My best approach yet is to distinguish if $n$ is even or odd. If it is odd, then surely $a$,$b$ and $c$ are odd. If it is even, from its prime factorization we can see that we have $n = 2^m\big(\frac{n}{2^m}\big)$ for a particular $m>0$, so either $a=\big(\frac{n}{2^m}\big)$ and $bc=2^m$, or $a=2^m$ and $bc=\big(\frac{n}{2^m}\big)$.
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I would start with $n=p^m$ where $p$ is a prime. – Marco Sep 18 '18 at 17:40
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The following MSE link includes a simple memoized recurrence to compute this statistic (non-ordered). – Marko Riedel Sep 18 '18 at 17:47