Let $n$ be a positive integer greater than zero. Let $a_n= {n(n+3)(n-1) \above 1.5pt 2}$ and set $$\rho_n = {a_n \above 1.5pt gcd(n, a_n)}$$ Computation show that $\rho_n$ is the sequence $$0,5,6,21,16,45,30,77,48,117,70,165,96,221,\ldots$$which is not in Sloan's database. However if you look at the terms that are strictly increasing in the sequence $$5,6,21,45,77,117,165,221,\ldots$$ they appear to form the sequence A276704. Now A276704 are the strictly increasing terms (or records) of the least common multiple of $n-2$ and $n+2$.
Is it possible to get to a closed fomrula for the records in $\rho_n$ ?
I can show that
$$\gcd(n,a_n) = \begin{cases} {n\above 1.5pt 2}, & \text{if $n$ is even} \\[2ex] n, & \text{if $n$ is odd} \end{cases}$$
So if $n$ is even we can write $n=2k$ and after substitution calculate $$\rho_n={2k(2k+3)(2k-1) \above 1.5pt 2}{1 \above 1.5pt k}=(2k+3)(2k-1)$$
We have:
$$\mathrm{lcm}(n-2,n+2)=\begin{cases}(n-2)(n+2)&n\text{ odd}\\ \frac{(n-2)(n+2)}2&n\text{ divisible by }4\\ \frac{(n-2)(n+2)}4&\text{otherwise} \end{cases}$$
Your sequence $\rho_n$ has the property that:
$$\rho_{n-1} = \begin{cases}(n-2)(n+2)&n\text{ odd}\\ \frac{(n-2)(n+2)}{2}&\text{otherwise} \end{cases}$$
The key is that only for small values of $n$ will we ever get the $\frac{1}{2}$ term be a "record." It is a mere coincidence of the shift that the one odd term $\rho_{3}$ which is a "record" has $3+1$ divisible by $4$, so it is also a "record" in the LCM sequence.
As noted in comments, there are several different ways to write $\rho_n$ in closed form, but they mostly involve tricks with $(-1)^n$, and are likely not very edifying.
For example:
$$\rho_n=(3+(-1)^n)\frac{(n-1)(n+3)}{4}$$