A function $f(n)$ is defined for $n=4^k$, $k\geq0$. $f(n)$ is given in a recursion format:
$$ f(n)= \begin{cases} &3f(\frac{n}{4})-2f(\frac{n}{16})+\frac{3}{8}n,\quad n\geq160 \\ &1,\quad n=1 \\ &3,\quad n=4 \end{cases} $$
I need to prove in a typical equation method only, that $f(n)=n-\sqrt{n}+1 $ is exist.
Now, I have tried this way:
1) $n=4^k$
$$ f(4^k)=3f(4^{k-1})-2f(4^{k-2})+\frac{3}{8}4^k $$
2) $ f(4^k)=g(k) $ for $ k\geq0 $
$$ g(k)=3g(k-1)-2g(k-2)+\frac{3}{8}4^k $$
$$ g(0)=1, g(1)=3 $$
3) Divide by $ 4^k $ and define $ h(k)=\frac{g(k)}{4^k} $
$$ h(k)=\frac{3}{4}h(k-1)-\frac{1}{8}h(k-2)+\frac{3}{8} $$
And this is where I get stack, I realy don't know if I did right or how to keep forward..
Please, HELP ?
Thanks!