Let F_n be [A000045][1] (i.e., Fibonacci numbers). Here

$$
F_n = F_{n-1} + F_{n-2}, \\
F_0 = 0, F_1 = 1
$$

Let $\operatorname{wt}(n)$ be [A000120][2] (i.e., number of ones in the binary expansion of $n$). Here

$$
\operatorname{wt}(2n+1)=\operatorname{wt}(n) + 1, \\
\operatorname{wt}(2n) = \operatorname{wt}(n), \\
\operatorname{wt}(0)=0
$$

Let $\nu_2(n)$ be [A007814][3] (i.e., number of trailing zeros in the binary expansion of $n$). Here

$$
\nu_2(2n+1)=0, \\
\nu_2(2n) = \nu_2(n) + 1
$$

Let $T(n,k)$ be an integer coefficients such that

$$
T(n,k) = [T(n,k-1)>0]\left\lfloor\frac{T(n,k-1)}{2^{\nu_2(T(n,k-1))+1}}\right\rfloor, \\
T(n,0) = n
$$

Let $b(n)$ be an integer sequence such that we start with a vector $\nu$ length $\operatorname{wt}(n)$ with elements $\nu_i = \operatorname{wt}(n) - i + 1$ and then for $i$ from $1$ to $\operatorname{wt}(n) - 1$ and for $j$ from $1$ to $\operatorname{wt}(n)-i$ consecutively apply

$$\nu_j = (\nu_2(T(n, i)) + 1)(\operatorname{wt}(n) - i - j + 1)(\nu_j - \nu_{j+1})$$

Then $b(n)=\nu_1$ after the whole transformation.

Let $s(n)$ be an integer sequence such that

$$
s(n) = \sum\limits_{i=1}^{2^n}b(2^n + i - 1)
$$

I conjecture that $$s(n)=F_{2n+1}.$$

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

    b(n) = my(A = n, B, C, v1, v2); v1 = []; while(A > 0, B = valuation(A, 2); v1 = concat(v1, B+1); A \= 2^(B+1)); A = #v1; v2 = vector(A, i, A - i + 1); for(i = 1, A-1, for(j=1, A-i, v2[j]=v1[i+1]*(A - i - j + 1)*(v2[j] - v2[j+1]))); v2[1]
    s(n) = my(A = 1 << n); sum(i=1, A, b(A + i - 1))
    test(n) = s(n) == fibonacci(2*n+1)

What do you think about

$$
R(n,k)=\sum\limits_{i=1}^{2^n}[b(2^n + i - 1)=k]
$$

for $k$ fixed and $n$ variable?

Here square bracket denotes [Iverson bracket][4].

Also $R(n,k)$ begins with

    [2]
    [3, 1]
    [4, 3, 1]
    [5, 6, 3, 2]
    [6, 10, 6, 7, 1, 2]
    [7, 15, 10, 16, 3, 9, 0, 3, 1]
    [8, 21, 15, 30, 6, 23, 1, 13, 4, 2, 0, 5]
    [9, 28, 21, 50, 10, 46, 3, 36, 10, 8, 0, 25, 0, 0, 2, 5, 0, 3]
    [10, 36, 28, 77, 15, 80, 6, 78, 21, 20, 0, 73, 0, 2, 8, 25, 0, 17, 0, 5, 0, 0, 0, 10, 0, 0, 1]
    [11, 45, 36, 112, 21, 127, 10, 146, 38, 41, 0, 165, 0, 8, 20, 78, 0, 53, 0, 23, 2, 0, 0, 59, 1, 0, 5, 0, 0, 6, 0, 8, 0, 0, 0, 9]

Is there a way to prove it? Is there a simple formula for $R(n,k)$?

  [1]: https://oeis.org/A000045
  [2]: https://oeis.org/A000120
  [3]: https://oeis.org/A007814
  [4]: https://en.wikipedia.org/wiki/Iverson_bracket