Simplification of the closed form for the A329369

Question

• Let $s(n,k)$ be a (signed) Stirling number of the first kind.

• Let ${n \brace k}$ be a Stirling number of the second kind.

• Let

$$ f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace j} $$

• Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k < m-1$) 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

$$ n = 2^m(2^{n_1}(2^{n_2}(\cdots(2^{n_{p-1}}(2^{n_p} - 1) - 1)\cdots) - 1) - 1) $$

I conjecture that

$$ a(n) = \sum\limits_{i_1=1}^{m+1} f(n_1, m, i_1) \sum\limits_{i_2=1}^{i_1+1} f(n_2-1, i_1, i_2) \sum\limits_{i_3=1}^{i_2+1} f(n_3-1, i_2, i_3) \cdots \sum\limits_{i_p=1}^{i_{p-1}+1} f(n_p-1, i_{p-1}, i_p) $$

Is there a way to simplify the last expression?

Answer 1

As it was noted in another answer, we have $$f(n,m,i) =[\tfrac{x^n}{n!}\tfrac{y^{m+1}}{(m+1)!}]\ \frac{\big(-\log(1+e^x(e^{-y}-1))\big)^i}{i!}.$$

The linked answer essentially establishes the same recurrence for $f(n,m,i)$ as satisfied by $a(n)$, and the conjecture then easily follows by induction on $p$.

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UPDATED. Now, let's simplify the given iterated sum expression.

Since $\log(1+e^x(e^{-y}-1))$ is a multiple of $y$, it follows that all summation bounds can be replaced with 0 and $+\infty$, respectively. Then $$S_p:=\sum_{i_p} f(n_p-1, i_{p-1}, i_p) = [\tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-1}+1}}{(i_{p-1}+1)!}]\ (1+e^{x_p}(e^{-y}-1))^{-1}$$ and correspondingly \begin{split} S_{p-1} &:= \sum_{i_{p-1}} f(n_{p-1}-1, i_{p-2}, i_{p-1}) \sum_{i_p} f(n_p-1, i_{p-1}, i_p) \\ &= [\tfrac{x_{p-1}^{n_{p-1}-1}}{(n_{p-1}-1)!} \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-2}+1}}{(i_{p-2}+1)!}\tfrac{t^{i_{p-1}+1}}{(i_{p-1}+1)!}]\ (1+e^{x_p}(e^{-t}-1))^{-1} \tfrac{\big(-\log(1+e^x(e^{-y}-1))\big)^{i_{p-1}}}{i_{p-1}!} \\ &= [\tfrac{x_{p-1}^{n_{p-1}-1}}{(n_{p-1}-1)!} \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-2}+1}}{(i_{p-2}+1)!}]\ \left.\tfrac{\partial}{\partial t} (1+e^{x_p}(e^{-t}-1))^{-1} \right|_{t=-\log(1+e^x(e^{-y}-1))} \\ &= [\tfrac{x_{p-1}^{n_{p-1}-1}}{(n_{p-1}-1)!} \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-2}+1}}{(i_{p-2}+1)!}]\ (e^{x_p}-1)g_{p-1}^{-2} + g_{p-1}^{-1}, \end{split} where $$g_k := 1+e^{x_k+\dots+x_p}(e^{-y}-1).$$ In general, $S_k$ (ie. the sum over $i_k,\dots,i_p$) is given by a linear combination of $g_k^{k-p-1}, g_k^{k-p}, \dots, g_k^{-1}$ with coefficients depending on $x_{k+1},\dots,x_p$ but not on $y$. Furthermore, $S_k$ can be obtained from $S_{k+1}$ by replacing each $g_{k+1}^{-\ell}$ with $\ell(e^{x_{k+1}+\dots+x_p}-1)g_k^{-\ell-1} + \ell g_k^{-\ell}$.

Then the iterated sum in question is obtained as the coefficient of $\tfrac{x_1^{n_1-1}}{(n_1-1)!}\cdots \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{m+1}}{(m+1)!}$ in $S_1$. It may be possible that $S_1$ has a simpler description but I have not found one yet.