A set of relative prime, positive integers $A = [a\_1, \dots, a\_d]$ describe the restricted partition function
$$ 
p\_A(n) = \# \{(m\_1,\dots,m\_d)\in\mathbb{Z}^d: \textrm{ all }m\_j \geq 0, \sum\_{j=1}^d m\_j a\_j = n \}
$$
I'm interested in a binary version of $p\_A(n)$, namely
$$
 q\_A(n) = 
\begin{cases} 
1 \textrm{ if } p\_A(n) \geq 1 \\ 
0 \textrm{ if } p\_A(n) = 0
\end{cases}
$$
Let $P\_A(n)$ and $Q\_A(n)$ be the polynomial parts of $p\_A(n)$ and $q\_A(n)$, respectively. In fact, $p\_A(n)$ is polynomial except the last coefficient term. Let us interpret $P\_A(n)$ as the expected number of solution to the restricted partition and $Q\_A(n)$ as the probability that there is a solution. *Without* the underlying structure, we could relate these quantities by:
$$
Q\_A(n) = 1 - \exp ( - P\_A(n) )
$$



> 
> Given a specified $\alpha$, how to choose $A$ with $\alpha \approx \prod\_{j=1}^N a\_j$ such that $Q\_A(n)$ is close to $1 - \exp ( - P\_A(n) )$?
> 
> 
> 


Example: $A = [49,51,100]$ is a bad choice, because $100$ gets "eaten" up by $49$ and $51$. Whereas $A = [49,63,81]$ is a better choice.


As is apparent, I am missing quite some terminology here, for example is there a name for $q\_A(n)$?



Relation to Ehrhart polynomial with Uniqueness
Your function $q\_A(n)$ is an indicator whether or not $n$ is representable as a nonnegative combination of $A$ and thus connected to the infamous *Frobenius problem* (also known as the *coin-exchange problem* and the *chicken nuggets progblem*): given $A$ (relatively prime), which is the largest integer that is not representable? Knowing $q\_A(n)$ or, equivalently, its generating function $f\_A(x) := \sum\_{n \ge 0} q\_A(n) \, x^n$, would mean you know everything needed for the Frobenius problem (and more): its solution is the degree of the polynomial $\frac 1 {1-x} - f\_A(x)$.


It's not hard to see that $f\_A(x)$ is a rational function with denominator $\prod\_{a \in A} (1-x^a)$, and it's a fun exercise to prove that $f\_{a\_1,a\_2}(x) = \frac{ 1 - x^{a\_1 a\_2} }{ (1-x^{a\_1})(1-x^{a\_2}) }$; you can then derive that $a\_1 a\_2 - a\_1 - a\_2$ is the solution to the Frobenius problem for $|A|=2$. There are beautiful (and almost as explicit) formulas for $f\_A(x)$ when $|A|=3$, due to Marcel Morales [J. Algebra 140 (1991), no. 1, 12–25] and Graham Denham [Electron. J. Combin. 10 (2003), Research Paper 36]; one amusing (and computationally important) consequence is that the numerator has either 4 or 6 terms, for any $|A|=3$. In sharp contrast, Henrik Bresinsky [Manuscripta Math. 17 (1975), no. 3, 205–219] proved that for $|A| \ge 4$ there is no absolute bound on the number of terms in the numerator of $f\_A(x)$.


