<s>[INST] Odd numbers as a sum in Generating Functions In combinatorics, I have to find (with the help of generating functions) in how many ways I can choose odd numbers from the numbers $[3..15]$ such that adding the numbers will give me $n$
So I said:
$$F(x) = (1+x^3+x^6...)(1+x^5+x^{10}+...) \cdot ... \cdot (1+x^{15} + x^{30} + ...)$$
It the number of options for $3$ multiplied by the number of options for $5$ and so on...
So I get them as a multiplication of sums:
$$F(x) = \sum_{n=0}^{\infty} {(x^3)^n} \cdot \sum_{n=0}^{\infty} {(x^5)^n} \cdot ... \cdot \sum_{n=0}^{\infty}{(x^{15})^n}$$
Which is equal to:
$$ F(x) = \frac{1}{(1-x^3)(1-x^5)\cdot...\cdot(1-x^{15})} $$
But how can I know the multiplier of $x^n$? or is it impossible? or did I do something really wrong? (perhaps the sums are finite to n)?
 [/INST] Using the method of partial fractions, rewrite $F(x)$ as
$$ F(x)=\frac{p_3(x)}{1-x^3}+\ldots +\frac{p_{15}(x)}{1-x^{15}}$$
where $\deg p_k<k$ (so, yes, $p_{15}$ might have degree $14$).
For each summand $\frac{p_k(x)}{1-x^k}$ the coefficient of $x^n$ is simply the coefficient of $x^{n\bmod k}$ in $p_k$.
Recall that the $p_k$ can be found from the polynomial identity obtained by multiplying the equation 
$$ \frac1{(1-x^3)(1-x^5)\cdots(1-x^{15})}=\frac{p_3(x)}{1-x^3}+\ldots +\frac{p_{15}(x)}{1-x^{15}}$$
with $(1-x^3)(1-x^5)\cdots(1-x^{15})$ and equating coefficients.
EDIT: I am afraid one should rather look for partial fractions based on the fcatorization 
$$(1-x)^8(1+x+x^2)^3(1+x+x^2+x^3+x^4)^2(1+x^3+x^6)\cdots $$
of the denominator, sigh.
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