4. First, prove a lemma using mathematical induction.

Lemma A non-negative integer appears in the sequence if and only if its ternary expansion contains only 0 and 1.
Proof It is obvious that the proposition holds for 0.
Assume the proposition holds for all non-negative integers less than $N$, and consider $N$.

If the ternary expansion of $N$ contains the digit 2, replace all occurrences of 2 with 0 to get the number $N_{0}$; replace all occurrences of 2 with 1 to get the number $N_{1}$. Thus, the ternary expansions of $N_{0}$ and $N_{1}$ do not contain the digit 2.
By the induction hypothesis, the numbers $N_{0}$ and $N_{1}$ are in the sequence.
Since $N_{0}$, $N_{1}$, and $N$ form an arithmetic sequence, $N$ does not appear in the sequence. If the ternary expansion of $N$ does not contain the digit 2, then we need to prove that $N$ must appear in the sequence. If not, then there exist terms $N_{0}$ and $N_{1}$ in the sequence such that $N_{0}$, $N_{1}$, and $N$ form an arithmetic sequence. Let the common difference be $d$, and $3^{k} \| d$. Then the ternary expansions of these three numbers have the same lowest $k-1$ digits, and the $k$-th digit is different for each pair, so one of the numbers must have a 2 in the $k$-th digit, which contradicts the assumption.
Returning to the original problem.
We only need to find the 2014th non-negative integer whose ternary expansion does not contain the digit 2.

Notice that under this restriction, the ternary carry-over method is equivalent to binary, so we just need to write 2013 (note $a_{1}=0$) in binary $(11111011101)_{2}$ and then convert it to ternary $a_{2014}=88327$.