斐波那契数是两个平衡数的乘积

IF 0.3 Q4 MATHEMATICS Annales Mathematicae et Informaticae Pub Date : 2019-06-26 DOI:10.33039/AMI.2019.06.001
F. Erduvan, R. Keskin
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引用次数: 2

摘要

当n≥2时,Fibonacci序列Fn定义为F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2。平衡数列Bn定义为:当n≥2时,B0 = 0, B1 = 1, Bn = 60 - 1 - Bn - 2。在本文中,我们找到了两个平衡数乘积的所有斐波那契数。我们还找到了所有的平衡数它们是两个斐波那契数的乘积。更一般地,取k,m,m为正整数,证明了Fk = BmBn意味着(k,m, n) = (1,1,1), (2,1,1), Bk = FmFn意味着(k,m, n) =(1,1,1),(1,1,2),(1,2,2),(2,3,4)。
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Fibonacci numbers which are products of two balancing numbers
The Fibonacci sequence (Fn) is defined by F0 = 0, F1 = 1 and Fn = Fn−1 + Fn−2 for n ≥ 2. The balancing number sequence (Bn) is defined by B0 = 0, B1 = 1 and Bn = 6Bn−1 − Bn−2 for n ≥ 2. In this paper, we find all Fibonacci numbers which are products of two balancing numbers. Also we found all balancing numbers which are products of two Fibonacci numbers. More generally, taking k,m,m as positive integers, it is proved that Fk = BmBn implies that (k,m, n) = (1, 1, 1), (2, 1, 1) and Bk = FmFn implies that (k,m, n) = (1, 1, 1), (1, 1, 2), (1, 2, 2), (2, 3, 4).
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