Time Complexity of a Loop when Loop variable “Expands or Shrinks” exponentially

For such cases, time complexity of the loop is O(log(log(n))).The following cases analyse different aspects of the problem. Case 1 : CPP

for (int i = 2; i <=n; i = pow(i, k)) 
{ 
    // some O(1) expressions or statements
}

In this case, i takes values 2, 2^k, (2^k)^k = 2^k2, (2^k2)^k = 2^k3, ..., 2^{klog_k(log(n))}. The last term must be less than or equal to n, and we have 2^{klog_k(log(n))} = 2^log(n) = n, which completely agrees with the value of our last term. So there are in total log_k(log(n)) many iterations, and each iteration takes a constant amount of time to run, therefore the total time complexity is O(log(log(n))). Case 2 : CPP

// func() is any constant root function
for (int i = n; i > 1; 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))). Refer below article for analysis of different types of loops. https://www.geeksforgeeks.org/dsa/how-to-analyse-loops-for-complexity-analysis-of-algorithms/