Introduce the auxiliary (sub)sequence $(S(k))$, defined for every $k\geqslant1$ by $$S(k)=4^{-k}\cdot T(3^{2^k-2}),$$ then the recursion on $(T(n))$ becomes $$S(k)=S(k-1)+4^{-k}\cdot(\log(3^{2^k-2}))^2,$$ for every $k\geqslant2$, that is, $$S(k)=S(k-1)+4^{-k}\cdot(\log3)^2\cdot(2^k-2)^2=S(k-1)+(\log3)^2\cdot(1-2^{-(k-1)})^2.$$ Since $(\log3)^2\cdot(1-2^{-(k-1)})^2\to(\log3)^2\ne0$ when $k\to\infty$, this yields $$S(k)\sim(\log3)^2\cdot k,$$ that is, $$T(3^{2^k-2})\sim(\log3)^2\cdot k\cdot4^k.$$
Let $n=3^{2^k-2}$, then $2^k=2+\log_3n$ hence $2^k\cdot\log3\sim\log n$, thus, $4^k\cdot(\log3)^2\sim(\log n)^2$, and $k\sim(\log\log n)/\log2$, hence $$T(n)\sim(\log\log n)(\log n)^2/(\log2),$$ in particular, $$T(n)=\Theta((\log\log n)(\log n)^2).$$ One sees that $T(n)$ is $O((\log\log n)(\log n)^2)$ and that $T(n)$ is not $O((\log n)^2)$.

Solving the recurrence relation $T(n)=4T\left(\frac{\sqrt{n}}{3}\right)+ \log^2n$
Given a tree, find a path on which every vertex has at most 4 leaves (can have 0 as well) and is the "biggest" (has the maximum amount of vertices possible - including the leaves).
Time complexity: $O(n)$.

I'm getting nowhere with this problem. I have tried BFS/DFS (seemed like such an obvious method), but even if I could modify them correctly, their time complexity is $O(|V|+|E|)$.

Any hint, thought is appreciated.

edit: so now I get that BFS/DFS is valid for this problem, but I'm still having difficulties with figuring out the algorithm itself. Help please.
