Generalizations of a Curious Family of MSTD Sets Hidden By Interior Blocks
aa r X i v : . [ m a t h . N T ] A ug GENERALIZATIONS OF A CURIOUS FAMILY OF MSTD SETSHIDDEN BY INTERIOR BLOCKSH`ung Viˆe.t Chu
Department of Mathematics, Washington and Lee University, Lexington, VA24450 [email protected]
Noah Luntzlara
Department of Mathematics, University of Michigan, Ann Arbor, MI 48109 [email protected]
Steven J. Miller
Department of Mathematics and Statistics, Williams College, Williamstown, MA01267 [email protected]
Lily Shao
Department of Mathematics and Statistics, Williams College, Williamstown, MA01267 [email protected]
Received: , Revised: , Accepted: , Published:
Abstract
A set A is MSTD (more-sum-than-difference) or sum-dominant if | A + A | > | A − A | ,and is RSD (restricted-sum dominant) if | A ˆ+ A | > | A − A | , where A ˆ+ A is the setof sums of distinct elements in A . We study an interesting family of MSTD setsthat have appeared many times in the literature (see the works of Hegarty, Martinand O’Bryant, and Penman and Wells). While these sets seem at first glance tobe ad hoc, looking at them in the right way reveals a nice common structure. Inparticular, instead of viewing them as explicitly written sets, we write them in termsof differences between two consecutive numbers in increasing order. We denote thisfamily by F and investigate many of its properties. Using F , we are able to generatemany sets A with high value of log | A + A | / log | A − A | , construct sets A with a fixed | A + A | − | A − A | more economically than previous authors, and improve the lowerbound on the proportion of RSD subsets of { , , , . . . , n − } to about 10 − (theprevious best bound was 10 − ). Lastly, by exhaustive computer search, we findsix RSD sets with cardinality 15, which is one lower than the smallest cardinalityfound to date, and find that 30 is the smallest diameter of RSD sets. NTEGERS: 19 (2019)
1. Introduction1.1. Background
Given a finite set of non-negative integers A , the sum set is defined to be A + A := { a i + a j : a i , a j ∈ A } and the difference set to be A − A := { a i − a j : a i , a j ∈ A } ; A is said to be sum-dominated or MSTD (more sums than differences) if | A + A | > | A − A | , balanced if | A + A | = | A − A | , and difference-dominated if | A + A | < | A − A | .Also, we define the restricted sum set to be A ˆ+ A := { a i + a j : a i , a j ∈ A and a i = a j } . We call a set A restricted sum-dominant (RSD) if | A ˆ+ A | > | A − A | . We couldsimilarly define a restricted difference set by only considering differences of distinctelements, but this would amount to removing the number 0 from the differenceset, decreasing the cardinality of A − A by one and not substantially changing thequestions about RSD sets. Thus we avoid this definition.Since Conway gave an early example of an MSTD set in 1969 , research onMSTDs has made incredible progress; see [2, 6, 10, 11, 13, 14, 15] for some of theearlier results and constructions. One of the most notable papers is by Martinand O’Bryant [7]. They proved the proportion of MSTD subsets of { , , , . . . , n − } is bounded below by a positive constant as n → ∞ . However, the proof isprobabilistic and does not give explicit constructions of MSTD sets. Later, Miller,Orosz and Scheinerman [8] gave an explicit construction of a dense family of MSTDsets (previous bounds were exponentially small). They showed that as n → ∞ , theproportion of MSTD subsets of { , , , . . . , n − } that are in their family is at least C/n for some constant C . The current record of a dense family belongs to Zhao[17] with a family of density
C/n .In this paper, we focus on a particular family of MSTD sets (which we denoteby F ) that has appeared many times in the literature. These sets appear to ariseat random and have no particular order, but if we look at them in the right way,they are very well-structured. In addition, our family F has many nice propertiesthat we will explore, despite not being dense.We first provide some examples of sets in F that have been discussed in theliterature. The following sets are found in [7]: S = { , , , , , , , , , , } ,S = { , , , , , , , , , , , , , } . see footnote 1 of [11] With a more refined analysis, the density can be improved to
