Latest Discussion:
$R = \left\lfloor log_{4}\left(\frac{n}{3}\right) \right\rfloor + 1$
$G = \left\lfloor\sqrt[R]{n}\right\rfloor$
S below describes a row
$$ \vec{S}=\begin{cases}[\{5,6\},6,6,...,\left\lceil\frac{n}{6^{R-1}}\right\rceil] &\text{if } G > 6\\ [\{4,5\},4,4,...,5?, 3] &\text{if } G < 4\\ [\{G,G+1\},G,...,G, G+1,...,G+1] &\text{if } 4\le G \le 6\end{cases} $$
The number of rounds with groups of size $G+1$ is $\left\lfloor \log_{(G+1)/G}\frac{n}{G^R} \right\rfloor$
Should the larger groups come earlier or later?
The largest possible group is $11$, which happens only when $n=11$.
When $n\in [43,47]$, the last group is size 8. For larger elections the maximum group size never exceeds 6.
The points at which the number of rounds increases are 12, 48, 192, ..., $3\times 4^k$.
Example groupings:
11 → 11