论两个 $ k $ 广义斐波那契数的并集

Alaa Altassan, Murat Alan
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引用次数: 0

摘要

设 $ k \geq 2 $ 为整数。$ k- $广义斐波那契数列是由递推关系 $ F_{n}^{(k)}=F_{n-1}^{(k)} +\cdots + F_{n-k}^{(k)}$ 定义的数列,对于所有 $ n \geq 2$,初始值为 $F_{i}^{(k)}=0$,对于 $ i=2-k, \ldots, 0 $ 和 $F_{1}^{(k)}=1。2020 年,班克斯和卢卡等人确定了所有斐波那契数,这些数都是两个斐波那契数的集合。在本文中,我们考虑了这一问题的类似问题,将 $ k-$ 的广义斐波那契数视为同一序列中两个项的集合。我们完全解决了所有 $ k \geq 3 的这一问题。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
On Concatenations of Two $ k $-Generalized Fibonacci Numbers
Let $ k \geq 2 $ be an integer. The $ k- $generalized Fibonacci sequence is a sequence defined by the recurrence relation $ F_{n}^{(k)}=F_{n-1}^{(k)} + \cdots + F_{n-k}^{(k)}$ for all $ n \geq 2$ with the initial values $ F_{i}^{(k)}=0 $ for $ i=2-k, \ldots, 0 $ and $ F_{1}^{(k)}=1.$ In 2020, Banks and Luca, among other things, determined all Fibonacci numbers which are concatenations of two Fibonacci numbers. In this paper, we consider the analogue of this problem by taking into account $ k-$generalized Fibonacci numbers as concatenations of two terms of the same sequence. We completely solve this problem for all $ k \geq 3.
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