<s>[INST] Ordinary Generating Function for the number of solution :$ x_1 + x_2 + \cdot\cdot\cdot + x_k = n$ Let $a_n$ be the number of solutions of the equation:
$$x_1 + x_2 + \cdot\cdot\cdot + x_k = n$$
where $x_i$ is the positive odd integer. 
Hereto I would like to find the ordinary generating function for the sequence $a_n$ 
ODG has is a function satisfies the below :
$$G(a_n;n) = \sum_0^\infty a_nx^n$$
But what is the typical algorithm to find the ordinary generating function that uniquely determines its coefficient that corresponds to the regarding combinatorial situation?
Anyone can guide me where to start from? 
 [/INST] The generating function is
$$
\begin{align}
&\left(\sum_{j=0}^\infty x^{2j+1}\right)^k\tag1\\
&=\left(\frac{x}{1-x^2}\right)^k\tag2\\
&=x^k\sum_{j=0}^\infty(-1)^j\binom{-k}{j}x^{2j}\tag3\\
&=\sum_{j=0}^\infty\binom{k+j-1}{j}x^{2j+k}\tag4\\
&=\sum_{n=k}^\infty\binom{\frac{n+k-2}2}{\frac{n-k}2}x^n\ [2|(n-k)]\tag5
\end{align}
$$
Explanation:
$(1)$: the exponent of $x$ chosen for each factor
$\phantom{(1)\text{: }}$represents the value chosen for each summand
$(2)$: sum the geometric series
$(3)$: apply the Binomial Theorem
$(4)$: $\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$ see this answer
$(5)$: $n=2j+k\implies j=\frac{n-k}2\ [2|(n-k)]$ where $[\cdots]$ are Iverson brackets
In step $(1)$, the exponent of $x$ chosen for each factor represents the value chosen for each summand. For example, if $k=4$,
$$
\scriptsize\left(\color{#CCC}{x^1+x^3+}x^5\color{#CCC}{+x^7+\cdots}\right)\left(\color{#CCC}{x^1+}x^3\color{#CCC}{+x^5+x^7+\cdots}\right)\left(\color{#CCC}{x^1+x^3+}x^5\color{#CCC}{+x^7+\cdots}\right)\left(\color{#CCC}{x^1+x^3+x^5+}x^7\color{#CCC}{+\cdots}\right)
$$
represents the sum $5+3+5+7=20$, and
$$
\scriptsize\left(\color{#CCC}{x^1+x^3+}x^5\color{#CCC}{+x^7+\cdots}\right)
\left(x^1\color{#CCC}{+x^3+x^5+x^7+\cdots}\right)
\left(\color{#CCC}{x^1+x^3+x^5+}x^7\color{#CCC}{+\cdots}\right)\left(\color{#CCC}{x^1+x^3+x^5+}x^7\color{#CCC}{+\cdots}\right)
$$
represents the sum $5+1+7+7=20$.
</s>