$t/3, t\in\mathbb{Z}$, perhaps?

take $ n=1,n=2 $ and you'll obtain $ n/3 $ where $ n $ in an integer.

yes, n=1, n=2 implying $15a\in\mathbb{Z}$ and $48a\in\mathbb{Z}$ and $3a=(48-3\cdot 15)a\in\mathbb{Z}$, so that $a=\frac{k}{3}$ for some $k\in\mathbb{Z}$. But one must also prove in the other direction that all such $a$ satisfy the required property, which is easy since one of $n$, $n+2$ and $n+4$ must be divisible by $3$. The grading guidelines say 4 pts if only the necessity is proved. If only sufficient condition - 2 pts.