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: 
Write $$ 2u\_{n+1}-u\_n = 6 = 2u\_n-u\_{n-1}$$
so we have $$ 2u\_{n+1}-3u\_n +u\_{n-1}=0$$

Then solving the characteristic equation $2x^2-3x+1=0$ which has a solution $x\_1=1$ ans $x\_2={1\over 2}$ we get a general solution $$ u\_n = a\cdot 1^n +b\cdot \Big({1\over 2} \Big)^n$$

Since $u\_1 = {19\over 6}$ by solving a system:

\begin{eqnarray\*}
 {1\over 3} &=& a+b\\
 {19\over 6} &=& a+{b\over 2}
\end{eqnarray\*}
we get the solution: 
 $$ u\_n = 6-{17\over 3\cdot 2^n}$$
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 $v\_0+6=u\_0\iff v\_0=?$