Simplifying iterated sums where the upper bound is the next index

Question

Consider first the iterated integral $I=m!\int_0^x\int_0^{x_1} ... \int_0^{x_{m-1}} \prod_{i=1}^m dx_i f(x_i)$.

By permuting the dummy indices, we can reorder $x_1$ to $x_m$ to find new expressions for $I$. By adding all of these expressions together and dividing by $m!$, we recover:

$$m!\int_0^x\int_0^{x_1} ... \int_0^{x_{m-1}}\prod_{i=1}^m dx_i f(x_i) = \left(\int_0^x dy f(y)\right)^m.$$

Now consider the sum: $$S = m! \sum_{n_1=0}^n \sum_{n_2=0}^{n_1} ... \sum_{n_m=0}^{n_{m-1}} \prod_{i=1}^m a_{n_i}$$

Using the same tactic used on the integrals for $S$ for the case of $m=2$, I find that $$2!\sum_{n_1=0}^n \sum_{n_2=0}^{n_1} a_{n_1} a_{n_2} = \left(\sum_{j=0}^n a_j\right)^2 + \left(\sum_{j=0}^n a_j^2\right)$$ so it is clear that there are necessarily extra terms relative to the integral case.

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How can I simplify the form of $S$? Since the choice of $a_n = 1$ simplifies nicely for general $m$ because of the hockey-stick identity, perhaps there is a straightforward combinatorical path.

Answer 1

Expressing $S_m$ in terms of $P_m$ where, for $a_0,\ldots,a_n$ considered fixed, $$S_m:=\sum_{0\leqslant k_1\leqslant\ldots\leqslant k_m\leqslant n}\prod_{j=1}^m a_{k_j},\qquad P_m:=\sum_{k=0}^n a_k^m,$$ can be done using generating functions. With $S_0:=1$, we have $$\sum_{m=0}^\infty S_m z^m=\prod_{k=0}^n\frac{1}{1-a_k z}=\exp\sum_{k=0}^n\big(-\log(1-a_k z)\big)=\exp\sum_{m=1}^\infty\frac{P_m}{m}z^m,$$ which gives the following formula: $$S_m=\sum_{\substack{n_1,n_2,\ldots,n_m\geqslant 0\\n_1+2n_2+\ldots+mn_m=m}}\prod_{k=1}^m\frac{(P_k/k)^{m_k}}{m_k!}.\tag{*}\label{result}$$

Here are particular cases, continuing yours: \begin{align*} 1!S_1&=P_1 \\\color{blue}{2!S_2}&=\color{blue}{P_1^2+P_2} \\3!S_3&=P_1^3+3P_1P_2+2P_3 \\4!S_4&=P_1^4+6P_1^2P_2+3P_2^2+8P_1P_3+6P_4 \\5!S_5&=P_1^5+10P_1^3P_2+15P_1P_2^2+20P_1^2P_3+20P_2P_3+30P_1P_4+24P_5 \end{align*}