The first is to precalculate lowest divisor of each number. It can be done with simple Eratosphenes sieve, which has time complexity $$$O(N \log N)$$$ if you do sieve loop for all numbers, and $$$O(N \log \log N)$$$ if you do loop only when number is prime. It can also be done with $$$O(N)$$$ time and memory using linear sieve.

Second part is the answer calculation, you need to add $$$2^k$$$, where $$$k$$$ -- number of different prime divisors. So if you haven't precalucated it on sieve step, it will be $$$O(\log x)$$$ per divisor, otherwise it will be $$$O(1)$$$.