Let 20 15 be in the nth group, then1+3+5+7+…+(2n-1) ≥1008,

That is (1+2n-1) n/2 ≥1008,

Solution: n≥3 1.7,

When n=32,1+3+5+7+…+63 =1024;

So the number 1008 is in the 32nd group.

Numbers1024: 2×1024-1= 2047,

The first number in the 32nd group is 2×962- 1= 1923.

Then 20 15 is (2015-1923)/2+1= 47.

Therefore, a2015 = (32,47).