We show that the number of subsets of [1, 2, ..., n] with no solution to x 1 +x 2 + } } } +x k = y 1 + y 2 + } } } + y l for k 4l&1 is at most c 2 %n where %=(k&l)รk.
1998 Academic Press
1. Introduction
A set S of positive integers is sum-free if x+ y=z has no solution in S. Similarly, a set S of positive integers is (k, l ) sum-free if x 1 + x 2 + } } } +x k = y 1 + y 2 + } } } + y l has no solution in S. Cameron and Erdo s have shown [5] that the number of sum-free sets contained in [(1ร3) n, ..., n] is c2 nร2 , and Alon [1], Calkin [3] and Erdo s and Granville (personal communication) have independently shown that the number of sum-free sets contained in [1, 2, ..., n] is o(2 n(1ร2+=) ) for every =>0. Calkin and Taylor [4] have shown that the number of (k, 1) sum-free sets in [1, 2, ..., n] is c 2 %n for k 3 and %=(k&1)รk. Bilu [2] has shown that the number of (4, 3) sum-free sets, and consequently, the number of (k+1, k) sum-free sets is at most c 2 (n+1)ร2 for k 3. We extend the results of Calkin and Taylor in a different direction by showing that for k 4l&1, the number of (k, l ) sum-free sets in [1, 2, ..., n] is at most c2 %n where %=(k&l)รk. Note that the number of (k, l ) sum-free sets in [1, 2, ..., n] is at least c 2 %n because any subset of [(lรk) n, ..., n] is (k, l ) sum-free, and there are 2 n&(lรk) n =2 %n such subsets. Our main result is the following theorem.
Theorem 1. For a fixed k, l with k 4l&1, let %=(k&l)รk. There exists a constant c depending on k such that the number of subsets of [1, 2, ..., n] containing no solution to x 1 +x 2 + } } } +x k = y 1 + y 2 + } } } + y l is at most c2 %n .
Article No. NT972191