On the linear diophantine problem of Frobenius in three variables.

On the linear diophantine problem of Frobenius in three variables. By Ernst S. Seltner and Öyvind Beyer at Bergen 1. Introduction Let 0 l 9 tf 2 > · · · » t f f c t>e relatively prime integers >1. The problem of Frobenius consists in determining the largest integer g(al, 02, . . . , ak) with no integral representation aixl+a2x2-i HGk x k> xi ^ 0· In later years, the number n(ai9a2, ..., ak) of positive integers with no such representation has also been studied. A closer description of Frobenius' problem, with an extensive list of references, can be found in Seimer [3]. In the present paper, we give a complete, explicit solution in the case k = 3. The solution was found when Beyer wrote his thesis [1] under the supervision of Seimer. The result was presented by Seimer at the 17. Scandinavian congress of mathematicians in Aabo (Turku), Finland in August 1976. Among the audience was Öystein Rödseth. He got interested in the method of solution by continued fractions, and managed to simplify the resulting formula. In particular, by using negative division remainders, he could replace our M-function in 2 m arguments by the minimum of two numbers only. Rödseth's solution will appear http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Journal für die reine und angewandte Mathematik (Crelle's Journal) de Gruyter

On the linear diophantine problem of Frobenius in three variables.

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Publisher
de Gruyter
Copyright
Copyright © 2009 Walter de Gruyter
ISSN
0075-4102
eISSN
1435-5345
D.O.I.
10.1515/crll.1978.301.161
Publisher site
See Article on Publisher Site

Abstract

By Ernst S. Seltner and Öyvind Beyer at Bergen 1. Introduction Let 0 l 9 tf 2 > · · · » t f f c t>e relatively prime integers >1. The problem of Frobenius consists in determining the largest integer g(al, 02, . . . , ak) with no integral representation aixl+a2x2-i HGk x k> xi ^ 0· In later years, the number n(ai9a2, ..., ak) of positive integers with no such representation has also been studied. A closer description of Frobenius' problem, with an extensive list of references, can be found in Seimer [3]. In the present paper, we give a complete, explicit solution in the case k = 3. The solution was found when Beyer wrote his thesis [1] under the supervision of Seimer. The result was presented by Seimer at the 17. Scandinavian congress of mathematicians in Aabo (Turku), Finland in August 1976. Among the audience was Öystein Rödseth. He got interested in the method of solution by continued fractions, and managed to simplify the resulting formula. In particular, by using negative division remainders, he could replace our M-function in 2 m arguments by the minimum of two numbers only. Rödseth's solution will appear

Journal

Journal für die reine und angewandte Mathematik (Crelle's Journal)de Gruyter

Published: Jan 1, 1978

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