How to determine the time complexity of this loop?

Question

// func() is any constant root function for (int i = n; i > 0; i = func(i)) {    // some O(1) expressions or statements }

"In this case, i takes values n, n^1/k, (n^1/k)^1/k = n^1/k2, n^1/k3, …, n^{1/klog_k(log(n))}, so there are in total log_k(log(n)) iterations and each iteration takes time O(1), so the total time complexity is O(log(log(n)))."

Can someone explain the above statement? How do we calculate that there are log_k(log(n)) iterations? 

Source: http://www.geeksforgeeks.org/time-complexity-loop-loop-variable-expands-shrinks-exponentially/

Answer 1

func() is given as root function. So, we write it as  n, n^1/k, n^1/k2, n^1/k3, …, n^1/kp  

Where we assume that n^1/kp  is the last term that is >0 (lets take it to be = e) according to the loop.(e = smallest positive root) Now  

n^1/kp  = e

taking log on both sides (1/k)^p log_e(n) = log_e(e)

(1/k)^p log_e(n) =  1

log(n) = k^p

again taking log on both the sides log(log_e(n)) = p log(k)

therefore p =  [log(log_e(n)) ]  /  [log(k)]

we know that [log (a)] / [log(b)] = log_ba

therfore p = log_k(log_e(n)