求逆斐波那契均值公式的推导

Q4 Multidisciplinary
Steven Elizalde, Romeo Patan
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引用次数: 0

摘要

反向斐波那契序列$\{J_n\}$由$n\geq2$的关系$J_n = 8(J_{n-1} - J_{n-2})$定义,以$J_0=0$和$J_1=1$作为初始项。在书籍和数学期刊中,已经推导出了一些公式来求解序列中缺失的项,但对于反向斐波那契序列却没有。因此,本文推导出了一个公式,可以演绎求解逆斐波那契数列的第一个缺失项$\{x_1\}$,公式为$x_1=\frac{b+8aJ_n}{J_{n+1}}$。通过使用$\{x_1\}$的导出公式,现在可以求解反向斐波那契数列的均值以及求解序列本身。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Deriving a Formula in Solving Reverse Fibonacci Means
Reverse Fibonacci sequence $\{J_n\}$ is defined by the relation $J_n = 8(J_{n-1} - J_{n-2})$ for $n\geq2$ with $J_0=0$ and $J_1=1$ as initial terms. A few formulas have been derived for solving the missing terms of a sequence in books and mathematical journals, but not for the reverse Fibonacci sequence. Thus, this paper derived a formula that deductively solves the first missing term $\{x_1\}$ of the reverse Fibonacci sequence and is given by the equation $x_1=\frac{b+8aJ_n}{J_{n+1}}$. By using the derived formula for $\{x_1\}$, it is now possible to solve the means of the reverse Fibonacci sequence as well as solving the sequence itself.
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来源期刊
CiteScore
0.70
自引率
0.00%
发文量
19
审稿时长
8 weeks
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