Conjecture: any factorial is a sum of two Jordan-Polya numbers.

Question

A Jordan-Polya number A001013 is a product of factorials $\prod^m_{k=1}a_k!$. For $n=2,\dots,12$ there are Jordan-Polya numbers $a,b$ such that $n!=a+b$.

2!=1+1
3!=2+4
4!=8+16
5!=48+72
6!=288+432
etc

Is this a coinsidence?

Answer 1

Writing a quick script to search for solutions, $n=15$ does not appear to work.

All other $n\le 21$ do; I may extend my program to search further than this by generating Jordan-Polya numbers beyond those in the OEIS b-file. Note that as Ty Jensen mentions in a comment, one obtains solutions for every number which is a sum of two Jordan-Polya numbers, so possible candidates for failure are $n=11,15,19,21,23,27,29,31,35,39,\ldots$.

Edit: After generating the $91,802$ Jordan-Polya numbers up to $30!$, I have confirmed that the values of $n$ up to $30$ for which $n!$ is not expressible as a sum of two Jordan-Polya numbers are $n=15,23,27,29$. Curiously, every other value of $n!$ has at least two such expressions (other than $n=2,3$).