11. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{2}=2, \frac{a_{n+2}}{a_{n}}=\frac{a_{n+1}^{2}+1}{a_{n}^{2}+1}(n=1,2, \cdots)$. Try to find $\left[a_{2017}\right]([a]$ represents the greatest integer not exceeding the real number $a$)
11. Analysis: From the recursive formula, we get: $\prod_{i=1}^{n-2} \frac{a_{i+2}}{a_{i}}=\prod_{i=1}^{n-2} \frac{a_{i+1}^{2}+1}{a_{i}^{2}+1}$, simplifying to: $\frac{a_{n} a_{n-1}}{a_{1} a_{2}}=\frac{a_{n-1}^{2}+1}{a_{1}^{2}+1}$. Given $a_{1}=1, a_{2}=2$, we know $a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$.

On one hand: $a_{n}^{2}=a_{n-1}^{2}+2+\frac{1}{a_{n-1}^{2}}>a_{n-1}^{2}+2>\ldots>a_{1}^{2}+2(n-1)$, it is easy to see that $a_{2017}^{2}>4033$, hence $a_{2017}>63$.

On the other hand: From $a_{n}^{2}=a_{n-1}^{2}+2+\frac{1}{a_{n-1}^{2}}=\ldots=a_{1}^{2}+2(n-1)+\sum_{i=1}^{n-1} \frac{1}{a_{i}^{2}}$, we have
$a_{2017}^{2}=4033+\sum_{i=1}^{2016} \frac{1}{a_{i}^{2}}$. Given the general term formula $a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$, we know $a_{n}>a_{n-1}$, thus $\sum_{i=1}^{2016} \frac{1}{a_{i}^{2}}=\sum_{i=1}^{10} \frac{1}{a_{i}^{2}}+\sum_{i=11}^{100} \frac{1}{a_{i}^{2}}+\sum_{i=101}^{2016} \frac{1}{a_{i}^{2}}<\frac{10}{a_{1}^{2}}+\frac{90}{a_{10}^{2}}+\frac{1916}{a_{100}^{2}}<10+\frac{90}{19}+\frac{1916}{199}<25$, hence $a_{2017}^{2}<4033+25<4096=64^{2}$, i.e., $a_{2017}<64$.

In summary, $\left[a_{2017}\right]=63$.