# What is the complexity of $i^i$?

What is the complexity of the following algorithm in Big O:

for(int i = 2; i < n; i = i^i) {     ...do somthing } 

I’m not sure if there is a valid operator to this type of complexity. My initial thought was as follows:

After $$k$$ iterations we want: (using tetration?)

$${^{k}i} = n \implies k=\log\log\log…_k\log{n}\implies\mathcal{O(\log\log\log…_k\log{n})}$$ (where we have k times the log function) but i’m not sure if this is evan a valid way of writing this. Anyway, we have a complexity that that includes $$k$$, which does not seems right to me.