Proving correspondence and partitions via generating functions, or at least I think so.

Question

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 $A$.

For example, for $n=6$, the partitions of the first type are $$6,~~~5+1,~~~4+2,~~~3+2+1,~~~2+2+2,$$ and the partitions of the second type are $$5+1,~~~4+1+1,~~~3+3,~~~3+1+1+1,~~~1+1+1+1+1+1,$$ and there are $5$ of each type.

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I believe that I could show the generating functions of the two are the same, but I do not know where to start. So I'm thinking to first set some variables and functions, so I decided to let $\operatorname{P}_0(n)$ be the set of partitions of $n$ in which no odd number appears more than once, and let $\operatorname{P}_1(n)$ be the set of partitions of $n$ containing no element of $A$. How do I move out from there? If there is anyone who would guide me to the correct proof, or give a proof, their help would be appreciated. Thanks!

Answer 1

Let $a_n$ be the number of partitions of $n$ of the first kind, and let $b_n$ be the number of partitions of $n$ of the second kind.

Then $$A(q)=\sum_{n=0}^\infty a_nq^n=\prod_{m=1}^\infty\frac{1+q^{2m-1}}{1-q^{2m}}$$ and $$B(q)=\sum_{n=0}^\infty b_nq^n=\prod_{m=1}^\infty\frac{1-q^{4m-2}}{1-q^m}.$$

I would try writing $1-q^{4m-2}=(1+q^{2m-1})(1-q^{2m-1})$ in the formula for $B(q)$.

Answer 2

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 [number which is twice an odd number].

Flip it round: it suffices to prove that the number of partitions of $n$ in which an odd number occurs twice is the same as the number of partitions of $n$ which contain a number of the form $4k+2$. Denote the set of partitions of the first type as $Q$ and the set of partitions of the second type as $R$.

It is possible that the intersection $Q \cap R$ is non-empty. We simplify things greatly by further reducing the problem to showing that $Q \setminus R$ and $R \setminus Q$ have the same cardinality.

Consider $f: (Q \setminus R) \to (R \setminus Q)$ which combines as many equal odd numbers as possible. E.g. $f(1^6) = 2^3$. Clearly this is bijective.