On 12/02/2015 03:29 PM, Timon Gehr wrote:
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⌋²It should actually be (⌊i/2⌋-1)² here, but this does not change the asymptotics.
Oops. No, it was actually right. Sorry for the noise. :-)
