It is required to find the number of partitions of a multiple of six into 3 positive terms.
Find $P(6n, 3)$ for $n \geq 0$. I'm want to know, what is the way to do it, if there is no formula depends of $n$. How to find it in combinatorics way? And how to do it with recurrent sequences?
For a certain formula, it’s easy, suppose the answer will be $3n^2$. Then $$P(6n, 3) = P(6n-3, 3) + P(6n - 1, 2) =$$ $$= P(6n-6, 3) + P(6n-4,2) + P(6n-1,2) =$$ $$= P(6n-6, 3) + \left[\frac{6n-4}{2}\right] + \left[\frac{6n-1}{2}\right] =$$ $$= P(6n-6, 3) + 6n - 3$$
And with induction finally $3n^2 = 3(n-1)^2 + 6n - 3$.
What if there is no formula?
There's a closed form formula for what you're looking for. The generalized formula for $p(n,3)$ is, $$p(n,3)=\frac{1}{72}\left(6n^{2}-7-9\left(-1\right)^{n}+16\cos\left(\frac{2}{3}\pi n\right)\right)$$ And in your generalized case, $p(6n,3)$ it is;
$$p(6n,3)=\frac{1}{72}\left(6^{3}n^{2}+16\left(\cos\left(4\pi n\right)-1\right)\right)$$