if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let 
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form  $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)}$$
then I feel very ugly,can you someone have good  partial fractions methods by hand?
because I take an hour to solve this problem. 
ago I have solve $$x+2y+3z=n$$ the number of the positive integer solution $a_{n}$ I found
$$a_{n}=\left[\dfrac{(n+3)^2}{12}\right]$$
Thank you

Here is an approach to compute $a_n$ using the observation that it is the number of ways to make change for $n$ using coins of denomination $2, 3, 4$.
Case: $n$ even
It may be noted that $3$ denomination coin can be used, but only as a multiple of $6$, so you have $n=2k=2p+4q+6r \implies k = p+2q+3r$.  Thus this is the number of ways you can make change for $k=n/2$ with coins of denominations $1, 2, 3$. So 
$$\sum_{k=0}^\infty a_{2k}x^k = \frac1{(1-x)(1-x^2)(1-x^3)} = \frac{(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4) (1+x^3)}{(1-x^6)^3} = \left(1+x+2 x^2+3 x^3+4 x^4+5 x^5+4 x^6+5 x^7+4 x^8+3 x^9+2 x^{10}+x^{11}+x^{12} \right)\left(\sum_{k=0}^\infty (-1)^k \binom{-3}{k} x^{6k} \right)$$ 
will help get $a_{2k}$.  
Case: n odd
Now we need to have one of the $3$ denominations used, and the rest is the even case, so we have $a_{2k+3} = a_{2k}$ in this case.
