ON A CLASS OF APERIODIC SUM-FREE SETS NEIL J. CALKIN PAUL ERD OS. that is that there are innitely many n not in S which are not a sum of two elements

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1 ON A CLASS OF APERIODIC SUM-FREE SETS NEIL J. CALKIN PAUL ERD OS Abstract. We show that certain natural aperiodic sum-free sets are incomplete, that is that there are innitely many n not in S which are not a sum of two elements of S.. Introduction An old question (ascribed to Dickson by Guy [6] and rediscovered in a more general setting by Cameron [4]) asks whether a sum-free set constructed in a greedy fashion from a nite initial set will be ultimately periodic. Cameron, Calkin and Finch have investigated this question, and it seems that the answer may be \no": in particular, the sum-free sets f; ; 8; 20; 26 : : : g and f2; 5; 6; 2; 27 : : : g are not known to be ultimately periodic. The approach taken by these authors is to compute a large number of elements of the respective sets, and to investigate the structure. Unfortunately there is currently no way to show that these sets are aperiodic (though there is good evidence [2, ] to suggest that they are). An alternative approach to answering Dickson's question in the negative would be to take a sum-free set already known to be aperiodic, and to show that it was complete. In this paper we shall show that this approach will fail for a natural set of aperiodic sum-free sets. 2. Definitions Given a set S of positive integers, we denote by S + S the set of pairwise sums, S + S = fx + y j x; y 2 Sg. A set S is said to be sum-free if S \ (S + S) = ;, that is if there do not exist x; y; z 2 S for which x + y = z. We call a sum-free set complete if there is an n 0 such that for all n > n 0, n 2 S [ (S + S), that is either n 2 S or there exist x; y 2 S with x + y = n. Equivalently, S is complete if and only if it is constructed greedily from a nite set. 99 Mathematics Subject Classication. B75. It seems that the problem as stated by Guy doesn't appear in the reference cited

2 2 NEIL J. CALKIN PAUL ERD OS We call S ultimately periodic with period m if its characteristic function is ultimately periodic with period m, that is if there exists a positive integer n o such that for all n > n 0, n 2 S if and only if n + m 2 S: otherwise we say that S is aperiodic. The following sets S are a natural set of aperiodic sum-free sets: for irrational, dene S = njfng 2 ( ; 2 ) where fxg denotes the fractional part of x. Clearly, S is aperiodic: indeed, if S were ultimately periodic with period m, then for suciently large k, km 62 S. However, since m is irrational, fkmg is dense on (0; ) as k runs through the positive integers. Since the sets S all have density and it is easily seen that S [ (S + S ) has density, it might be hoped that amongst the S we might nd a complete set, thus answering Dickson's question in the negative (it is clear that not every S can be complete, since the set of complete sum-free sets is countable, and there are uncountably many S ). We prove below that this is not the case.. Preliminaries We need an understanding of when n 62 S [ (S + S ). Lemma. n 2 S + S if and only if either (i) fng =? ; where 0 < < and there exists an m < n with 2? fmg = < or (ii) fng = 2 +; where 0 < < and there exists an m < n with fmg? = <. Further, we may choose m n or. 2 2 Proof: Suppose that n 2 S + S. Then either fng 2 (0; ) or fng 2 ( 2 ; ). Suppose the former. Let m ; m 2 2 S, with m + m 2 = n, and fm g = 2? ; and fm 2 g = 2? 2: Then f(m + m 2 )g = f 4?? 2 g = f? ( + 2 )g and since fng =?, we have + 2 =. Since m + m 2 = n, the smaller of m and m 2 is at most n, and 2 since + 2 =, the smaller of and 2 is at most. 2 Conversely, if there is an m with 2? fmg = < then clearly (n? m) 2 S, and so n 2 S + S. The case where fng 2 ( 2 ; ) is handled similarly. We also require the following basic facts about continued fractions (cf [7]): Lemma 2. If is irrational, with continued fraction convergents a i fraction expansion and continued

3 ON A CLASS OF APERIODIC SUM-FREE SETS then for every i, (i) a 2i b 2 < < a 2i+ b 2i+ (ii) a i +? a i+ = (iii) a i+ = q i a i + a i?, + = q i +? (iv) a i? < + q 0 + q + q 2 + q + q 4 + : : : (v) If a; b 2 Zwith 0 < b < b n+, and b 6= b n, then ja? bj > ja n? b n j. 4. The Result Theorem. For every irrational, the set S is incomplete. Proof: Consider the continued fraction convergents a i to =. Suppose i is even: then a i < ; so > a i : Hence if a i 2 (mod ), then 0 <? a i < + so 0 < f g? 2 < ; + and by Lemma 2(v), we see that there are no a; b; with b < and jb?aj < j?a i j: Hence 62 S [ (S + S ): Similarly, if i is odd, and a (mod ) then a i > ; so < a i and 0 <? fg < + and so 62 S [ (S + S ): In the remainder of the proof we may thus assume that for all i suciently large, a 2i 6 2 (mod ) and a 2i+ 6 (mod ): Further, since a i +? a i+ =, we have (a i ; a i+ ) =, and so a i ; a i+ cannot both be divisible by. Hence one of the following situations occurs: (i) a 2i (mod ); a 2i+ 2 (mod ) for innitely many i,

