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Write down the generating function for the number of partitions of $n$ in which each odd part can occur any number of times but each even part can occur at most twice. Apply algebra to this generating function to complete the following theorem:

The number of partitions of n in which each odd part can occur any number of times but each even part can occur at most twice is the number of partitions of n in which...

The generating function I've found for the first part is $$\prod_{k\geq1}\frac{(1+x^{2k})^2}{1-x^{2k-1}}$$ but I'm unsure how to start the second part, any help would be appreciated.

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The generating function for the first part is \begin{eqnarray*} \prod_{n=1}^{\infty} \frac{ (1+x^{2n}+x^{4n})}{1-x^{2n-1}}. \end{eqnarray*} The terms in the numerator deal with the even numbers occuring at most twice & the terms in the denominator deal with the odd numbers.

Now observe that \begin{eqnarray*} 1+x^{2n}+x^{4n} = \frac{ 1-x^{6n} }{ 1-x^{2n} }. \end{eqnarray*} The generating function can be rewritten as \begin{eqnarray*} \prod_{n=1 \text{ and } 6 \nmid n }^{\infty} \frac{ 1}{1-x^{n}}. \end{eqnarray*} So it is the same as partitioning numbers whose parts are not multiples of $6$.

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    $\begingroup$ +1 Love it when generating functions yield a non-intuitive / obvious result. $\endgroup$ – Calvin Lin Dec 1 '19 at 22:45
  • $\begingroup$ @CalvinLin Really tough challenge ... set up a bijection between these two types of partitions ... Thank goodness for generating functions? $\endgroup$ – Donald Splutterwit Dec 1 '19 at 22:48
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    $\begingroup$ This is awesome! $\endgroup$ – Hendrix Dec 2 '19 at 1:10

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