Q: Integer partitions under divisibility constraint Consider integer partitions of $x \in \mathbb{N}$ of size $k$ under the constraint that the partition elements are distinct and the ratio of any element to each smaller element is a natural number.  
Example $f(x,k)$:


*

*$f(17,3) = \{ 1, 4, 12\}$

*$f(14,3) = \{ 2, 4, 8 \}$

*$f(101,4) = \{ 1, 2, 14, 84 \}$

*$f(4,3) = \emptyset$
As pointed out by @Henrik, one can express one constraint as:  $x = a_1 + a_1 a_2 + \ldots + a_1\cdots a_k$ and then factor $x$, $x-a_1$, $x - a_1 - a_1 a_2$, etc. to find candidates for successive $a_i$.  (In the special case of $x$ being prime and $k>1$, then $a_1 = 1$.) However, because there are typically several choices for each successive $a_i$, perhaps one must use some sophisticated search (with backtracking) or variant on linear programming.
Have these partitions (or series) been studied?  Is there an efficient algorithm or method for finding them, given $x$ and $k$?  When might $f(x,k)$ be empty or not unique?
Influenced by the comment from @Gerhard "always has a clever middle name" Paseman, I thought I'd plot candidates for the case $f(28,2)$.  The abscissa is $a_1$ (whose value must be a factor of $28$, i.e., $1, 2, 4, 7, 14, 28$), and the ordinate $a_1 a_2$.  Because $a_1 + a_1 a_2$ must ultimately equal $n = 28$, we need consider the smallest partition in the range $0 < a_1 \leq \lfloor {n \over k+1} \rfloor = 9$ (marked by the non-gray region).  Because the $a_2 \geq 2$, the region of candidates for $a_1 a_2$ must be $\geq 2 a_1$, as shown in yellow.
Thus we seek a point $(a_1, a_1 a_2)$ on the line $a_1 + a_1 a_2 = n$, as shown by the black line.
For this case, the only solution is $(1, 27)$.
I'm not quite sure how this helps in finding an efficient algorithm, but my problem-solving style is to visualize or graph as much as possible, so perhaps this will shed light for someone else.

Here's a table of candidates for $f(25,3)$, showing there are four solutions:  $\{ 1, 6, 18\}$, $\{ 1, 2, 22\}$, $\{ 1, 4, 20\}$, $\{ 1, 8, 16 \}$.  (Actually, we can eliminate the case $a_1 = 5$ a priori, but I include it for completeness.)
$$
\begin{array}{|r|c|l|} \hline
a_1 & a_1 a_2 & a_1 a_2 a_3 \\ \hline
1 & 2 & 4,8, 12, 16, 20, {\bf 22} \\
  & 4 & 8, 12, 16, {\bf 20} \\
  & 6 & 12, {\bf 18} \\
  & 8 & {\bf 16} \\ \hline
5 & 10 & 20 \\ \hline
\end{array}
$$
