12. Let $r, s, t$ be integers, and the set $\left\{a \mid a=2^{r}+\right.$ $2^{s}+2^{t}, 0 \leqslant t<s<r \mid$ consists of numbers arranged in ascending order to form the sequence $\left\{a_{n}\right\}: 7,11,13,14, \cdots$. Then $a_{34}=$ $\qquad$
12.131.

Since $r, s, 1$ are integers $\mathrm{HL}, 0 \leqslant 1<s<r$, then $r$ takes the smallest value 2. At this time, the numbers that meet the condition are $C_{2}^{2}=1$.

When $r=3$, $s, 1$ can be chosen from $0,1,2$, the numbers that meet the condition are $C_{i}^{3}=3$.
Similarly, when $r=4$, the numbers that meet the condition are $\mathrm{C}_{4}^{2}=6$; when $r=6$, the numbers that meet the condition are $C_{0}^{3}=15$; therefore, $a_{x}$ is the smallest value when $r=7$, i.e., $a_{x_{0}}=2^{\prime \prime}+2^{1}+2^{7}=131$.