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.
Timon Gehr via Digitalmars-d Wed, 02 Dec 2015 06:31:40 -0800
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.