13. Let $N_{0}$ be the set of non-negative integers, and $f: N_{0} \rightarrow N_{0}$ be a function such that $f(0)=0$, and for any $n \in N_{0}$, we have $[f(2 n+1)]^{2}-[f(2 n)]^{2}=6 f(n)+1, f(2 n)>f(n)$. How many elements in $f\left(N_{0}\right)$ are less than $2004 ?$
13. From the given, we have $f(2 n+1)=f(2 n)+1, f(2 n)=3 f(n)$.

From the above, it is easy to prove by mathematical induction that $f(n)$ is strictly increasing. Since $f(128)=f\left(2^{7}\right)=3 f\left(2^{6}\right)=\cdots=3^{7} f(1)=$ $2187>2004$, while $f(127)=f(126)+1=3 \cdot f(63)+1=3 \cdot f(62)+4=9 \cdot f(31)+4=9 \cdot f(30)+13=$ $27 \cdot f(15)+13=27 \cdot f(14)+40=81 \cdot f(7)+40=81 \cdot f(6)+121=243 \cdot f(3)+121=243 \cdot$ $f(2)+364=729 \cdot f(1)+364=1093<2004$.
Therefore, there are 128 elements $f(0), f(1), f(2), \cdots, f(127)$ that do not exceed 2004.