Krdytkyu

A starting point is this problem: $P subset Z_n = {1, 2, dots, n}$ is binary, if there exists a number $k$ such that both $k in P$ and $2k in P$, otherwise it is antibinary. One is supposed, for given $n$, to show antibinary set with maximum amount of elements (there might be many sets satisfying this property). One can show that set $P_n = {4^m(2k + 1) mid m, k in mathbb{N}} cap Z_n$ is a proper solution. My question is: how big is this set? I'm looking for $displaystylelim_{n to infty} p_n$, where $p_n = dfrac{|P_n|}{n}$. Simple program I wrote in Python suggests that the answer is $dfrac{2}{3}$, but I don't know, how to prove it.

Since $4^m (2k+1)leq n $ if and only if $kleqfrac {n/4^m-1}{2}$, we have
begin{align}
frac {P (n)}n
&=frac 1nsum_{m=0}^{lfloorlog_4(m)rfloor}leftlfloor 1+frac {n/4^m-1}{2}rightrfloor\
&=frac 1nsum_{m=0}^{lfloorlog_4(m)rfloor}leftlfloorfrac {n+4^m}{2cdot 4^m}rightrfloor\
&=frac 1nsum_{m=0}^{lfloorlog_4(m)rfloor}frac {n+4^m}{2cdot 4^m}+Oleft(frac {log (n)}nright)\
&=frac 12sum_{m=0}^{lfloorlog_4(m)rfloor}frac 1{4^m}+Oleft(frac {log (n)}nright)\
&=frac 23left(1-4^{-1-lfloorlog_4 (n)rfloor}right)+Oleft(frac {log (n)}nright)\
&xrightarrow {ntoinfty}frac 23
end{align}

Roughly $frac 12$ of the numbers are of the form $2k+1$, $frac 18$ of the numbers are of the form $8k+4$, $frac 1{32}$ of the numbers are of the form $32k+16$, $frac 1{128}$ are of the form $128k+64$, so the density is
$$frac 12left(1+frac 14+frac 1{16}+ldotsright)=frac 18 cdot frac 1{1-frac 14}=frac 23$$