Consider the non-decreasing sequence of odd integers
$\left(a_{1}, a_{2}, a_{3}, \cdots\right)=(1,3,3,5,5,5,5,5, \cdots)$, where each positive odd integer $k$ appears $k$ times. It is known that there exist integers $b$, $c$, $d$ such that for all positive integers $n$, $a_{n}=b[\sqrt{n+c}]+d$. Find $b, c, d$.
Solution: Group the sequence $\left\{a_{n}\right\}$ as follows:
$$
\left(a_{1}\right),\left(a_{2}, a_{3}, a_{4}\right),\left(a_{5}, a_{6}, a_{7}, a_{8}, a_{9}\right), \cdots,
$$

where the $k$-th group has $2 k-1$ elements. From the properties of the original sequence, it is known that each term in the $k$-th group is $2 k-1, k=1,2, \cdots$, and that $a_{n}$ is an element of the $k$-th group if and only if $(k-1)^{2}<n \leqslant k^{2}$. Therefore, $a_{n}$ is an element of the $k$-th group if and only if $(k-1)^{2} \leqslant n-1 \leqslant k^{2}-1$, which means that when $k-1=[\sqrt{n-1}]$, the value of $a_{n}$ is $2 k-1$.

In summary, the general term of the sequence $\left\{a_{n}\right\}$ is $a_{n}=2[\sqrt{n-1}]+1, n=1,2, \cdots$, so $b=2, c=-1$, $d=1$.