7、Sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=t, a_{n+1}=4 a_{n}\left(1-a_{n}\right)\left(n \in N_{+}\right)$ and $a_{2017}$ is the first term in the sequence that is 0. The number of real numbers $t$ that satisfy the condition is $\qquad$

The number of real numbers $t$ that satisfy the condition is $\qquad$
7. $2^{2015}-1$ solutions

Analysis: It is easy to see that if $t \in (1,+\infty) \cup (-\infty, 0)$, then $a_{n}<0, \forall n \geq 2$; if $t \in \{0,1\}$, then $a_{2}=0$, which is a contradiction. Therefore, $t \in (0,1)$. Suppose $a_{1}=t=\sin ^{2} \theta$. By induction, we can show that $a_{2017}=\sin ^{2}\left(2^{2016} \theta\right)$. Hence, $a_{2017}=0 \Leftrightarrow \theta=\frac{k \pi}{2^{2016}}$, so $\mathrm{k}$ has $2^{2015}-1$ solutions, meaning $\mathrm{t}$ has $2^{2015}-1$ different solutions.