So I'm given a set: [10] x [2] x $\mathbb N$ with a weight function: $w(a, b, c) = 4a + 2b + c$
And i'm asked to determine the generating series of this, but I'm confused due to the 3 variables..
I understand that $f(x) = \sum_{n=0}^\infty a_n x^n $ and that $a_n$ is the number of $(a, b, c)$ such that $n = 4a + 2b + c$. But I'm confused as to how I would get this, given the set.
Would this be acceptable for example?
We can pick $a\in{1, 2, 3, ... , 10}$ so that there are 10 possibilities of this. Then $b$ only has two possibilities (since it's either 1 or 2), so there are now 20 possibilities where $c$ is fixed at $c = n - 4a - 2b$. But where would I go from here?
EDIT --- adding on to Brain M. Scott's answer:
picking $a\in[10]$ gives me
$$x^4+x^8+\ldots+x^{40}=\sum_{a\in[10]}x^{4a} = \frac{1-(x^4)^{11}}{1-x^4}.$$ by geometric series.
$b\in[2]$ gives me
$$x^2+x^4=\sum_{b\in[2]}x^{2b}$$
and $c\in\mathbb N$ gives
$$x+x^2+x^3+\ldots=\sum_{c\in\Bbb N}x^c\; = \frac{x}{1-x}$$
I removed the $1$ at the beginning since this set excludes $0$.
So I multiplied these 3, giving me:
${x^3+x^5-x^{45}-x^{49}}\over{1-x-x^4+x^5}$
Would this be my final rational expression?
Suppose that you were just picking $a\in[10]$; then your series would be
$$x^4+x^8+\ldots+x^{40}=\sum_{a\in[10]}x^{4a}\;.$$
Similarly, $b$ would give you
$$x^2+x^4=\sum_{b\in[2]}x^{2b}\;,$$
and $c$ would give you
$$1+x+x^2+x^3+\ldots=\sum_{c\in\Bbb N}x^c\;.$$
What happens if you multiply those three series together?