TAILIEUCHUNG - Báo cáo toán học: "Linear Recurrence Relations for Sums of Products of Two Terms"

Tuyển tập các báo cáo nghiên cứu khoa học ngành toán học tạp chí Department of Mathematic dành cho các bạn yêu thích môn toán học đề tài: Linear Recurrence Relations for Sums of Products of Two Terms. | Linear Recurrence Relations for Sums of Products of Two Terms Yan-Ping Mu College of Science Tianjin University of Technology Tianjin 300384 . China Submitted Dec 27 2010 Accepted Aug 16 2011 Published Aug 26 2011 Mathematics Subject Classification 68W30 33F10 05A19 Abstract For a sum of the form Yk F n k G n k we set up two systems of equations involving shifts of F n k and G n k . Then we solve the systems by utilizing the recursion of F n k and the method of undetermined coefficients. From the solutions we derive linear recurrence relations for the sum. With this method we prove many identities involving Bernoulli numbers and Stirling numbers. 1. Introduction Finding recurrence relations is a basic method for proving identities. Fasenmyer 4-6 proposed a systematic method to find linear recurrence relations for hypergeometric sums. Wilf and Zeilberger 10-13 provided an efficient algorithm called Zeilberger s algorithm to construct linear recurrence relations for hypergeometric sums. Chyzak 2 extended Zeilberger s algorithm to holonomic systems. Kauers 8 presented algorithms for sums involving Stirling-like sequences. Chyzak Kauers and Salvy 3 further considered non-holonomic systems. In this paper we focus on deriving a linear recurrence relation for the sum tt f n F n k G n k O k - x where n ni . nr and F n k G n k are two functions of n and k. Our approach can be described as follows. We first take a finite subset S of Zr 1 as the set of shifts of Supported by the National Natural Science Foundation of China Project 11001198 . THE ELECTRONIC JOURNAL OF COMBINATORICS 18 2011 P170 1 the variables n and k. Then we make an ansatz Xa 3 n k F n - a k - ft 0 a eS where Xa p n k are functions to be determined. Denote AS a G Zr there exists ft G Z such that a ft G S and Sa ft G Z a ft G S . For each a G AS we take a finite subset S a of Zr and make an ansatz Aa n k ft F n a k G n k ft eSa Ca Y n F n Y k G n Y k C1-3 yeS a where caY n are .

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• báo cáo của tạp chí Department of Mathematic
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