Rounding logarithm to next integer — Potential function
The analysis there is slightly wrong, though the conclusions are correct. Here is the correct calculation:


$i$ is not a power of 2. In that case $\lfloor \log_2 i \rfloor = \lfloor \log_2 (i-1) \rfloor$ and so $\Phi(i) - \Phi(i-1) = (2i+1) - (2(i-1)+1) = 2$.
$i$ is a power of 2. In that case $2^{\lfloor \log_2 i \rfloor} = i$ and $2^{\lfloor \log_2 (i-1) \rfloor} = i/2$, and so $\Phi(i) - \Phi(i-1) = (2i - 2i + 1) - (2(i-1) - i + 1) = 2-i$.