【Analysis】For $p \mid n, p$ being a prime, consider $a_{1}, a_{2}, \cdots, a_{p-1}$.
If some of the $a_{i} \equiv 0,1(\bmod p)$, then
$a_{i+1}=2 a_{i}^{2}-1 \equiv \pm 1(\bmod p)$,
$a_{i+2} \equiv 1(\bmod p)$.
It can be deduced that $a_{p} \equiv \pm 1(\bmod p)$.
Thus, $p$ † $a_{p}$.
Also, $a_{k p} \equiv 1(\bmod p)$, so $a_{n} \equiv 1(\bmod p)$.
Therefore, $p \nmid a_{n}$.
If $a_{1}, a_{2}, \cdots, a_{p-1}$ have no remainders 0 and 1 modulo $p$, then there must be
$a_{i} \equiv a_{j}(\bmod p)(i \neq j)$.
Thus, $a_{i+1} \equiv a_{j+1}(\bmod p)$, and $a_{i}, a_{i+1}, \cdots, a_{p-2}, a_{p-1}$ are not divisible by $p$, with a congruence period of $j-i$.
Hence, $a_{n} \equiv a_{k(j-i)+r} \equiv a_{r}(\bmod p)(0 \leqslant r \leqslant j-i-1)$.
If $r=0$, then $a_{n} \equiv a_{j-i}(\bmod p)$.
Therefore, $a_{n} \neq \equiv(\bmod p)$.
If $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$, then $p_{i} \nmid a_{n}$.
Thus, $\left(n, a_{n}\right)=1$.