C/n . NTEGERS: 19 (2019) A = { , , , , , , , , } ,A = S ,A = { , , , , , , , , , , , , , , , , , , , , , , } . Last but not least, the following sets are found in [12]: T ′ j = { , } ∪ { , , . . . , j } ∪ { , , . . . , j }∪ { , , . . . , j } ∪ { j, j + 1) } (Theorem 1) ,T j = { , } ∪ { , , . . . , j + 1) } ∪ { , , . . . , j }∪ { , , . . . , j } ∪ { j, j + 1) } (Theorem 4) ,R j = { , } ∪ { , , . . . , j } ∪ { , , . . . , j }∪ { , , . . . , j } ∪ { , , . . . , j } ∪ { j, j } (Theorem 6) . These sets play important roles in the papers which initially described them. Forexample, A is used to prove Theorem 8 in [2], which states that there exists apositive constant lower bound for the proportion of sets with fixed cardinalities ofsum sets and difference sets. The sets T j and T ′ j give explicit construction of RSDsets. The set R j gives a set A with the highest known value of log | A + A | / log | A − A | .The study of our family F was motivated by trying to describe a common patternamong these remarkable MSTD sets. Although they arose in somewhat differentsituations, and were presented ad hoc for the purposes of each of the papers whichdescribed them, it turned out that all these sets have something in common, andall belong to F .The members of the family F have many nice properties, including (1) sets A with large values of log | A + A | / log | A − A | ; (2) economical construction of sets A with fixed | A + A | − | A − A | ; (3) demonstration of Spohn’s conjecture (1973);(4) compactness; (5) more constructions of RSD subsets; and (5) examples of smallfringes. Let nonnegative numbers a ≤ b be chosen. Let [ a, b ] = { x | a ≤ x ≤ b } and[ a, b ] q = { x | x ≡ a (mod q ) and a ≤ x ≤ b } .We use a different notation to represent sets of integers; it was first introducedby Spohn [16] (1973): Given a set S = { a , a , . . . , a n } , we arrange its elements inincreasing order and form the sequence of differences between consecutive elements.Suppose that a < a < · · · < a n , then our sequence is a − a , a − a , a − a , . . . , a n − a n − and we represent S = ( a | a − a , a − a , a − a , . . . , a n − a n − ) . The highest value of log | A + A | / log | A − A | for a set A in F is about 1.03059. NTEGERS: 19 (2019) S = { , , , , } , we would arrange the elements in increasingorder to get 2 < < < <
10, then write S = (2 | , , , a − a , a − a , a − a , . . . , a n − a n − the sequence ofconsecutive differences (SCD) . The advantage of this notation is that differencesbetween elements of S correspond to sums of consecutive runs in the SCD. Forexample, look at the SCD 1 , , ,
1. We know that 7 is in the difference set becausethe run 1 , , F family and interior blocks . Definition 1.1.
Let M k denote the sequence , , . . . , | {z } k -times , . We define F to be thefamily of sets with SCD , , , , M k , M k , . . . , M k ℓ , M where ℓ, k , . . . , k ℓ are positive integers, and M is either , or , , or , , , . Remark 1.2.
It can be verified that all the sets S , S , A , A , A , T ′ j , T j , R j arein F . Conjecture 1.3.
All sets in F are MSTD. Example 1.4.
The set S = (0 | , , , , , , , , , , , , , , , , , , { , , , , , , , , , , , , , , , , , , , } has | S + S | − | S − S | = 86 −
83 = 3 . In this paper, we prove that the conjecture is true for a periodic subfamily of F . Definition 1.5. [Interior block] Consider a set S with its SCD. Let B denote aconsecutive subsequence of the SCD. Suppose there exists N ∈ N such that for k ≥ N , the sets S k with SCD constructed by repeating B for k times in place of B are MSTD sets. Then we call B an interior block and let | B | denote the length ofthe interior block. A natural question to ask is what are the possible values of | B | , which is addressedby the following theorem: Theorem 1.6.
Let k and ℓ be arbitrary positive integers. The following threesubfamilies of F consist of MSTD sets; we specify the exact values of | S + S |−| S − S | .1. The family { S k,ℓ | k, ℓ ∈ N } , where S k,ℓ = (0 | , , , , , . . . , | {z } k -times , , , , . . . , | {z } k -times , , . . . , , , . . . , | {z } k -times , | {z } ℓ -times , , , , has | S k,ℓ + S k,ℓ | − | S k,ℓ − S k,ℓ | = 2 ℓ . NTEGERS: 19 (2019)
2. The family { S ′ k,ℓ | k, ℓ ∈ N } , where S ′ k,ℓ = (0 | , , , , , . . . , | {z } k -times , , , , . . . , | {z } k -times , , . . . , , , . . . , | {z } k -times , | {z } ℓ -times , , , has | S ′ k,ℓ + S ′ k,ℓ | − | S ′ k,ℓ − S ′ k,ℓ | = 2 ℓ − .3. The family { S ′′ k,ℓ | k, ℓ ∈ N } , where S ′′ k,ℓ = (0 | , , , , , . . . , | {z } k -times , , , , . . . , | {z } k -times , , . . . , , , . . . , | {z } k -times , | {z } ℓ -times , , has | S ′′ k,ℓ + S ′′ k,ℓ | − | S ′′ k,ℓ − S ′′ k,ℓ | = ℓ .We call { S k,ℓ | k, ℓ ∈ N } ∪ { S ′ k,ℓ | k, ℓ ∈ N } ∪ { S ′′ k,ℓ | k, ℓ ∈ N } the F per family, which isa periodic subfamily of the larger family F . Example 1.7.
1. The set S , = (0 | , , , , , , , , , , , , , , , , , , { , , , , , , , , , , , , , , , , , , , } has | S , + S , | − | S , − S , | = 85 −
79 = 6 = 2 · .2. The set S ′ , = (0 | , , , , , , , , , , , , , , , { , , , , , , , , , , , , , , , , } has | S ′ , + S ′ , | − | S ′ , − S ′ , | = 72 −
69 = 3 = 2 · − .3. The set S ′′ , = (0 | , , , , , , , , , , , , , , { , , , , , , , , , , , , , , , } has | S ′′ , + S ′′ , | − | S ′′ , − S ′′ , | = 67 −
65 = 2 . Remark 1.8.
