If I can easily state the number of steps for an algorithms for certain values of $n$, e.g. for $n = 2^k$, where $k$ is a whole number, the number of steps is $n\log n$, is this enough to allow me to state the complexity is amortized $O(n\log n)$?
I have a tree building algorithm - when adding elements a rebalancing operation occurs every so often, with those operations getting progressively more expensive as the tree size grows close to a power of two elements (before becoming cheaper again once we pass such a power). At every power of two elements its easy to state the number of steps done so far but in between it's not so easy.
I'm slightly embarrassed by this question as I currently assume it is OK to state an amortized time of $O(n\log n)$ for this situation but my conclusions, from reading a discussion elsewhere (see next part), seemed to contradict this.