Question 4: Call an $n$-tuple $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ "intermittently periodic" if there exist non-negative integers $i$ and positive integers $p$ such that $i+2 p \leqslant n$, and for all $j=1,2, \cdots, p$,
$$
a_{i+j}=a_{i+p+j}.
$$

Let $k$ be a given positive integer. Find the smallest positive integer $n$ such that the $n$-tuple $\left(a_{1}, a_{2}, \cdots, a_{n}\right)\left(a_{i} \in\{1,2, \cdots, k\}\right)$ is not intermittently periodic, but its extension $\left(a_{1}, a_{2}, \cdots, a_{n+1}\right)$ is intermittently periodic for any $a_{n+1} \in\{1,2, \cdots, k\}$.
(2015, Estonian National Team Selection Exam)
Solve for the smallest positive integer $n=2^{k}-1$.
Let $p$ be the period length of the array.
Suppose the period length of $a_{n+1}=i(i=1,2, \cdots, k)$ is $p_{i}$. If $i_{1} \neq i_{2} \in\{1,2, \cdots, k\}$, then $p_{i_{1}} \neq p_{i_{2}}$. Otherwise, $i_{1}=a_{n+1}=a_{n+1-p_{n_{1}}}=a_{n+1-p_{n}}=a_{n+1}^{\prime}=i_{2}$, which is a contradiction.
Therefore, without loss of generality, assume $p_{i_{1}}p_{j}$, we have $p_{i} \geqslant 2 p_{j}(1 \leqslant i \neq j \leqslant k)$. It is easy to see that the minimum period length
\[
\min \left\{p_{1}, p_{2}, \cdots, p_{k}\right\}=1 \text{. }
\]

At this point, $a_{n+1}=a_{n}$.
Thus, $p^{\prime}=\max \left\{p_{1}, p_{2}, \cdots, p_{k}\right\} \geqslant 2^{k-1}$.
Therefore, $n+1 \geqslant 2 p^{\prime} \geqslant 2 \times 2^{k-1} \Rightarrow n \geqslant 2^{k}-1$.
We will use mathematical induction to provide a specific construction for $n=2^{k}-1$.
When $k=1$, the arrays $(1)$ and $(1,1)$ satisfy the conditions.
Assume that when $k-1$, the array $\left(a_{1}, a_{2}, \cdots, a_{n_{0}}\right)$ meets the requirements.
Then $n_{0}=2^{k-1}-1$.
At this point, denote this array as $L_{k-1}$.
Now consider the case for $k$.
Consider the array $M_{k}=\left(L_{k-1}, k, L_{k-1}\right)$.
Clearly, $M_{k}$ is not periodically intermittent, and for any $i \in\{1,2, \cdots, k-1\}$, by the induction hypothesis,
\[
\left(M_{k}, i\right)=\left(L_{k-1}, k, L_{k-1}, i\right)
\]

is periodically intermittent;
When $i=k$,
\[
\left(M_{k}, k\right)=\left(L_{k-1}, k, L_{k-1}, k\right)
\]

is the periodically intermittent array with the largest period length.
Thus, the array $M_{k}$ meets the requirements.
At this point, $n=\left(2^{k-1}-1\right) \times 2+1=2^{k}-1$.
This provides a specific construction for $n=2^{k}-1$.
In summary, the minimum value of $n$ is $2^{k}-1$.