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I've been having a few issues coming up with a generating function for an integer partition such that at least one part is even. What I have got so far is:

The generating function with no restrictions$$\begin{align*} \\&= even~parts * odd~parts \\ &= (1+x^2+x^4+\ldots)(1+x^4+x^8+\ldots)(1+x^6+x^{12}+\ldots)etc * (1+x+x^2+\ldots)(1+x^3+x^6+\ldots)(1+x^5+x^{10})+\ldots)etc \\&= \prod_{k= 1}^\infty1/(1-x^k)\end{align*}$$

Assuming that the situation where no even numbers are present in the partition is represented by the 1 in the expanded multiplication of the "even" side, we must remove this case by subtracting 1 from these terms.

Thus the generating function $ even~parts*odd~parts$ becomes $$\begin{align*}\\& (even~parts-1)*odd~parts\\ &= (even~parts*odd~parts) - odd~parts\\&=all~parts-odd~parts\\ &=\prod_{k= 1}^\infty1/(1-x^k) -\prod_{k= 1}^\infty1/(1-x^{2k+1})\end{align*}$$

Is this the right idea or am I completely on the wrong track? If it is the right idea how do I combine both the product signs $\prod$ into one? Thanks in advance.

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    You have the right idea here. I don't think it's possible to write the difference of the two products as a single product.2012-06-03
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    Thanks for that Michael, its good to know I'm not too far off beam. However, number 18 and the ensuing possibilities on this [page](http://garsia.math.yorku.ca/~zabrocki/math5020fw1112/handouts/Npartitionsids.pdf) seem to indicate that there is an answer for this question which does not contain 2 product signs. Is there another way of approaching this problem whereby the 2 signs are avoided from the beginning?2012-06-03
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    The sequence is http://oeis.org/A047967 where a generating function is given that looks pretty much like the one you got, which suggests there's nothing better. There are some links there that may be of interest.2012-06-03
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    Cheers for that @Gerry. That oeis entry is very helpful.2012-06-03

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