Let $$A = \begin{pmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$$, $$a, b \in \mathbb{R}$$. If for some $$n \in \mathbb{N}$$, $$A^n = \begin{pmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{pmatrix}$$ then $$n + a + b$$ is equal to ______.

Given $$A = \begin{pmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$$ and $$A^n = \begin{pmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{pmatrix}$$.

Write $$A = I + N$$ where $$N = \begin{pmatrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}$$. Since $$N^2 = \begin{pmatrix} 0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ and $$N^3 = 0$$, we have

$$A^n = I + nN + \binom{n}{2}N^2 = \begin{pmatrix} 1 & na & na + \binom{n}{2}ab \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{pmatrix}$$

Matching entries gives $$na = 48$$ so $$a = \frac{48}{n}$$, and $$nb = 96$$ so $$b = \frac{96}{n}$$. Also, from the $$(1,3)$$ entry, $$na + \binom{n}{2}ab = 2160$$.

Substituting yields

$$48 + \frac{n(n-1)}{2} \cdot \frac{48}{n} \cdot \frac{96}{n} = 2160$$

which simplifies to

$$48 + \frac{(n-1) \cdot 48 \cdot 96}{2n} = 2160$$

$$\frac{2304(n-1)}{n} = 2112$$

$$2304n - 2304 = 2112n$$

$$192n = 2304 \implies n = 12$$

Thus $$a = \frac{48}{12} = 4$$ and $$b = \frac{96}{12} = 8$$, and hence $$n + a + b = 12 + 4 + 8 = 24$$.

The answer is $$\boxed{24}$$.