How to find upper bound for recurrence relation

[ciphone]

Homework Statement

Find a tight upper bound for the recurrence relation using a recursion tree argument

Homework Equations

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

The Attempt at a Solution

I don't know how to do this problem because the tree doesn't have symmetry. One side of the tree can keep going because of the lack of symmetry plus at the end there is no way that you get T(1). How do you solve this problem using a recursion tree?

 

[geoffrey159]

I don't understand.
If one consider that the indexation of sequence $T(n)$ is in $\mathbb{N}-\{0\}$, then the index $n$ must be a multiple of $6$, since 2 and 3 are mutually prime divisors of $n$.

At first sight you must consider a sequence of type $T(6n) = T(3n) + T(2n) + c$. But for 6 to divide $3n$ and $2n$, it must divide $n$. It leads you to consider the sequence $T(36n) = T(18n) + T(12n) + c$.

With that, what sense do you give to $T(n)$ for indexes that are not multiple of 12 and 18 ?