5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a_{2}=\frac{1}{2}, a_{n+1}=2 a_{n}+a_{n-1}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Then $\left[\sum_{k=2}^{2016} \frac{1}{a_{k-1} a_{k+1}}\right]=$ $\qquad$ .
5.1.

Given, the sequence $\left\{a_{n}\right\}$ is an increasing sequence of positive numbers.
From $a_{k+1}-a_{k-1}=2 a_{k}$, we know
$$
\begin{array}{l}
\frac{1}{a_{k+1} a_{k-1}}=\frac{1}{2}\left(\frac{1}{a_{k-1} a_{k}}-\frac{1}{a_{k} a_{k+1}}\right) \\
\Rightarrow \sum_{k=2}^{2016} \frac{1}{a_{k-1} a_{k+1}}=\frac{1}{2} \sum_{k=2}^{2016}\left(\frac{1}{a_{k-1} a_{k}}-\frac{1}{a_{k} a_{k+1}}\right) \\
=\frac{1}{2}\left(\frac{1}{a_{1} a_{2}}-\frac{1}{a_{2016} a_{2017}}\right) \\
=2-\frac{1}{2 a_{2016} a_{2017}}\frac{1}{a_{1} a_{3}}=\frac{4}{3} \text {, hence } \\
{\left[\sum_{k=2}^{2016} \frac{1}{a_{k-1} a_{k+1}}\right]=1 .} \\
\end{array}
$$