A partition of $n$ is said to be ‘3-phobic’ if none of the parts is a multiple of 3. Show that the number of 3-phobic partitions of $n$ is equal to the number of partitions of n that have no double-gaps.
I get what a double gap is: A partition of $n$ is said to have ‘no double-gaps’ if it is not missing two consecutive integers in the sequence $1, 2, . . . , k_{max}$ where $k_{max}$ is the size of the largest part. For example $10=5+3+2$ doesn't have a double gap but $10=5+2+2+1$ does have a double gap.
I'm really lost on how to show this.