Suppse $n > m$ are positive integers. Then it is an elementary exercise to show that the number of $m$-tuples of non-negative integers $(x\_1, \cdots, x\_m)$ such that $x\_1 + \cdots + x\_m = n$ (order matters) is $\binom{n+m-1}{m-1}$. My question is a variant of this, where instead of considering the sum $x\_1 + \cdots + x\_m = n$ we are consider the weighted sum $w\_1 x\_1 + \cdots + w\_m x\_m = n$ for some weight vector $(w\_1, \cdots, w\_m)$, where $w\_i \geq 1, 1 \leq i \leq m$ are co-prime positive integers. 


Likely there is no explicit formula as in the case $(w\_1, \cdots, w\_m) = (1, \cdots, 1)$. However, one might expect that there will be an asymptotic formula. Is there a quick and dirty way to derive such an asymptotic formula? 



A weighted sum of non-negative integers
The quick and dirty way is to regard the weighted sum as counting the number of lattice points $(x\_2, ... x\_m)$ satisfying $w\_2 x\_2 + ... + w\_m x\_m \le n$ and $w\_1 | (n - w\_2 x\_2 - ... - w\_m x\_m)$. The second condition is satisfied $\frac{1}{w\_1}$ of the time. The number of lattice points satisfying the first condition can be approximated by the volume of the corresponding polytope, which is $\frac{n^{m-1}}{w\_2 ... w\_m (m-1)!}$, so the final asymptotic is
$$\frac{n^{m-1}}{w\_1 w\_2 ... w\_m (m-1)!} + O(n^{m-2}).$$


Here is a more rigorous derivation which also provides an "explicit" formula. The number of solutions $a\_n$ has generating function
$$A(z) = \sum a\_n z^n = \frac{1}{(1 - z^{w\_1})...(1 - z^{w\_m})}$$


and so the asymptotics of $a\_n$ are controlled by the behavior of $A(z)$ at the dominant pole $z = 1$, which has multiplicity $m$. The dominant term at this pole is (by l'Hopital's rule for example)
$$\frac{1}{w\_1 ... w\_m (1 - z)^m}$$


and the asymptotic follows. The point here is that one can also sum the contributions from the other poles and, for fixed $w\_1, ... w\_m$, the result is a completely explicit formula (polynomial-exponential) in $n$. The condition that the $w\_i$ are coprime implies that the only pole with a multiplicity of greater than $1$ is $z = 1$, so in fact to get within $O(1)$ one only needs to sum the contributions from $z = 1$ (which, I should add, are very easy to compute in any CAS that can handle power series: one just needs to substitute $x = 1 - z$, multiply by $x^m$, and then compute the first $m$ terms in the resulting series). 