Theorem 1.6 answers a question raised by Spohn [16], on whetherthe interior block must contain at least 3 elements. The answer is no: set ℓ = 1 andchoose k ≥ for sets in Theorem 1.6. Then, we can use either the interior block or , . In other words, we have shown that for all i ∈ N , there exists an interiorblock B with | B | = i . Also, this theorem provides an infinite family of sets whichdemonstrate Conjecture 6 in [16]. The repetition of certain interior blocks can cause the number of sums to be increased by agreater constant than that by which the number of differences is increased.
NTEGERS: 19 (2019) A with large values of the ratiolog | A + A | / log | A − A | . An early high ratio of about 1 . . . A with highratios from [2] and [12] belong to F . We offer examples of several sets in F thatgive higher ratios than the ones given in [1] and [2]; there are at least 22 sets A in F with log | A + A | / log | A − A | > . F gives an economical way (in the sense of having arelatively small width between its minimum and maximum element) to construct aset A with any specific value of | A + A | − | A − A | . Martin and O’Bryant proved thatfor a given x ∈ N , there exists a set A ⊆ [0 , | x | ] such that | A + A | − | A − A | = x , which is significantly more efficient than the base expansion method. Withsubfamilies of F , we further improve this. Theorem 1.9.
Given x ∈ N , there exists a set A ⊆ [0 ,
12 + 4 x ] such that | A + A | − | A − A | = x . Furthermore, it is impossible to construct A ⊆ [0 , f ( x )] such that | A + A | − | A − A | = x , where f ( x ) is sub-linear. This means that a linear growth ofthe interval containing A is the best we can do. Finally, we improve the lower bound for the proportion of RSD subsets of [0 , n − n goes to infinity. RSD implies MSTD, and compared to MSTD sets, RSD setsare much less common: exhaustive computer search shows that there are no RSDsubsets of [0 , . · MSTD sets in the same interval.In [12], the lower bound on the proportion of RSD subsets of [0 , n −
1] as n → ∞ isabout 10 − ; we improve this bound to 4 . · − by a better fringe formed byusing F . Theorem 1.10.
For n ≥ , the proportion of RSD subsets of [0 , n − is at least . · − . This work was supported by NSF Grants DMS1561945 and DMS1659037, theUniversity of Michigan, Washington and Lee University, and Williams College. Wethank the referee for helpful comments on an earlier draft. We can generate an infinite family of MSTD sets from a given MSTD set through the baseexpansion method. Let A be an MSTD set, and let A k,m = { P ki =1 a i m i − : a i ∈ A } . If m issufficiently large, then | A k,m ± A k,m | = | A ± A | k and | A k,m | = | A | k . There are exactly 6 RSD subsets of [0 ,
30] and 16 RSD subsets of [0 , , n −
1] as n → ∞ to be about3 · − . NTEGERS: 19 (2019)
2. Proof of Theorem 1.6
We only prove (1) for conciseness, since the proofs of (2) and (3) are similar. For(1), the case when k = 1 is Theorem 4 in [12], so we prove the case k ≥ Lemma 2.1.
Fix i ≥ and consider i . For some fixed k and ℓ , if there is arun that sums up to i in the SCD , , , , , . . . , | {z } k -times , , . . . , , , . . . , | {z } k -times , | {z } ℓ -times , , , , , then the run is one of the forms(A) , , , , , . . . , | {z } k -times , , . . . , , , . . . , | {z } k -times , | {z } ℓ -times , , (B) , , , . . . , | {z } k -times , , . . . , , , . . . , | {z } k -times , | {z } j -times (C) , , , . . . , | {z } k -times , , . . . , , , . . . , | {z } k -times , | {z } ℓ -times , , , (D) , , . . . , | {z } k -times , , . . . , , , . . . , | {z } k -times , | {z } j -times , , for some j ≥ .Proof. We consider possible cases for where the run that sums up to 6 + 4 i canstart.1. Case I : the run starts at the first 1. Since 1 + 1 + 2 + 1 = 5 <
6, the runmust contain 1 , , , Subcase 1 : If the run ends at 4, the run sums up to 1 + 4 m for some m ≥ Subcase 2 : If the run ends at 3, we have 4 m for some m ≥ Subcase 3 : If the run ends with 3 ,
1, we have 1 + 4 m for some m ≥ Subcase 4 : If the run ends with 3 , ,
1, we have 6 + 4 m for some m ≥ NTEGERS: 19 (2019)
Subcase 5 : If the run ends with 3 , , ,
2, we have 4 m for some m ≥ Subcase 6 : If the run ends with 3 , , , ,
1, we have 1 + 4 m for some m ≥ Case II : the run starts at the second 1. As above, the run must contain1 , ,
1. Using the same argument, we see that there are no such runs that sumup to 6 + 4 i .3. Case III : the run starts at the first 2. As above, the run must contain 2 , i , the run must either endat 3 or end with 3 , , ,
2. We have form (B) and form (C).4.
Case IV : the run starts with 1 ,
4; it must end with 3 , , Case V : the run starts at 4; there are no such runs.6.
