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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. – Michael Lugo Jun 3 '12 at 3:04
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 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? – Johann Jun 3 '12 at 4:38
The sequence is 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. – Gerry Myerson Jun 3 '12 at 5:06
Cheers for that @Gerry. That oeis entry is very helpful. – Johann Jun 3 '12 at 8:38

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