任意阶的线性矢量递推

IF 1 Q1 MATHEMATICS
Bernadette Faye, L´aszl´o N´emeth, L'aszl'o Szalay
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引用次数: 0

摘要

解决各种组合问题往往需要高阶(即阶数大于 1)的向量递归。假设给定矩阵 A 1 , A 2 , ., A s,均来自 C k × k。这些矩阵允许我们定义矢量 ¯ v n ∈ C k , n ≥ s 的矢量递推关系 ¯ v n = A 1 ¯ v n - 1 + A 2 ¯ v n - 2 + - - + A s ¯ v n - s 。 本文的重要成果是,我们可以分离矢量的分量序列,并找到一个共同的线性递推关系来描述它们。我们这种方法的主要优点是处理方法统一,而且有可能实现自动化。我们可以应用主要结果来回答一个问题,这个问题涉及参数 { 4 , 5 } 的双曲帕斯卡三角形的行。.我们还验证了文献中的另外两种说法,以说明该方法的威力。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Linear vector recursions of arbitrary order
Solution of various combinatorial problems often requires vector recurrences of higher order (i.e., the order is larger than 1). Assume that there are given matrices A 1 , A 2 , . . . , A s , all from C k × k . These matrices allow us to define the vector recurrence ¯ v n = A 1 ¯ v n − 1 + A 2 ¯ v n − 2 + · · · + A s ¯ v n − s for the vectors ¯ v n ∈ C k , n ≥ s . The paramount result of this paper is that we could separate the component sequences of the vectors and find a common linear recurrence relation to describe them. The principal advantage of our approach is a uniform treatment and the possibility of automatism. We could apply the main result to answer a problem that arose concerning the rows of the modified hyperbolic Pascal triangle with parameters { 4 , 5 } . We also verified two other statements from the literature in order to illustrate the power of the method.
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来源期刊
Discrete Mathematics Letters
Discrete Mathematics Letters Mathematics-Discrete Mathematics and Combinatorics
CiteScore
1.50
自引率
12.50%
发文量
47
审稿时长
12 weeks
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