Question: I'm having trouble with writing this sequence as a function of $n$ because it's neither geometric nor arithmetic. 

> 
> $$\begin{cases}
> u\_{n+1} = \frac12 u\_{n} + 3\qquad \forall n \in \mathbb N\\
> u\_{0} = \frac13
> \end{cases}$$
> 
> 
>

Answer: 
Hint. Note that for some real number $a$ (which one?),
$$u\_{n+1}-a = \frac12 \left(u\_{n}-a\right).$$
Hence, the sequence $(u\_n-a)\_n$ is of geometric type:
$$u\_{n}-a=\frac12 \left(u\_{n-1}-a\right)=\frac{1}{2^2} \left(u\_{n-2}-a\right)=\dots =\frac{1}{2^{n}} \left(u\_{0}-a\right).$$
Similarly to what RobertZ did, turning the arithmetico-geometric recurrence to a purely geometric one, you can get rid of the multiplicative coefficient.

Let $u\_n:=\dfrac{v\_n}{2^n}$. Then

$$\dfrac{v\_{n+1}}{2^{n+1}}=\frac12\dfrac{v\_n}{2^n}+3$$ or

$$v\_{n+1}=v\_n+3\cdot2^{n+1}.$$

This recurrence is now easily solved as a geometric summation

$$v\_n=6\,(2^n-1)+\frac13.$$