3Ask1 * The sequence $a_{1}, a_{2}, \cdots$ is defined as follows: $a_{n}=2^{n}+3^{n}+6^{n}-1(n=1,2,3, \cdots)$. Find all positive integers that are coprime to every term of this sequence.
The only positive integer that satisfies the condition is 1.
The following is the proof: For any prime number $p$, it must be a factor of some term in the sequence $\left\{a_{n}\right\}$.
For $p-2$ and $p-3$, they are divisors of $a_{2}=48$.
For every prime number $p$ greater than 3, since $(2, p)=1,(3, p)=1,(6, p)=1$, by Fermat's Little Theorem, $2^{p-1} \equiv 1(\bmod p), 3^{p-1} \equiv 1(\bmod p), 6^{p-1}=1(\bmod p)$, so
$$
3 \cdot 2^{p-1}+2 \cdot 3^{p-1}+6^{p-1} \equiv 3+2+1=6(\bmod p),
$$

which means $6 \cdot 2^{p-2}+6 \cdot 3^{p-2}+6 \cdot 6^{p-2} \equiv 6(\bmod p)$, hence $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv$ $0(\bmod p)$, that is, $p \mid a_{p-2}$.

For any positive integer $n$ greater than 1, it must have a prime factor $p$. If $p \in\{2,3\}$, then $\left(n, a_{2}\right)>1$; if $p \geqslant 5$, then $\left(n, a_{p-2}\right)>1$, so all positive integers greater than 1 do not meet the requirement.
Since 1 is coprime with all positive integers, the only positive integer that satisfies the condition can only be 1.