$\begingroup$

Here's a proof of the first conjecture.

Preliminary remarks

Let's call "balanced strings" those described by that sequence.

Let $s$ be a balanced binary string of length $2n-1$. We denote $s_i$ its $i$-th bit (indexed from $1$), and $s_i^j$ the substring of bits from $i$ to $j$ (inclusive). The number of $1$ is denoted $|s|$.

Note that another way to say that two binary strings are permutations of each other is that they have the same numbers of $0$ and $1$: $s$ being balanced is equivalent to saying that for every $i < n$, $|s_1^n| = |s_n^{2n-1}|$.

This implies that every odd prefix is also balanced. Conversely, we can generate balanced strings by appending two bits at a time. This is also suggested by the binary tree you drew above.

How many ways are there to extend $s$ into a balanced string of length $2(n+1)-1$? We enumerate pairs of bits, $s_{2n}$ and $s_{2n+1}$, such that $|s_1^{n+1}| = |s_{n+1}^{2n+1}|$. We must consider four cases of the possible values for the middle bits $s_n$ and $s_{n+1}$. Here's one:

If $s_n = 0$ and $s_{n+1} = 1$, then $$\begin{aligned} |s_1^{n+1}| &= |s_1^n| + 1 \\ &= |s_n^{2n-1}| + 1 \quad \text{(since $s$ is balanced)}\\ &= |s_{n+1}^{2n-1}| + 1 \end{aligned}$$ For that to be equal to $|s_{n+1}^{2n+1}|$, the new bits can be either $01$ or $10$ (two choices).

The conclusion is that if $s_n

eq s_{n+1}$, then there are two ways to extend $s$ (with $01$ or $10$), otherwise there is only one.

Main result

$A297789(2^k+1)=2⋅A297789(2^{k})−1$ for $k>0$

As a consequence of the previous intermediate result, that is equivalent to claiming that every balanced string of length $2^{k+1}-1$ can be extended in exactly two ways, except the one string made of all ones.

That, in turn, amounts to saying that in every balanced string $s$ of length $2^{k+1}-1$, the middle bits $s_{2^k}$ and $s_{2^k+1}$ are distinct. In that case, the new bits we append at the end are $01$ or $10$, i.e., also distinct. Interesting coincidence, we can thus prove that claim by induction on $k$.