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.
On first thought, my mind is blank. I simply do not know how to approach this problem. Solutions are greatly appreciated. Thanks in advance!