Number of solutions of $ax+by+cz=n$ in nonnegative integers Let $a,b,c,n\in\mathbb{Z}_0^+$. The number of solutions of
$$ax+by+cz=n$$
(i.e. the number of ordered triplets $(x,y,z)$ satisfying the equation) in $\mathbb{Z}_0^+$ as a function of $n$ is
$$\frac{1}{n!}\lim_{w\to 0}\frac{d^n}{dw^n}\frac{1}{(1-w^a)(1-w^b)(1-w^c)}=\frac{1}{2\pi i}\oint \frac{dw}{(1-w^a)(1-w^b)(1-w^c)w^{n+1}}$$
by the theory of generating functions and Cauchy's integral formula.
If $a,b,c$ are small, then the $n$th derivative/contour integral is amenable for computation using partial fraction decomposition. But for large $a,b,c$, the partial fraction decomposition becomes very tedious.
Are there any faster methods of computing this particular integral, or faster methods of obtaining the number of solutions in general?

The topic is standardly known as the Frobenius coin problem, and
it is quite complicated in general.
In the 2D case $ax+by=n$ there is an interesting theorem, the Popovicius' theorem
which tells that the number of non-negative solutions $ p_{\left\{ {a,b} \right\}} (n)$ is given by
$$
\eqalign{
  & p_{\left\{ {a,b} \right\}} (n) = \left| {\,\left\{ \matrix{
  0 \le x,y,a,b,n \in Z \hfill \cr 
  \gcd (a,b) = 1 \hfill \cr 
  ax + by = n \hfill \cr}  \right.\;} \right| =   \cr 
  &  = {n \over {ab}} - \left\{ {{{b^{\,\left( { - 1} \right)} n} \over a}} \right\}
 - \left\{ {{{a^{\,\left( { - 1} \right)} n} \over b}} \right\} + 1 \cr} 
$$
where
$$
\left\{ x \right\} = x - \left\lfloor x \right\rfloor
 \quad b^{\,\left( { - 1} \right)} b \equiv 1\;\left( {\bmod a} \right)
\quad a^{\,\left( { - 1} \right)} a \equiv 1\;\left( {\bmod b} \right)
$$
For the 3D case, like the present, we can apply the above to $ax+by=n-cz$ and sum over $z$.
$$
\eqalign{
  & p_{\left\{ {a,b,\,c} \right\}} (n)\quad \left| {\; \gcd(a,b) = 1} \right.\quad
  = \sum\limits_{z = 0}^{\left\lfloor {n/c} \right\rfloor } {p_{\left\{ {a,b} \right\}} (n - cz)}  =   \cr 
  &  = \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor } {\left( {{{n - ck} \over {ab}}
 - \left\{ {{{b^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over a}} \right\}
 - \left\{ {{{a^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over b}} \right\} + 1} \right)}  =   \cr 
  &  = {n \over {ab}}\left( {\left\lfloor {{n \over c}} \right\rfloor  + 1} \right)
 - {c \over {2ab}}\left\lfloor {{n \over c}} \right\rfloor \left( {\left\lfloor {{n \over c}} \right\rfloor
  + 1} \right) + \left( {\left\lfloor {{n \over c}} \right\rfloor  + 1} \right) +   \cr 
  &  - \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor } {\left( {\left\{ {{{b^{\,\left( { - 1} \right)}
 \left( {n - ck} \right)} \over a}} \right\} + \left\{ {{{a^{\,\left( { - 1} \right)}
 \left( {n - ck} \right)} \over b}} \right\}} \right)}  =   \cr 
  &  = \left( {{n \over {ab}} - {c \over {2ab}}\left\lfloor {{n \over c}} \right\rfloor  + 1} \right)
\left( {\left\lfloor {{n \over c}} \right\rfloor  + 1} \right)
 - \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor }
 {\left( {\left\{ {{{b^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over a}} \right\}
 + \left\{ {{{a^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over b}} \right\}} \right)}  \cr} 
$$
Example:
$$n=13, \, a=3, \, b=4, \, c=5$$
the solutions are
$$
\left( {x,y,z} \right) \in
 \left\{ {\left( {0,2,1} \right),\left( {1,0,2} \right),\left( {3,1,0} \right)} \right\}
$$
The modular inverses of $a$ and $b$ are
$$
\eqalign{
  & 3 \cdot 3 \equiv 1\left( {\bmod 4} \right) \Rightarrow a^{\,\left( { - 1} \right)}  = 3  \cr 
  & 1 \cdot 4 \equiv 1\left( {\bmod 3} \right) \Rightarrow b^{\,\left( { - 1} \right)}  = 1 \cr} 
$$
and $p_{\left\{ {a,b,\,c} \right\}} (n)$ becomes
$$
\eqalign{
  & p_{\left\{ {3,4,5} \right\}} (13) =   \cr 
  &  = \left( {{{13} \over {12}} - {5 \over {24}}\left\lfloor {{{13} \over 5}} \right\rfloor  + 1} \right)
\left( {\left\lfloor {{{13} \over 5}} \right\rfloor  + 1} \right)
 - \sum\limits_{k = 0}^2 {\left( {\left\{ {{{\left( {13 - 5k} \right)} \over 3}} \right\}
 + \left\{ {{{3\left( {13 - 5k} \right)} \over 4}} \right\}} \right)}  =   \cr 
  &  = 5 - \sum\limits_{k = 0}^2 {\left( {\left\{ {{{13} \over 3}} \right\}
 + \left\{ {{{39} \over 4}} \right\} + \left\{ {{8 \over 3}} \right\} + \left\{ {{{24} \over 4}} \right\}
 + \left\{ {{3 \over 3}} \right\} + \left\{ {{9 \over 4}} \right\}} \right)}  =   \cr 
  &  = 5 - \sum\limits_{k = 0}^2 {\left( {{1 \over 3} + {3 \over 4} + {2 \over 3} + {1 \over 4}} \right)}  = 3 \cr} 
$$
