Example 6 In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5,5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$ such that for all integers $n$, $a_{n} = b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$. Then $b+c+d$ equals ( ).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(1980, American Mathematics Competition)
Explanation: Divide the sequence $\{1,3,3,3,5,5,5,5,5, \cdots\}$ into groups:
(1), $(3,3,3),(5,5,5,5,5), \cdots,(2 k-1, 2 k-1, \cdots, 2 k-1), \cdots$,
where the $k$-th group consists of $2 k-1$ occurrences of $2 k-1$. If $a_{n}$ is in the $k$-th group, then $a_{n}=2 k-1$. Therefore,
$$
\begin{array}{l}
1+3+\cdots+(2 k-3)+1 \\
\leqslant n0$, solving this gives
$$
\sqrt{n-1}<k \leqslant \sqrt{n-1}+1 \text {. }
$$

Since $k \in \mathbf{Z}_{+}$, and
$$
\sqrt{n-1}+1-\sqrt{n-1}=1 \text {, }
$$

we have $k=[\sqrt{n-1}+1]=[\sqrt{n-1}]+1$.
Thus, $a_{n}=2 k-1=2[\sqrt{n-1}]+1$.
Hence, $b=2, c=-1, d=1$.
Therefore, $b+c+d=2+(-1)+1=2$.
Note: The 1981 Austrian Mathematical Competition, the second part of the second question, is another form of this problem:

Given $\left\{a_{1}, a_{2}, \cdots\right\}=\{1,3,3,3,5,5,5,5, 5, \cdots\}$, i.e., each odd number $k$ appears $k$ times. Prove: There exist integers $b, c, d$ such that $a_{n}=b[\sqrt{n+c}]+d(n \geqslant 1)$, and there is only one set of values for $b, c, d$ that satisfies the equation.