1. Let the sets be
$$
\begin{aligned}
A & =\{n(n+1) \mid n=1,2, \cdots\}, \\
B & =\{3 m-1 \mid m=1,2, \cdots\} .
\end{aligned}
$$

If the elements of the set $A \cap B$ are arranged in ascending order to form a sequence $\left\{a_{k}\right\}$, then the general term formula for the sequence $\left\{a_{k}\right\}$ is $a_{k}=$ $\qquad$
$$
-1.9 k^{2}-9 k+2 \text {. }
$$

It is easy to see that, $A=\{2,6,12,20,30,42, \cdots\}$,
$$
B=\{2,5,8,11,14,17,20, \cdots\} \text {. }
$$

If $n(n+1)=3 m-1$
$$
\Rightarrow(2 n+1)^{2}=3(4 m-1) \text {. }
$$

Thus, $4 m-1$ is three times some odd square number.
Let $4 m-1=3(2 k-1)^{2}$. Then
$$
\begin{array}{l}
2 n+1=3(2 k-1) \\
\Rightarrow n=3 k-2, m=3 k^{2}-3 k+1(k=1,2, \cdots) .
\end{array}
$$

Therefore, $a_{k}=n(n+1)=3 m-1=9 k^{2}-9 k+2$.