Let $$f(x)=c_0+c_1x+c_2x^2+\cdots +c_kx^k$$

and $c_i\in\{0,1,2,3\}$ where $i=0, 1, 2,\cdots, k$. Then, $f(2)=n$ is equivalent to $$c_0 + 2c_1 + 2^2c_2 +\cdots +2^kc_k=n$$

Note that every $2^ic_i$ can only take a value of $0$, $2^i$, $2\cdot 2^{i}$, or $3\cdot 2^i$. Thus, $a_n$ is the coefficient of the term $t^n$ in the following polynomial $$P(x)=\prod_{i=0}^{\infty}\left(1+t^{2^i} + t^{2\cdot 2^{i}} + t^{3\cdot 2^i}\right)$$

Applying the sum of geometric sequence formula gives $$\begin{align*} P(x)=\ &\frac{1-t^4}{1-t}\frac{1-t^8}{1-t^2}\frac{1-t^{16}}{1-t^4} \frac{1-t^{32}}{1-t^8}\cdots\\ =\ &\frac{1}{1-t}\frac{1}{1-t^2} \\=\ &\left(1+t+t^2+t^3+\cdots\right)(1+t^2+t^4+t^6+\cdots)\\=\ & 1+t +2t^2 + 2t^3 + 3t^4 + 3t^5 + 4t^6 +\cdots\end{align*}$$

Therefore, the final answer is $$a_n=\boxed{\left\lfloor\frac{n}{2}\right\rfloor + 1}$$