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n-th root of a continued fraction of Lucas numbers

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Tolaso J Kos
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n-th root of a continued fraction of Lucas numbers

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Post by Tolaso J Kos » Fri Jan 06, 2017 10:09 pm

Let us denote the $n$ -th Lucas number as $L_n$. It is known that $L_0=2$ , $L_1=1$ as well as

$$L_{n+2}=L_{n+1} + L_{n} \quad \text{forall} \; n \geq 0$$

Calculate the value of

$$\mathcal{R}=\sqrt[n]{L_{2n} - \frac{1}{L_{2n}- \dfrac{1}{L_{2n}-\ddots}}}$$

I don't have a solution.
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