$14 \cdot 37$ 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 integers $b, c, d$ are to be determined. Find the value of $b+c+d$.
(Advanced Math Test for Science Experimental Class Admission, 1989)
[Solution] Since
$$
a_{n+1}-a_{n}=b[\sqrt{n+1+c}]-b[\sqrt{n+c}] \text {, }
$$

and $a_{n}$ is odd, $a_{n+1} \geqslant a_{n}$, then
$$
a_{n+1}-a_{n} \in\{0,2\} \text {. }
$$

That is,
$$
b[\sqrt{n+1+c}]-b[\sqrt{n+c}]=0 \text { or } 2 \text {. }
$$

For any natural number $n$, we always have
$$
[\sqrt{n+1+c}]-[\sqrt{n+c}] \in\{0,1\} .
$$

Clearly, $b \neq 0$.
When $a_{n+1}-a_{n}=2$, that is,
$$
b([\sqrt{n+1+c}]-[\sqrt{n+c}])=2
$$

by (1), we can only have
Therefore
$$
\begin{array}{c}
{[\sqrt{n+1+c}]-[\sqrt{n+c}]=1 .} \\
b=2 .
\end{array}
$$

Since $b$ is a constant, $b=2$.
Next, we find $c$.
By
$$
\begin{aligned}
a_{1} & =b[\sqrt{1+c}]+d \\
& =2[\sqrt{1+c}]+d
\end{aligned}
$$

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
$$
a_{(k+1)^{2}}-a_{1}=[2(k+1)-1]-1=2 k .
$$

That is, $2\left[\sqrt{(k+1)^{2}+c}\right]-2[\sqrt{1+c}]=2 k$.
$$
\left[\sqrt{(k+1)^{2}+c}\right]-[\sqrt{1+c}]=k .
$$

By inequality (2), we have
and
$$
[\sqrt{1+c}]<2,
$$

therefore
$$
c<3 \text {. }
$$

Thus, $\quad-1 \leqslant c<3, \quad c=-1,0,1,2$.
If $c=0,1,2$, then in all cases
$$
\begin{array}{c}
a_{5}-a_{4}=[\sqrt{5+c}]-[\sqrt{4+c}]=2-2=0 \\
a_{5}-a_{4}=2
\end{array}
$$

which is a contradiction.
Therefore, we can only have $c=-1$.
Next, we find $d$.
By
$$
a_{1}=2[\sqrt{1+c}]+d=1
$$

we know that
$$
d=1 \text {. }
$$

Therefore, $\quad b+c+d=2+(-1)+1=2$.