Fibonacci numbers in binary system

We get the sequence of Fibonacci numbers by slightly changing the binary number system – let’s do the conversion. Example:
$(n)_{notation}$
$(5)_{10}=(101)_2$
How we will transform: we will replace zeros with ones, ones with twos, then we sum them up. $101\rightarrow 212 \rightarrow 5$
$(6)_{10}=(011)_2 \rightarrow 122 \rightarrow 5$
$(7)_{10}=(111)_2 \rightarrow 222 \rightarrow 6$
Let’s write in two columns the numbers before the conversion, and after:
$0\rightarrow 1$ ; $1\rightarrow 2$ ; $2\rightarrow3$ ; $3\rightarrow4$ ; $4\rightarrow4$ ; $5\rightarrow5$ ; $6\rightarrow5$ ; $7\rightarrow6$ ; $8\rightarrow5$ ; $9\rightarrow6$ ; $10\rightarrow6$ ; $11\rightarrow7$
The numbers on the right are repeated, let’s count their repetitions:
$1\rightarrow1$ ; $2\rightarrow1$ ; $3\rightarrow1$ ; $4\rightarrow2$ ; $5\rightarrow3$ ; $6\rightarrow5$ ; $7\rightarrow8$ ; $8\rightarrow13$ ;
We get a sequence of Fibonacci numbers.
To derive a formula for Fibonacci numbers, you need to count the number of repetitions of the number. I note that each repeated number $n$ (converted from a binary number) occurs in the natural series of numbers in the interval $\min=floor \left( \frac{n+1}{2} \right)$ ; $\max=2^{n-1}$ ;
Let’s take the repeating number $8$ as an example. In how many ways can we write the number $8$ as a sum of twos and ones?
$1+1+1+1+1+1+2$
$1+1+1+1+2+2$
$1+1+2+2+2$
$2+2+2+2$
Subtract one deuce from each of the options:
$1+1+1+1+1+1$
$1+1+1+1+2$
$1+1+2+2$
$2+2+2$
For each case, we calculate the number of combinations from $n$ (the number of numbers) by $k$ (the number of twos), and sum them up. We get:
$\sum_{k=0}^{floor \left( \frac{n-2}{2} \right) } ( \frac{(n-(k+2))!}{(n -(2k+2))!\cdot k!} )$
Wikipedia only mentions diagonal summation in Pascal’s triangle.