tailieunhanh - Báo cáo toán học: "Reconstructing integer sets from their representation functions"

Tuyển tập các báo cáo nghiên cứu khoa học trên tạp chí toán học quốc tế đề tài: Reconstructing integer sets from their representation functions. | Reconstructing integer sets from their representation functions Vsevolod F. Lev Department of Mathematics University of Haifa at Oranim Tivon 36006 Israel seva@ Submitted Oct 5 2004 Accepted Oct 27 2004 Published Nov 3 2004 Mathematics Subject Classifications 11B34 05A17 11B13 Abstract We give a simple common proof to recent results by Dombi and by Chen and Wang concerning the number of representations of an integer in the form a1 a2 where a1 and a2 are elements of a given infinite set of integers. Considering the similar problem for differences we show that there exists a partition N g 1Ak of the set of positive integers such that each Ak is a perfect difference set meaning that any non-zero integer has a unique representation as a1 a2 with ai a2 2 Ak . A number of open problems are presented. 1 Introduction and summary For a set A c Z and an integer n 2 Z consider the representation functions R n a1 a2 2 A X A a1 a2 ng RA n ai a2 2 A X A ai a2 n ai a2g and RA3 n a1 a2 2 A X A a1 a2 n a1 a2g. To what extent do Rj n determine the set A Problems of this sort were to our knowledge first studied by Nathanson in N78 . Let N denote the set of all positive integers. In his research talks and private communications Sarkozy has raised the following question Jr u . o 1 jj do there exist A B c N with the infinite symmetric difference such that Ra n Rb n for all but finitely many n 2 N Dombi noticed in D02 that the answer is negative for j 1 by the simple observation that rA n is odd if and only if n 2a for some a 2 A. On the other hand he has shown that for j 2 the answer is positive and indeed there is a partition N A B such that rA n rB n for all n 2 N. THE ELECTRONIC JOURNAL OF COMBINATORICS 11 2004 R78 1 Theorem 1 Dombi D02 Define the mapping T N 1 1 by T 1 1 T 2n T 2n 1 T 2n 1 T n 1 n 2 N and let A n 2 N T n 1 B n 2 N T n 1 . Then rA n rB n for all n 2 N. For the function rA n the problem was solved by Chen and Wang in CW03 . Theorem 2 Chen and Wang CW03