Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\cdots,m\}$ 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 $n$). Here $$ a(2^m(2k+1)) = \sum\limits_{j=0}^{m}\binom{m+1}{j}a(2^jk), \\ a(0) = 1 $$
Let $\nu_2(n)$ be A007814 (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 $\operatorname{wt}(n)$ be A000120 (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 $b(n)$ be an integer sequence such that we start with $A=2n+1, B$ and vector $s$ of length $\operatorname{wt}(n)+1$ and for $i$ from $1$ to $\operatorname{wt}(n)+1$ consecutively apply $B := \nu_2(A)$, $s_i := B+1$, $A := \left\lfloor\frac{A}{2^{B+1}}\right\rfloor$. After that, we reverse vector $s$ and create vector $t$ of length $\operatorname{wt}(n)+1$ with elements $t_i=1$. Then we start with $A = \operatorname{wt}(n)+1, B, C$ and for $i$ from $1$ to $A-1$ with $B := A - i$ and for $j$ from $1$ to $B$ consecutively apply $C := B-j+2$, $t_j := t_jC^{s_B} - t_{j+1}(C-1)^{s_B}$. Finally, $b(n) = t_1$ after the whole transformation.
I conjecture that $$b(n)=a(n).$$
Here is the PARI/GP program to check it numerically:
a(n) = if(n == 0, 1, my(A = valuation(n, 2), B = n >> (A+1)); sum(j=0, A, binomial(A+1, j)*a(B << j)))
b(n) = my(A = 2*n+1, B, C, v1, v2); v1 = []; while(A > 0, B = valuation(A, 2); v1 = concat(v1, B+1); A \= 2^(B+1)); v1 = Vecrev(v1); A = #v1; v2 = vector(A, i, 1); for(i=1, A-1, B = A-i; for(j=1, B, C = B-j+2; v2[j] = v2[j]*C^v1[B] - v2[j+1]*(C-1)^v1[B])); v2[1]
test(n) = b(n) == a(n)
Note that this algorithm has similarities to the Gandhi polynomials used to calculate the Genocchi numbers (subsequece $a\left(\frac{4^n-1}{3}\right)$).
Is there a way to prove it?
