, 1997

We have $a_{2015} = 4531$.
We claim that
$$a_{4k+1} = 9k + 1,\quad a_{4k+2} = 9k + 3,\quad a_{4k+3} = 9k + 4,\quad a_{4k+4} = 9k + 7$$
for any integer $k \ge 0$. The base cases $k = 0, 1$ can be verified directly. Indeed, the first 8 terms are
1, 3, 4, 7, 10, 12, 13, 16.
Now, assume the claim holds for $k = 0, 1, \dots, n-1$. Observe that none of the previous terms is congruent to 2 modulo 3, and all numbers congruent to 1 modulo 3 less than $9n$ has appeared. Consider the case $k = n$.
* Since $(9n-1)+4 = 3(3n+1)$ and $9n+3 = 3(3n+1)$, we have $a_{4n+1} \ne 9n-1, 9n$. If $(9n+1)+a_i = 3a_j$, then $a_i \equiv 2 \pmod 3$, contradiction. Thus, $a_{4n+1} = 9n+1$.
* Since $(9n+2)+1 = 3(3n+1)$, we have $a_{4n+2} \ne 9n+2$. If $(9n+3)+a_i = 3a_j$, then $3 \mid a_i$. By the inductive hypothesis, we must have $a_i = 9t+3$ for some integer $t$. Then we have $a_j = 3n+3t+2 \equiv 2 \pmod 3$, contradiction. Thus, $a_{4n+2} = 9n+3$.
* If $(9n+4)+a_i = 3a_j$, then $a_i \equiv 2 \pmod 3$, contradiction. Thus, $a_{4n+3} = 9n+4$.
* Since $(9n+5)+7 = 3(3n+4)$ and $(9n+6)+(9n+6) = 3(6n+4)$, we have $a_{4n+1} \ne 9n+5, 9n+6$. If $(9n+7)+a_i = 3a_j$, then $a_i \equiv 2 \pmod 3$, contradiction. Thus, $a_{4n+4} = 9n+7$.
This proves our claim by induction. Therefore,
$$a_{2015} = a_{4 \times 503 + 3} = 9 \times 503 + 4 = 4531.$$