Let $q(n)$ be A007814, i.e., the number of trailing zeros in the binary representation of $n$. Here $$q(2n+1)=0, q(2n)=q(n)+1$$
Let $a(n)$ be A329369. Here $$a(2n+1)=a(n), a(2n)=a(n)+a(n-2^{q(n)})+a(2n-2^{q(n)}), a(0)=1$$
We have $$a(2^m(2^n-1))=\sum\limits_{i=1}^{n+1}i!i^mS(n+1,i)(-1)^{n-i+1}$$ Here $n\geqslant 0$, $m\geqslant 0$ and $S(n,k)$ is a Stirling number of the second kind.
You can see the proof by R. Ehrenborg and E. Steingrimsson (equivalent to proposition 6.5 on page 297, see the link in A329369).
I tried to solve this question in general, so at least I conjecture that $$a(2^m(2^n-2^p-1))=\sum\limits_{i=1}^{n}i!i^m(-1)^{n-i}\left((i-p+1)S(n,i)-S(n-p,i-p)+\sum\limits_{j=0}^{p-2}\frac{p-j-1}{j!}S(n-p,i-j)\sum\limits_{k=0}^{j}(i-k)^p\binom{j}{k}(-1)^k\right)$$ Here $n>2$, $m\geqslant 0$, $0<p<n-1$ and also we consider that $S(n,k)=0$ for $n\geqslant 0$, $k<0$.
Here is the PARI/GP program to check it numerically:
a(n) = my(n = (n + 1)/2^valuation(n + 1, 2), A = n, B, C, v = [], v1); while(A > 0, B = valuation(A, 2); v = concat(v, B + 1); A \= 2^(B + 1)); v = Vecrev(v); A = #v; if(n == 1, 1, v1 = vector(A, i, A - i + 1); for(i = 1, A - 1, B = A - i; for(j = 1, B, C = B - j + 2; v1[j] = v1[j] * C ^ v[B] - v1[j + 1] * (C - 1) ^ v[B])); v1[1])
S(n, k) = if(k < 0, 0, stirling(n, k, 2))
b1(n, m, p) = sum(i = 1, n, i! * i ^ m * (-1) ^ (n - i) * ((i - p + 1)*S(n, i) - S(n - p, i - p) + sum(j = 0, p - 2, (p - j - 1) / j! * S(n - p, i - j)*sum(k = 0, j, (i - k) ^ p * binomial(j, k) * (-1) ^ k))))
test(n, m, p) = b1(n, m, p) == a(2 ^ m * (2 ^ n - 2 ^ p - 1))
Is there a way to prove it? Is there any way to simplify the last expression?
