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As shown here, the (rounded) index $n$ of a given Fibonacci number $F$ is calculated with

$$ n(F) = \left\lfloor \log_\varphi (F \cdot \sqrt{5} + \frac12) \right\rfloor, $$

where

$$ \log_\varphi(x) = \frac{\ln(x)}{\ln(\varphi)} = \frac{\log_{10}(x)}{\log_{10}(\varphi)}. $$

Now my question is: How do I get there, e. g. from this formula?

$$ F_n = \left\lfloor \frac{\varphi^n}{\sqrt{5}} + \frac12 \right\rfloor, n \geq 0 $$

KLJ
  • 23

1 Answers1

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From Binet’s formula we have

$\sqrt{5}F_n=\phi^n-(-\phi)^{-n}$

and since $\phi^2 = \left(\frac{1+\sqrt5}{2}\right)^2 = \frac{1+2\sqrt5+5}{4} > \frac{10}{4} > 2$ we have

$\sqrt{5}F_n + \frac12 > \phi^n \space \forall n\ge 2$

But $\sqrt{5}F_n$ cannot be as great as $\phi^{n+1}$ so

$n<\log_{\phi}(\sqrt{5}F_n+\frac12)<n+1 \space \forall n \ge2$

hkBst
  • 455
gandalf61
  • 15,326
  • Why cannot $\sqrt5 F_n (+1/2?)$ be as great as $\phi^{n+1}? – hkBst Dec 24 '21 at 12:21
  • That does not hold for even $n$. For example $n=0 \implies \sqrt5 < \phi = (1+\sqrt5)/2 < \sqrt5$. And you are missing a $(-1)^n$ in your Binet's formula. – hkBst Dec 24 '21 at 15:08