On 12/02/2015 12:26 PM, Timon Gehr wrote: > ... > ∑ᵢ[2≤i≤m]·log(i)·∑ₖ[0≤k≤⌊i/2⌋-1]·(2·k+1) > = > ∑ᵢ[2≤i≤m]·log(i)·⌊i/2⌋² > It should actually be (⌊i/2⌋-1)² here, but this does not change the asymptotics.