Density of some intervals whose lengths are powers of 2 Let $\mathbb{N} = \{0,1,2,\ldots\}$.  Define the upper power density of a set $A\subseteq \mathbb{N}$ to be $$ \bar{d_p}(A) = \limsup_{n\to\infty} \frac{\sum \{2^k\;|\; k\in A\cap\{0,1,\ldots,n\}\}}{1 + 2 + \cdots + 2^n} $$  In other words, $\bar{d_p}(A)$ is the upper density of the set $B$ obtained by replacing each $n\in\mathbb{N}$ with an interval of length $2^n$.  Clearly, for each $A\subseteq\mathbb{N}$, either $\bar{d_p}(A) \ge 1/2$ or $\bar{d_p}(A^C) \ge 1/2$.
Question: Does there exist a set $A\subseteq\mathbb{N}$ such that $\bar{d_p}(A) = \bar{d_p}(A^C) = 1/2$?  If not, what is the maximal $\alpha\in [0,1]$ such that for every $A\subseteq\mathbb{N}$, either $\bar{d_p}(A)\ge \alpha$ or $\bar{d_p}(A^C) \ge \alpha$?
Note that $\bar{d_p}(evens) = \bar{d_p}(odds) = 2/3$.
 A: Almost right after posting this question, I found what I think is an easy answer.  Oh well.
Suppose that for infinitely-many $n$, both $n$ and $n+1$ are in $A$.  Then, for such an $n$, we have $$ \frac{\sum\{2^k\;|\; k\in A\cap \{1,\ldots,n+1\}\}}{1 + 2 + \cdots + 2^{n+1}} \ge \frac{2^n + 2^{n+1}}{2^{n+2} - 1} \approx 3/4, $$ so $\bar{d_p}(A) \ge 3/4$.
Of course if there are infinitely-many $n$ such that neither $n$ nor $n+1$ are in $A$ then we have $\bar{d_p}(A^C) \ge 3/4$.
The only other case is where, for large enough $n$, we have $n\in A$ if and only if $n+1\not\in A$.  In this case $A$ has the same upper power density as the evens (or the odds), i.e. 2/3.  So it seems the answer to the first question is no, and the $\alpha$ from the second question is $2/3$.
What is interesting here is that there are essentially only two witnesses to the minimal $\bar{d_p}$, and every other set has $\bar{d_p}(A) \ge 3/4$ or $\bar{d_p}(A^C) \ge 3/4$.
A: In order for $\overline d_p(A) = \overline d_p(A^C) = 1/2$, we must have the lim inf of $A$ to also be $1/2$; which is equivalent to saying the total limit is $1/2$.
Suppose this was possible and we had a set $A$ for which it was true. Then we can fix a small $\varepsilon$ (say less than $1/8$) and find $N$ large for which the quantity
$$s_n = \frac{\sum \{2^k \mid k \in A \cap \{0, 1, \dots, n\}\}}{1 + 2 + \dots + 2^n}$$
is in $[1/2-\varepsilon, 1/2+\varepsilon]$ for all $n \geq N$. So suppose $s_n$ is in the interval for some large $n$. 
For $n+1$, if $n+1 \not\in A$ then $s_{n+1} < \frac{1}{2} s_n < 1/2 - \varepsilon$, so it is not in the interval.
For $n+1$, if $n+1 \in A$ then 
$$s_{n+1} = \dfrac{2^{n+1}-1}{2^{n+2} - 1} s_n + \frac{2^{n+1}}{2^{n+2} - 1} \approx \frac{1}{2} (s_n + 1) > 1/2 + \varepsilon$$
hence there is no set $A$ for which $\overline d_p(A) = \overline d_p(A^C) = 1/2$.
