$5 \cdot 85$ If all the coefficients of a polynomial are equal to $0, 1, 2$, or 3, the polynomial is called compatible. For a given natural number $n$, find the number of all compatible polynomials that satisfy the condition $p(2)=n$.


[Solution] Let $0 \leqslant m \leqslant\left[\frac{n}{2}\right]$, and
$$\begin{array}{l}
n-2 m=a_{0}+a_{1} \cdot 2+a_{2} \cdot 2^{2}+\cdots+a_{n} \cdot 2^{n} \\
m=b_{0}+b_{1} \cdot 2+b_{2} \cdot 2^{2}+\cdots+b_{n} \cdot 2^{n}
\end{array}$$

where $a_{i}, b_{i}$ all belong to $\{0,1\}$. Thus, for each value $m=0,1,2, \cdots,\left[\frac{n}{2}\right]$, there is a polynomial $p_{m}(x)=\sum_{i=0}^{n}\left(a_{i}+2 b_{i}\right) x^{i}$ corresponding to it, and clearly, each polynomial in $p_{m}(x)$ is compatible.

Conversely, each compatible polynomial is identical to one of the polynomials in $p_{m}(x)$.
Therefore, the number of all compatible polynomials that satisfy the conditions is $\left[\frac{n}{2}\right]+1$.