13. If the function $f$ satisfies $f\left(n^{2}\right)=f(n)+2, n \geqslant 2$ and $f(2)=1$, then, $f(256)=(\quad)$.
(A) 3
(B) 5
(C) 7
(D) 9
$(\mathrm{E}) *$
13. (C).

From $f\left(n^{2}\right)=f(n)+2$, we have $f\left(n^{2^{4}}\right)=f\left(n^{2^{t-1}}\right)+2$, for $\forall k \in \mathbf{N}$. Then $f\left(n^{2^{2}}\right)=f(n)+2 k$. Therefore, $f(256)=f\left(2^{2^{3}}\right)=f(2)+2 \times 3=7$.