将数字表示为两个斐波那契数或卢卡斯数之差

Q3 Mathematics Matematychni Studii Pub Date : 2021-12-26 DOI:10.30970/ms.56.2.124-132
P. Ray, K. Bhoi
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引用次数: 0

摘要

在本研究中,我们研究了所有表示为两个Fibonacci或Lucas数之差的重数位。我们证明,如果$F_{n}-F_{m} $是一个重数位,其中$F_{n}$表示第$n$个斐波那契数,然后$(n,m)\in\{此外,如果$L_{n}$表示第$n$个Lucas数,则$L_{n}-L_{m} $是$(n,m)\in\{(6,4),(7,4)、(7,6),(8,2)\}的重复数字,其中$n>m$也就是说,唯一可以表示为两个斐波那契数之差的repdigits是$11,33,55,88$和$555$;他们的陈述是$11=F_{7}-F_{3} ,\ 33=F_{9}-F_{1} =F_{9}-F_{2} ,\55=F_{11}-F_{9} =F_{12}-F_{11} ,\88=F_{11}-F_{1} =F_{11}-F_{2} ,\555=F_{15}-F_{10} $(定理2)。两个Lucas数差的相似结果:唯一可以表示为两个Luca数差的重复数字是$11,22$和$44;$他们的陈述是$11=L_{6}-L_{4} =L_{7}-L_{6} ,\22=L_{7}-L_{4} ,\4=L_{8}-L_{2} $(定理3)。
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Repdigits as difference of two Fibonacci or Lucas numbers
In the present study we investigate all repdigits which are expressed as a difference of two Fibonacci or Lucas numbers. We show that if $F_{n}-F_{m}$ is a repdigit, where $F_{n}$ denotes the $n$-th Fibonacci number, then $(n,m)\in \{(7,3),(9,1),(9,2),(11,1),(11,2),$ $(11,9),(12,11),(15,10)\}.$ Further, if $L_{n}$ denotes the $n$-th Lucas number, then $L_{n}-L_{m}$ is a repdigit for $(n,m)\in\{(6,4),(7,4),(7,6),(8,2)\},$ where $n>m.$Namely, the only repdigits that can be expressed as difference of two Fibonacci numbers are $11,33,55,88$ and $555$; their representations are $11=F_{7}-F_{3},\33=F_{9}-F_{1}=F_{9}-F_{2},\55=F_{11}-F_{9}=F_{12}-F_{11},\88=F_{11}-F_{1}=F_{11}-F_{2},\555=F_{15}-F_{10}$ (Theorem 2). Similar result for difference of two Lucas numbers: The only repdigits that can be expressed as difference of two Lucas numbers are $11,22$ and $44;$ their representations are $11=L_{6}-L_{4}=L_{7}-L_{6},\ 22=L_{7}-L_{4},\4=L_{8}-L_{2}$ (Theorem 3).
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来源期刊
Matematychni Studii
Matematychni Studii Mathematics-Mathematics (all)
CiteScore
1.00
自引率
0.00%
发文量
38
期刊介绍: Journal is devoted to research in all fields of mathematics.
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