Big O (time complexity) calculation

The outer loop for (int i = 0; i < N; i++) is O(N), after a constant threshold (1024 in your example) number of iterations are halved but this does not change complexity.

The important part of this loop for (int j = 0; j < N; j += j+1) is j += j+1 which can be evaluated to 2j + 1, after t iterations j = 2^t giving us O(logN).

Here for (int k = 0; k < N; k +=(i+j+5)) we look at k += (i+j+5)) which gives us O(N/(i+j+5)) however since j is growing exponentially the number of iterations this loop hits shrinks exponentially giving us O(logN) again this is only with taking into account values of j, by itself it is O(N/(i+j)).

So, overall the algorithm is O(N logN logN) = O(Nlog^{^2}N)

For each value of i and j, the value of k will increment approximately N/(i+j+5) times. Since values of j increase as powers of 2, altogether this gives ∑_i=0^N(∑_j=0^{log N} N/(i+2^j+5)) iterations. The sums over i and j can be interchanged, which gives ∑_j=0^{log N} (N(∑_i=0^N 1/(i+2^j+5))). The inner sum is a partial sum of the harmonic series, so it can be approximated by log(N + 2^j) - log(2^j) = log(1 + N/2^j) = log(1+2^{(log N - j)}) ≈ log(2^{(log N - j)}) = log N -j. The double sum is then approximately equal to N(∑_j=0^{log N} (log N -j)) ≈ N(log N)(log N + 1)/2 ≈Nlog(N)²/2. Which gives complexity O(Nlog(N)²).