So the divide and concur recurrence has to be of the form H($n$) = $a$H($n$/$b$) + $cn^d$. I already figured out that $a$ = 4 and $b$ = 2. I am really stuck on how to find $cn^d$ however. I understand $cn^d$ is the number of operations it takes to combine the sub-problems. How can I figure $cn^d$ out however if I am not even sure what kind of problem we are solving?
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While $a * H(\frac{n}{b})$ counts the number of nodes within your first-order trees, the other term counts the number of nodes needed to connect those trees. You probably already figured this out. The next step I would do is go through some examples for small n and see if $c$ and $d$ come out to be constants to be proved later.
The "problem" in this case appears to be, "How many nodes are in the tree?" Trying to find "number of operations to be performed" is indeed confusing, since no operations on the tree have been specified, but if you replace "number of operations to combine the sub-problem" with "number of nodes to connect all the sub-trees" then the recurrence relation is just as valid a way to count nodes as it is to count operations of some algorithm.
If you must count operations, you can make up an operation that takes one step to find an uncounted node and "count" it. Then the number of operations is exactly equal to the number of nodes in the tree.
I think the term you are looking for is there in the diagram: the nodes that are counted in the "$cn^d$" term are all colored black.