Let $B \subset \mathbb{Z}^+$. Define $r\_{B,h}(n)$ to be the number of ways of writing $n$ as the sum of $h$ elements of $B$ and $R\_{B,h}(n)$ the number of ways to write $n$ as the sum of $h$ DISTINCT elements of $B$. In many applications, such as the Erdos-Tetalli theorem which finds a set $B$ such that $R\_{B,h}(n) = \Theta(\log(n))$ it is more convenient to work with $R\_{B,h}(n)$. The reason why such convenience does not affect the results is because in general, the number of elements in $B^h$ that sum to $n$ counted by $r\_{B,h}(n)$ but not by $R\_{B,h}(n)$, namely those with repeat summands, is negligible. 


To illustrate with a fairly concrete example, consider the famous Goldbach Conjecture which asserts that every positive even integer larger than 2 can be written as the sum of two primes. In other words if $B$ is the set of primes, then Goldbach Conjecture is the assertion that $r\_{B,2}(2n) > 0$ for all $n > 1$. But the truth is that one expects $r\_{B,2}(2n)$ to tend to infinity. In this case the number of sums with repeat summands are precisely to write $2p = p + p$ for some prime $p$, and if $r\_{B,2}(2n)$ does indeed tend to infinity, then this is a complete triviality to replace $r\_{B,2}(2n)$ with $R\_{B,2}(2n)$, since $0 \leq r\_{B,2}(2n) - R\_{B,2}(2n) \leq 1$ for all $n$. 


So my question is, is there some general argument for this observation? That is, in a sufficiently general setting, one can essentially assume that the summands are distinct.


I note that in some very trivial cases this assumption is not appropriate at all. For example, consider $B =$ {$3k : k \in \mathbb{N}$} $\cup$ {$0,1$}. Then $B$ is an asymptotic basis of order 3, but it would NOT be a basis at all if we demanded that the summands be unique. This of course is a somewhat contrived example, since for a set of positive density the number of representations is very small. So of course our criteria for a 'sufficiently general setting' would have to exclude such trivial cases.



Any rigorous way to claim that sums with repeat summands are few?
Here is a fairly crude first attempt.


First note that for any basis $B$, when $h=2$, we have


$$0\leq r\_{B,2}(n)-R\_{B,2}(n)\leq 1$$


for all n, as you note in your question. Hence assume $h\geq 3$. In this case, if $r'\_{B,h}(n)=r\_{B,h}(n)-R\_{B,h}(n)$ counts the number of representations where some elements are identical, then


$$ r'\_{B,h}(n)=\sum\_{m\in 2\cdot(B\cap[1,n])}r\_{B,h-2}(n-m)\ll\lvert B\_n\rvert r\_{B,h-2}(n).$$


Hence if we can show that this upper bound is $o(r\_{B,h}(n))$, then we have $R\_{B,h}(n)\sim r\_{B,h}(n)$ as required. 


For example, note that this shows that distinct summands dominate in the case of Waring bases, where $\lvert B\_n\rvert\approx n^{1/k}$ and $r\_{B,h}(n)\approx n^{h/k-1}$.


Also note that it deals with e.g. the ternary Goldbach case, where $\lvert B\_n\rvert\approx n/\log n$ and $r\_{B,3}\gg n^2$. 


This does not, however, deal with thin bases, where $r\_{B,h}(n)\approx\log n$ and $\lvert B\_n\rvert\approx n^{1/k}$. For such cases you may need to rely on probabilistic arguments, as in the proof of the Erdos-Tetalli theorem.