Case VI : the run starts at 3; there are no such runs.We have iterated through all possible cases and thus the proof is complete.
Lemma 2.2.
Let k and ℓ ∈ N be chosen. Consider S = (0 | , , , , . . . , | {z } k -times , , . . . , , . . . , | {z } k -times , | {z } ℓ -times , , , , . Then the set [1 , max( S )] \ ( S − S ) of missing positive differences is exactly T = [6 , k − ∪ [14 + 4( k − ,
14 + 8( k − ∪· · · ∪ [6 + 8( ℓ −
1) + 4( ℓ − k − , ℓ −
1) + 4 ℓ ( k − . (Recall from Section 1.2 that [ a, b ] := { x | x ≡ a (mod 4) and a ≤ x ≤ b } .) Example 2.3.
We use S , as an example. We have S , − S , = [ − , \{± , ± , ± , ± , ± , ± } . Note that { , , , , , } = [6 , ∪ [18 , ∪ [30 , , which is the set T in Lemma 2.2.Proof. Pick 1 ≤ i ≤ ℓ . We show that S − S misses[6 + 8( i −
1) + 4( i − k − , i −
1) + 4 i ( k − ; NTEGERS: 19 (2019) i −
1) + 4( i − k −
1) + 4 m for all 0 ≤ m ≤ k −
1. We prove this by contradiction. Pick some 0 ≤ m ≤ k − i −
1) + 4( i − k −
1) + 4 m ≤ i −
1) + 4 i ( k − ≤ ℓ −
1) + 4 ℓ ( k − . Since both form (A) and form (C) in Lemma 2.1 gives 6 + 8 ℓ + 4( k − ℓ , our runmust be of the form (B) or (D). We consider these two cases.1. Case I : the run is of form (B). Then it sums up to 2 + 4( k + 1) j for some j ≥
1. We have:2 + 4( k + 1) j = 6 + 8( i −
1) + 4( i − k −
1) + 4 m ( k + 1)( j − i + 1) = 1 + m. (1)So, 1 ≤ ( k + 1)( j − i + 1) = 1 + m ≤ k and so, 0 < j − i + 1 <
1, which is acontradiction.2.
Case II : the run is of the form (D). Then it sums up to 4( k + 1) j + 2 for some j ≥
1. As above, we find a contradiction. We have shown that S − S misses T .To complete the proof, we show that S − S contains [0 , k + 1) ℓ ] \ T . Note thatclose to the beginning of the SCD, we have 1+2+1 = 4 and after that, the sequenceimplicitly contains consecutive differences of 4 (because 3 + 1 = 4 and 1 + 2 + 1 = 4 . )So, S − S contains all numbers in [0 , k + 1) ℓ ] that are 0 mod 4. Similarly, it isnot hard to see that S − S contains all numbers that are either 1 mod 4 or 3 mod 4.Next, we show that all numbers that are 2 mod 4 and not in T are in S − S . We have2 ∈ S − S and 10+4( k − ∈ S − S because 10+4( k −
1) = (2+1)+4( k − { k − k +3) i | ≤ i ≤ ℓ − } ⊆ S − S . Lastly, 6+4( k +1) ℓ ∈ S − S . because we have the run 2 , , . . . , | {z } k -times , , . . . , , . . . , | {z } k -times , | {z } ℓ -times , , , Corollary 2.4.
Choose k and ℓ ∈ N , and let S = (0 | , , , , . . . , | {z } k -times , , . . . , , . . . , | {z } k -times , | {z } ℓ -times , , , , . Then | S − S | = 19 + ℓ (6 k + 8) . NTEGERS: 19 (2019) Lemma 2.5.
Choose k ≥ and ℓ ∈ N , and let S = (0 | , , , , . . . , | {z } k -times , , . . . , , . . . , | {z } k -times , | {z } ℓ -times , , , , . Then S + S contains [0 , k + 1) ℓ + 18] \ T , where T = [ i ∈ [1 ,ℓ ] [12 + 12( i −
1) + 4( i − k − ,
12 + 12( i −
1) + 4 i ( k − ∪ [ i ∈ [1 ,ℓ ] [12 ℓ + 4 ℓ ( k −
2) + 16 + 4( i − k −
2) + 12( i − , ℓ + 4 ℓ ( k −
2) + 16 + 4 i ( k −
2) + 12( i − ! . Proof.
