Unique decomposition of $c$ sums of products of $k$ prime numbers, allowing duplicates? Suppose that there are $n$ different prime numbers. Define procedure a) as following ($k \leq n$ and $k$ fixed): procedure a) for each time, we select one number out of $n$ possible cases and multiply it to the product (number may be repeated). For the first time, we choose a number and that is the product. We do this $k$ times.
Then, we do the following for $c \leq n^n$ times where $c$ is fixed: procedure b) we define variable $z$. For the first time procedure b) is started, $z = 0$. Now call procedure a) and get a result into variable $y$. Then we add $y$ to $z$ - and the sum replaces the original value of $z$ into $z$. We repeat this $c$ times.
So, step is: we call procedure b) (which then recursively calls procedure a) ) and see the result $z$.
Now the question is, is this $z$ unique in its decomposition, given $k$ and $c$? (So, there would be $c$ sums of product of $k$ prime numbers, allowing duplicates.)
Or if not, what would be the general rule when this holds?
For every step, the set of $n$ prime numbers remains the same - that is, it does not change.
 A: If $c=1$ then the decomposition is unique (up to ordering), as the selection of primes gives the prime factorization of $z$.
If $c>1$ then in general the decomposition is not unique. In the simplest case $k=1,c=2$ you consider the number of ways $z$ can be written as the sum of two primes (Goldbach partitions), e.g. let your primes be $\{5,11,13,19\}$ then $z=24=5+19=11+13$.
In general procedure a) has $N=\binom{n+k-1}{k}\simeq n^k/k!$ possible results (by stars-and-bars), and as multisets procedure b) has $$M=\binom{N+c-1}{c}\simeq \frac{n^{kc}}{(k!)^cc!}$$
If $L$ and $U$ are your smallest and largest primes, then
$$
cL^k\le z \le cU^k
$$
and a counting argument shows in general they can't all be distinct. For example, let your primes be $\{29,31,37,41,43,47,53,59,61,67,71\}$, $n=10,k=5,c=5$.
Then
$$
N=\binom{14}{5}=2002 \\
M=\binom{2006}{5}=2.69\times 10^{14} \\
cL^k=5\cdot 29^5 = 1.026\times 10^8, \quad cU^k=5\cdot 71^5 = 9.02\times 10^9
$$
So procedure b) has $>10^{14}$ possible selections from procedure a), but can only output $<10^{10}$ possible values, so some values must have a large number of decompositions.
