Example 21 (Mathematics Retest Question for Science Experimental Class Admission in 1989) The general term of the sequence is $a_{n}=b[\sqrt{n+c}]+d$, and the terms are calculated successively as
$$
1,3,3,3,5,5,5,5,5, \cdots
$$

where each positive odd number $m$ appears exactly $m$ times consecutively. The values of $b, c, d$ are to be determined. Find the value of $b+c+d$.
Given that $a_{n+1}-a_{n}=b[\sqrt{n+1+c}]-b[\sqrt{n+c}]$, and $a_{n}$ is an odd number, $a_{n+1} \geqslant a_{n}$, then $a_{n+1}-a_{n} \in\{0,2\}$, i.e., $b[\sqrt{n+1+c}]-b[\sqrt{n+c}]=0$ or 2.
For any natural number $n$, we always have
$$
[\sqrt{n+1+c}]-[\sqrt{n+c}] \in\{0,1\} \text {. }
$$

Clearly, $b \neq 0$.
When $a_{n+1}-a_{n}=2$, i.e., $b([\sqrt{n+1+c}]-[\sqrt{n+c}])=2$, by (1), we can only have $[\sqrt{n+1+c}]-[\sqrt{n+c}]=1$.
Therefore, $b=2$.
Since $b$ is a constant, $b=2$.
Next, we find $c$.
From $a_{1}=b[\sqrt{1+c}]+d=2[\sqrt{1+c}]+d$, we know that $c \geqslant-1$.
By the characteristics of the sequence given in the problem, we have $a_{k}^{2}=2 k-1$.
Taking a sufficiently large $k$ such that $2 k+3>c$, at this time
$(k+1)^{2}+c<(k+1)^{2}+2 k+3=(k+2)^{2}$, thus
$\left[\sqrt{(k+1)^{2}+c}\right]<k+2$.
On the other hand, we also have $a_{(k+1)^{2}}=2(k+1)-1$, then
$a_{(k+1)^{2}}-a_{1}=[2(k+1)-1]-1=2 k$, i.e.,
$2\left[\sqrt{(k+1)^{2}+c}\right]-2[\sqrt{1+c}]=2 k$, i.e.,
$\left[\sqrt{(k+1)^{2}+c}\right]-[\sqrt{1+c}]=k$.
By inequality (2), we have $[\sqrt{1+c}]<2$, thus $c<3$.
Therefore, $-1 \leqslant c<3, c=-1,0,1,2$.
If $c=0,1,2$, then in all cases,
$a_{5}-a_{4}=[\sqrt{5+c}]-[\sqrt{4+c}]=2-2=0$, which contradicts $a_{5}-a_{4}=2$.
Thus, we can only have $c=-1$.
Next, we find $d$.
From $a_{1}=2[\sqrt{1+c}]+d=1$, we know that $d=1$.
Therefore, $b+c+d=2+(-1)+1=2$.