Observe that S = { i | ≤ i ≤ k + 1) ℓ + 9 and i ≡ } ⊆ S . So, allof the following sets are in S + S :0 + S = { i | ≤ i ≤ k + 1) ℓ + 9 and i ≡ } , S = { i | ≤ i ≤ k + 1) ℓ + 9 and i ≡ } , S = { i | ≤ i ≤ k + 1) ℓ + 9 and i ≡ } , k + 1) ℓ + 9 + S = { i | k + 1) ℓ + 10 ≤ i ≤ k + 1) ℓ + 18 and i ≡ } , k + 1) ℓ + 8 + S = { i | k + 1) ℓ + 9 ≤ i ≤ k + 1) ℓ + 17 and i ≡ } , k + 1) ℓ + 6 + S = { i | k + 1) ℓ + 7 ≤ i ≤ k + 1) ℓ + 15 and i ≡ } Thus S + S contains all numbers that are either 1 , , k + 1) ℓ + 18]. Now, we focus on numbers that are divisible by 4. Observe that S = { , , k + 1) , k + 1) , . . . , ℓ ( k + 1) , ℓ ( k + 1) } ⊆ S . We write S = { , ℓ ( k + 1) } ∪ { i ( k + 1) | ≤ i ≤ ℓ } . We show that all numbers divisible by 4 that are not in T are in S + S . The set ofall numbers divisible by 4 that are not in T is { , , ,
12 + 4 ℓ ( k + 1) , ℓ + 8 kℓ + 16 } ∪ (4 + {
12 + 12( i −
1) + 4 i ( k − | ≤ i ≤ ℓ } ) ∪ (8 + {
12 + 12( i −
1) + 4 i ( k − | ≤ i ≤ ℓ } ) ∪ (20 + { ℓ + 4( ℓ + i )( k −
2) + 12( i − | ≤ i ≤ ℓ } ) ∪ (24 + { ℓ + 4( ℓ + i )( k −
2) + 12( i − | ≤ i ≤ ℓ } ) . We know the following:
NTEGERS: 19 (2019) , , k + 1) ℓ ∈ S , { , , ,
12 + 4 ℓ ( k + 1) , ℓ + 8 kℓ + 16 } ⊆ S + S .2. For each 1 ≤ i ≤ ℓ , we have 4+(12+12( i − i ( k − i ( k +1)) ∈ S + S .3. For each 1 ≤ i ≤ ℓ , we have 8+(12+12( i − i ( k − i ( k +1)) ∈ S + S .4. For each 1 ≤ i ≤ ℓ , we have 20 + 12 ℓ + 4( ℓ + i )( k −
2) + 12( i −
1) = (4 + 4 ℓ ( k +1)) + (4 + 4 i ( k + 1)) ∈ S + S .5. For each 1 ≤ i ≤ ℓ , we have 24 + 12 ℓ + 4( ℓ + i )( k −
2) + 12( i −
1) = (8 + 4 ℓ ( k +1)) + (4 + 4 i ( k + 1)) ∈ S + S .We have shown that all numbers divisible by 4 that are not in T are in S + S , andthis completes the proof. Lemma 2.6.
Choose k ≥ and ℓ ∈ N , and let S = (0 | , , , , . . . , | {z } k -times , , . . . , , . . . , | {z } k -times , | {z } ℓ -times , , , , . Then S + S contains none of the elements in T = [ i ∈ [1 ,ℓ ] (cid:2)
12 + 12( i −
1) + 4( i − k − ,
12 + 12( i −
1) + 4 i ( k − (cid:3) ! ∪ [ i ∈ [1 ,ℓ ] (cid:2) ℓ + 4 ℓ ( k −
2) + 16 + 4( i − k −
2) + 12( i − , ℓ + 4 ℓ ( k −
2) + 16 + 4 i ( k −
2) + 12( i − (cid:3) ! . Proof.
To complete the proof, we prove that none of the numbers in T are in S + S .We write out S explicitly: S = { , , , , } ∪ { j − k + 1) + 4 i | ≤ j ≤ ℓ, ≤ i ≤ k }∪ { i ( k + 1) | ≤ i ≤ ℓ } ∪ { i ( k + 1) | ≤ i ≤ ℓ }∪ { ℓ ( k + 1) , ℓ ( k + 1) , ℓ ( k + 1) } . We consider elements in [ i ∈ [1 ,ℓ ] (cid:2)
12 + 12( i −
1) + 4( i − k − ,
12 + 12( i −
1) + 4 i ( k − (cid:3) . NTEGERS: 19 (2019) ≤ m ≤ ℓ and 0 ≤ n ≤ k −
2. Consider12 + 12( m −
1) + 4( m − k −
2) + 4 n = 4 m ( k + 1) − k + 8 + 4 n ≤ ℓ + 4 kℓ − k + 8 + 4( k − ℓ ( k + 1) . Because S contains no numbers that are 3 mod 4, for a pair whose sum is 4 m ( k +1) − k +8+4 n , we cannot use numbers that are 1 mod 4. Also, because 4 m ( k +1) − k + 8 + 4 n ≤ ℓ ( k + 1), we can ignore all numbers that are greater than 4 ℓ ( k + 1).Hence, our set of concern is { , , } ∪ { i ( k + 1) | ≤ i ≤ ℓ } . If a pair that sums to 4 m ( k + 1) + 4( n − k ) + 8 is in { i ( k + 1) | ≤ i ≤ ℓ } , thenthere exists m ′ and n ′ such that8 + 4( m ′ + n ′ )( k + 1) = 4 m ( k + 1) + 4( n − k ) + 8 , ( m ′ + n ′ − m + 1)( k + 1) = n + 1 . Thus, 0 < ( m ′ + n ′ − m +1)( k +1) = n +1 ≤ k − < m ′ + n ′ − m +1 < { , , } . Let 4 + 4 m ′ ( k + 1) bethe number used in { i ( k + 1) | ≤ i ≤ ℓ } . We consider three cases correspondingto the three elements in { , , } .1. We have 0 + (4 + 4 m ′ ( k + 1)) = 4 m ( k + 1) − k + 8 + 4 n . So, 0 < ( k + 1)( m ′ − m + 1) = n + 2 ≤ k , which implies 0 < m ′ − m + 1 <
