Given $n,k\in\mathbb Z$ what is the minimum $r$ needed such that following iteration is $\leq2$?
$$n_0=n$$ $$n_1=\lceil n_0^{1/k}\rceil$$ $$\vdots$$ $$n_{i+1}=\lceil n_i^{1/k}\rceil$$ $$\vdots$$ $$n_r\leq2$$
What is the gap $r-\frac{\log_2\log_2 n}{\log_2k}$?
If $n_0\le 2$ then $r=0$, so we assume that $n_0\ge 3$.
If $k\le 0$ then $n_1\le 2$ and $r=1$, so we assume that $k\ge 1$.
If $k=1$, then $n_1=n_0$ and $n_i$ newer is not bigger than $2$, so we assume that $k\ge 1$.
By an induction with respect to $j$ we can check that $n_r\le 2$ iff $n_{r-j}\le 2^{k^j}$ for each $j$. Thus $2^{k^{r-1}}<n_0\le 2^{k^r}$ that is $r=\lceil \log_k \log_2 n\rceil$ and the gap is less than $1$.