Non-linear recurrence relation $T(n)=nT^2\left(\frac{n}{2}\right)$ with $T(1)=6$

Question

I am concerned to solve the recurrence relation $T(n)=nT^2\left(\frac{n}{2}\right)$ with initial condition $T(1)=6$.

I have tried to do some transformation like $N=2^k$ and some other things like that but all of them was meaningless.

Answer 1

As usual, this only defines $T$ on the powers of $2$ hence consider $$ t_k=2^kT(2^k). $$ Then $t_k=4t_{k-1}^2$ for every $k\geqslant1$. Iterating, one gets $$ t_k=4\cdot4^2\cdots4^{2^{k-1}}\cdot (t_0)^{2^k}, $$ that is, $$ 2^k\cdot T(2^k)=4^{2^k-1}\cdot T(1)^{2^k}, $$ or $$ T(\color{red}{2^k})=\frac{(4T(1))^{\color{red}{2^k}}}{4\cdot\color{red}{2^k}}. $$ If one wants to extend this to every $n$ (which is not a consequence of the hypothesis), probably the most natural formula is $$ T(\color{blue}{n})=\frac{(4T(1))^\color{blue}{n}}{4\cdot\color{blue}{n}}. $$