Summing over set of sets I am having trouble trying to understand an expression based on the multivariate Taylor series which is used to generate a system of residual equations. The expression is from Chauhan (2017) Benchmarking Approaches for the Multidisciplinary Analysis of Complex Systems Using a Taylor Series-Based Scalable Problem and the expression is
\begin{align}
R_i(v_1,\dots,v_n)=\sum\limits_{r=1}^{d_i}\frac{1}{r!}\sum\limits_{(j_1,\cdots,j_r)\\1\leq j_k\leq n\\j_1,\cdots,j_r \in A(i)} \frac{\partial^rR_i}{\partial v_{j_i}\cdots\partial v_{j_r}} \prod\limits_{k=1}^{r}v_{j_k}
\end{align}
where $n$ is the number of variables, $d_i$ is the degree of the polynomial, and $A(i)$ are "arguments of the $i^{th}$ equation; $A:\{1,\cdots,n\}\rightarrow P(\{1,\cdots,n\})$ where $P()$ is the power set". From this description I assume $A(i)$ is a function that returns the power set for a given list of elements, e.g. $A([1,2])=\{\{\},\{1\},\{2\},\{1,2\}\}$. 
My point of confusion is the summation term and its three conditions. Is it a single sum over all elements $(j_1,\cdots,j_r)$ in the power set, giving four terms in the sum? Or is it a nested sum $\sum\limits_{j_1}\sum\limits_{j_2}\cdots\sum\limits_{j_r}$ where each sum is performed over the sets in the power set? How would the empty set or a set like $j_k=\{1,2\}$ be substituted as a single index into $v_{j_k}$?
 A: Here   are  some   hints  which might be useful.

The following representations   are  equivalent
  \begin{align*}
\sum_{1\leq  j_1,j_2\leq n}a_{j_{1},j_{2}}&=\sum_{j_1=1}^n\left(\sum_{j_2=1}^n   a_{j_1,j_2}\right)\\
&=\sum_{j_2=1}^n\left(\sum_{j_1=1}^na_{j_{1},j_{2}}\right)\\
&=\sum_{{(j_1,j_2)}\atop{1\leq j_k\leq n}}a_{j_1,j_2}\tag{1}\\
&=\sum_{{1\leq  j_k\leq n}\atop  {1\leq  k\leq 2}}a_{j_1,j_2}\tag{2}
\end{align*}
  
  
*
  
*Note that (1)  tells  us  to sum up over all  tupels  $(j_1,j_2)$ which gives  us  just the information how many different summation indices are in use.
  
*Precisely the same is stated in (2) when we list the valid indices $j_k$ via $1\leq k\leq 2$.

Let's  have a  look at  the inequality chain $1\leq    j_k\leq n$.


*

*The inequality chain implies that $j_k$ are positive integer values fulfilling an order relation. So, the $j_k$ cannot be sets.

*The relation $j_1,\ldots,j_r\in A(i)$ tells us due to the element-operator $\in$ that $A(i)$ is a set containing integer values. It is not a set of sets.

Example: We take $n=5,r=3, i=1, A(1)=\{1,3\}$.
We obtain
  \begin{align*}
\sum\limits_{{{(j_1,\cdots,j_r)}\atop{1\leq j_k\leq n}}\atop{j_1,\cdots,j_r \in A(i) }}a_{j_1,j_2,\ldots,j_r}
&=\sum\limits_{{{(j_1,j_2,j_3)}\atop{1\leq j_k\leq 5}}\atop{j_1,j_2,j_3\in \{1,3\} }}a_{j_1,j_2,j_3}\\
&=a_{1,1,1}+a_{1,1,3}+a_{1,3,1}+a_{1,3,3}\\
&\qquad+a_{3,1,1}+a_{3,1,3}+a_{3,3,1}+a_{3,3,3}
\end{align*}
Note that   although $1\leq j_k\leq  5$     with  $1\leq k\leq 3$ we have  only a small number of terms due    to $j_k\in\{1,3\}$. 

