What is a general formula $n(n)$ for this? We know that starting from below, we can see how many numbers a certain $n$ generates by counting the number of numbers contained in the column $n$ is in, along with the numbers in the next column BUT NOT the numbers in the same row as $n$, and so on. for example,
$\begin{align}n(1) = 1\\ n(2) = 2\\ n(3) = 3\\ n(4) = 5 = (2 + 2 + 1)\\ n(5) = 7 = (2 + 2 + 2 + 1)\\ n(6) = 11 = (2 + 2 + 2 + 4 + 1)\\ n(7) = 15 = (2 + 2 + 2 + 4 + 4 + 1)\\ n(8) = 21 = (2 + 2 + 2 + 4 + 4 + 6 + 1)\\ n(9) = 27 (2 + 2 + 2 + 4 + 4 + 6 + 6 + 1)\\ n(10) = 35 = (2 + 2 + 2 + 4 + 4 + 6 + 6 + 8 + 1)\\ n(11) = 43 = (2 + 2 + 2 + 4 + 4 + 6 + 6 + 8 + 8 + 1)\\ \cdots \end{align}$
Firstly, don't ever name a function $n$ and then make its argument $n$ as well (like when you wrote $n(n)$).
That said, it appears you want the partition function, which is typically written $p(n)$. As mentioned in the comments, one of the best ways to compute $p(n)$ is to use Euler's Pentagonal Formula. One of our users, Paramanand Singh, wrote up a really nice introduction as a blog entry on our community blog.
If you want an approximation, one may consider the Hardy-Ramanujan asymptotic formula, which is given by: $$p(n) \sim \frac{1}{4n\sqrt 3}\exp\left(2\pi\sqrt{n/6}\right)$$