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

$$$\begin{array}{l}\mathrm{f}:\{1,2,3,4\}\to \{\mathrm{a}\in \mathrm{z},|\mathrm{a}|\le 8\}\\ \mathrm{f}\left(\mathrm{n}\right)+\frac{1}{n}\mathrm{f}(\mathrm{n}+1)=1,\mathrm{n}\in \{1,2,3\}\end{array}$

$f\left(n+1\right)\text{ must be divisible by }\mathrm{n}$

$\begin{array}{l}\mathrm{f}\left(4\right)\Rightarrow -6,-3,0,3,6\\ \mathrm{f}\left(3\right)\Rightarrow -8,-6,-4,-2,0,2,4,6,8\\ \mathrm{f}\left(2\right)\Rightarrow -8,\dots \dots \cdots \cdots \dots ,8\\ \mathrm{f}\left(1\right)\Rightarrow -8,\dots \dots \cdots \cdots \cdots ,8\end{array}$

$\frac{\mathrm{f}\left(4\right)}{3}\text{ must be odd since }\mathrm{f}\left(3\right) \mathrm{should} \mathrm{be} \mathrm{even}$

$$\begin{gathered} \text{ therefore }2\text{ solution possible. } \\ \mathrm{f}=\left\{\left(1,1\right)\left(2,0\right)\left(3,2\right)\left(4,-3\right)\right\} \end{gathered}$$

$\mathrm{or} \mathrm{f}=\left\{\left(1,0\right),\left(2,1\right),\left(3,0\right),\left(4,3\right)\right\}$