scholarly journals Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers

1997 ◽  
Vol 73 (7) ◽  
pp. 140-142 ◽  
Author(s):  
Daniel Duverney ◽  
Keiji Nishioka ◽  
Kumiko Nishioka ◽  
Iekata Shiokawa
2007 ◽  
Vol 130 (1) ◽  
pp. 37-60 ◽  
Author(s):  
Carsten Elsner ◽  
Shun Shimomura ◽  
Iekata Shiokawa

2020 ◽  
Vol 1 (3) ◽  
pp. 112-122
Author(s):  
Agung Prabowo

Fibonacci numbers, Lucas numbers and Mulatu numbers are built in the same method. The three numbers differ in the first term, while the second term is entirely the same. The next terms are the sum of two successive terms. In this article, generalizations of Fibonacci, Lucas and Mulatu (GFLM) numbers are built which are generalizations of the three types of numbers. The Binet formula is then built for the GFLM numbers, and determines the golden ratio, silver ratio and Bronze ratio of the GFLM numbers. This article also presents generalizations of these three types of ratios, called Metallic ratios. In the last part we state the Metallic ratio in the form of continued fraction and nested radicals.


2007 ◽  
Vol 17 (3) ◽  
pp. 429-446 ◽  
Author(s):  
C. Elsner ◽  
S. Shimomura ◽  
I. Shiokawa

2013 ◽  
Vol 2013 ◽  
pp. 1-5
Author(s):  
Daniel Fishman ◽  
Steven J. Miller

We derive closed form expressions for the continued fractions of powers of certain quadratic surds. Specifically, consider the recurrence relation with , , a positive integer, and (note that gives the Fibonacci numbers). Let . We find simple closed form continued fraction expansions for for any integer by exploiting elementary properties of the recurrence relation and continued fractions.


Mathematics ◽  
2021 ◽  
Vol 9 (2) ◽  
pp. 178
Author(s):  
Younseok Choo

This paper concerns the properties of the generalized bi-periodic Fibonacci numbers {Gn} generated from the recurrence relation: Gn=aGn−1+Gn−2 (n is even) or Gn=bGn−1+Gn−2 (n is odd). We derive general identities for the reciprocal sums of products of two generalized bi-periodic Fibonacci numbers. More precisely, we obtain formulas for the integer parts of the numbers ∑k=n∞(a/b)ξ(k+1)GkGk+m−1,m=0,2,4,⋯, and ∑k=n∞1GkGk+m−1,m=1,3,5,⋯.


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