Tuyển tập các báo cáo nghiên cứu khoa học trên tạp chí toán học quốc tế đề tài: A note on the number of (k, )-sum-free sets. | A note on the number of k -sum-free sets Tomasz Schoen Mathematisches Seminar Universitat zu Kiel Ludewig-Meyn-Str. 4 24098 Kiel Germany tos@ and Department of Discrete Mathematics Adam Mickiewicz University Poznan Poland Abstract A set A c N is k -sum-free for k 2 N k if it contains no solutions to the equation X1 - xk y1 - y . Let p p k be the smallest natural number not dividing k and let r rn 0 r p be such that r n mod p . The main result of this note says that if k is small in terms of p then the number of k -sum-free subsets of 1 n is equal to p r p o 1 2Ln J where r x denotes the number of positive integers m r relatively prime to x and x x x . Submitted February 15 1999 Accepted May 23 2000. 1991 Mathematics Subject Classification 11B75 11P99. A set A of positive integers is k -sum-free for k Ế 2 N k if there are no solutions to the equation X1 xk y yt in A. Denote by SFn the number of k -sum-free subsets of 1 n . Since the set of 1 THE ELECTRONIC .JOURNAL OF COMBINATORICS 7 2000 R30 2 odd numbers is 2 1 -sum-free we have SFn 1 2b n 1 2C. In fact Erdos and Cameron 6 conjectured SFn1 O 2n 2 . This conjecture is still open and the best upper bounds for SFn 1 given independently by Alon 1 and Calkin 3 say that for 1 SFỊ ự SFn 1 O 2n 2 o n . For 3 this bound was recently improved by Bilu 2 who proved that in this case SFn 1 1 o 1 2L n 1 2J. The case of k being much larger than was treated by Calkin and Taylor 4 . They showed that for some constant ck the number of k 1 -sum-free subsets of 1 n is at most ck2 k n provided k 3. Furthermore Calkin and Thomson proved 5 that for every k and with k 4 1 SFnk Ck2 k n k. In order to study the behaviour of SFn i let us observe first that there are two natural examples of large k -sum-free subsets of the interval 1 n n k 1 . n SF I t max 2n p- and m 2 1 2 . n m r mod p where gcd r p 1 and p p k min s 2 N s does not divide k . Thus 2Ĩ k . n k In this note we study the case k so that 2 p- 2d k- n k and we
