We can see that all numbers in given sequence are distinct. since all numbers in sequence are of the form . Hence if ai = aj then = = 2x - m = 3y - n which is not possible because any power of 2 will be an even number and any power of 3 will be an odd number.
if there exists in sequence then 2 * ai can not be in sequence and vice versa. We can prove it using contradiction. Suppose there is a number ai such that both and 2 * ai exists in sequence. by little bit of tricks this = , this again is not possible by same argument as above, we just have to change the order of exponents.
Hence for each ai in sequence we see if or 2 * ai is present(remember that only one of them can be present). Now if there is any ai such that both and 2 * ai is not in sequence then this should be an. if there is any such ai that for all 0 ≤ j ≤ n: ≠ ai AND 2 * ai ≠ ai, then this is a1.
Once you have got a1, you keep on producing sequence just by doing binary search for and 2 * ai (remember only one of them is present so once you find it you print it).