3. For an integer $m$, its unit digit is denoted by $f(m)$, and let $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$. Then $a_{2006}$ $=$ . $\qquad$
3.7.

It is known that $f\left(2^{m}\right)=f\left(2^{m+4}\right)$. Therefore,
$$
a_{2 \omega 0}=f\left(2^{2 \omega 1}-1\right)=f\left(2^{2 \omega 07}\right)-1=f(8)-1=7 .
$$