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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: $1-x-x^3+x^4+x^5+x^6-x^7+2x^8-2x^9+2x^{10}-2x^{11}+3x^{12}-3x^{13}+3x^{14} ...$

jimjim
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Mathy Person
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    What is the pattern here? – vadim123 Dec 21 '14 at 00:52
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    Would you mind writing out a few more terms to make a pattern more apparent – ClassicStyle Dec 21 '14 at 00:53
  • @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 00:53
  • @TylerHG: Sure, I don't mind. :) You can also reference my comment below. – Mathy Person Dec 21 '14 at 00:54
  • @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 01:06
  • @TylerHG: Actually, never mind... the recent terms I just added are a counterexample to what I originally thought. – Mathy Person Dec 21 '14 at 01:11
  • It might be worthwhile to see of the terms end up looking like $$a_n=(-1)^nnx^n$$ as this can be summed directly. Then to get a formula for the full series one could subtract off the first few terms. Edit: nevermind for my comment too after yours came up – ClassicStyle Dec 21 '14 at 01:12
  • @TylerHG: I did find this: http://math.stackexchange.com/questions/543561/partitions-of-n-into-distinct-odd-and-even-parts-proof . see 1. – Mathy Person Dec 21 '14 at 01:20
  • This seems to be the number of partitions minus twice the number of distinct partitions of $n$. It also seems to be number of partitions of $n$ into distinct odd parts multiplied by a sign based on parity. – Henry Dec 21 '14 at 01:23

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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.

vadim123
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