On a problem of Chen and Fang related to infinite additive complements
aa r X i v : . [ m a t h . N T ] D ec On a problem of Chen and Fang related toinfinite additive complements
Sándor Z. Kiss ∗ , Csaba Sándor † Abstract
Two infinite sets A and B of nonnegative integers are called additivecomplements if their sumset contains every nonnegative integer. In1964, Danzer constructed infinite additive complements A and B with A ( x ) B ( x ) = (1 + o (1)) x as x → ∞ , where A ( x ) and B ( x ) denote thecounting function of the sets A and B , respectively. In this paper wesolve a problem of Chen and Fang by extending the construction ofDanzer. Keywords and phrases: additive number theory, additive comple-ment counting function, sumset.
Let N be the set of nonnegative integers and let A and B be infinite sets ofnonnegative integers. We define their sum by A + B = { a + b : a ∈ A, b ∈ B } .We say A and B are infinite additive complements if their sum contains allnonnegative integers i.e., A + B = N . Let A ( x ) be the number of elementsof A up to x i.e., A ( x ) = X a ∈ A a ≤ x . Since A and B are infinite additive complements, every nonnegative integer x can be written in the form a + b = x , where a ∈ A , b ∈ B . Then clearly ∗ Institute of Mathematics, Budapest University of Technology and Economics, H-1529B.O. Box, Hungary; [email protected]; This author was supported by the National Re-search, Development and Innovation Office NKFIH Grant No. K115288 and K129335.This paper was supported by the János Bolyai Research Scholarship of the HungarianAcademy of Sciences. Supported by the ÚNKP-20-5 New National Excellence Programof the Ministry for Innovation and Technology from the source of the National Research,Development and Innovation Fund. † Institute of Mathematics, Budapest University of Technology and Economics, MTA-BME Lendület Arithmetic Combinatorics Research Group H-1529 B.O. Box, Hungary,[email protected]. This author was supported by the NKFIH Grants No. K129335.Research supported by the Lendület program of the Hungarian Academy of Sciences(MTA), under grant number LP2019-15/2019.
17] we have A ( x ) B ( x ) ≥ x + 1 , which implies that lim sup x →∞ A ( x ) B ( x ) x ≥ lim inf x →∞ A ( x ) B ( x ) x ≥ . According to a conjecture of H. Hanani [3], the above result can be sharpenedin the following way.
Conjecture 1 (Hanani, 1957) If A and B are infinite additive comple-ments, then lim sup x →∞ A ( x ) B ( x ) x > . Later, Danzer [2] disproved the above conjecture of Hanani.
Theorem 1 (Danzer, 1964)
There exist infinite additive complements A and B such that lim x →∞ A ( x ) B ( x ) x = 1 . Let A , . . . , A r be infinite sets of nonnegative integers. We define theirsum by A + A + . . . + A r = { a + a + . . . + a r : a i ∈ A i , ≤ i ≤ r } .Chen and Fang extended the notion of additive complements to more thantwo sets in the following way [1]. The infinite sets A , . . . , A r of nonnegativeintegers are said to form infinite additive complements if their sum containsall nonnegative integers. Again, it is easy to see that A ( x ) · · · A r ( x ) ≥ ( A + . . . + A r )( x ) = x + 1 , thus lim inf x →∞ A ( x ) · · · A r ( x ) x ≥ . Furthermore, they posed the following problem.
