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.

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