Let $$R_1$$ and $$R_2$$ be relations on the set $$\{1, 2, \ldots, 50\}$$ such that $$R_1 = \{(p, p^n) : p$$ is a prime and $$n \geq 0$$ is an integer$$\}$$ and $$R_2 = \{(p, p^n) : p$$ is a prime and $$n = 0$$ or $$1\}$$. Then, the number of elements in $$R_1 - R_2$$ is ______

$$R_1 = \{(p, p^n) : p \text{ is prime}, n \geq 0 \text{ integer}, p^n \in \{1, 2, \ldots, 50\}\}$$

$$R_2 = \{(p, p^n) : p \text{ is prime}, n = 0 \text{ or } 1, p^n \in \{1, 2, \ldots, 50\}\}$$

$$R_1 - R_2$$ consists of pairs $$(p, p^n)$$ where $$n \geq 2$$, $$p$$ is prime, and $$p^n \leq 50$$.

For each prime $$p$$, we list $$p^n$$ with $$n \geq 2$$ and $$p^n \leq 50$$:

$$p = 2$$: $$4, 8, 16, 32$$ (i.e., $$2^2, 2^3, 2^4, 2^5$$) — 4 elements

$$p = 3$$: $$9, 27$$ (i.e., $$3^2, 3^3$$) — 2 elements

$$p = 5$$: $$25$$ (i.e., $$5^2$$) — 1 element

$$p = 7$$: $$49$$ (i.e., $$7^2$$) — 1 element

$$p \geq 11$$: $$p^2 \geq 121 > 50$$ — no elements

Total elements in $$R_1 - R_2 = 4 + 2 + 1 + 1 = 8$$.

Hence the answer is $$\boxed{8}$$.