Problem 1
For each integer r ≥ find additive complements A , . . . , A r such that lim x →∞ A ( x ) · · · A r ( x ) x = 1 . In this paper we solve this problem. Note that our construction is the exten-sion of Danzer’s result to r > . Theorem 2
For each integer h ≥ there exist infinite sets of nonnegativeintegers A , . . . , A h with the following properties:(1) A + . . . + A h = N , A ( x ) · · · A h ( x ) = (1 + o (1)) x as x → ∞ . Let R A + B ( n ) be the number of representations of the integer n in the form a + b = n , where a ∈ A , b ∈ B . W. Narkiewicz [4] proved the followingtheorem. Theorem 3 (Narkiewicz, 1960) If R A + B ( n ) ≥ C for every sufficientlylarge integer n , where C is a constant and lim sup x →∞ A ( x ) B ( x ) x ≤ C, then lim x →∞ A (2 x ) x = 1 , or lim x →∞ B (2 x ) x = 1 . Additive complements A , B are called exact if A ( x ) B ( x ) = (1 + o (1)) x as x → ∞ . For any h ≥ integer let us define the system of sets A h by A h = { A ⊂ N : there exist A , . . . , A h ⊂ N ,A + A + . . . + A h = N , A ( x ) · A ( x ) · · · A h ( x ) = (1 + o (1)) x as x → ∞} . Theorem 3 implies that A h = ∅ for every h ≥ . We prove that the A h ’sform an infinite chain. Theorem 4
We have A ⊇ A ⊇ . . . According to Theorem 3, if A ∈ A , then A ( x ) = x o (1) or A ( x ) = x o (1) as x → ∞ . Then for any h ≥ , A ∈ A h implies that A ( x ) = x o (1) or A ( x ) = x o (1) as x → ∞ . If the sets A , . . . , A h ⊂ N satisfy A + . . . + A h = N and A ( x ) · · · A h ( x ) = (1+ o (1)) x as x → ∞ , then A i ( x ) = x o (1) or A i ( x ) = x o (1) for every ≤ i ≤ h while x → ∞ . As a corollary, one can get from Theorem4 that Corollary 1
Let A , . . . , A h be infinite sets of nonnegative integers such that A + . . . + A h = N and A ( x ) · · · A h ( x ) = (1 + o (1)) x as x → ∞ . Then there exists an index i such that A i ( x ) = x o (1) and A j ( x ) = x o (1) for every ≤ j ≤ h with j = i as x → ∞ .
3e pose the following problems for further research.
Problem 2
Does A h = A h +1 hold for every h ≥ ? Problem 3
Assume that A + . . . + A h = N and A ( x ) · · · A h ( x ) = (1+ o (1)) x hold as x → ∞ . Does there exist a permutation i , . . . , i h of the indices , . . . , h such that A i j ( x ) = ( A i j − ( x )) o (1) for every ≤ j ≤ h as x → ∞ ? The statement in Problem 2 holds for h = 2 .The exact complemets have been investigated by many authors in the lastfew decades. In particular, they studied what kind of sets A of nonnegativeintegers with A ( x ) = x o (1) as x → ∞ have exact additive complement. It wasproved in [2] that the sequence a n = ( n !) + 1 has an exact complement. In[5] Ruzsa showed that the set of the powers of an integer a ≥ has an exactcomplement. Furthermore, in [6] he proved that the set of powers of hasan exact complement. Moreover, he also proved in [6] that A = { a , a , . . . } with ≤ a < a < . . . has an exact complement if lim n →∞ a n +1 na n = ∞ . Inview of these results, it is natural to ask Problem 4
