6. Let $r, s, t$ be integers, and the set $\left\{a \mid a=2^{r}+2^{s}+2^{t}, 0 \leqslant t<s<r\right\}$ consists of numbers that form the sequence $\left\{a_{n}\right\}: 7,11,13,14, \cdots$, then $a_{36}=$ $\qquad$
6. 131 Detailed Explanation: $\because r, s, t$ are integers and $0 \leqslant t<s<r, \therefore r$ can take the smallest value of 2, at which point the number of valid combinations is $C_{2}^{2}=1$.
$r=3, s, t$ can be chosen from $0,1,2$, the number of valid combinations is $C_{3}^{2}=3$. Similarly, when $r=4$, the number of valid combinations is $C_{4}^{2}=6$, when $r=5$, the number of valid combinations is $C_{5}^{2}=10$, when $r=6$, the number of valid combinations is $C_{6}^{2}=15$, and when $r=7$, the number of valid combinations is $C_{7}^{2}=21$. Therefore, $a_{36}$ is the smallest value in the case of $r=7$, i.e., $a_{36}=2^{0}+2^{1}+2^{7}=131$.