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A simple proof of a remarkable continued fraction identity


Authors: P. G. Anderson, T. C. Brown and P. J.-S. Shiue
Journal: Proc. Amer. Math. Soc. 123 (1995), 2005-2009
MSC: Primary 11A55
DOI: https://doi.org/10.1090/S0002-9939-1995-1249866-4
MathSciNet review: 1249866
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Abstract | References | Similar Articles | Additional Information

Abstract: We give a simple proof of a generalization of the equality

$\displaystyle \sum\limits_{n = 1}^\infty {\frac{1}{{{2^{[n/\tau ]}}}} = [0,{2^0},{2^1},{2^1},{2^2},{2^3},{2^5}, \ldots ],} $

where $ \tau = (1 + \sqrt 5 )/2$ and the exponents of the partial quotients are the Fibonacci numbers, and some closely related results.

References [Enhancements On Off] (What's this?)

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Additional Information

DOI: https://doi.org/10.1090/S0002-9939-1995-1249866-4
Article copyright: © Copyright 1995 American Mathematical Society

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