simple but (seemingly?) tricky recurrence
Let us attempt to solve the easier recurrence $T(2n) = T(n) \log T(n)$. Define $S(m) = \log T(2^m)$. Then $S(m+1) = S(m) + \log S(m)$. Let us heuristically convert this into a differential equation: $S' = \log S$. The solution to this equation is $S = \mathrm{li}^{-1} + C$, where $\mathrm{li}$ is the logarithmic integral. Roughly speaking, $\mathrm{li}(x) \approx x/\log x$, and so $\mathrm{li}^{-1}(x) \approx x\log x$. This heuristic implies that $S(m) \approx m \log m$, and so $T(n) \approx \exp \Theta(\log n \log \log n)$. Of course, at this point this is just an educated guess.

As a sanity check, suppose that $T(n) = \exp (\log n \log \log n)$, where all logarithms are base&nbsp;2. Then
$$
\begin{align*}
T(n) \log T(n) &amp;= \exp (\log n \log \log n) \log n \log \log n \\ &amp;=
\exp (\log n \log \log n + \log \log n + \log \log \log n) \\ &amp;=
\exp (\log (2n) \log \log n + \log \log \log n).
\end{align*}
$$
This is quite similar to $T(2n) = \exp (\log (2n) \log \log (2n))$ (though not the same).