<s>[INST] 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
 [/INST] 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 $$(\lfloor 10/4\rfloor+1)+(\lfloor 4/4\rfloor +1)=3+2=5$$
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