How to calculate the time complexity of a Catalan-like recurrence by substitution? I was given the following problem:


  Calculate the computational time complexity of the recurrence $$P(n) = \begin{cases} 1 &amp; \text{if } n = 1 \\ \sum_{k=1}^{n-1} P(k) P(n-k) &amp; \text{if } n \geq 2 \end{cases}$$ using the substitution method. Answer: $\Omega(2^n)$.


I have seen there is a closed-form for this recurrence but I am unsure whether it can be used out of the blue or not. I don't know if I have to prove it by deriving it from the recurrence or if there are simpler ways of solving such a problem by mere substitution.

On the other hand, I think substitution is not the right tool to use here. I have tried with both reverse and forward substitution, but in neither approach I could determine the function as with other problems (e.g. QuickSort).

Is it possible to solve this problem with only substitution at all? If so, how can I devise it?
You can use induction. First, check directly that $P(n) \geq 2^{n-2}$ for all $n \leq 4$ (the relevant values are $1,1,2,5$). Now suppose that $P(n) \geq 2^{n-2}$ holds for all $n \leq m$, where $m \geq 4$. Then
$$
P(m+1) = \sum_{k=1}^m P(k) P(m+1-k) \geq \sum_{k=1}^m 2^{k-2} 2^{m+1-k-2} \geq m 2^{m+1-4} \geq 2^{m+1-2},
$$
since $m \geq 4$.