Let $ k$ be an even number. For a $ k$ -dimensional cube (http://mathworld.wolfram.com/HypercubeGraph.html) $ Q_k$ , let $ G$ be a subgraph of $ Q_k$ with $ 2^{k-1}+s$ vertices, for $ 1\le s\le 2^{k-1}-1$ . I am wondering what is maximum number of edges $ G$ can have?

A theorem says average degree of $ G$ is at most $ v_G\log_2 v_G$ , so the number of edges of $ G$ is at most $ \lfloor\frac{v_G\log_2 v_G}{2}\rfloor$ . But is this the best bound? For example when $ s=1$ , in my mind, to maximize the number of edges on $ 2^{k-1}+1$ vertices, it should contains a $ (k-1)$ -dim subcube, say $ \{(x_1,\dots,x_{n-1},0):x_i\in\{0,1\}\}$ . But then the remaining vertex $ (y_1,\dots,y_{n-1},1)$ can have only one edge with the subcube. So this $ G$ has only $ (k-1)2^{k-2}+1$ edges. But for even $ k$ , $ $ \lfloor\frac{v_G\log_2 v_G}{2}\rfloor=\lfloor\frac{(2^{k-1}+1)\log_2 (2^{k-1}+1)}{2}\rfloor\ge \lfloor\frac{(2^{k-1}+1)(k-1)}{2}\rfloor=\frac{(2^{k-1}+1)(k-1)-1}{2}=(k-1)2^{k-2}+\frac{k-2}{2},$ $ which is much larger than $ (k-1)2^{k-2}+1$ when $ k$ is large. Therefore I am asking when $ k$ is even and $ 2^{k-1}+1\le v_G\le 2^{k-1}+2^{k-1}-1$ , is the bound $ \lfloor\frac{v_G\log_2 v_G}{2}\rfloor$ tight, or is there any other better result?