Recurrence of T(n) = T(n/3) + T(2n/3) I've searched online for this but I only seem to find answers for a similar equation: 

T(n) = T(n/3) + T(2n/3) + cn


But the one I'm trying to solve is:

T(n) = T(n/3) + T(2n/3)


Base case: We can assume T(a) = Theta(1) for any constant a.

I've succeeded in proving (by induction) that T(n) = O(n*log(n)). I thought the answer should be Theta(n*log(n)), but I cannot prove that T(n) = Omega(n*log(n)).

So my question is - am I correct that the answer is O(n*log(n)), and NOT Theta(n*log(n))? IF that's true that would really be great...

If I'm wrong I will of course explain where I'm stuck in the induction process...

Thanks!

P.S. If you need to, please try to explain using induction, because I haven't learned all methods for solving these problems yet.
Hint: Suppose that $T(m) \leq Cm$ for all $m &lt; n$, and that $n$ is divisible by $3$. Then
$$ T(n) = T(n/3) + T(2n/3) \leq C(n/3) + C(2n/3) = Cn. $$
Use this to obtain a good bound on $T(n)$ when $n$ is a power of $3$.