Generating Function for the number of ways representing positive integer with odd numbers I had an exam and this question struck me out of nowhere, making me sad :)
Let $f(n)$ denote the number of possibilities representing $n$ using odd positive single-digit numbers $[1,3,...,9]$
For example, $f(6)=4$ because:
$6=1+1+1+1+1+1$, $6=3+3$, $6=5+1$, $6=1+1+1+3$
Note that there is no difference between $1+5$ and $5+1$ and both of the possibilities count as one.
Find the generating function representing $f(n)$ 

Since you are not making a difference based on the order of the numbers, the only important thing here is how many of each numbers is present in a partition. That means we can model 1's with this generating function:
$$1+x+x^2+x^3+\cdots=\frac{1}{1-x}$$
$1=x^0$ is the case when we don't take any ones, $x$ is the case when we take 1 one, $x^2$ when we take 2 ones and so on... The same can be done for other odd integers, for example:
for 3: $$1+x^3+x^6+x^9+\cdots=\frac{1}{1-x^3}$$
for 5: $$1+x^5+x^{10}+x^{15}+\cdots=\frac{1}{1-x^5}$$
and so on...
That means that the whole generating function will be:
$$\frac{1}{(1-x)\left(1-x^3\right)\left(1-x^5\right)\left(1-x^7\right)\left(1-x^9\right)}$$
In general, if you have a set $S$ of numbers you want to partition into, you can use the generating function
$$\frac{1}{\prod_{s\in S}\left(1-x^s\right)}$$
