4. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{2}=10$, for all positive integers $n$ we have $a_{n+2}=a_{n+1}-a_{n}$. Then $a_{2015}=$
4. -10 .

Upon inspection, the sequence $\left\{a_{n}\right\}$ has a period of 6. Therefore,
$$
a_{2015}=a_{335 \times 6+5}=a_{5}=-10 \text {. }
$$