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.

Without the additive term, the (exact) solution will be of the form $T(n)=\alpha n$ for some constant $\alpha$ depending on the first terms.
By substitution (or induction if you prefer):
$$T(n) = T\left(\frac{n}{3}\right)+T\left(\frac{2n}{3}\right) = \alpha \frac{n}{3}+\alpha \frac{2n}{3} = \alpha n$$
