5. 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
5.2.

Divide the known sequence into groups as follows:
$$
\begin{array}{l}
(1),(3,3,3),(5,5,5,5,5), \cdots, \\
(\underbrace{2 k-1,2 k-1, \cdots, 2 k-1}_{2 k-1 \uparrow}),
\end{array}
$$

Let $a_{n}$ be in the $k$-th group, where $a_{n}=2 k-1$. Then we have
$$
\begin{array}{l}
1+3+5+\cdots+2 k-3+1 \\
\leqslant n0$, solving this gives $\sqrt{n-1}<k \leqslant \sqrt{n-1}+1$.
Therefore, $k=[\sqrt{n-1}+1]=[\sqrt{n-1}]+1$.
Thus, $a_{n}=2[\sqrt{n-1}]+1$.
Hence, $b+c+d=2+(-1)+1=2$.