Asymptotic bound of $\sum^M_{x=1}{x \log_2(\frac{M}{x})}$, $x=1,2,4,8,..,M$

Question

What is the big-O bound for the following?

$$\sum^{ N}_{k=0}{2^k \log_2\left(\frac{2^N}{2^k}\right)}$$.

where $N = \log M$

My math is too weak to analyze it.

Answer 1

Note that $$\sum\limits_{x=1}^Nx\log_2\left(\frac Nx\right)=\sum\limits_{x=1}^N\sum\limits_{y=x}^N\log_2\left(\frac Ny\right)$$$$=\sum\limits_{x=1}^N\log_2\left(\prod\limits_{y=x}^N\frac Ny\right)=\sum\limits_{x=1}^N\log_2\left(\frac{(x-1)!}{N!}\cdot N^{N-x+1}\right)$$$$=\sum\limits_{x=1}^N\log_2\left(\frac{(x-1)!}{N!}\right)+\sum\limits_{x=1}^N(N-x+1)\cdot\log_2N$$$$N\cdot(N+1)\cdot(\log_2N-1)-N\log_2\left(N!\right)+\sum\limits_{x=1}^N\log_2\left((x-1)!\right)$$You can finish with Stirling's Approximation.

Answer 2

$$S_N=\sum^{N}_{k=0}{2^k \log_2\left(\frac{2^N}{2^k}\right)}=N\sum^{N}_{k=0}2^k-\sum^{N}_{k=0}k\,2^k$$ that is to say $$S_N=\left(2^{N+1}-1\right)N-2 \left(2^N N-2^N+1\right)=2^{N+1}-(N+2)$$