- Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\dotsc,m\}$ 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 $$ a(2^m(2k+1)) = \sum\limits_{j=0}^{m} \binom{m+1}{j}a(2^jk), \\ a(0) = 1. $$ We have $$ a(2^m(2^n-1))=\sum\limits_{i=1}^{n+1}i!i^m{n+1 \brace i}(-1)^{n-i+1}. $$ You can see the proof by R. Ehrenborg and E. Steingrimsson (equivalent to proposition 6.5 on page 297, see the link in A329369).
- 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 $T(n,k)$ be an integer coefficients (A358612) such that $$ T(2n+1, k) = kT(n, k) + T(n, k-1), \\ T(2n, k) = kT(n, k) + T(n, k-1) - \frac{T(2n, k-1) + T(n, k-1)}{k-1}, \\ T(n, 1) = T(0, 2) = 1. $$
- Let $s(m,k)$ be the family of integer sequences such that $$ s(m,k)=\sum\limits_{i=1}^{\operatorname{wt}(k)+2}i!i^mT(k,i)(-1)^{\operatorname{wt}(k)-i}. $$
I conjecture that $$ s(m,k) = a(2^m(2k+1)). $$
Note that for $2k+1=2^n-1$ we have $\operatorname{wt}(k)+2=n+1$, $T(k,i)={n+1 \brace i}$, so it completely agrees with observation given above.
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)))
row1(n) = my(A, v1, v2); v1 = [1, 1]; if(n==0, v1, forstep(i=logint(n,2), 0, -1, A = bittest(n,i); v2 = vector(#v1+A, i, 0); v2[1] = 1; for(j=2, #v2, v2[j] = j*if(j==#v1+1, 0, v1[j]) + v1[j-1] - if(A, 0, (v2[j-1] + v1[j-1])/(j-1))); v1 = v2); v1)
s(m,k) = my(v1); v1 = row1(k); sum(i=1, #v1, i!*i^m*v1[i]*(-1)^(#v1-i))
test1(m,k) = s(m,k) == a(2^m*(2*k+1))
Is there a way to prove it?
