A simple proof of a remarkable continued fraction identity
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- by P. G. Anderson, T. C. Brown and P. J.-S. Shiue
- Proc. Amer. Math. Soc. 123 (1995), 2005-2009
- DOI: https://doi.org/10.1090/S0002-9939-1995-1249866-4
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Abstract:
We give a simple proof of a generalization of the equality \[ \sum \limits _{n = 1}^\infty {\frac {1}{{{2^{[n/\tau ]}}}} = [0,{2^0},{2^1},{2^1},{2^2},{2^3},{2^5}, \ldots ],} \] where $\tau = (1 + \sqrt 5 )/2$ and the exponents of the partial quotients are the Fibonacci numbers, and some closely related results.References
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Bibliographic Information
- © Copyright 1995 American Mathematical Society
- 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