Clever use of mutual recursion, almost self-explained! I reduce the base case a little bit and add some comments for easier understand.

public int lastRemaining(int n) {
return leftToRight(n);
}
// eliminate [1...n] first from left to right, then alternate
private int leftToRight(int n) {
if (n == 1) return 1;
// scan from left to right is simple, the length of array doesn't matter
// [1, 2, 3, 4] -> 2 * [1, 2]
// [1, 2, 3, 4, 5] -> 2 * [1, 2]
return 2 * rightToLeft(n / 2);
}
// eliminate [1...n] first from right to left, then alternate
private int rightToLeft(int n) {
if (n == 1) return 1;
// if the length of array is even, we will get only odd number
// [1, 2, 3, 4] -> [1, 3] = 2 * [1, 2] - 1
if (n % 2 == 0) return 2 * leftToRight(n / 2) - 1;
// else if the length of array is odd, we will get only even number
// [1, 2, 3, 4, 5] -> [2, 4] = 2 * [1, 2]
else return 2 * leftToRight(n / 2);
}