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.
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!
Best Answer
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 $B(q)$.