1, a contradiction.2. We have 2 + (4 + 4 m ′ ( k + 1)) = 4 m ( k + 1) − k + 8 + 4 n . So, 0 < m ′ − m +1)( k + 1) = 3 + 2 n ≤ k −
1, which implies 0 < m ′ − m + 1 <
1, a contradiction.3. We have 4 + (4 + 4 m ′ ( k + 1)) = 4 m ( k + 1) − k + 8 + 4 n . So, 0 < ( m ′ − m +1)( k + 1) = n + 1 ≤ k −
1, which implies 0 < m ′ − m + 1 <
1, a contradiction.Next, we consider elements in [ i ∈ [1 ,ℓ ] (cid:2) ℓ +4 ℓ ( k − i − k − i − , ℓ +4 ℓ ( k − i ( k − i − (cid:3) . Pick 1 ≤ m ≤ ℓ and 0 ≤ n ≤ k −
2. Consider12 ℓ + 4 ℓ ( k −
2) + 16 + 4( m − k −
2) + 12( m −
1) + 4 n = 4( ℓ + m )( k + 1) + 12 − k + 4 n ≥ ℓ + 1)( k + 1) + 12 − k = 4( k + 1) ℓ + 16 . NTEGERS: 19 (2019) , , { i ( k + 1) | ≤ i ≤ ℓ } ∪ { ℓ ( k + 1) , ℓ ( k + 1) } . If a pair that sums to 4( ℓ + m )( k + 1) + 12 − k + 4 n is in { i ( k + 1) | ≤ i ≤ ℓ } ,then for some 1 ≤ m ′ , n ′ ≤ ℓ we have:4( ℓ + m )( k + 1) + 12 − k + 4 n = 8 + 4( k + 1)( m ′ + n ′ )0 < ( − ℓ − m + m ′ + n ′ + 1)( k + 1) = n + 2 ≤ k So, 0 < − ℓ − m + m ′ + n ′ + 1 <
1, a contradiction. Therefore, a number in the pairmust be in { ℓ ( k + 1) , ℓ ( k + 1) } . Since 4( ℓ + m )( k + 1) + 12 − k + 4 n ≤ ℓ ( k + 1) + 4, both numbers cannot be in { ℓ ( k + 1) , ℓ ( k + 1) } . We considertwo cases corresponding to the two elements of { ℓ ( k + 1) , ℓ ( k + 1) } :1. We have (6 + 4 ℓ ( k + 1)) + (4 + 4 m ′ ( k + 1)) = 4( ℓ + m )( k + 1) + 12 − k + 4 n .Equivalently, 0 < k + 1)( m ′ − m + 1) = 2 n + 3 ≤ k −
2) + 3 = 2 k − < m ′ − m + 1 <
1, a contradiction.2. We have (8 + 4 ℓ ( k + 1)) + (4 + 4 m ′ ( k + 1)) = 4( ℓ + m )( k + 1) + 12 − k + 4 n .Equivalently, 0 < ( k + 1)( m ′ − m + 1) = n + 1 ≤ ( k −
2) + 1 = k − < m ′ − m + 1 <
1, a contradiction.This completes our proof.
Example 2.7.
We use S , as an example to illustrate Lemma 2.5 and Lemma 2.6.We have S , + S , = [0 , \{ , , , , , } , and the set { , , , , , } is exactly the set T in Lemmas 2.5 and Lemma2.6. Corollary 2.8.
Let k ≥ and ℓ ∈ N be chosen. Let S = (0 | , , , , . . . , | {z } k -times , , . . . , , . . . , | {z } k -times , | {z } ℓ -times , , , , . Then | S + S | = 19 + ℓ (6 k + 10) .Proof of Theorem 1.6, Item 1. The proof follows immediately from Corollaries 2.4and 2.8, because | S k,ℓ + S k,ℓ | − | S k,ℓ − S k,ℓ | = [19 + ℓ (6 k + 10)] − [19 + ℓ (6 k + 8)] = 2 ℓ. NTEGERS: 19 (2019) Remark 2.9.
Theorem 1.6 is a generalization of Theorem 2 and Theorem 3 in [12].With a little more work, we can show that sets in F per are RSD for ℓ sufficientlylarge. We also offer another family of MSTD sets formed by repeating certain interiorblocks. We do not prove the theorem since it is not in the focus of the currentpaper. However, the proof is very similar to the proof of Theorem 1.6 but replacing“modulo 4” by “modulo k ” throughout. Theorem 2.10.
