Linear vector recursions of arbitrary order

IF 1 Q1 MATHEMATICS Discrete Mathematics Letters Pub Date : 2024-04-05 DOI:10.47443/dml.2024.029
Bernadette Faye, L´aszl´o N´emeth, L'aszl'o Szalay
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引用次数: 0

Abstract

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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任意阶的线性矢量递推
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。
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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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