I am trying to find a $\Theta$ bound for the following recurrence equation:
$$ T(n) = 2 T(n/2) + T(n/3) + 2n^2+ 5n + 42 $$
I figure Master Theorem is inappropriate due to differing amount of subproblems and divisions. Also recursion trees do not work since there is no $T(1)$ or rather $T(0)$.
Yes, recursion trees do still work! It doesn't matter at all whether the base case occurs at $T(0)$ or $T(1)$ or $T(2)$ or even $T(10^{100})$. It also doesn't matter what the actual value of the base case is; whatever that value is, it's a constant.
Seen through big-Theta glasses, the recurrence is $T(n) = 2T(n/2)+T(n/3)+n^2$.
The root of the recursion tree has value $n^2$.
The root has three children, with values $(n/2)^2$, $(n/2)^2$, and $(n/3)^2$. Thus, the total value of all children is $(11/18) n^2$.
Sanity check: The root has nine grandchildren: four with value $(n/4)^2$, four with value $(n/6)^2$, and one with value $(n/9)^2$. The sum of those values is $(11/18)^2 n^2$.
An easy induction proof implies that for any integer $\ell\ge 0$, the $3^\ell$ nodes at level $\ell$ have total value $(11/18)^\ell n^2$.
The level sums form a descending geometric series, so only the largest term $\ell=0$ matters.
We conclude that $T(n) = \Theta(n^2)$.
You can use the more general Akra-Bazzi method.
In your case, we would need to find $p$ such that
$$ \frac{1}{2^{p-1}} + \frac{1}{3^p} = 1$$
(which gives $p \approx 1.364$)
and we then have
$$T(x) = \Theta(x^p + x^p\int_{1}^{x} t^{1-p} \text{d}t) = \Theta(x^2)$$
Note that you don't really need to solve for $p$. All you need to know is that $1 \lt p \lt 2$.
A simpler method would be to set $T(x) = x^2 g(x)$, and try proving that $g(x)$ is bounded.
Let $f(n) = 2T(n/2) + T(n/3) + 2n^2 + 5n + 42$ be a shorthand for the right-hand side of the recurrence. We find an lower and upper bound for $f$ by using $T(n/3) \leq T(n/2)$:
$$ 3 T(n/3) + 2n^2+ 5n + 42 \quad \le\quad f(n) \quad\leq \quad 3 T(n/2) + 2n^2+ 5n + 42 \quad$$
If we use the lower resp. upper bound as right-hand side of the recurrence, we get $T'(n) \in \Theta(n^2)$ in both cases by the Master theorem. Thus, $T(n)$ is bounded from above by $O(n^2)$ and from below by $\Omega(n^2)$ or, equivalently, $T(n) \in \Theta(n^2)$.
