Find the coefficient of $x^n$ in $\frac{x^4}{(1-x^2)(1-x)^2}$

Question

I'm solving a combinatorics problem using generating function and reached the following expression: $$\frac{x^4}{(1-x^2)(1-x)^2}$$

How can I expand it to find a general expression for the coefficient of $x^n$?

Any help would be appreciated.

Answer 1

Partial fraction: $$\frac{1}{(1-x^2)(1-x)^2}=\frac1{8(1+x)}+\frac1{8(1-x)}+\frac1{4(1-x)^2}+\frac1{2(1-x)^3}$$

Taylor series: $$\frac1{1+x}=1-x+x^2-x^3+...$$$$\frac1{1-x}=1+x+x^2+x^3+...$$ for $|x|<1$.

Hence $$\frac1{8(1+x)}+\frac1{8(1-x)}=\frac18(2+2x^2+2x^4+...)=\frac14\sum_{n=0}^\infty x^{2n}$$$$\frac1{4(1-x)^2}+\frac1{2(1-x)^3}=\frac14((1+x+x^2+x^3+...)^2+2(1-x+x^2-x^3+...)^3)=\frac14\left(\sum_{n=0}^\infty (1+n)x^n+\sum_{n=0}^\infty (1+n)(2+n)x^n\right)=\frac14\sum_{n=0}^\infty (1+n)(3+n)x^n$$ So $$\frac{1}{(1-x^2)(1-x)^2}=\frac14\sum_{n=0}^\infty\left[x^{2n}+(1+n)(3+n)x^n\right]$$ This means that $$\boxed{\frac{x^4}{(1-x^2)(1-x)^2}=\frac14\sum_{n=0}^\infty\left[x^{4+2n}+(1+n)(3+n)x^{4+n}\right]\, \text{for} \,|x|<1}$$ Can you finish?

Answer 2

Let $$\frac{x^4}{(1-x^2)(1-x)^2}=\sum_{n=0}^\infty a_n x^n. $$ Since the denominator has the form $$ (1-x^2)(1-x)^2=-{x}^{4}+2\,{x}^{3}-2\,x+1, $$ then $a_n$ satisfied the following recurrence relation $$ a_n=2a_{n-1}-2a_{n-3}+a_{n-4},a_1=0,a_2=0,a_3=0,a_4=1. $$ By solving it we get $$ a_n=\frac 3 8-\frac 7 4\,n+\frac 1 8\, \left( -1 \right) ^{n}+\frac 1 2\, \left( n+1 \right) \left( \frac 1 2\,n+1 \right). $$