, 2018

Let $S$ be the set of positive integers which can be written as $s=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ for some positive integers $a, b, c, d$.
Since $s>0$, we can assume that $a>b, c>d$ and write $s=2^{b-d} \frac{2^{a-b}-1}{2^{c-d}-1}$. It's now clear that $b-d=v_{2}(x)$.
So if we set $x=2^{v_{2}(x)} y$ then $y=\frac{2^{u}-1}{2^{v}-1}$ ($u, v$ are positive integers). In particular, the number we are looking for is odd. Clearly the numbers of the form $2^{u}-1$ ($v=0$) belong to $S$. More generally, we know that $2^{v}-1 \mid 2^{u}-1$ if and only if $v \mid u$. Set $u=d v$ then
$$y=\frac{2^{u}-1}{2^{v}-1}=\frac{2^{d v}-1}{2^{v}-1}=2^{(d-1) v}+2^{(d-2) v}+\cdots+2^{v}+1$$
This means that the base 2 representation of $y$ is of the form
$$y=1 \underbrace{0 \ldots 0}_{v-1} 1 \underbrace{0 \ldots 0}_{v-1} 1 \ldots \underbrace{0 \ldots 0}_{v-1} 1$$
($d$ digits 1, every two consecutive 1's are separated by $v-1$ 0's). So the number we are looking for is the smallest odd number not of this form (for some $d \geq 1, v \geq 1$). Note that
$$1=\overline{1},\ 3=\overline{11},\ 5=\overline{101},\ 7=\overline{111},\ 9=\overline{1001},\ 11=\overline{1011}$$
We can conclude that 11 is the smallest odd number whose base 2 representation is not of the desired form. So the answer is $n=11$.