Let $m \geqslant 1$ be a fixed integer.
Let $\operatorname{wt}(n)$ be A000120, $1$'s-counting sequence: number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Let $f(n)$ be A007814, exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Let $\operatorname{tm}(n)$ be A010060, Thue-Morse sequence: let $A_k$ denote the first $2^k$ terms; then $A_0 = 0$ and for $k \geqslant 0$, $A_{k+1} = A_k B_k$, where $B_k$ is obtained from $A_k$ by interchanging $0$'s and $1$'s.
Then we have an integer sequence given by \begin{align} a(0)& = 1\\ a(n)& = (m+\operatorname{wt}(n))a(\left\lfloor\frac{n}{2}\right\rfloor) \end{align} Let $$b(n) = \sum\limits_{k=0}^{n}(-1)^{\operatorname{tm}(n-k)}(\binom{n}{k}\operatorname{mod} 2)a(k)$$ Then I conjecture that \begin{align} b(0)& = 1\\ b(2n+1)& = b(n) + (m-1)(-1)^{\operatorname{tm}(n)}\\ b(2n)& = b(n) + b(n-2^{f(n)}) + b(2n-2^{f(n)}) - (m-1)(m+1)^{f(n)+1}(-1)^{\operatorname{tm}(n)} \end{align}
Is there a way to prove it?
