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I'm interested in asymptotically minimizing the maximum between the height of a tree of degree $k$ with $n$ leaves, and $k$, i.e. minimizing $\max(k, \log_kn)$ asymptotically.

If I set $k = \frac {\log n}{\log \log n}$, then the height of the tree is $\log_\frac{\log n}{\log \log n}(n) = \frac{\log n}{\log \frac {\log n}{\log \log n}}$.

What is the asymptotic height of that tree, is $$\frac{\log n}{\log \frac {\log n}{\log \log n}} = \Theta(\frac{\log n}{\log \log n})$$ and if so, is this optimal?

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Yes, it is true that $\dfrac{\log n}{\log \frac {\log n}{\log \log n}} = \Theta(\dfrac{\log n}{\log \log n}).$ In fact, we have the following more accurate approximation, $$\frac{\log n}{\log \frac {\log n}{\log \log n}} \sim \frac{\log n}{\log \log n}.\tag{app}$$

Proof. $$\begin{aligned} \lim_{n\to\infty}\frac{\ \frac{\log n}{\log \frac {\log n}{\log \log n}}\ }{\ \frac{\log n}{\log \log n}\ }&=\lim_{n\to\infty}\frac{\log\log n}{\log \frac {\log n}{\log \log n}} =\lim_{n\to\infty}\frac{\log\log n}{\log\log n-\log\log \log n}\\ &=\frac{1}{1-\lim_{n\to\infty}\frac{\log\log \log n}{\log\log n}} =\frac{1}{1-0}=1. \end{aligned}$$

So, if we set $k = \frac {\log n}{\log \log n}$, then the height of the tree, $\frac{\log n}{\log \frac {\log n}{\log \log n}}$ is $\frac{\log n}{\log \log n}$ asymptotically and, hence, so is the larger one between the height of the tree and $k$.


Now, the question is what is the asymptotic behavior of the minimum of the larger one between $\log_kn $ and $k$, i.e. $h(n)=\min_k\max(\log_kn,k)$.

Since $k$ is increasing with respect to $k$ while $\log_k n$ is decreasing with respect to $k$, the minimum of the larger one between $k$ and $\log_k n$ for a given $n$ is obtained when $k=\log_k n$, assuming $k,n>1$.

What is the solution to $k=\log_k n$ for a given $n$? $$k=\log_kn\Leftrightarrow k= \frac{\log n}{\log k}\Leftrightarrow \log k\cdot k= \log n\Leftrightarrow \log k\cdot e^{\log k}= \log n$$ By the definition of the Lambert $W$ function, we should have $$\log k = W_0(\log n),$$ where $W_0(\cdot)$ is the principal branch of the Lambert $W$ function, as we are only interested in the real positive values. So, $$k = e^{W_0(\log n)}.$$ Accordingly, $$h(n)=\left.\max(\log_kn,k)\right|_{k=e^{W_0(\log n)}}=e^{W_0(\log n)}.$$

The same article on Lambert $W$ function tells that $$ W_0(x) = \log x - \log\log x + o(1).$$ Substituting $\log n$ for $x$, we have $$ W_0(\log n) = \log\log n - \log\log\log n + o(1).$$ So, $$ h(n) =e^{W(\log n)}= e^{\log\log n} e^{-\log\log\log n} e^{ o(1)}=\frac{\log n}{\log\log n}\ e^{o(1)},$$ Since $e^{o(1)}$ goes to 1 when $n$ goes to infinity, $$h(n)\sim{\frac{\log n}{\log\log n}}.$$


Combining the above conclusions, we see that if we set $k = \frac {\log n}{\log \log n}$, then the larger one between the height of the tree and $k$ approaches its minimum $h(n)$ as well as ${\frac{\log n}{\log\log n}}$ asymptotically.


To be nit picky, it should be pointed out that we have been ignoring the expected requirement that $k$ should be an integer. It would require cumbersome analysis to adapt the above analysis and conclusion for integer $k$. In the end, we have still $$\min_{1\lt k\le n,\ k\in\mathbb N}\max(k, \log_kn)\sim\frac{\log n}{\log\log n}.$$

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