Is it true that if A ∈ A , A ( x ) = x o (1) as x → ∞ , then A ( x ) = O (log x ) ? For any nonnegative integers a < b , let us define [ a, b ] = { x ∈ N : a ≤ x ≤ b } .For any nonnegative integer k , we define [ a, b ] · k = { kx : x ∈ [ a, b ] ∩ N } . Thefollowing lemma plays the key role in the proof of Theorem 2. Lemma 1
Assume that A , . . . , A h ⊂ N infinite subsets with the followingproperties:(1) A + . . . + A h = N ,(2) there exists a monotone increasing arithmetic function f h ( n ) ≥ with lim n →∞ f h ( n ) = ∞ such that the equation a + . . . + a h = n , a i ∈ A i has a solution with a i ≥ f h ( n ) ,(3) A ( x ) · · · A h ( x ) = (1 + o (1)) x as x → ∞ . or m ∈ N , let g ( m ) be a strictly increasing function such that g ( f h ( n )) ≥ n for every n ∈ N . Furthermore, for ≤ i ≤ h , let B i = { g ( a )! + a : a ∈ A i } and define the sets of integers B h +1 = [0 , g (6)! + h ( g (6) − − ∪ ( [ n ≥ (cid:20) g ( n )! n − ⌈√ n ⌉ − , g ( n + 1)! + h ( g ( n + 1) − n − ⌈√ n ⌉ (cid:21) · ( n − ⌈√ n ⌉ ) ) . Then(i) B + . . . + B h +1 = N ,(ii) there exists a monotone increasing arithmetic function f h +1 ( n ) ≥ with lim n →∞ f h +1 ( n ) = ∞ such that the equation b + . . . + b h +1 = n , b i ∈ B i has a solution with b i ≥ f h +1 ( n ) ,(iii) B ( x ) · · · B h +1 ( x ) = (1 + o (1)) x as x → ∞ . Now we prove that for any N ≥ , B + . . . + B h + (cid:20) g ( N )! N − ⌈√ N ⌉ − , g ( N + 1)! + h ( g ( N + 1) − N − ⌈√ N ⌉ (cid:21) · ( N −⌈√ N ⌉ ) ⊇ [ g ( N )! − N − ⌈√ N ⌉ ) + h ( g ( N ) − N, g ( N + 1)! + h ( g ( N + 1) − . Consider an element from the interval on the right hand side i.e., let y be g ( N )! − N − ⌈√ N ⌉ ) + h ( g ( N ) − N ≤ y ≤ g ( N + 1)! + h ( g ( N + 1) − . It is clear that there exists an ⌈√ N ⌉ ≤ m ≤ N − with y ≡ m (mod N −⌈√ N ⌉ ) . By (2), there exist a , . . . , a h integers with a i ∈ A i such that m = a + . . . + a h and a i ≥ f h ( m ) . Since f h ( m ) is a monotone increasing functionand g ( m ) is a strictly increasing function, then we have g ( a i ) ≥ g ( f h ( m )) ≥ g ( f h ( ⌈√ N ⌉ )) ≥ ( ⌈√ N ⌉ ) ≥ N and so g ( a i )! ≡ N − ⌈√ N ⌉ ) . Let b i = g ( a i )! + a i . Then b i ∈ B i forevery ≤ i ≤ h . It follows that h X i =1 b i = h X i =1 ( g ( a i )! + a i ) ≡ h X i =1 a i ≡ m ≡ y (mod N − ⌈√ N ⌉ ) , y − ( b + ... + b h ) N −⌈√ N ⌉ is an integer and clearly y = b + . . . + b h + y − ( b + . . . + b h ) N − ⌈√ N ⌉ · ( N − ⌈√ N ⌉ ) . In view of these facts, it is enough to show that g ( N )! N − ⌈√ N ⌉ − ≤ y − ( b + . . . + b h ) N − ⌈√ N ⌉ ≤ g ( N + 1)! + h ( g ( N + 1) − N − ⌈√ N ⌉ . Since g ( n ) is a strictly increasing function, we have ≤ b i = g ( a i )! + a i ≤ g ( m ) + m ≤ g ( N − N − < ( g ( N ) − N and so ≤ h X i =1 b i < h (( g ( N ) − N ) . It follows that y − ( b + . . . + b h ) N − ⌈√ N ⌉ ≥ y − h (( g ( N ) − N ) N − ⌈√ N ⌉≥ g ( N )! − N − ⌈√ N ⌉ ) + h (( g ( N ) − N ) − h ( g ( N ) − N ) N − ⌈√ N ⌉ = g ( N )! N − ⌈√ N ⌉ − and y − ( b + . . . + b h ) N − ⌈√ N ⌉ ≤ yN − ⌈√ N ⌉ ≤ g ( N + 1)! + h ( g ( N + 1) − N − ⌈√ N ⌉ . Thus for N ≥ , we have B + . . . + B h +1 ⊇ [ g ( N )! − N − ⌈√ N ⌉ ) + h ( g ( N ) − N, g ( N + 1)! + h ( g ( N + 1) − ⊇ [ g ( N )! + h ( g ( N ) − , g ( N + 1)! + h ( g ( N + 1) − . This implies that B + . . . + B h +1 ⊇ [ N ≥ [ g ( N )! + h ( g ( N ) − , g ( N + 1)! + h ( g ( N + 1) − [ g (6)! + h ( g (6) − , + ∞ ) . Moreover, for ≤ i ≤ h , ∈ B i and B h +1 ⊇ [0 , g (6)! + h ( g (6) − − ,thus [0 , g (6)! + h ( g (6) − − ⊆ B + . . . + B h +1 and so B + . . . + B h +1 = N ,which proves (i).If g ( N )!+ h ( g ( N ) − ≤ n ≤ g ( N +1)+ h ( g ( N +1) − , then there existsa representation n = b + . . . + b h +1 , where b i = g ( a i ) + a i ≥ a i ≥ f h ( ⌈√ N ⌉ ) and b h +1 ≥ g ( N )! − N − ⌈√ N ⌉ ) ≥ N ! − N − ⌈√ N ⌉ ) , which proves (ii)with a suitable function f h +1 ( n ) .To prove (iii) we assume that g ( N )! + h ( g ( N ) − ≤ x ≤ g ( N + 1)! + h ( g ( N + 1) − . Since g ( N ) is strictly increasing, g ( N + 2 h ) ≥ g ( N + 1) + h .This implies that x ≤ g ( N + 1)! + h ( g ( N + 1) − g ( N + 1) + h )( g ( N + 1) − ≤ g ( N + 2 h )( g ( N + 1) − < g ( N + 2 h )! + N + 2 h and x ≥ g ( N )! + h ( g ( N ) − > g ( N )! + h ( N − ≥ g ( N − N − . Therefore, by the definition of the sets B i , we have A i ( N ) ≤ B i ( x ) ≤ A i ( N +2 h ) for every ≤ i ≤ h . Thus we have, B i ( x ) = A i ( N ) + O (1) = (1 + o (1)) A i ( N ) as x → ∞ for every ≤ i ≤ h . Now, we have B ( x ) · · · B h ( x ) =(1 + o (1)) A ( N ) · · · A h ( N ) = (1 + o (1)) N as x → ∞ . It remains to provethat B h +1 ( x ) = xN (1 + o (1)) as x → ∞ . It follows from the definition of B h that for x ≥ g (6)! + h ( g (6) − we have B h ( x ) = g (6)! + h ( g (6) − N − X n =6 (cid:18) g ( n + 1)! + h ( g ( n + 1) − n − ⌈√ n ⌉ − g ( n )! n − ⌈√ n ⌉ + 3 (cid:19) + (cid:22) xN − ⌈√ N ⌉ − g ( N )! N − ⌈√ N ⌉ + 3 (cid:23) = O ( N ) + N − X n =6 (cid:18) g ( n + 1)! + h ( g ( n + 1) − n − ⌈√ n ⌉ − g ( n )! n − ⌈√ n ⌉ (cid:19) + (cid:18) xN − ⌈√ N ⌉ − g ( N )! N − ⌈√ N ⌉ (cid:19) . By x ≥ g ( N )! + h ( g ( N ) − ≥ N ! , we have O ( N ) = o (cid:0) xN (cid:1) as x → ∞ . Itfollows from (2) in Lemma 1 that n ≥ f h ( n ) . Then by the definition of g ( n ) ,7e have g ( n ) ≥ g ( f h ( n ) ≥ n . Applying this observation, a straightforwardcomputation shows that g ( n + 1)! + h ( g ( n + 1) − n − ⌈√ n ⌉ − g ( n )! n − ⌈√ n ⌉ = (cid:18) O (cid:18) n (cid:19)(cid:19) · g ( n + 1)! + h ( g ( n + 1) − n − ⌈√ n ⌉ = (cid:18) O (cid:18) √ n (cid:19)(cid:19) · g ( n + 1)! n + 1 . Hence, N − X n =6 g ( n + 1)! + h ( g ( n + 1) − n − ⌈√ n ⌉ − g ( n )! n − ⌈√ n ⌉ = N − X n =6 (cid:18) O (cid:18) √ n (cid:19)(cid:19) · g ( n + 1)! n + 1 . In the next step, we show that N − X n =6 (cid:18) O (cid:18) √ n (cid:19)(cid:19) · g ( n + 1)! n + 1 = (1 + o (1)) · g ( N )! N as N → ∞ . Since g ( m ) is strictly increasing, g ( N + 1)! g ( N )! ≥ ( g ( N ) + 1)! g ( N )! = g ( N ) + 1 ≥ N + 1 ≥ N + 1 N , which implies that g ( N )! N is monotone increasing. By g ( m ) ≥ m , we have g ( N − ≤ N g ( N )! . On the other hand, g ( N − N − ≤ g ( N )! /N N − O (cid:18) g ( N )! N (cid:19) . By using the above observations, we have N − X n =6 (cid:18) O (cid:18) √ n (cid:19)(cid:19) · g ( n + 1)! n + 1 = N − X n =7 (cid:18) O (cid:18) √ n (cid:19)(cid:19) · g ( n )! n + g ( N )! N (1+ o (1)) N − X n =7 O (cid:18) g ( N − N − (cid:19) + (1 + o (1)) g ( N )! N = O (cid:18) N g ( N )! N (cid:19) + g ( N )! N (1 + o (1)) = g ( N )! N (1 + o (1)) as x → ∞ . It is clear that xN − ⌈√ N ⌉ − g ( N )! N − ⌈√ N ⌉ = (cid:18) O (cid:18) √ N (cid:19)(cid:19) (cid:18) x − g ( N )! N (cid:19) = (1 + o (1)) x − g ( N )! N as x → ∞ . Then it follows that B h +1 ( x ) = o (cid:16) xN (cid:17) + (1 + o (1)) g ( N )! N + x − g ( N )! N (1 + o (1)) = (1 + o (1)) xN as x → ∞ , which proves (iii). The proof of Lemma 1 is completed. Now, we prove Theorem 2 by induction on h . We show that there existinfinite sets A , . . . , A h ⊂ N with the following properties:(1) A + . . . + A h = N ,(2) there exists a monotone increasing arithmetic function f h ( n ) ≥ with lim n →∞ f h ( n ) = ∞ such that the equation a + . . . + a h = n , a i ∈ A i has a solution with a i ≥ f h ( n ) .(3) A ( x ) · · · A h ( x ) = (1 + o (1)) x as x → ∞ .For h = 1 consider the set of natural numbers and the function f ( n ) = n ,which gives the result. Assume that the statement of Theorem 2 holds for h . For h + 1 the result follows from Lemma 1. (Actually, for h = 2 ourconstruction is the same as the construction of Danzer [2]). The proof ofTheorem 2 is completed. 9 Proof of Theorem 4
Let h ≥ . We will prove that A h +1 ⊆ A h . Let A ∈ A h +1 . Thenthere exist A , . . . , A h +1 ⊆ N such that A + A + . . . + A h +1 = N and A ( x ) A ( x ) · · · A h +1 ( x ) = (1 + o (1)) x as x → ∞ . Let A ∗ h = A h + A h +1 .It is clear that A ∗ h ( x ) ≤ A h ( x ) · A h +1 ( x ) . Then we have A + A + . . . + A h − + A ∗ h = N and so A ( x ) A ( x ) · · · A h − ( x ) A ∗ h ( x ) ≥ x + 1 . On the otherhand, A ( x ) A ( x ) · · · A h − ( x ) A ∗ h ( x ) ≤ A ( x ) A ( x ) · · · A h +1 ( x ) = (1 + o (1)) x as x → ∞ , thus we have A ( x ) A ( x ) · · · A h − ( x ) A ∗ h ( x ) = (1 + o (1)) x as x → ∞ ,which implies that A ∈ A h . The proof of Theorem 4 is completed. References [1]
Y.-G. Chen, J.-H. Fang . On a conjecture of Sárközy and Szemerédi ,Acta Arith., , (2015) 47-58.[2]
L. Danzer . Über eine Frage von G.Hanani aus der additiven Zahlen-theorie , J. Reine Angew. Math, , (1964) 392-394.[3]
P. Erdős . Some unsolved problems , Michigan Math., J. (1957) 291-300.[4] W. Narkiewicz . Remarks on a conjecture of Hanani in additive num-ber theory , Colloq. Math., , (1959/60), 161-165.[5] I. Z. Ruzsa . An asymptotically exact additive completion , Studia Sci.Math. Hungar., , (1996) 51-57.[6] I. Z. Ruzsa . Additive completion of lacunary sequences , Combinatorica, , (2001) 279-291.[7] A. Sárközy, E. Szemerédi . On a problem in additive number theory ,Acta Math. Hungar.,64