Since ~a_{1 \cdot 1} = a_1+a_1~, ~a_1~ has to be equal to ~0~, so if the first element is changed the sequence cannot become logarithmic again. It is easily shown that ~a_{xy} = a_x+a_y~ for every pair of positive integers ~x, y~ is equivalent to ~a_m = \alpha_1 a_{p_1} + \alpha_2 a_{p_2} + \dots + \alpha_k a_{p_k}~ for positive integer ~m = p_1^{\alpha_1} \dots p_k^{\alpha_k}~. We conclude that logarithmic sequence is uniquely defined by values at prime indices. If the index ~p~ of changed element is prime one possible way to "fix" the sequence is to change all of the elements at indices which are multiples of ~p~, there are ~\lfloor \frac n p \rfloor-1~ of them. If that is the minimal number of changes necessary, then for any index ~x~ of changed element the minimal number of changes necessary will be ~\lfloor \frac{n}{\text{greatest prime factor of }x} \rfloor-1~ (since it is necessary to change at least one prime factor of ~x~).

We will prove that is the solution and we can implement it with standard algorithm for finding prime factorization in ~\mathcal O(q \sqrt n)~.

Proof: Using induction by ~n~ let's prove a stronger claim for odd primes ~p~: "Changing multiples of ~p~ is the only minimal solution".

If ~n~ is not divisible by ~p~ the number of changes stays the same and the solution is unique.

If ~n~ is divisible by ~p~ then clearly the solution for ~n-1~ needs to be expanded by changing the ~n~-th element and we get that ~\lfloor \frac n p \rfloor-1~ is the least number of changes. Assuming that there exists another minimal solution that solution doesn't change the ~n~-th element and it is necessary to change another odd prime index ~q~ (for ~q = 2~ it can be seperately shown it requires more than minimal changes) for which by induction hypothesis has unique solution for first ~n-~ element. Since ~2q~ is not divisible by ~p~ that solution would have at least ~\lfloor \frac n p \rfloor~ changes so it is not minimal.

For ~p = 2~ it doesn't necessarily hold that minimal solution is unique (for prime ~\frac n 3 < q \le \frac n 2~ it is possible to change ~q~ instead of ~2q~), however it still holds that the minimal number of changes necessary is ~\lfloor \frac n 2 \rfloor-1~. The proof similar to the previous is left for the reader as exercise as well as finding implementation which runs in time complexity of ~\mathcal O(n+q)~.