7. The general term of the sequence is $a_{\mathrm{n}}=b[\sqrt{n+c}]+d$, and the terms are calculated successively as
$$
1,3,3,3,5,5,5,5,5, \cdots \text {. }
$$

where each positive odd number $m$ appears exactly $m$ times consecutively. The above $b, c, d$ are undetermined integers. Then, the value of $b+c+d$ is $\qquad$ where $[x]$ denotes the greatest integer not exceeding $x$.
Solve: First determine $b$.
Since $a_{n}$ is odd, and $a_{n}+1 \geqslant a_{n}$, we know that $a_{n}+1-a_{n}$ $\in\{0,2\}$,
i.e., $b[\sqrt{n+1+c}]-b[\sqrt{n+c}]$ is 0 or 2.
For any natural number $n$, it always holds that
$[\sqrt{n+1+c}]-[\sqrt{n+c}] \in\{0,1\}$.
Clearly, $b \neq 0$.
When $-[\sqrt{n+c}])=2$, by (1), it can only be that $[\sqrt{n+1+c}]-[\sqrt{n+c}]=1, b=2$.
Since $b$ is a constant, we have $b=2$.
Next, find $c$.
From $=2[\sqrt{u+c}]+d$ we know $c \geqslant-1$.
Based on the characteristics of the sequence $a_{\mathrm{n}}$, we have $a_{k}^{2}=2 k-1$.
Taking a sufficiently large $k$, ensuring $2 k+3>c$, at this time $(k+1)^{2}+c<(k+1)^{2}+2 k+3=(k+2)^{2}$; we get $\left[\sqrt{(k+1)^{2}+c}\right]<k+2$. $-1=2 k$ or
$$
\left[\sqrt{(k+1)^{2}+c}\right]-[\sqrt{1+c}] \equiv k \text {. }
$$

Substituting (2) gives: $[\sqrt{1+c}]<2$.
Therefore, $c<3$.

If $c=0,1,2$, then in each case $[\sqrt{5+c}]-[\sqrt{4+c}]$
$=2-2=0$ which contradicts $a_{5}-a_{4}=2$.
In summary, $c=-1$.
Finally, $d=a_{1}=1$.
Thus, $b+c+a'=2$.