4. The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$
a_{n}=2^{n}+3^{n}+6^{n}-1(n=1,2,3, \cdots) \text {. }
$$

Find all positive integers that are coprime with every term of this sequence.
(Poland provided)
4. The only positive integer that satisfies the condition is 1.

The following is the proof: For any prime number $p$, it must be a divisor 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
$$
(2, p)=1,(3, p)=1,(6, p)=1 \text {, }
$$

by Fermat's Little Theorem, we have
$$
\begin{array}{l}
2^{p-1} \equiv 1(\bmod p), \\
3^{p-1} \equiv 1(\bmod p), \\
6^{p-1} \equiv 1(\bmod p) .
\end{array}
$$

Then $3 \times 2^{p-1}+2 \times 3^{p-1}+6^{p-1}$
$$
\equiv 3+2+1=6(\bmod p) \text {, }
$$

i.e., $6 \times 2^{p-2}+6 \times 3^{p-2}+6 \times 6^{p-2} \equiv 6(\bmod p)$.
Thus, $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0(\bmod p)$.
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 $\left(n, a_{p-2}\right)>1$.
Therefore, 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 is 1.
Note: The average score for this problem is 3.78.