How can I unfold the recurrence: T(n)=2T((n+2)/3)

Question

I\'m trying to solve this recurrence, but I don\'t know how to unfold it.

T(n)=2T((n+2)/3) + 1

Can I ignore that \"+2\" and solve it as it was

Answer 1

Sort of. The rigorous version of this trick is to set U(n) = T(n+1) and write

U(n) = T(n+1)
     = 2T((n+1+2)/3) + 1
     = 2T(n/3 + 1) + 1
     = 2U(n/3) + 1.

Then solve for U (e.g., U(n) = O(n^log3(2))) and then you should be able to find an asymptotic expression for T of the same order.