Let $p, q \in \mathbb{Z}$.

Let $\operatorname{wt}(n)$ is A000120, number of $1$'s in binary expansion of $n$ (or the binary weight of $n$) and

$$n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_{wt(n)}+1}))\dots)$$

Then we have an integer sequence given by \begin{align} a(0, m)& = 1\\ a(n, m)& = \sum\limits_{j=0}^{2^{\operatorname{wt}(n)}-1}m^{\operatorname{wt}(n)-\operatorname{wt}(j)}\prod\limits_{k=0}^{\operatorname{wt}(n)-1}(1+\operatorname{wt}(\left\lfloor\frac{j}{2^k}\right\rfloor))^{t_{k+1}+1} \end{align} I conjecture that $$a(n, -1) = \sum\limits_{j=0}^{2^{\operatorname{wt}(n)}-1}(-1)^{\operatorname{wt}(n)-\operatorname{wt}(j)}a(f(n,j),0)$$ and $$a(n, 0) = \sum\limits_{j=0}^{2^{\operatorname{wt}(n)}-1}a(f(n,j),-1)$$ where $a(n,-1)$ is A329369, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)...b(1)b(0)$ ($0 \leqslant k < m-1$) is the binary expansion of $n$. The excedance set of a permutation $p$ of ${1,2,...,m}$ is the set of indices $i$ such that $p(i) > i$; it is a subset of ${1,2,...,m-1}$.

and $a(n,0)$ is A284005, \begin{align} a(0, 0)& = 1\\ a(n,0)& = (1+\operatorname{wt}(n))a(\left\lfloor\frac{n}{2}\right\rfloor,0) \end{align}

and finally $f(n,k)$ is A295989, irregular triangle $T(n, k)$, read by rows, $n \geqslant 0$ and $0 \leqslant k <$ A001316$(n)$: $T(n, k)$ is the $(k+1)$-th nonnegative number $m$ such that $n \operatorname{AND} m = m$ (where $\operatorname{AND}$ denotes the bitwise $\operatorname{AND}$ operator).

\begin{align} T(n, 0)& = 0\\ T(2n, k)& = 2T(n,k)\\ T(2n+1, 2k)& = 2T(n,k)\\ T(2n+1, 2k+1)& = 2T(n,k) + 1 \end{align}

I also conjecture that

$$a(n, p) = \sum\limits_{j=0}^{2^{\operatorname{wt}(n)}-1}(p-q)^{\operatorname{wt}(n)-\operatorname{wt}(j)}a(f(n,j),q)$$

Is there a way to prove it?