Let $$\alpha, \beta$$ ($$\alpha > \beta$$) be the roots of the quadratic equation $$x^2 - x - 4 = 0$$. If $$P_n = \alpha^n - \beta^n$$, $$n \in \mathbb{N}$$, then $$\frac{P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}}{P_{13}P_{14}}$$ is equal to _____

We have $$\alpha, \beta$$ as roots of $$x^2 - x - 4 = 0$$ with $$\alpha > \beta$$, and $$P_n = \alpha^n - \beta^n$$. Since $$\alpha$$ and $$\beta$$ satisfy $$x^2 = x + 4$$, we get the recurrence $$P_n = P_{n-1} + 4P_{n-2}$$ for all $$n \geq 2$$.

We simplify the given expression $$\frac{P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}}{P_{13}P_{14}}$$. Factoring the numerator: $$P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14}) = (P_{15} - P_{14})(P_{16} - P_{15})$$.

From the recurrence $$P_n = P_{n-1} + 4P_{n-2}$$, we get $$P_n - P_{n-1} = 4P_{n-2}$$. Hence $$P_{15} - P_{14} = 4P_{13}$$ and $$P_{16} - P_{15} = 4P_{14}$$.

The expression becomes $$\frac{4P_{13} \cdot 4P_{14}}{P_{13}P_{14}} = \frac{16\,P_{13}P_{14}}{P_{13}P_{14}} = 16$$.

Hence, the correct answer is $$\boxed{16}$$.