An identity relating Fibonacci and Lucas numbers of order k

Q2 Mathematics

Electronic Notes in Discrete Mathematics Pub Date : 2018-12-01 DOI:10.1016/j.endm.2018.11.006

Spiros D. Dafnis, Andreas N. Philippou, Ioannis E. Livieris

引用次数: 3

Abstract

The following relation between Fibonacci and Lucas numbers of order k, $\sum_{i = 0}^{n} m^{i} [l_{i}^{(k)} + (m - 2) F_{i + 1}^{(k)} - \sum_{j = 3}^{k} (j - 2) F_{i - j + 1}^{(k)}] = m^{n + 1} F_{n + 1}^{(k)} + k - 2,$ is derived by means of colored tiling. This relation generalizes the well-known Fibonacci-Lucas identities, $\sum_{i = 0}^{n} 2^{i} L_{i} = 2^{n + 1} F_{n + 1}, \sum_{i = 0}^{n} 3^{i} (L_{i} + F_{i + 1}) = 3^{n + 1} F_{n + 1}$ and $\sum_{i = 0}^{n} m^{i} (L_{i} + (m - 2) F_{i + 1}) = m^{n + 1} F_{n + 1}$ of A.T. Benjamin and J.J. Quinn, D. Marques, and T. Edgar, respectively.

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关于k阶的斐波那契数和卢卡斯数的恒等式

利用彩色平铺法导出了k阶Lucas数与Fibonacci数之间的关系式∑i=0nmi[li(k)+(m−2)Fi+1(k)] -∑j=3k(j−2)Fi−j+1(k)]=mn+1Fn+1(k)+k−2。此关系推广了A.T. Benjamin、J.J. Quinn、D. Marques和T. Edgar分别提出的著名Fibonacci-Lucas恒等式∑i=0n2iLi=2n+1Fn+1、∑i=0n3i(Li+Fi+1)=3n+1Fn+1和∑i=0nmi(Li+(m−2)Fi+1)=mn+1Fn+1。

本文章由计算机程序翻译，如有差异，请以英文原文为准。

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来源期刊

Electronic Notes in Discrete Mathematics Mathematics-Discrete Mathematics and Combinatorics

CiteScore

1.30

自引率

0.00%

发文量

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