Let $a(n)$ be A344960 (i.e., position of binary complement of $n$-th word in A341258).

Let $s(n)$ be A000201, $t(n)$ be A001950, $p(n)$ be A005206 and $q(n)$ be A060144.

Let $w(n)$ be $n$-th word in A341258. To reproduce the sequence from itself, start with $w(1)=0$, $w(2)=1$ and apply $w(n)=0w(p(n-1))$ if $n=s(p(n))$, $w(n)=1w(q(n-1))$ otherwise.

Let $b(n)$ be A345253 (i.e., maximal Fibonacci tree: Arrangement of the positive integers as labels of a complete binary tree).

Let $c(n)$ be A353654 (i.e., numbers whose binary expansion has the same number of trailing $0$ bits as other $0$ bits).

Let $d(n)$ be A030109 (i.e., write $n$ in binary, reverse bits, subtract $1$, divide by $2$).

Sequence Machine conjectures (1, 2) that $$a(n) = b(d(c(n+2))) - 1.$$

Here is the PARI/GP program to check it numerically:

s(n) = (n + sqrtint(5*n^2))\2
t(n) = s(n) + n
p(n) = s(n+1) - n - 1
q(n) = n - p(n)
w(n) = if(n < 3, [n - 1], if(n==s(p(n)), concat(0, w(p(n-1))), concat(1, w(q(n-1)))))
a(n) = my(A = 0, v1); v1 = w(n); for(i=1, #v1, v1[i] = !v1[i]); until(w(A)==v1, A++); A
b(n) = my(x=0, y=0); for(i=0, logint(n, 2), [x, y]=[y+1, x+y]; if(bittest(n, i), [x, y]=[y, x+y])); y;
d(n) = fromdigits(Vecrev(binary(n)), 2)\2
isok(k) = if (k==1, 1, (logint(k-1, 2)-hammingweight(k-1) == valuation(k, 2)));
my(z=4); for(k=1,299, while(!(isok(z)), z++); print(b(d(z))-1==a(k)); z++;);

Is there a way to prove it?