Let $$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$ $$f(n)=n-2^{\ell(n)}$$ $$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$ $$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \operatorname{wt}(2n)=\operatorname{wt}(n), \operatorname{wt}(0)=0$$ $$\operatorname{tr}(2n+1)=0, \operatorname{tr}(2n)=\operatorname{tr}(n)+1$$ Here $f(n)$ is the distance to largest power of $2$ less than or equal to $n$, $T(n,k)$ is the $(k+1)$-th bit from the right side in the binary representation of $n$, $\operatorname{wt}(n)$ is the binary weight of $n$ and $\operatorname{tr}(n)$ is the number of trailing zeros in the binary representaton of $n$.
Let $a(n)$ be a sequence of positive integers such that $$a(2n+1)=a(n), a(2n)=a(n)+a(n-2^{\operatorname{tr}(n)})+a(2n-2^{\operatorname{tr}(n)}), a(0)=1$$ Here $a(n)$ is A329369, i.e., number of permutations of $\left\lbrace1,2,\cdots,m\right\rbrace$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b_{i-1} = 1$ where $b_{k}b_{k-1}\cdots b_{1}b_{0}$ ($0 \leqslant k < m-1$) is the binary expansion of $n$. The excedance set of a permutation $\pi$ of $\left\lbrace1,2,\cdots,m\right\rbrace$ is the set of indices $i$ such that $\pi_i > i$; it is a subset of $\left\lbrace1,2,\cdots,m-1\right\rbrace$.
I conjecture that there exist recurrence such that for $n>1$ $$a(n)=(\ell(n)-\operatorname{wt}(n)+2)\cdot a(f(n))+\sum\limits_{k=0}^{\ell(n)-1} (1-T(n,k))\cdot a(f(n) + 2^{k}(1 - T(n,k)))$$ There are no counterexamples up to $10^6$.
Is there a way to prove it?