For k ≥ , the following is a MSTD set: A k, = (0 | , , . . . , | {z } k − -times , , , k k + 1 , k + 1 , . . . , k + 1 | {z } k − -times , , , , . . . , | {z } k − -times , , , and | A k, + A k, | − | A k, − A k, | = 2 . Define A k,ℓ to be a similarly built set with thesequence , k, k + 1 , k + 1 , . . . , k + 1 | {z } k − -times , repeated k times, then | A k,ℓ + A k,ℓ | − | A k,ℓ − A k,ℓ | = 2 ℓ Remark 2.11.
If we consider Theorem 1.6 to be a generalization of Theorem 2 andTheorem 4 in [12], then Theorem 2.10 is another generalization from a differentperspective. Notice that A ,ℓ = S ,ℓ . Sets A k,ℓ also form a family of MSTD setsand RSD sets with interior blocks.
3. Good Properties of the Family F A with Large log | A + A | / log | A − A | The first application of our family F is that the family produces many sets A with large value of log | A + A | / log | A − A | . For convenience, we define f ( A ) :=log | A + A | / log | A − A | . An early example of a set A with high f ( A ) is given byHegarty [2]. The set is A = { , , , , , , , , , , , , , , , , , , , , , , } . In our notation, A = (0 | , , , , , , , , , , , , , , , , , , , , , , which is very close to a set in our family F per , namely S , = (0 | , , , , , , , , , , , , , , , , , , . It turns out that f ( S , ) > f ( A ). NTEGERS: 19 (2019) F per defined in Theorem 1.6, we find theset S , with the property of f ( S , ) = 1 . . . . , which is larger than previousresults in [2] (1.0208. . . ) and [1] (1.0213. . . ) but smaller than the current record in[12] (1.03059. . . ). It is worth noting, however, the set with the highest known valueof f ( A ) (1 . . . . ) exhibited by Penman and Wells [12] is a member of the family F :(0 | , , , , , , , , , , , , . . ., | {z } , , . . . , , , . . . , | {z } , | {z } , , , , , , , , , , , . In fact, there are at least 22 sets A in F with f ( A ) > .
03: these sets are of theform:(0 | , , , , , , , , , , , , . . ., | {z } , , . . . , , , . . . , | {z } , | {z } ℓ -times , , , , , , , , , , , , (0 | , , , , , , , , , , , , . . ., | {z } , , . . . , , , . . . , | {z } , | {z } ℓ -times , , , , , , , , , , . A with Fixed | A + A | − | A − A | We show another application of our large family F of MSTD sets, which is toconstruct sets with a fixed difference | A + A | − | A − A | economically, i.e. with arelatively small width between their maximum and minimum elements. Proof of Theorem 1.9.
Fix x ∈ N . If x is even, pick k = 1 and ℓ = x/ A = S k,ℓ with | S k,ℓ + S k,ℓ | − | S k,ℓ − S k,ℓ | = 2 ℓ = x and max A = 9 + 8 ℓ = 9 + 4 x , min A = 0.If x is odd, pick k = 1 and ℓ = ( x + 1) / A = S ′ k,ℓ with | S ′ k,ℓ + S ′ k,ℓ | − | S ′ k,ℓ − S ′ k,ℓ | = 2 ℓ − x and max A = 8 + 8 ℓ = 12 + 4 x , min A = 0.Hence, for any positive integer x , there exists A ⊆ [0 ,
12 + 4 x ] such that | A + A | − | A − A | = x . Remark 3.1.
Linear growth of the interval containing A is the best we can do. Tosee this, assume that the theorem is true for A x ⊆ [0 , φ ( x )] , where φ ( x ) is sub-linear. NTEGERS: 19 (2019) We have: lim x →∞ | A x + A x | − | A x − A x | φ ( x ) = lim x →∞ xφ ( x ) = ∞ , (2) which is a contradiction, since for all sets A x ⊆ [0 , φ ( x )] for large enough x , | A x + A x | − | A x − A x | φ ( x ) ≤ | A x + A x | φ ( x ) < φ ( x ) + 1 φ ( x ) ≤ . Many classes of MSTD sets can be generated by finding a good fringe pair, i.e.the two sets of elements on the leftmost and rightmost sides of the interval [0 , n ].Examples can be found the proofs of Theorem 8 in [2], Theorem 1.4 in [1], Theorem1 in [7], Theorem 1.1 in [8] and Theorem 17 in [12]. Often, when shifted close toeach other, two sets in a fringe pair form an MSTD set. However, these fringe pairshave been found by brute force and there has not been a systematic way to generatefringes. It turns out that F can be a good fringe generator; we demonstrate thisby improving the lower bound for the proportion of RSD sets of { , , . . . , n − } mentioned in Theorem 17 [2].In particular, Pennman and Wells used a fringe pair of size 120 generated by thefringe pair used in [7]. The method is to repeat blocks of sets, which inefficientlycreates a small lower bound of about 10 − . The authors mentioned that Zhao’stechniques can be modified to improve the result; however, this task requires asubstantial computation. We believe that this is true since RSD sets are muchrarer than MSTD sets . As Zhao’s technique relies on extensive search for fringepairs, the technique is much less effective when applied to RSD sets. Therefore, afeasible and simple way to improve the bound is to find a better fringe pair. Hereis a fringe pair generated by F (we use L and U to match the notations with [12]): L = (0 | , , , , , , , , , , , , , , , , , { , , , , , , , , , , , , , , , , , , } ,U = ( n −
41) + (0 | , , , , , , , , , , , , , , , , , , n −
41) + { , , , , , , , , , , , , , , , , , , , } = n − { , , , , , , , , , , , , , , , , , , , } . Observe that the fringe pair is formed by the MSTD set S , . We have: L ˆ+ L = [0 , \{ , , } ,U ˆ+ U = [ n − , n + 38] ,U ˆ+ U = [2 n − , n − \{ n − , n − , n − , n − } . Exhaustive computer search shows that there are no RSD subsets of [0 , . · MSTD sets in the same interval.
