Time Complexity increasing by n^(1/3)log(n)

Question

int j = 2
while (j < n) {
    int k = j
    while (k < n) {
        Sum += a[k] * b[k]
        k += n^(1/3) log(n)
    }
    j = j * sqrt(5)
}

From the midterm practice

Answer 1

Outer loop Analysis (for j)

j goes from 2 to n, multiplied by a constant at ever stype

n = (2 * sqrt(5)) * sqrt(5)).... = 

Therefore, number of steps = 2log[5,(n/2)] = O(log(n))

Inner loop Analysis (for k) 

n = k * n^(1/3)* log(n) 

Number of steps: O(n^(2/3)/log(n))

Overall:

O(log(n) * n^(2/3)/log(n)) 

= O(n^(2/3))

Comments

Sorry for the poor formatting.  This site doesn't seem to work very well with super / subscripts and added a new line whenever i attempted to use one