Let $\operatorname{wt}(n)$ be A000120, i.e. the number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Also let's consider $$\ell(n)=\left\lfloor\log_{2} n\right\rfloor$$ and $$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2.$$ Here $T(n,k)$ is the $(k+1)$-th bit from the right side in binary expansion of $n$.
Now let's start with $T(n,\ell(n)-\operatorname{wt}(n)+m+1)$ where $m=0$ and apply $m:=m+1$ until $T(n,\ell(n)-\operatorname{wt}(n)+m+1)=1$.
Let $a(n)$ be the number of iterations needed to reach $1$ in operation above.
Let $b(n)=a(2n+1)$: the sequence begins with $$1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 3, 2$$
Let $c(n)=\operatorname{wt}(n)-b(n)$: the sequence begins with $$0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 2, 1, 1, 2, 3, 0, 0, 0, 0, 0$$
Let $d(n)$ be the sequence of numbers $k$ such that $c(k)>0$ and $c(k-1)=c(k+1)=0$: the sequence begins with $$3, 39, 75, 143, 147, 279, 283, 291, 543, 551, 555, 563, 579, 1071, 1079, 1083, 1095, 1099, 1107, 1123$$
Let $e(n)=\ell(d(n))$: the sequence begins with $$\underbrace{1}_{1}, \underbrace{5}_{1}, \underbrace{6}_{1}, \underbrace{7, 7}_{2}, \underbrace{8, 8, 8}_{3}, \underbrace{9, 9, 9, 9, 9}_{5}, \underbrace{10, 10, 10, 10, 10, 10, 10, 10}_{8}$$ Conjecture: the numbers of occurrence of different values of $e(n)$ is the sequence of Fibonacci numbers.
Let $$f(n)=\frac{d(n)-2^{e(n)}+1}{4}$$ with $f(1)=1$: the sequence begins with $$1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 17, 12, 14, 15, 18, 19, 21, 25$$
Conjecture: $f(n)$ is the sequence of permutation of natural numbers.
UPD: If we start with $T(n,m)$ instead of $T(n,\ell(n)-\operatorname{wt}(n)+m+1)$, then for $e'(n)$ we have
$$\underbrace{3}_{1}, \underbrace{4}_{1}, \underbrace{5}_{1}, \underbrace{6, 6}_{2}, \underbrace{7, 7, 7}_{3}, \underbrace{8, 8, 8, 8}_{4}, \underbrace{9, 9, 9, 9, 9, 9}_{6}$$
Conjecture: the numbers of occurrence of different values of $e'(n)$ is A000930, Narayana's cows sequence.
It also looks like $f(n)-1$ is A243571.
Is there a way to prove it?
