# An interesting Fibonacci discovery

If $\displaystyle a_{m} = \sum_{n=2}^{m} \left(\frac{1}{2^{n-1}} + \sum_{k=1}^{n-1} \frac{L_{k}}{2^{n-k}}\right)$ for $m ≥ 2$, find $\displaystyle \left \lfloor \left(\sum_{m=2}^{\infty} \frac{1}{2+a_{m}} \right)^{-1}\right \rfloor$.

Notation: $L_n$ denotes the $n$th Lucas number, where $L_0 = 2$, $L_1=1$, and $L_n = L_{n-1} + L_{n-2}$ for $n \ge 2$.

Largely the same as this problem.

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