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Let $A=\{2,6,10,14,\ldots\}$ be the set of integers that are twice an odd number.

Prove that, for every positive integer $n$, the number of partitions of $n$ in which no odd number appears more than once is equal to the number of partitions of $n$ containing no element of $A$.

I can't seem to find the generating function for either of these.

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    $\begingroup$ Do you have a guess for either generating function? or do you know the generating function for partitions with similar restrictions? $\endgroup$ – Greg Martin Sep 27 '14 at 0:58
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The generating function for partitions where no odd number appears more than once is $$\prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 0} (1+z^{2k+1}).$$

The number of partitions containing no element of $A$ is $$\prod_{k\ge 1} \frac{1}{1-z^k} \prod_{k\ge 0} (1-z^{4k+2}).$$ Re-write this as $$\prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 0} \frac{1}{1-z^{2k+1}} \prod_{k\ge 0} (1-z^{4k+2}).$$

Finally observe that $$\frac{1-z^{4k+2}}{1-z^{2k+1}} = 1 + z^{2k+1}$$ so this becomes $$\prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 0} (1+z^{2k+1})$$ which is the same as the first generating function.

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  • $\begingroup$ Can you explain how you re-wrote the generating function for "number of partitions containing no element of A" into $\prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 0} \frac{1}{1-z^{2k+1}} \prod_{k\ge 0} (1-z^{4k+2}). $ $\endgroup$ – qs13 Oct 16 '20 at 4:11
  • $\begingroup$ @qs13 it is just splitting up the index set of positive integers into evens and odds. $\endgroup$ – RobPratt Nov 13 '20 at 4:34

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