NTEGERS: 19 (2019) U − L misses ± ( n − , ± ( n − , ± ( n − , ± ( n − S − S misses at least 8 numbers. If we can guarantee that S ˆ+ S misses only 7 elements in { , , , ± (2 n − , ± (2 n − , ± (2 n − , ± (2 n − } , then S is RSD. Followingthe proof of Theorem 17 in [12], we find a lower bound of(1 − − + 2 − )) · − (40+41) = 4 . · − . (3)This improvement comes from the reduction in fringe size from 120 to 81. Canwe find a better bound for the proportion of RSD subsets of { , , , . . . , n − } as n → ∞ ? Since there are no RSD subsets in [0 , − ≈ − .
4. Observation: Interior Block Sizes and The Growth of | A + A | −| A − A | Spohn [16] was the first to share the concept of and raise several questions aboutinterior blocks existing within MSTD sets. He noted that the repetition of interiorblocks may increase the cardinality of the sum set by more than that of the differenceset. For a set A having an interior block B A , let T A be the value that the sum setincrease by more than the difference set when B A is repeated. We observe therelationship between T A and | B A | . Theorem 1.6 gives us the following: Theorem 4.1.
The following results about T A / | B A | are true.1. There exists a set A such that T A / | B A | = 0 .2. For any ε > , there exists a set A with < T A / | B A | < ε .3. For any ≤ ε < . , there exists a set A such that T A / | B A | > ε .Proof.
1. Consider the set S , in Theorem 1.6. Notice that | S , + S , | − | S , − S , | = | S , + S , | − | S , − S , | = 2 and so, T S , = 0. In other words, repeating 4does not change | S , + S , | − | S , − S , | . This proves (1).2. Pick ε > k such that 2 / ( k + 2) < ε . Consider the set S k, inTheorem 1.6. We have ( | S k, + S k, | − | S k, − S k, | ) − ( | S k, + S k, | − | S k, + S k, | ) = 2 and so, T S k, = 2. Hence, 0 < T S k, / | B S k, | = 2 / ( k + 2) < ε . Thisproves (2).3. Finally, Theorem 12 in [12] shows that ( | Q + Q | − | Q − Q | ) − ( | Q + Q | − | Q − Q | ) = 6, while the interior block is 1,4,4,4,3 (of size 5 . ) Hence, T Q / | B Q | = 6 / .
2. This proves (3).
NTEGERS: 19 (2019) A with large f ( A ). The previous record A in [2] has T A / | B A | = 2 /
3, the highest known at that time. The new record Q in [12] has T Q / | B Q | = 6 /
5, which is much higher and this explains why the current record(1 . . . . ) is much higher than the old record of (1 . . . . ).
5. Smallest Cardinality for RSD sets
Hegarty proved that the smallest MSTD sets have size 8, and there is exactly onesuch set up to affine transformation. The method is to reduce the problem to finitecomputations and run through all possible cases by computers. As commented in[2], this method is not feasible in finding all possible MSTD sets of cardinality 9since there are many pair of possible equal differences for a set of 9 random numbers.However, Penman and Wells [2] proved that the list of size-9 MSTD sets given by[2] is exhaustive (up to affine transformation). They also observed that the smallestcardinality of RSD sets must be in the interval [10 , Theorem 5.1.
The smallest cardinality of RSD sets is in the interval [10 , .Furthermore there are no RSD subsets of [0 , and the smallest diameter of anRSD set is 30. There are exactly 6 RSD sets in [0 , C = { , , , , , , , , , , , , , , } ,C = { , , , , , , , , , , , , , , } ,C = { , , , , , , , , , , , , , , } ,C = 30 − C ,C = 30 − C ,C = 30 − C . For all 1 ≤ i ≤ | C i + C i | − | C i − C i | = 1.
6. Open Questions
We end with these open questions:1. What are the possible values of T A / | B A | over all sets A ? Is there a set A suchthat T A / | B A | > .
2? This may lead to an increase in the highest known valueof log | A + A | / log | A − A | . NTEGERS: 19 (2019) T A / | B A | > | B A | ≥ , n −
1] as n → ∞ ?6. What is the size of the smallest RSD sets? Is there a better way to find outthis number than Hegarty’s method (which requires large computing power)? Acknowledgement.
The authors were supported by NSF grants DMS1659037and DMS1561945, the Finnerty Fund, Washington and Lee University and WilliamsCollege. We thank the participants from the 2018 SMALL REU program for manyhelpful conversations.
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