TITLE: Coefficients of (generating) function
QUESTION [7 upvotes]: If I have the generating function 
\begin{equation*}
A(x)= \frac{1}{(1-x^{10})\cdot(1-x^5)\cdot(1-x) }\,, 
\end{equation*}
what is a clean way to find the coefficients of $x^{n}$. This coefficient would tell me, in how many ways I can combine an element of the first, the second and the third to get $x^n$.
I am pretty new to generating functions and I only now how to set them up from a series, but not how to get the series from a function :-(
Thanks in advance.

REPLY [4 votes]: We can apply the geometric  series expansion to derive the coefficients of
\begin{align*}
A(x)&=\frac{1}{\left(1-x^{10}\right)\left(1-x^5\right)\left(1-x\right)}\\
&=1 + x + x^2 + x^3 + x^4 + 2 x^5 + 2 x^6 + 2 x^7 + 2 x^8 + 2 x^9\\
&\qquad + 4 x^{10} + \cdots+ 4 x^{14} + 6 x^{15} + \cdots + 6 x^{19}\\
&\qquad + 9 x^{20} + \cdots + 9 x^{24} + 12 x^{25} + \cdots + 12 x^{29}\\
&\qquad + 16 x^{30} + \cdots + 16 x^{34} + 20 x^{35} + \cdots + 20 x^{39}\\
&\qquad + 25 x^{40} + \cdots + 25 x^{44} + 30 x^{45} + \cdots + 30 x^{49}\\
&\qquad + \cdots
\end{align*}

We start with
  \begin{align*}
\color{blue}{A(x)}&\color{blue}{=\frac{1}{\left(1-x^{10}\right)\left(1-x^5\right)\left(1-x\right)}}\\
&=\frac{1}{1-x}\left(\sum_{j=0}^\infty x^{10j}\right)\left(\sum_{k=0}^\infty x^{5k}\right)\\
&=\frac{1}{1-x}\left(\sum_{n=0}^\infty\left(\sum_{{10j+5k=n}\atop{j,k\geq 0}}1\right)x^n\right)\tag{1}\\
&=\frac{1}{1-x}\left(\sum_{n=0}^\infty\left(\sum_{{2j+k=n}\atop{j,k\geq 0}}1\right)x^{5n}\right)\tag{2}\\
&=\frac{1}{1-x}\left(\sum_{n=0}^\infty\left(\sum_{j=0}^{\lfloor{n/2}\rfloor}1\right)x^{5n}\right)\tag{3}\\
&=\left(\sum_{m=0}^\infty  x^m\right)\left(\sum_{n=0}^\infty\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)x^{5n}\right)\tag{4}\\
&=\sum_{t=0}^\infty\left(\sum_{{m+5n=t}\atop{m,n\geq 0}}\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)\right)x^t\tag{5}\\
&=\sum_{t=0}^\infty\left(\sum_{n=0}^{\lfloor t/5\rfloor}\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)\right)x^t\tag{6}\\
\end{align*}
  and denoting with $[x^t]$ the coefficient of $x^t$ in a series we observe from (6)
  \begin{align*}
\color{blue}{[x^t]A(x)=\sum_{n=0}^{\lfloor t/5\rfloor}\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{t}{5}\right\rfloor+1}\tag{7}
\end{align*}
  which is already a nice formula for the coefficients of $A(x)$.

Comment:

In (1) we apply the Cauchy-multiplication of series.
In (2) we note that only multiples of $5n$ of the index $n$ do contribute.
In (3) we rewrite the sum by eliminating the index $k$ and using the floor function $x\leq \lfloor x\rfloor < x+1$.
In (4) we evaluate the inner sum and expand the factor $\frac{1}{1-x}$.
In (5) we do again a Cauchy-multiplication.
In (6) we eliminate the index $m$ from the inner sum.


If we like we can derive a closed formula from (7) by noting that
  \begin{align*}
\color{blue}{\sum_{n=0}^N\left\lfloor\frac{n}{2}\right\rfloor}
&=\sum_{{n=0}\atop{ n\equiv 0(2)}}^N\frac{n}{2}+\sum_{{n=0}\atop{ n\equiv 1(2)}}^N\frac{n-1}{2}\\
&=\sum_{n=0}^N\frac{n}{2}\left(\frac{1+(-1)^n}{2}\right)+\sum_{n=0}^N\frac{n-1}{2}\left(\frac{1-(-1)^n}{2}\right)\\
&=\frac{1}{2}\sum_{n=0}^N n-\frac{1}{2}\sum_{n=0}^N\left(\frac{1-(-1)^n}{2}\right)\\
&\,\,\color{blue}{=\frac{1}{4}N(N+1)-\frac{1}{2}\left\lfloor\frac{N+1}{2}\right\rfloor} \tag{8}
\end{align*}
  We finally obtain from (7) and (8) by setting $N=\lfloor t/5\rfloor$:
  \begin{align*}
\color{blue}{[x^t]A(x)}&=\sum_{n=0}^{\lfloor t/5\rfloor}\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{t}{5}\right\rfloor+1\\
&=\frac{1}{4}\left\lfloor\frac{t}{5}\right\rfloor\left(\left\lfloor\frac{t}{5}\right\rfloor+1\right)-\frac{1}{2}\left\lfloor\frac{\left\lfloor\frac{t}{5}\right\rfloor+1}{2}\right\rfloor+\left\lfloor\frac{t}{5}\right\rfloor+1\\
&\,\,\color{blue}{=\frac{1}{4}\left\lfloor\frac{t}{5}\right\rfloor^2+\frac{5}{4}\left\lfloor\frac{t}{5}\right\rfloor
-\frac{1}{2}\left\lfloor\frac{t}{10}+\frac{1}{2}\right\rfloor+1}
\end{align*}