Let $n=2^k$ and $T(2)=1$. Then, if we define $T(n)=\sqrt{n}(T(\sqrt{n}))+n$ for $n>2$, then what would be the exact solution of the recurrence?
The master theorem might not work here, and guessing a solution seems hard. In addition, I suspect $n=2^{2^k}$, for otherwise, we get irrational values. I think $T(n)=O(n\log n)$. But, is a better bound possible? Thanks beforehand.
$$ T(e^{\ln n})=\sqrt n T(e^{\frac 12\ln n})+n $$
or calling $T'(u) = T(e^u)$
$$ T'(\ln n) = \sqrt n T'(\frac 12\ln n) + n $$
or
$$ T'(2^{\log_2(\ln n)}) = \sqrt n T'(2^{\log_2(\frac 12\ln n)})+n $$
now calling $z = \log_2(\ln n)$ and $T''(z) = T'(2^z)$ we have
$$ T''(z) = e^{\frac 12 2^z}T''(z-1)+e^{2^z} $$
This is a linear recurrence with solution
$$ T''(z) = C_1 e^{2^z-2}+ze^{2^z} $$
and finally
$$ T(n) = C_1\frac{n}{e^2}+n\log_2(\ln n) $$
