12. Let $a$ be the largest positive root of the equation $x^{3}-3 x^{2}+1=0$, then among the 2006 numbers $\left[a^{n}\right](1 \leqslant n \leqslant 2006)$, there are $\qquad$ numbers that can be divided by 17.
12. 251 Let the three roots of the equation $x^{3}-3 x^{2}+1=0$ be $\alpha, \beta, a$, respectively. Then, from $f(-1) < 0, f\left(\frac{1}{2}\right)>0, f(1) < 0$ (where $f(x)=x^{3}-3 x^{2}+1$), we can assume $-1 < \alpha < \frac{1}{2} < \beta < 1 < a$. Since $0 < \alpha < \frac{1}{2} < \beta < 1$, we have $0 < \alpha + \beta < 2$ and $0 < \alpha \beta < \frac{1}{2}$. Therefore, $-\alpha(2 \sqrt{2})^{2}=8$, so $\alpha^{2}+\beta^{2}<1$.

Define $\mu_{n}=\alpha^{n}+\beta^{n}+a^{n}(n=0,1,2, \cdots)$, then $\mu_{0}=3, \mu_{1}=2+\beta+a=3, \mu_{2}=\alpha^{2}+\beta^{2}+a^{2}=9$, $\mu_{n+3}=3 \mu_{n+2}-\mu_{n}(n \geqslant 0)$.

Since $0<\alpha^{2}+\beta^{2}<1$, we have $a^{n}=\mu_{n}-\left(\alpha^{n}+\beta^{n}\right)=\mu_{n}-1-\left[1-\left(\alpha^{n}+\beta^{n}\right)\right]$. Therefore, $\left[a^{n}\right]=\mu_{n}-1 .(n=1,2, \cdots)$, so $17\left|\left[a^{n}\right] \Leftrightarrow 17\right| \mu_{n}-1$.

Now consider the sequence $\left\{\mu_{n}\right\}$ modulo 17: $\mu_{0} \equiv 3, \mu_{1} \equiv 3, \mu_{2} \equiv 9, \mu_{3} \equiv 7, \mu_{4} \equiv 1, \mu_{5} \equiv 11, \mu_{6} \equiv 9, \mu_{7} \equiv 9, \mu_{8} \equiv 16, \mu_{9} \equiv 5, \mu_{10} \equiv 6, \mu_{11} \equiv 2, \mu_{12} \equiv 1, \mu_{13} \equiv 14, \mu_{14} \equiv 6, \mu_{15} \equiv 0, \mu_{16} \equiv 3, \mu_{17} \equiv 3, \mu_{18} \equiv 9, \cdots$ It is clear that modulo 17, $\left\{\mu_{n}\right\}$ is a periodic sequence with a period of 16, i.e., $\mu_{n+16} \equiv \mu_{n}(\bmod 17)$, hence $17!\left[a^{n}\right] \Leftrightarrow n \equiv 4$ or $12(\bmod 16)$. In the range $1 \leqslant n \leqslant 2006$, there are $\left[\frac{2006}{16}\right] \times 2+1=251$ values of $n$ that satisfy $n \equiv 4$ or $12(\bmod 16)$.