tailieunhanh - Báo cáo toán học: "A counterexample to a conjecture of Erd˝s, Graham and Spencer o"

Tuyển tập các báo cáo nghiên cứu khoa học về toán học trên tạp chí toán học quốc tế đề tài: A counterexample to a conjecture of Erd˝s, Graham and Spencer o. | A counterexample to a conjecture of Erdos Graham and Spencer Song Guo Department of Mathematics Huaiyin Teachers College Huaian 223300 The People s Republic of China guosong77@ Submitted Oct 6 2008 Accepted Dec 2 2008 Published Dec 9 2008 Mathematics Subject Classihcation 11B75 Abstract It is conjectured by Erdos Graham and Spencer that if 1 a1 a2 as with s 11 ai n 1 30 then this sum can be decomposed into n parts so that all partial sums are 1. In this note we propose a counterexample which gives a negative answer to this conjecture. Keywords Erdos-Graham-Spencer conjecture Erdos problem Partition. 1 Introduction Erdos 2 p. 41 asked the following question is it true that if ai s are positive integers with 1 a1 a2 as and s 1 1 ai 2 then there exists a subset A of 1 2 . sg such that X a 1 X . a 1 i2A a i 1 . s A a Sándor 3 gave a simple construction to show that the answer is negative let aig divisors of 120 with the exception of 1 and 120 g. Furthermore Sándor 3 proved the following results Theorem 1. For every n 2 there exist integers 1 a1 a2 as with E 1 1 ai n and this sum cannot be split into n parts so that all partial sums are 1. This author is supported by Natural Science Research Project of Ordinary Universities in Jiangsu Province 08KJB110002 . THE ELECTRONIC JOURNAL OF COMBINATORICS 15 2008 N43 1 Theorem 2. Let n 2. If 1 a1 a2 as with 22s 1 1 ai n - en T then this sum can be decomposed into n parts so that all partial sums are 1. If we allow repetition of integers it is conjectured by Erdos Graham and Spencer 2 p. 41 that if 1 a1 a2 as with 22s 1 1 ai n 1 30 then this sum can be decomposed into n parts so that all partial sums are 1. This is not true for Xs I 1 ai n 1 30 as shown by a1 2 a2 a3 3 a4 . a5n_3 5. Sandor 3 proved a weaker assertion when the n 1 30 was replaced by n 1 2. Let o n denote the least real number such that for any integers 1 a1 a2 as with n 2 and 22s 1 1 ai n o n this sum can be decomposed into n parts so that all