A divide-and-conquer algorithm creates one recursive call on $\lceil n/3 \rceil$ elements and another on $\lfloor 2n/3 \rfloor$ elements, plus linear time to combine: $T(n) = T(\lceil n/3 \rceil) + T(\lfloor 2n/3 \rfloor) + n$ Using the Akra–Bazzi theorem, we need to find $p$ satisfying $(1/3)^p + (2/3)^p = 1$. Which value of $p$ solves this?