑁+17=0, (33) or 𝜀_16𝑁+25=𝜀_16𝑁+23=𝜀_16𝑁+17=1, (34) which contradicts (32).
This completes the proof of Theorem 1.

3. Proof of Theorem 2

Proof. Suppose that there exist integer 𝑛₀ and sets 𝐴,𝐵⊂ℕ with 𝐴∪𝐵=ℕ and 𝐴∩𝐵={8𝑘∶𝑘∈ℕ} such that 𝑅₃(𝐴,𝑛)=𝑅₃(𝐵,𝑛) for all 𝑛≥𝑛₀. Without loss of generality, we may assume that 𝑛₀=16𝑁−1, 𝑁 is a positive integer then there exists a polynomial 𝑞(𝑥) of degree at most 16𝑁−2 such that ^∞_𝑛=0𝑅₃(𝐴,𝑛)−𝑅₃(𝐵,𝑛)𝑥^𝑛=𝑞(𝑥). (35)
Define 𝑓(𝑥),𝑔(𝑥), and 𝜀_𝑖 as in the proof of Theorem 1. Then we have ^∞_𝑛=0𝑅₃(𝐴,𝑛)𝑥^𝑛=12𝑓²(𝑥)+𝑓𝑥²,_∞𝑛=0𝑅₃(𝐵,𝑛)𝑥^𝑛=12𝑔²(𝑥)+𝑔𝑥²=121+11−𝑥1−𝑥⁸−𝑓(𝑥)²+11−𝑥²+11−𝑥¹⁶−𝑓𝑥²=1(1−𝑥)1−𝑥²+11−𝑥⁸1−𝑥¹⁶+1(1−𝑥)1−𝑥⁸−𝑓(𝑥)−𝑓(𝑥)1−𝑥1−𝑥⁸+12𝑓²(𝑥)−12𝑓𝑥². (36) Therefore, ^∞_𝑛=0𝑅₃(𝐴,𝑛)−𝑅₃(𝐵,𝑛)𝑥^𝑛=𝑓(𝑥)+𝑓(𝑥)1−𝑥1−𝑥⁸+𝑓𝑥²−1(1−𝑥)1−𝑥²−11−𝑥⁸1−𝑥¹⁶−1(1−𝑥)1−𝑥⁸. (37)
Write 𝑞(𝑥)1−𝑥⁸=^16𝑁+6_𝑖=0𝛽_𝑖𝑥^𝑖. (38) By (35) and (37), we have ⁷_𝑖=0𝑥^𝑖+1𝑓(𝑥)=𝑥⁸−1𝑓𝑥²+2+𝑥²+𝑥⁴+𝑥⁶+11−𝑥1−𝑥¹⁶+𝑞(𝑥)1−𝑥⁸. (39) We write both sides of (39) in power series; then we have ^∞_𝑖=02𝜀_𝑖𝑥^𝑖+^∞_𝑖=0𝜀_𝑖𝑥^𝑖+1+⋯+^∞_𝑖=0𝜀_𝑖𝑥^𝑖+7=^∞_𝑖=0𝜀_𝑖𝑥^2𝑖+8−^∞_𝑖=0𝜀_𝑖𝑥^2𝑖+^∞_𝑖=02𝑥^𝑖+^∞_𝑖=0𝑥^𝑖+2+^∞_𝑖=0𝑥^𝑖+4+^∞_𝑖=0𝑥^𝑖+6+^∞_𝑖=0𝑥^16𝑖+^16𝑁+6_𝑖=0𝛽_𝑖𝑥^𝑖. (40)
Comparing the coefficients of 𝑥^16𝑁+𝑡,𝑡=7,9,11,13,15,17,19,21,23,25 on the both sides of (40), we have 2𝜀_16𝑁+𝑡+𝜀_{16𝑁+𝑡−1}+⋯+𝜀_{16𝑁+𝑡−7}=5,𝑡=7,9,…,25. (41)
The remainder of the proof is the same as that of the proof of Theorem 1. We omit it here.

Acknowledgments

This work was supported by the National Natural Science Foundation of China, Grant no. 10901002 and Anhui Provincial Natural Science Foundation, Grant no. 1208085QA02.