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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 $A$.


Not sure where to start.Generating Functions?

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$${1+x\over1-x^2}{1\over1-x^4}{1+x^3\over1-x^6}{1\over1-x^8}\dots={1\over1-x}{1\over1-x^3}{1\over1-x^4}{1\over1-x^5}{1\over1-x^7}\dots$$

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Hint: Create an explicit bijection between the two sets of partitions.

A partition that satisfies both conditions will map to itself. A partition that satisfies only one of them can map to one that satisfies the other.

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  • $\begingroup$ What exactly is this bijection? $\endgroup$ – user118161 Jul 10 '20 at 8:22

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