* Let $a(n)$ be [A344960][1] (i.e., position of binary complement of $n$-th word in [A341258][2]). By definition, in order to calculate $a(n)$, we need to know [A341258][2]. Below we will correspond this sequence with the function $w(n)$.

* To compute [A341258][2], we also need to know a few additional features listed below.

* Let $s(n)$ be [A000201][3], $t(n)$ be [A001950][4], $p(n)$ be [A005206][5] and $q(n)$ be [A060144][6].

* Let $w(n)$ be $n$-th word in [A341258][2]. 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. Compare it with definition in [A341258][2]:

> Let $s = (s(n))$ be a strictly increasing sequence of positive integers with infinite complement, $t = (t(n))$. For $n\geqslant1$, let $s'(n)$ be the number of $s(i)$ that are $\leqslant n-1$ and let $t'(n)$ be the number of $t(i)$ that are $\leqslant n-1$. Define $w(1) = 0$, $w(t(1)) = 1$, and $w(n) = 0w(s'(n))$ if $n$ is in $s$, and $w(n) = 1w(t'(n))$ if $n$ is in $t$. Then $(w(n))$ is the "s-induced ordering" of all $01$-words.
$s$ = A000201; $t$ = A001950; $s'$ = A005206; $t'$ = A060144;

What does it mean? This means that we must first choose $s(n)$. In our case it is [A000201][3]. Then $t(n)$ is a complement of $s(n)$. Also $p(n)$ and $q(n)$ are closely related to $s'(n)$ and $t'(n)$, where we just change condition $\leqslant n-1$ to $\leqslant n$. So obviously $n$ is in $s$ if $s(p(n))=n$ and $n$ is in $t$ if $t(q(n))=n$. Since $t(n)$ is a compement of $s(n)$ we define second case as *otherwise* to first case.


* Let $b(n)$ be [A345253][7] (i.e., maximal Fibonacci tree: Arrangement of the positive integers as labels of a complete binary tree).

* Let $c(n)$ be [A353654][8] (i.e., numbers whose binary expansion has the same number of trailing $0$ bits as other $0$ bits).

* Let $d(n)$ be [A030109][9] (i.e., write $n$ in binary, reverse bits, subtract $1$, divide by $2$).

Sequence Machine conjectures ([1][10], [2][11]) 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?

  [1]: https://oeis.org/A344960
  [2]: https://oeis.org/A341258
  [3]: https://oeis.org/A000201
  [4]: https://oeis.org/A001950
  [5]: https://oeis.org/A005206
  [6]: https://oeis.org/A060144
  [7]: https://oeis.org/A345253
  [8]: https://oeis.org/A353654
  [9]: https://oeis.org/A030109
  [10]: https://sequencedb.net/s/A348366
  [11]: https://sequencedb.net/s/A344960