Geometric/arithmetic sequences: $u_{n+1} = \frac12 u_{n} + 3$ I'm having trouble with writing this sequence as a function of $n$ because it's neither geometric nor arithmetic. 

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


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 $v_0+6=u_0\iff v_0=?$
