Closed form for coefficient of $x^N$ of $\frac{1}{(1-x^2)(1-x^3)(1-x^5)}$ I'm trying to get a closed form for the generating function which counts the number of non-negative integer solutions to 
$$2e_1+3e_2+5e_3 = N$$
I note that we have something like: 
$$\frac{1}{(1-x^2)(1-x^3)(1-x^5)}$$ 
but can't get something to translate nicely. I've thought of using $e^{2x}  e^{3x}  e^{5x}$, but is it valid?

According to Maple:

convert(1/(1-x^2)/(1-x^3)/(1-x^5),parfrac,x);

$$-1/30\, \left( x-1 \right) ^{-3}+1/8\, \left( x+1 \right) ^{-1}-{
\frac {77}{360}}\, \left( x-1 \right) ^{-1}+1/5\,{\frac {2+x+{x}^{2}+{
x}^{3}}{{x}^{4}+{x}^{3}+{x}^{2}+x+1}}+{\frac {7}{60}}\, \left( x-1
 \right) ^{-2}+1/9\,{\frac {-x+1}{{x}^{2}+x+1}}$$

map(convert, %, FPS, x);

$$\sum _{k=0}^{\infty } \left( \sum _{{\it \alpha}={\it RootOf} \left( 
{{\it \_Z}}^{4}+{{\it \_Z}}^{3}+{{\it \_Z}}^{2}+{\it \_Z}+1 \right) }-
1/25\,{\frac {{x}^{k} \left( -2\,{\it \alpha}+1+{{\it \alpha}}^{2}
 \right) }{{{\it \alpha}}^{k+1}}} \right) +\sum _{k=0}^{\infty }
 \left( -{\frac {1}{54}}\, \left( -\sqrt {3}+3\,i \right) \sqrt {3}
 \left( -1/2-1/2\,i\sqrt {3} \right) ^{k}-{\frac {1}{54}}\,i\sqrt {3}
 \left( -3+i\sqrt {3} \right)  \left( -1/2+1/2\,i\sqrt {3} \right) ^{k
} \right) {x}^{k}+\sum _{k=0}^{\infty }{\frac {77}{360}}\,{x}^{k}+
\sum _{k=0}^{\infty } \left( {\frac {7}{60}}+{\frac {7}{60}}\,k
 \right) {x}^{k}+\sum _{k=0}^{\infty }1/8\, \left( -1 \right) ^{k}{x}^
{k}+\sum _{k=0}^{\infty } \left( 1/30+1/20\,k+{\frac {1}{60}}\,{k}^{2}
 \right) {x}^{k}$$
