Example 5 "Sieve" the sequence $\left\{a_{n}\right\}$ from the set of natural numbers $\mathbf{N}$: Let $a_{1}=1$, remove $b_{1}=$ $a_{1}+k$, where $k$ is some natural number; then take $a_{2}$ as the smallest number in $\mathbf{N}$ except for $a_{1}, b_{1}$, remove $b_{2}=a_{2}+2 k$; take $a_{3}$ as the smallest number in $\mathbf{N}$ except for $a_{1}, a_{2}, b_{1}, b_{2}$, remove $b_{3}=a_{3}+3 k$, and so on, to obtain the sequence $\left\{a_{n}\right\}$, find the expression for $a_{n}$.
Given the construction of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, they are known to be $\mathbf{N}-$ complementary. The uniqueness of $\left\{a_{n}\right\}$ is evident. Using a constructive method, we can give it a form and find its general term. Let $a_{n} = [n \alpha] (\alpha > 0, \alpha \notin \mathbf{Q})$, then $b_{n} = a_{n} + n k = [n \alpha] + n k = [n (\alpha + k)]$. Let $\beta = \alpha + k \notin \mathbf{Q}$. From $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, we find $\alpha = \frac{1}{2} \left(2 - k + \sqrt{k^2 + 4}\right) > 0$. Since $k > 0$ is a natural number, $k^2 + 4$ is not a perfect square, thus $\alpha$ is a positive irrational number. By Example 1, $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are complementary. $a_{n} = [n \alpha]$ is the desired sequence.

From the theorem, we can formulate and solve the following problems:
(1) Any arithmetic sequence with a common difference $d = k - 1$ and $a_{1} \in \mathbf{N}$ has a $\mathbf{N}$-complementary sequence of the form $b_{n} = n + [\varphi(n)]$ (take $\varphi(x) = (k - 1) x + a_{1} - k + \alpha$, where $\alpha > 0$ is an irrational number).
(2) Given the sequence $\left\{a_{n}\right\}$ with the general term $a_{n} = n + \left[\csc \frac{\pi}{2(n+1)}\right]$, and the sequence $\left\{b_{n}\right\}$ with the general term $b_{n} = n - 1 + \left\lceil \frac{2}{\pi} \arcsin \frac{1}{n} \right\rceil$, then $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are $\mathbf{N}$-complementary.
(3) The sequence $\left\{a_{n}\right\}$ has the general term $a_{n} = 3n^2 - 2n$, and the sequence $\left\{b_{n}\right\}$ has the general term $b_{n} = n + \left[\sqrt{\frac{n}{3}} + \frac{1}{2}\right] (n \in \mathbf{N})$, then $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are $\mathbf{N}$-complementary (take $\varphi(x) = \sqrt{\frac{x}{3}} + \frac{1}{2}$).
(4) Given the sequence $\left\{a_{n}\right\}$ with the general term $a_{n} = n + \left[\sqrt{3n} + \frac{1}{2}\right]$, and the sequence $\left\{b_{n}\right\}$ with the general term $b_{n} = \left[\frac{n^2 + 2n}{3}\right]$, then $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are $\mathbf{N}$-complementary.
(Hint: Take $\varphi(x) = \sqrt{3x} + \frac{1}{2}$, then $\varphi^{-1}(x) = \frac{1}{3} \left(x - \frac{1}{2}\right)^2$. At this time, $b_{n} = n + \left[\varphi^{-1}(n)\right] = \left[n + \frac{n^2 - n}{3} + \frac{1}{12}\right] = \left[\frac{n^2 + 2n}{3} + \frac{1}{12}\right] = \left[\frac{n^2 + 2n}{3}\right]$. It is clear that the constant $\frac{1}{2}$ in $\varphi(x)$ can be replaced by any positive number less than 1.)