5. 85 If all coefficients of a polynomial are equal to $0,1,2$, or 3, then 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. 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 $\left[\frac{n}{2}\right]+1$.