The number of functions $$f : \{1, 2, 3, 4\} \to \{a \in \mathbb{Z} : |a| \leq 8\}$$ satisfying $$f(n) + \frac{1}{n}f(n+1) = 1$$, $$\forall n \in \{1, 2, 3\}$$ is

Given: $$f : \{1, 2, 3, 4\} \to \{a \in \mathbb{Z} : |a| \leq 8\}$$ satisfying $$f(n) + \frac{1}{n}f(n+1) = 1$$ for all $$n \in \{1, 2, 3\}$$.

From the recurrence relation, we express all values in terms of $$f(1)$$:

For $$n = 1$$: $$f(1) + f(2) = 1 \implies f(2) = 1 - f(1)$$

For $$n = 2$$: $$f(2) + \frac{1}{2}f(3) = 1 \implies f(3) = 2(1 - f(2)) = 2(1 - (1 - f(1))) = 2f(1)$$

For $$n = 3$$: $$f(3) + \frac{1}{3}f(4) = 1 \implies f(4) = 3(1 - f(3)) = 3(1 - 2f(1)) = 3 - 6f(1)$$

All function values must be integers with $$|f(i)| \leq 8$$. Since $$f(1)$$ is an integer, we need:

• $$|f(1)| \leq 8$$
• $$|1 - f(1)| \leq 8 \implies -7 \leq f(1) \leq 9$$
• $$|2f(1)| \leq 8 \implies -4 \leq f(1) \leq 4$$
• $$|3 - 6f(1)| \leq 8 \implies -\frac{5}{6} \leq f(1) \leq \frac{11}{6}$$

The most restrictive constraint gives $$f(1) \in \{0, 1\}$$ (integers in $$[-0.83, 1.83]$$).

The two valid functions are:

• $$f(1) = 0$$: $$f = (0, 1, 0, 3)$$
• $$f(1) = 1$$: $$f = (1, 0, 2, -3)$$

The number of such functions is 2.

Therefore, the correct answer is Option D: $$\mathbf{2}$$.