Solving recurrence equation $T(n)=T(n^{2/3})+17$
This post answers the original version of the question, where the recurrence relation is

$$T(n)=T\left(\frac{n^2}{3}\right)+17$$



The recurrence relation is somewhat unconventional. Here is an outline to solve it.

Suppose $n\ge 6$. Let $n=3\cdot2^{2^{m}}$, where $m=\log_2(\log_2\frac n3)\ge0$.

$$\begin{align}
T(n)
&amp;=T\left(3\cdot2^{2^{m}}\right)
 =T\left(\sqrt{3\cdot\left(3\cdot2^{2^{m}}\right)}\right)-17\\
&amp;=T\left(3\cdot2^{2^{m-1}}\right)-17
 =T\left(\sqrt{3\cdot3\left(\cdot2^{2^{m-1}}\right)}\right)-2\cdot17\\
&amp;=T\left(3\cdot2^{2^{m-2}}\right)-2\cdot17
 =T\left(\sqrt{3\cdot\left(3\cdot2^{2^{m-2}}\right)}\right)-3\cdot17\\
&amp;=\cdots\\
&amp;=T\left(3\cdot2^{2^{m-\lceil m\rceil}}\ \right)-\lceil m \rceil\cdot 17
\end{align}$$

Since $-1\lt m-\lceil m\rceil\le0$, $3\sqrt2\lt3\cdot2^{2^{m-\lceil m\rceil}}\le6.$ So 
$T(n)\sim -17\log_2(\log_2 n)$ when $n$ goes to $\infty$. 





Here are two related exercises.

Exercise 1. What is the recurrence relation for function $S$ such that $S(n)=T(3n)$?

Exercise 2. What is the asymptotic behavior of $T(n)$ if $n$ goes to 3 from above, assuming $T$ is continuous?