How do I express this recursive function in implicit form? (Please show working out)
$$K_{(n)}=2K_{(n-1)}+\frac{4^{n-1}-1}3$$
$$K_{(0)}=1$$
This is my proposed upper bound for the Koch Art Gallery Problem (https://file.scirp.org/pdf/OJDM20120400004_11740065.pdf), where 'n' is equal to the iteration of the Koch curve and the value returned is the maximum number of guards required for full vision inside the koch curve polygon. I can find the recursive function, however as I was never formally taught how to solve first order difference equations with exponentional functions, I am unaware how to progress.
THIS IS THE SOLUTION BEFORE OP EDITED THE QUESTION ON SEPT 15.
Let $K_{(n)} = \frac23 4^n + R_{(n)}$, then we have $$ K_{(n)}=2K_{(n-1)}+\frac{4^n-1}3 \quad \rightarrow \quad R_{(n)}=2R_{(n-1)}-\frac{1}3 $$
Note that in here, the original condition as posted by the OP was $\frac{4^{\color{Red}{ n}}-1}3$ instead of the new edit $\frac{4^{\color{Red}{ n-1}}-1}3$
The recursive equation for the remainder $R_{(n)}$ can be solved by setting $R_{(n)} = c \cdot a^n + b$ which gives $$ c \; a \; a^{n-1} + b=2 \; c \; a^{n-1} + 2 b -\frac{1}3 $$ i.e. $a=2$ and $b = \frac{1}3$, with arbitrary $c$. In total, we obtain a general solution
$K_{(n)} = \frac23 4^n + c\; 2^n + \frac{1}3$
Satisfying $K_{(0)} =1$ demands $ \frac23 + c + \frac{1}3= 1$, i.e. $c=0$, hence
$K_{(n)} = \frac23 4^n + \frac{1}3$
For other initial conditions, we obtain different $c$. If $K_{(-1)} =1$ (see the OPs comment), we have $1 = \frac23 4^{-1} + c\; 2^{-1} + \frac{1}3$, i.e. $c=1$, hence in this case
$K_{(n)} = \frac23 4^n + 2^n + \frac{1}3$