Formula for computing the coefficients of Bell polynomial I'm working on Bell polynomials and have learned some of its properties, but I've never seen any formula for calculating the coefficient in Bell polynomials. My trying to find these coefficients was useless.
For example consider the following incomplete Bell polynomial:
$$B_{4,3}(x_1,x_2)$$
Here we are going to partition a set with $4$ distinguished objects into $3$ nonempty parts. Also this $x_i$ is called a block.
Notice that this can be done such that: two blocks of size $1$ and one block of size $2$, also there exist 
$6$ cases for doing that, hence we have $$B_{4,3}(x_1,x_2)=6{x_1}^{2}x_2$$
I always do the partition using Stirling numbers of the second kind, because I know the sum of the 
coefficients in $B_{n,k}$ are $S(n,k)$, but I still can't find a formula for computing the coefficient, and 
the way I'm using Stirling numbers of the second kind is not useful always and does not give me the 
coefficient foe each term in incomplete Bell polynomials.
Any help is really highly appreciated.
 A: A relation of a complete Bell polynomial $B_n(x_1,x_2,\ldots,x_n)$ in terms of incomplete Bell polynomials $B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})$ is 
\begin{align*}
B_n(x_1,x_2,\ldots,x_n)=\sum_{k=1}^nB_{n,k}(x_1,x_2,\ldots,x_{n-k+1})\qquad n\geq 1
\end{align*}
where we conveniently set $B_0=B_{0,0}=1$.
A formula for incomplete Bell polynomials is
\begin{align*}
&B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})\\
&\qquad=\sum_{{{j_1+j_2+\ldots+j_{n-k+1}=k}\atop{j_1+2j_2+\ldots+(n-k+1)j_{n-k+1}=n}}\atop{j_1,j_2,\ldots,j_{n-k+1}\geq  0}}
\frac{n!}{j_1!j_2!\cdots j_{n-k+1}!}\left(\frac{x_1}{j_1!}\right)^{j_1}\left(\frac{x_2}{j_2!}\right)^{j_2}\cdots \left(\frac{x_{n-k+1}}{j_{n-k+1}!}\right)^{j_{n-k+1}}\tag{1}
\end{align*}
according to this Wiki-page. 
Combinatorial reasoning: $B_{n,k}$  is a multivariate polynomial which describes  the number of ways to partition an $n$-element set into $k$ non-empty partitions. The relations
\begin{align*}
j_1+j_2+\cdots+j_{n-k+1}&=k\\
j_1+2j_2+\cdots+(n-k+1)j_{n-k+1}&=n\\
j_1,j_2,\ldots,j_{n-k+1}&\geq 0
\end{align*}
stated in the the index region show that we consider precisely the number of ways with $j_1$ one-element partitions, $j_2$ two-element partitions up to $j_{n-k+1}$ $(n-k+1)$-element partitions, where the numbers $j_1,j_2,\ldots j_{n-k+1}$ are non-negative integers. This situation is marked by the $n-k+1$ variables
\begin{align*}
x_1^{j_1}x_2^{j_2}\cdots x_{n-k+1}^{j_{n-k+1}}
\end{align*}
where the index indicates the number of elements of the addressed partition and the exponent  the number of partitions with the   corresponding   size. The  number  of ways  with this structure  is given by
\begin{align*}
\frac{n!}{j_1!1!^{j_1}j_2!2!^{j_2}\cdots j_{n-k+1}!(n-k+1)!^{j_{n-k+1}}}
\end{align*}


*

*We have $n!$ ways to permute the $n$ elements of the set.

*There are $j_m, 1\leq m\leq n-k+1$ partitions with $m$ elements each. Since the order of these $j_m$ partitions is not relevant we identify these cases by dividing by $j_m!, 1\leq m\leq n-k+1$.

*There are $m, 1\leq m\leq n-k+1$ elements in each of the $j_m$ partitions. Since the order of these $m$ elements is not relevant we identity these $m$ elements for each $j$ by dividing by $m!^{j_m}$.

We obtain from (1)
  \begin{align*}
\color{blue}{B_{4,3}(x_1,x_2)}&=\sum_{{{j_1+j_2=3}\atop{j_1+2j_2=4}}\atop{j_1,j_2\geq 0}}\frac{4!}{j_1!j_2!}\left(\frac{x_1}{1!}\right)^{j_1}\left(\frac{x_2}{2!}\right)^{j_2}\\
&=\frac{4!}{2!1!}\left(\frac{x_1}{1!}\right)^2\left(\frac{x_2}{2!}\right)^1\\
&\,\,\color{blue}{=6x_1^2x_2}
\end{align*}
  We observe there is only one pair $(j_1,j_2)=(2,1)$ which fulfills the conditions of the index region. This corresponds to the following $6$ structures of a set $\{a_1,a_2,a_3,a_4\}$ partitioned into three blocks with two of size $1$ and one of size $2$:
  \begin{align*}
&\{\{a_1\},\{a_2\},\{a_3,a_4\}\}, \qquad\{\{a_2\},\{a_3\},\{a_1,a_4\}\},\\
&\{\{a_1\},\{a_3\},\{a_2,a_4\}\}, \qquad\{\{a_2\},\{a_4\},\{a_1,a_3\}\},\\
&\{\{a_1\},\{a_4\},\{a_2,a_3\}\}, \qquad\{\{a_3\},\{a_4\},\{a_1,a_2\}\}\\
\end{align*}

Note: The Stirling numbers of the second kind ${n\brace k}$ are not fine-grained enough to derive the coefficients of the incomplete partial Bell polynomials. They give us the number of partitions of an $n$-element set into $k$ non-empty blocks, but they don't tell us how many elements are in each of these blocks.
When we look for instance at
\begin{align*}
B_4(x_1,x_2,x_3,x_4)=x_1^4+6x_1^2x_2+\color{blue}{4}x_1x_3+\color{blue}{3}x_2^2+x^4
\end{align*}
then the blue marked coefficients $4$ and $3$ correspond to the Stirling number ${4 \brace 3}=\color{blue}{7}$, which tells us there are $7$ partitions of a $4$-element set into $3$ non-empty blocks. But ${4\brace 3}=7$ don't recognise the split into two blocks of type $x_1x_3$ and of type $x_2x_2$ respectively.
