For what $n$ do we have $ F_{n}=\lfloor \frac{1}{\sqrt{5}} ( \frac{1+\sqrt{5}}{2})^n +\frac{1}{2} \rfloor $

Asked Oct 17 '13 at 09:19

Active Oct 17 '13 at 09:33

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For what $n$ do we have $ F_{n}=\lfloor \frac{1}{\sqrt{5}} ( \frac{1+\sqrt{5}}{2})^n +\frac{1}{2} \rfloor $ where $F_{n}$ is the nth element of the Fibonacci sequence.

Naturally I have to use the formula $F_{n}= \frac{1}{\sqrt{5}}( ( \frac{1+\sqrt{5}}{2})^n -(\frac{1-\sqrt{5}}{2})^n) $, but I have trouble imagining the floor above...

EDIT:

Here's my trying.

Since $F_{n}$ is an integer for all $n$ than we can enter the floor by $F_{n}-F'_{n}$ where $F'_{n}$ is the equation with the floor.

We get $F_{n}-F'_{n}=\lfloor \frac{1}{\sqrt{5}}(\frac{1-\sqrt{5}}{2})^{n}+\frac{1}{2} \rfloor$

And we get that for all odd numbered $n$'s. Is this in any way correct?

edited Oct 17 '13 at 09:33

asked Oct 17 '13 at 09:19

Arek Krawczyk

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Well, now you need only see that $F_n - F_n'$ is always $0$. That's an easy consequence of $\left\lvert\dfrac{1-\sqrt{5}}{2}\right\rvert < 1$ and $\sqrt{5} > 2$. – Daniel Fischer Oct 17 '13 at 09:29
Very cool, thanks :) – Arek Krawczyk Oct 17 '13 at 09:41

For what $n$ do we have $ F_{n}=\lfloor \frac{1}{\sqrt{5}} ( \frac{1+\sqrt{5}}{2})^n +\frac{1}{2} \rfloor $

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