I have given an algorithm that solves a system depending on the parameters $m$ and $n$ in time $O(m)^{(m+1)^2}O(n)^{m(m+1)}log^2(mn)$. I am interested in the running time of the algorithm for all $n$ when we restrict ourselves to the case where $m=2$. How do I compute the new runningtime of the algorithm?
I was thinking about the following simplifications: $O(m)^{(m+1)^2}=O(2)^9=O(1)^9=O(1)$
$O(n)^{m(m+1)}=O(n)^6$ and
$log^2(mn)=log^2(2n)$
Which yields a running time of $O(1)O(n)^6log^2(2n)$.
Is this line of thought correct?