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
 A: The $n$-th term is obtained from $(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)(1+x^4+x^8+\cdots)$ which is the number of triples $(i,jk)$ that are solutions to $2i+3j+4k=n$. Hence,  $$a_n=\sum_{2i+3j+4k=n}1=\sum_{j=0}^{\lfloor n/3\rfloor }\sum_{2i+4k=n-3j}1$$
Now we want to obtain $$b_m=\sum_{2i+4k=m}1$$
It is clear $b_{2m+1}=0$, so we look at $$b_{2m}=\sum_{i+2k=m}1$$ Now $$\sum_{2k=l}1=  [2\mid l]$$ hence $$b_{2m}=\left\lfloor\frac m2\right \rfloor+1$$
For example, $2i+4k=10$ has three solutions, $(5,0),(3,1),(1,2)$, i.e $\lfloor 5/2\rfloor+1=3$.
Now $2\mid n-3j\iff 2\mid n-j$, so we get $$a_n= \sum_{j=0}^{\lfloor n/3\rfloor}\left(\left\lfloor \frac{n-3j}4\right\rfloor+1 \right)[2 \mid n-j]$$
For example, $2k+3j+4k=10$ has solutions $(5,0,0), (0,2,1), (2,2,0),(1,0,2),(3,0,1)$, and the formula gives $$(\lfloor 10/4\rfloor+1)+(\lfloor 4/4\rfloor +1)=3+2=5$$
A: Incomplete answer (maybe it can be used):
If $\left|z\right|<1$ then $\frac{1}{1-z}=1+z+z^{2}+\cdots$
Applying this for $z=x^{4},x^{3},x^{2}$ you find:
$$\frac{1}{\left(1-x^{4}\right)\left(1-x^{3}\right)\left(1-x^{2}\right)}=\left(1+x^{4}+x^{8}+\cdots\right)\left(1+x^{3}+x^{6}+\cdots\right)\left(1+x^{2}+x^{4}+\cdots\right)$$
Then $a_{n}$ must equalize the cardinality of $\left\{ \left(p,q,r\right)\in\left\{ 0,1,2,\dots\right\} \mid4p+3q+2r=n\right\} $ (as Macavity comments). 
A: Hints :


*

*First, prove that 
$$\frac{1}{(1-x^2)(1-x^3)(1-x^4)} = \frac{7}{32(x+1)}-\frac{59}{288(x-1)}+\frac{1}{8(x-1)^2}+\frac{1}{16(x+1)^2}-\frac{1}{24(x-1)^3}+\frac{x+2}{9(x^2+x+1)}+\frac{1-x}{8(x^2+1)}.$$

*Then, use that
$$ \frac{1-x}{x^2+1} = -\frac{1+i}{2(x-i)}+\frac{-1+i}{2(x+i)}.$$
and
$$ \frac{x+2}{x^2+x+1} = \frac{j}{x+j}+\frac{\overline{j}}{x+\overline{j}}$$ where $j = \frac{1+i\sqrt{3}}{2}$.

*Finally, use the classical series
$$ \sum_{n=0}^{+\infty} x^k = \frac{1}{1-x},$$
$$ \sum_{n=0}^{+\infty} (k+1)x^k = \frac{1}{(1-x)^2},$$
and
$$ \sum_{n=0}^{+\infty} \frac{1}{2}(k+1)(k+2)x^k = \frac{1}{(1-x)^3}.$$

A: Note that $\left(1-x^2\right)\left(1-x^3\right)\left(1-x^4\right)$ divides $\left(1-x^{12}\right)^3$ and $\displaystyle\left(1-z\right)^{-3}=\sum_{n=0}^{\infty}{n+2 \choose 2}z^n$.
A: 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.
