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: 
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

$$v\_n=6\,(2^n-1)+\frac13.$$
Let $u\_m=v\_m+a\_0+a\_1m+\cdots$

$$6=2u\_{n+1}-u\_n=2v\_{m+1}-v\_m+a\_0(2-1)+a\_1(2(m+1)-m)+\cdots$$

Set $a\_0=6,a\_r=0\forall r>0$

to find $$v\_{n+1}=\dfrac{v\_n}2=\cdots=\dfrac{v\_{n-p}}{2^{p+1}}$$

Now $v\_0+6=u\_0\iff v\_0=?$