Problem 846

Solution:
Answer: $\frac{1}{6^{n}}$

Let $f_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ denote the $n$-variable version of the function. We will prove that $f_{n}(5, \ldots, 5)=\frac{1}{6^{n}}$ by induction.

The base case was done in the two previous problems. Suppose we know that $f_{n-1}(5,5, \ldots, 5)=\frac{1}{6^{n-1}}$. Let $g\left(x_{1}, \ldots, x_{n-1}\right)=3 f_{n}\left(x_{1}, \ldots, x_{n-1}, 3\right)$. We have that $g$ is linear in $x_{1}, \ldots, x_{n-1}$ and $g\left(x_{1}, \ldots, x_{n-1}\right)=\frac{1}{x_{1} \cdots x_{n-1}}$ for all $x_{1}, \ldots, x_{n-1} \in\{3,4\}$.

By the inductive hypothesis, we have $g(5, \ldots, 5)=\frac{1}{6^{n-1}}=f_{n-1}(5, \ldots, 5)$. Therefore, $f_{n}(5, \ldots, 5,3)=\frac{f_{n-1}(5, \ldots, 5)}{3}$. Similarly, $f_{n}(5, \ldots, 5,4)=\frac{f_{n-1}(5, \ldots, 5)}{4}$.

$$\begin{aligned} f_{n}(5,5, \ldots, 5,5) & =2 f_{n}(5,5, \ldots, 5,4)-f_{n}(5,5, \ldots, 5,3) \\ & =2 \cdot \frac{f_{n-1}(5, \ldots, 5)}{4} - \frac{f_{n-1}(5, \ldots, 5)}{3} \\ & =\frac{1}{2} f_{n-1}(5,5, \ldots, 5)-\frac{1}{3} f_{n-1}(5,5, \ldots, 5) \\ & =\frac{1}{6} f_{n-1}(5,5, \ldots, 5) \\ & =\frac{1}{6 \cdot 6^{n-1}} \\ & =\frac{1}{6^{n}} \end{aligned}$$
and this proves our conjecture by induction.