Trivial Fibonacci 2

If am=n=2m(12n1+k=1n1Lk2nk)\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 m2m ≥ 2, find (m=21(2+am)2)1\displaystyle \left \lfloor \left(\sum_{m=2}^{\infty} \frac{1}{(2+a_{m})^2} \right)^{-1}\right \rfloor.

Notation: LnL_n denotes the nnth Lucas number, where L0=2L_0 = 2, L1=1L_1=1, and Ln=Ln1+Ln2L_n = L_{n-1} + L_{n-2} for n2n \ge 2.

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