Imagine if you have a line from 0 to 1, and you begin partitioning it dyadically. The first point will be at 0, the second at 1 and the third at 0.5, the fourth at 0.25, the fifth at 0.75 etc. Let's call this the count of the points.
Now, if we label them from left to right from 0 to $2^n$, what will 1 on this label map to the count of points?
In general, what is the mapping between the counts and the labels?
There are several sequences/functions/implicit ideas in your question, and I'll try to clear them up as best as I can.
There is an enumeration of the dyadic fractions $d_c$ where $d_1=0,d_2=1,d_3=\frac12$ that continues $\frac14,\frac34,\frac18,\frac38,\frac58,\frac78,\ldots$. Now, for fixed $n\ge0$, we can look at $d_1$ through $d_{2^n+1}$ and assign them labels based on how they are ordered in $[0,1]$, so there is a corresponding sequence of integers $\ell_{c,n}$. For example, if $n=2$ then $d_1$ through $d_5$ are $0,1,\frac12,\frac14,\frac34$ and $\ell_{1,2}$ through $\ell_{5,n}$ are $0,4,2,1,3$. Note that in general, $\ell_{c,n}=d_c*2^n$, so that it's easy enough to find these labels once we have a formula for $d_c$.
Putting $d_1$ and $d_1$ aside for a moment we can gain some insight by looking at $d_c$ in binary, starting with $d_3$. They are: $.1,.01,.11,.001,.011,.101,.111,\ldots$ They group nicely into the first $1$, the next $2$, the next $4$, the next $8$,... Furthermore, each group has a nice pattern: the group of $4$ is $.00+.001$, $.01+.001$, $.10+.001$, $.11+.001$, where $.00$ through $.11$ are just $0$ through $4$ with the binary point shifted.
We can figure out which "group" we're in by using floor and the base-$2$ logarithm: $d_c$ is in the group with $\lfloor\log_2(c-2)\rfloor+1$ bits. Now, we also need to figure out how far along we are in the group. Indexing from $0$, we can take $c-2$ and subtract the relevant power of $2$: For example, $c=6$ and $c=7$ (corresponding to $d_6=.001$ and $d_7=.011$) have $c-2-2\text{^}\lfloor\log_2(c-2)\rfloor$ equal to $0$ and $1$. Putting this together with the "binary point shifted" and the "add $.001$" idea, we get the following formula for $c>2$: $d_c=\dfrac{c-2-2\text{^}\lfloor\log_2(c-2)\rfloor}{2\text{^}\lfloor\log_2(c-2)\rfloor}+\dfrac{1}{2\text{^}\left(\lfloor\log_2(c-2)\rfloor+1\right)}$. This simplifies to $d_c=\dfrac{2(c-2)+1}{2\text{^}\left(\lfloor\log_2(c-2)\rfloor+1\right)}-1$.
Therefore, $d_c=\begin{cases} 0 & \text{ if }c=1\\ 1 & \text{ if }c=2\\ \dfrac{2(c-2)+1}{2\text{^}\left(\lfloor\log_2(c-2)\rfloor+1\right)}-1 & \text{ if }c>2 \end{cases}$ and the mapping from count to label is $\ell_{c,n}=2^nd_c=\begin{cases} 0 & \text{ if }c=1\\ 2^n & \text{ if }c=2\\ -2^n+(2(c-2)+1)*2\text{^}\left(n-1-\lfloor\log_2(c-2)\rfloor\right) & \text{ if }c>2\text{ and }n>0 \end{cases}$
Since this gives the label as a function of the count, it doesn't easily answer the part of the question that asks what the count of the label $1$ is. For the first $2^n+1$ dyadics, the last $2^{n-1}$ of these have the highest denominator of $2^n$ when written in simplest form, so the count corresponding to label $1$ is $\left(2^n+1-2^{n-1}\right)+1=2^n-2^{n-1}+2$. For example, with $n=3$, label $1$ occurs at count $6$: $0,1,\frac12,\frac14,\frac34,\boxed{\frac18}$.
If you want the general function from the label to the count (the inverse of $c\mapsto\ell_{c,n}$), that will be a bit trickier to write down, as it depends on the binary representation of the label. For $n=3$, you can see a pattern in the binary representations of the labels for $c$ from $3$ to $8+1$ (skipping the exceptional $c=1,2$: $100,010,110,001,011,101,111$. First is the label ending in $100$, then are the two labels ending in $10$ in increasing order, then are the four labels ending in $1$ in increasing order. A map from labels to counts would involve the highest power of $2$ dividing the label (to see which group it's in).