Example 9 The sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies $a_{0}=1, a_{1}=1, a_{n+1}=2 a_{n}+(a-1) a_{n-1}(n \in \mathbf{N}), a \in \mathbf{N}$ is a parameter. Let $p_{0}>2$ be a given prime. Find the smallest value of $a$ that satisfies the following two conditions.
(i) If $p$ is a prime and $p \leqslant p_{0}$, then $p \mid a_{p}$.
(ii) If $p$ is a prime and $p>p_{0}$, then $p \nmid a_{p}$.
(23rd IMO Shortlist)
Given the conditions of the problem,
$$
a_{n}=\frac{1}{2 \sqrt{a}}\left[(1+\sqrt{a})^{n}-(1-\sqrt{a})^{n}\right] \text {. }
$$

Noting that $p$ is a prime number, then
$$
p \mid \mathrm{C}_{p}^{k}(k=1,2, \cdots, p-1) .
$$

By the binomial theorem, we have
$$
\begin{aligned}
a_{p} & =\frac{1}{2 \sqrt{a}}\left[(1+\sqrt{a})^{p}-(1-\sqrt{a})^{p}\right] \\
& =\sum_{i=0}^{p-1} C_{p}^{2 i+1} a^{i} \\
& \equiv a^{\frac{p-1}{2}}(\bmod p) .
\end{aligned}
$$

When $p=2$,
$$
\begin{aligned}
a_{2} & =\frac{1}{2 \sqrt{a}}\left[(1+\sqrt{a})^{2}-(1-\sqrt{a})^{2}\right] \\
& =2 \equiv 0(\bmod 2) .
\end{aligned}
$$

Therefore, the necessary and sufficient condition for (i) and (ii) to hold is that $a$ is divisible by every prime $p$ satisfying $0 < p \leqslant p_{0}$, but not by any prime $p$ greater than $p_{0}$. The natural number $c$ that satisfies this property is the product of all primes $p$ satisfying $3 \leqslant p \leqslant p_{0}$.