6. The sequence $\left\{x_{n}\right\}: 1,3,3,3,5,5,5,5,5, \cdots$ is formed by arranging all positive odd numbers in ascending order, and each odd number $k$ appears consecutively $k$ times, $k=1,3,5, \cdots$. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$, then $a+b+c$ $+d=$
6. 3 Detailed Explanation: From $x_{k}^{2}+1=x_{k}^{2}+2=\cdots=x_{(k+1)^{2}}=2k+1$, that is, when $k^{2}+1 \leqslant n \leqslant (k+1)^{2}$, $x_{n}=2k+1$.
$k=[\sqrt{n-1}]$, so $x_{n}=2[\sqrt{n-1}]+1$, thus, $(a, b, c, d)=(2,1,-1,1), a+b+c+d=3$.