15. Let $\left\{a_{n}\right\}$ be a sequence of real numbers,
$$
a_{1}=t, a_{n+1}=4 a_{n}\left(1-a_{n}\right) \text {. }
$$

Find how many different real numbers $t$ make $a_{2006}=0$
15. When $t>1$, $a_{2}<0$.

By mathematical induction, for any $n \geqslant 2$, we have $a_{n}<0$. Therefore, $a_{2006} \neq 0$.

When $t<0$, similarly, for any $n \in \mathbf{N}$, we have $a_{n}<0$. Thus, we only need to consider the case $0 \leqslant t \leqslant 1$.
When $0 \leqslant t \leqslant 1$, let $t=\sin ^{2} \theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$, then $a_{2}=4 \sin ^{2} \theta \cdot \cos ^{2} \theta=\sin ^{2} 2 \theta$.
Using mathematical induction, we get $a_{n}=\sin ^{2} 2^{n-1} \theta$.
From $a_{2006}=0$, we can get $\sin 2^{2005} \theta=0$.
Thus, $\theta=\frac{k \pi}{2^{200}}(k \in \mathbf{Z})$, and $0 \leqslant k \leqslant 2^{2004}$.
Since the sine function is monotonically increasing on $\left[0, \frac{\pi}{2}\right]$, for different $\theta \in\left[0, \frac{\pi}{2}\right]$, the value of $a_{1}$ is different.
Therefore, the number of $t$ that satisfy the condition is $2^{2004}+1$.