So, I'm working on a proof of a personal problem that I solved years ago unrigorously, and am now trying to rigorously prove. The problem I have is as follows:
Given some $v\in\Bbb Z$, $v\ge2$, find a formula giving $m\in\Bbb Z$, $m>0$ such that
$$\frac{(2+\sqrt{5})^m-(2-\sqrt{5})^m}{\sqrt{5}}\le v<\frac{(2+\sqrt{5})^{m+1}-(2-\sqrt{5})^{m+1}}{\sqrt{5}}$$
Now, if - as I wholly suspect - the $(2-\sqrt{5})^m$ and $(2-\sqrt{5})^{m+1}$ terms can be ignored due to their lack of effect on the final result (as $|2-\sqrt{5}|<\frac{1}{2}$ and $\lim_{m\to\infty}|(2-\sqrt{5})^m|=0$), then we have:
$$m\le \frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}<m+1$$
which implies that:
$$m=\lfloor\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}\rfloor$$
which certainly seems to work empirically. However, the difficulty I'm having is in proving that that $(2-\sqrt{5})^m$ term can be dropped since we're taking the floor of the final result. Boiling it down, I need to prove that:
$$\lfloor\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}\rfloor=\lfloor\frac{\ln{(v\sqrt{5}+(2-\sqrt{5})^m)}}{\ln(2+\sqrt{5})}\rfloor$$
for all $v\in\Bbb Z$, $v\ge2$ and $m\in\Bbb Z$, $m>0$ (as above), and I'm having a heck of a time figuring out how best to go about it. Any pointers in the correct direction would be appreciated!
Let $$v = \frac{(2+\sqrt{5})^m-(2-\sqrt{5})^m}{\sqrt{5}}$$ with even $m$. Then $$ (2+\sqrt{5})^{m-1}<(2+\sqrt{5})^m-1<v\sqrt{5}< (2+\sqrt{5})^m. $$ which implies $$m-1\le \frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}<m.$$ Therefore, $$\left\lfloor\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}\right\rfloor=m-1,$$ and your claim is false, unfortunately.
If, however, both inequalities are strict, $$ F_m:=\frac{(2+\sqrt{5})^m-(2-\sqrt{5})^m}{\sqrt{5}}< v<\frac{(2+\sqrt{5})^{m+1}-(2-\sqrt{5})^{m+1}}{\sqrt{5}},$$ then (knowing that $F_m$ is integer) $F_m+1\le v\le F_{m+1}-1$. Therefore, $$ (2+\sqrt5)^m < \sqrt{5}F_{m+1} + \sqrt{5}\le v\sqrt{5}\le F_{m+1}-\sqrt{5} <(2+\sqrt5)^{m+1}, $$ whence $$\left\lfloor\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}\right\rfloor=m.$$