7. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{n}=\left[(2+\sqrt{5})^{n}+\frac{1}{2^{n}}\right]\left(n \in \mathbf{N}^{*}\right)$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. Let $C$ be a real number, and for any positive integer $n$, we have $\sum_{k=1}^{n} \frac{1}{a_{k} a_{k+2}} \leqslant C$, then the minimum value of $C$ is $\qquad$
Let $x_{1}=2+\sqrt{5}, x_{2}=2-\sqrt{5}, b_{n}=x_{1}^{n}+x_{2}^{n}$. Note that $x_{1}+x_{2}=4, x_{1} \cdot x_{2}=-1$, so $x_{1}, x_{2}$ are the roots of the equation $x^{2}-4 x-1=0$. Substituting $b_{n}=x_{1}^{n}+x_{2}^{n}$ into $b_{n+2}=4 b_{n+1}+b_{n}$, we get $x_{1}^{n+2}+x_{2}^{n+2}=4\left(x_{1}^{n+1}+x_{2}^{n+1}\right)+x_{1}^{n}+x_{2}^{n}$. Thus, the sequence $\left\{b_{n}\right\}$ satisfies $b_{n+2}=4 b_{n+1}+b_{n}$, and $b_{1}=4, b_{2}=x_{1}^{2}+x_{2}^{2}=18$.

Since $|2-\sqrt{5}|<\frac{1}{2} \Rightarrow x_{1}^{n}+x_{2}^{n}<(2+\sqrt{5})^{n}+\frac{1}{2^{n}}<x_{1}^{n}+x_{2}^{n}+|2-\sqrt{5}|^{n}+\frac{1}{2^{n}}$ $<x_{1}^{n}+x_{2}^{n}+1 \Rightarrow\left[(2+\sqrt{5})^{n}+\frac{1}{2^{n}}\right]=x_{1}^{n}+x_{2}^{n} \Rightarrow a_{n}=b_{n}$.

Therefore, $\frac{1}{a_{k} a_{k+2}}=\frac{4 a_{k+1}}{4 a_{k} a_{k+1} a_{k+2}}=\frac{a_{k+2}-a_{k}}{4 a_{k} a_{k+1} a_{k+2}}=\frac{1}{4}\left(\frac{1}{a_{k} a_{k+1}}-\frac{1}{a_{k+1} a_{k+2}}\right)$
$$
\Rightarrow \sum_{k=1}^{n} \frac{1}{a_{k} a_{k+2}}=\frac{1}{4}\left(\frac{1}{a_{1} a_{2}}-\frac{1}{a_{n+1} a_{n+2}}\right) \leqslant \frac{1}{4 a_{1} a_{2}}=\frac{1}{288} \text {, }
$$

with equality holding if and only if $n=1$. Therefore, the minimum value of $C$ is $\frac{1}{288}$.