4 4 NEIL J. CALKIN PAUL ERD OS (ii) a 2i? 0 (mod ); a 2i (mod ) for innitely many i, (i) a 2i 0 (mod ); a 2i+ 2 (mod ) for innitely many i. Case (i): We will show that b 2i+? b 2i 62 S [ (S + S ). Let fb 2i g = + 2i Then fb 2i+ g = 2? 2i+: f(b 2i+? b 2i )g =? 2i+? 2i ; so b 2i+? b 2i 62 S. Further, for every a; b with b < b 2i+ ; b 6= b 2i we have 2? fbg b? a Thus there is no b < b 2i+? b 2i with = jb? aj > jb 2i? a 2i j = 2i > 2 ( 2i+ + 2i ) 2? 2 ( 2i+ + 2i ) < fbg < 2 : Hence by Lemma, b 2i+? b 2i 62 S + S. Case (ii): we will show that b 2i + b 2i? 62 S [ (S + S ) Let fb 2i? g =? 2i? fb 2i g = + 2i: Then f(b 2i? + b 2i )g =? ( 2i?? 2i )

5 ON A CLASS OF APERIODIC SUM-FREE SETS 5 so b 2i + b 2i? 62 S. Further, for every a; b with b < b 2i ; b 6= b 2i? we have fbg? 2 b? a = jb? aj > jb 2i?? a 2i? j = 2i? : Now, since 2 (b 2i + b 2i? ) < b 2i there is no b < 2 (b 2i + b 2i? ) with 2? ( 2i?? 2i ) < fbg < 2 : Hence, by Lemma, b 2i? + b 2i 62 S + S. Case (iii) is proved in a similar fashion to case (ii), obtaining b 2i +b 2i+ 62 S [(S +S ), completing the proof of the theorem. 5. Open Questions In this section we present some open questions in related areas. () The sets considered above show that there are aperiodic maximal sum-free sets for which S [ (S + S) omits about log n elements up to n. Are there maximal aperiodic sum-free sets which omit o(log n) elements? (2) If there are no aperiodic complete sum-free sets, then there is a function M = M(n) so that if S is a sum-free set, and if S [ (S + S) contains all integers greater than n, then S has period M. How fast must M(n) grow? () Erdos [5] and Alon and Kleitman [] have shown using similar ideas that if S is a set of n integers, then there is a sum-free subset of S of cardinality greater than n. Is n best possible? (4) By considering chosen uniformly at random from the interval ( ; m ) m 2? m 2? we see that any set of S of n positive integers contains a subset of cardinality n with the property that it has no solutions to the equation x m+ + x x m = y. Is this best possible? In particular, if c m is such that any set S of cardinality n contains a subset of cardinality c m n with no solutions to x + x x m = y then what is the behaviour of c m as m!? Does c m!? To 0? (5) Similarly, we see that if S = fs ; s 2 ; s ; : : : g then there is a subset S 0 = fs i ; s i2 ; s i ; : : : g having no solution to x + x x m = y for which i n lim n! n = m + :

6 6 NEIL J. CALKIN PAUL ERD OS Is m+ best possible here? (6) Call a set strongly m-sum-free if it contains no solution to x +x 2 + +x j = y for 2 j m. Given a set S of n positive integers, how large a strongly m- sum-free set must it contain? Let d m be such that S must contain a set of size d m n. What is the behaviour of d m as m!? Is it possible that d m is bounded away from 0? (7) Which sets S having no solutions to x + x x m = y have the greatest density? In particular, if k is the least integer not dividing m? then the set of integers congruent to (mod k) has no solutions. Is this best possible? This is easy to show when m = 2, and has been shown by Luczak (personal communication) in the case m =. The rst interesting case is thus m = 7. (8) Cameron [4] has considered a probability measure on the set of all sum-free sets: does fsj9 s.t. S S g have positive measure? References. N Alon and D.J. Kleitman. Sum-free subsets. In A. Baker, B Bollobas, and A. Hajnal, editors, A tribute to Paul Erdos, pages {26. Cambridge University Press, Neil J. Calkin and Steven R. Finch. Necessary and sucient conditions for periodicity of sum-free sets. In preparation.. Neil J. Calkin and Steven R. Finch. Dierence densities of sum-free sets. In preparation. 4. P. J. Cameron. Portrait of a typical sum-free set. In C. Whitehead, editor, Surveys in Combinatorics 987, volume 2 of London Mathematical Society Lecture Notes, pages {42. Cambridge University Press, P. Erdos. Extremal problems in number theory. Proc. Symp. Pure Maths., VIII:8{89, Richard K. Guy. Unsolved Problems in Number Theory. Springer Verlag, 980. Problem C9. 7. G. H. Hardy and E. M. Wright An Introduction to the Theory of Numbers. Oxford University Press, Oxford, England, 979. School of Mathematics, Georgia Institute of Technology, Atlanta, Ga 02, Hungarian Academy of Sciences, Mathematical Institute, Budapest, Hungary address: calkin@math.gatech.edu

ALMOST ODD RANDOM SUM-FREE SETS NEIL J. CALKIN AND P. J. CAMERON. 1. introduction

ALMOST ODD RANDOM SUM-FREE SETS NEIL J. CALKIN AND P. J. CAMERON Abstract. We show that if S 1 is a strongly complete sum-free set of positive integers, and if S 0 is a nite sum-free set, then with positive