Sequence of N operations Amortized Analysis
How about this. Assume we maintain a binary counter $C$ that counts the operations executed so far. Let $D_i$ be the data structure after the $i$-th operation. We then define as potential $\Phi(D_i)$ number of $\tt 1$-digits in $C$. As usual $c_i$ denotes the actual costs, and $\hat c_i$ the amortized costs.

If $i$ is not a power of two, then we have
$$ \hat c_i = c_i + \Phi(D_i) - \Phi(D_{i-1}) \le 2 .$$
This is true since the addition of a $\tt 1$ on $C$ will turn one digit from $\tt 0$ to $\tt 1$ while possibly turning some digits from $\tt 1$ to $0$.

If $i$ is a power of two then clearly
$$ \hat c_i = c_i + \Phi(D_i) - \Phi(D_{i-1})= i + 1 - (i-1) =2 .$$ 

As a final comment, the amortized costs do always depend on the potential function. So another potential function yields different amortized costs.