# How to write $1-x-x^3+x^4+x^5+x^6-x^7 \cdots$ as a power series representation

How can I write $1-x-x^3+x^4+x^5+x^6-x^7 ....$ as a power series representation (i.e., a neat fraction such as $\frac{1}{1-x}$.

This stems from $\binom{\text{number of partitions of }n}{\text{into an even number of parts}}-\binom{\text{number of partitions of }n}{\text{into an odd number of parts}}$.

I've been pondering this for a while, yet can't seem to think of any ways to solve this. Any hints?

EDIT: The polynominal with a few extra terms i: $1-x-x^3+x^4+x^5+x^6-x^7+2x^8-2x^9+2x^{10}-2x^{11}+3x^{12}-3x^{13}+3x^{14} ...$

• What is the pattern here? – vadim123 Dec 21 '14 at 0:52
• Would you mind writing out a few more terms to make a pattern more apparent – ClassicStyle Dec 21 '14 at 0:53
• @vadim123: I'm not sure if there is any specific pattern. It was originally $\binom{\text{number of partitions of }n}{\text{into an even number of parts}}-\binom{\text{number of partitions of }n}{\text{into an odd number of parts}}$, but I found some terms. – Mathy Person Dec 21 '14 at 0:53
• @TylerHG: Sure, I don't mind. :) You can also reference my comment below. – Mathy Person Dec 21 '14 at 0:54
• @TylerHG Ok. Added some more. It seems to alternate (positive, negative). Also, for the first 8 terms, the coefficient is 1. For the next 4 terms, the coefficient is 2. I'm assuming that for the next 2 terms after that, the coefficient is 3. – Mathy Person Dec 21 '14 at 1:06

This appears in the oeis, wherein it is given that the generating function is $$\prod_{k>0}1-x^{2k-1}=\prod_{k>0}\frac{1}{1+x^k}$$ There are also many references there, I highly recommend that link.
• That's correct; the products are each from $k=1$ on up. – vadim123 Dec 21 '14 at 1:23
• @vadim123: After using that: how would I turn $\frac{1}{(1+x)(1-x)(1+x^2)(1-x^2)(1+x^3)(1-x^3)....}$ into a generating function power series again (for powers past 15)? – Mathy Person Dec 21 '14 at 1:26