I'm somewhat stuck on a problem involving using generating functions to determine the number of possible solutions to an equation. I've attempted to solve the problem by following an example in my textbook, however I am unsure still if I have correctly answered the problem.
The problem follows:
Find the generating function for the number of integer solutions of:
a) $2w + 3x +5y + 7z = n$ where $0 \leq w, x, y, z$
b) $2w + 3x + 5y + 7z = n$ where $ 0 \leq w, 4 \leq x, y, 5 \leq z$
Now, the work I've so far attempted to do is to assign the following functions to each of the terms:
$2w = 1+a^2+a^4+a^6+ ..$
$3x = 1 + a^3 + a^6 + a^9+..$
$5y = 1 + a^5 + a^{10} +a^{15} +..$
$7z = 1+a^7+a^{14}+a^{21}+..$
Again, I've only just followed the example in the textbook as I am unsure of what to do. Could anyone explain what this means? (if it is correct?)
Then, I've come up with the generating function by multiplying these values together:
$f(x)=(1+a^2+a^4+..)(1+a^3+a^6+..)(1+a^5+a^10+..)(1+a^7+a^14+..) $
Now as far as I know, this is a valid generating function. However, my textbook takes one more step, and I've followed suit:
$f(x)=\frac{1}{1-x^2}*\frac{1}{1-x^3}*\frac{1}{1-x^5}*\frac{1}{1-x^7}$
I don't understand the step that is taken here, but this is what I have come up with for my final answer to a).
Now for b), I've taken the equations that I lined out and simply eliminated the earlier terms according to the restrictions on $w,x,y,z$
$2w = 1+a^2+a^4+a^6+ ..$
$3x = a^9+a^{12}+..$
$5y = a^{15} + a^{20}+..$
$7z = a^{28}+a^{35}+..$
If anyone can offer any insight here I would greatly appreciate it. I'm having quite a bit of trouble with my combinatorics class..
Cheers everyone