Sum-free Sets of Integers

نویسندگان

  • H. L. ABBOTT
  • E. T. H. WANG
چکیده

A set S of integers is said to be sum-free if a, b e 5 implies a + b 6 S. In this paper, we investigate two new problems on sum-free sets: (1) Let f(k) denote the largest positive integer for which there exists a partition of (1, 2,... ,f(k)) into k sum-free sets, and let h(k) denote the largest positive integer for which there exists a partition of {1, 2, . . . , h(k)) into k sets which are sum-free mod h(k) + 1. We obtain evidence to support the conjecture that f(k) = h(k) for all k. (2) Let g(n, k) denote the cardinality of a largest subset of {1, 2,. . ., n] that can be partitioned into k sum-free sets. We obtain upper and lower bounds for g(n, k). We also show that g(n, 1) = [(ii + l)/2] and indicate how one may show that for all n < 54, g(n, 2) = n [n/5]. A set S of integers is said to be sum-free if a, b E S implies a + b $ S. The case a = b is not excluded, so that a E S implies 2a G S. A well-known theorem of I. Schur [14] states that if the set {1, 2, . . . , [k\e]} is partitioned arbitrarily into k sets, at least one of the sets fails to be sum-free. Thus we may define f(k) to be the largest positive integer for which there exists some way of partitioning {1, 2, . . . ,f(k)} into k sum-free sets. The determination of the numbers/(/c) is a notoriously difficult problem. It is easy to verify that/(l) = l,/(2) = 4 and with a little effort one can show that /(3) = 13. L. D. Baumert [4], with the aid of a computer, showed that /(4) = 44. The value of /(5) seems to be out of reach at the present time. Recently H. Fredericksen [9] proved that/(5) > 138 and this seems to be the record. Schur proved that f(k + 1) > 3f(k) + 1 and this, together with his theorem mentioned in the first paragraph, shows that (1) (3* l)/2 < f(k) <[k\e] 1. Abbott and Hanson [3], improving on an earlier inequality of Abbott and Moser [2], proved that for all positive integers k and /, (2) f(k + l)>2f(k)f(l)+f(k)+f(l). From (2), or the earlier inequality of Abbott and Moser, it follows, via a well-known argument, that a = hmk^x f(k)l/k exists, although a may be Received by the editors December 20, 1976 and, in revised form, April 29, 1977. AMS (MOS) subject classifications (1970). Primary 10L10.

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تاریخ انتشار 2010