Consider the partitions of $n$. For $n = 5,7,9,\ldots$, it appears as if the number of pairwise partitions $\{a,b\}$, where both $a$ and $b$ are composite, equals the total number of individual odd composite parts of $n-2$. Take $n=13$ for example. The partitions of $13$ are $\{\{12, 1\}, \{11, 2\}, \{10, 3\}, \{9, 4\}, \{8, 5\}, \{7, 6\}\}$ and $\{9,4\}$ is the only partition where both a and b are composite. The partitions of $(n=13)-2 = 11$ are $\{\{10, 1\}, \{9, 2\}, \{8, 3\}, \{7, 4\}, \{6, 5\}\}$ and $9$ is the only odd composite part.
Is it obvious to anyone why this is true?
Well, any composite odd $p<n-2$ will produce a pair of composite parts for $n$, since $n-p$ will be even and greater than $2$. The only remaining odd number $n-2$ will not contribute a new pair since $2$ is prime.