Recently I presented the topic Fibonacci numbers in my class. It was massively successful and it has motivated me to conduct the part $2$ of the presentation. But unfortunately I have not sufficient data to proceed. I shall be thankful if you guys can present here some interesting or so called mind blowing identities including Fibonacci numbers.

I used following identities in the previous presentation:

$1.$ As observed by Johannas Kepler, $\frac{F_{n+1}}{F_n}\to \phi$ as $n\to\infty$.

$2.$ Binet's formula: $F_n=\frac{\phi^n-(\phi^{-n})}{\sqrt{5}}=\frac{\phi^n-(\phi^{-n})}{2\phi-1}$.

$3.$ $\phi^n=F_n\phi+F_{n-1}$

$4.$ Number of binary strings of length $n$ without consecutive $1's$ is the Fibonacci number $F_{n+2}$. For instance if $n=2${length of string} then there are total $2^2=4$ strings $00,01,10,11$. Note that $\underbrace{00,01,10}_{3 \text{times 1 is not being repeated}}$ and $3=F_4$.

$5.$ Number of ordered ways of writing a number $n$ in terms of sum of $1's$ and $2's$ is $F_{n+1}$. For instance $2=2=1+1\to $ two ways $=F_3$ ways.

$6.$ Pascal triangle and Fibonacci numbers:

This image is quite self explainatory, the Fibonacci numbers are sum of shallow diagonals of Pascal triangle. We represent it mathematically as $F_n={\sum_{k=0}^{[\frac{n}{2}]}}$ ${n-k+1}\choose{k}$ where $[ \ ]$ is floor function.

Side note: Whatever result you are providing, please link the source and try to give the proof if it can be understood by general audience (does not require much knowledge).

Almost all of the content I have given is available on Wikipedia.


Update: Thanks to @Claude, this gives a lot of instances where Fibonacci numbers are seen.

Update: As the answers are not of the form I expected(though answer by Frpzzd has good content), so let me show you a masterpiece. This is something which we can call interesting


closed as too broad by Jack, Namaste, Shailesh, C. Falcon, Daniel W. Farlow Jun 27 '17 at 1:04

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  • $\begingroup$ Notice that $\phi$ is Golden ratio $\endgroup$ – Vidyanshu Mishra Jun 9 '17 at 15:52
  • $\begingroup$ oeis.org/A000045 lists lots of ways of getting $F_n$ $\endgroup$ – Claude Jun 9 '17 at 16:11
  • $\begingroup$ I remember few facts about these numbers, but I know a journal dedicated to these numbers, I say it if you don't know it: The Fibonacci Quarterly. Here should be interesting results involving Fibonacci numbers that you can read. $\endgroup$ – user243301 Jun 9 '17 at 19:59

Here are a few ideas.

1.Talk about the relationship between the Fibonacci numbers and the Lucas numbers. The Lucas numbers are defined recursively in the same way as the Fibonacci numbers, but it starts with $2$ and $1$ instead of $1$ and $1$. There exist a few interesting relationships between the two sequences, such as $$F_{2n}=L_nF_n$$

  1. The Fibonacci numbers have some interesting properties related to divisibility such as:

    • for each positive integer, $k$, the Fibonacci number $F_n$ evenly divides the Fibonacci number $F_{kn}$.
    • $GCD(F_n,F_m)=F_{GCD(n,m)}$.
    • The function $M(x)=F_x \bmod m$ is periodic for any given $m$.
  2. Talk about the extension of $F_n$ to negative $n$. These are given by the formula $$F_{-n}=(-1)^{n+1}F_n$$

  3. Prove that the number of ways to cover a $2$ by $n$ checkerboard with $2$ by $1$ dominoes is $F_{n+1}$.

  • $\begingroup$ Proving that two consecutive Fibonacci numbers are coprime is also a good idea, but it is fairly easy too. +1 $\endgroup$ – Vidyanshu Mishra Jun 9 '17 at 16:29
  • $\begingroup$ That's true; however, proving the second one of my divisibility identities isn't so easy. $\endgroup$ – Franklin Pezzuti Dyer Jun 9 '17 at 16:30
  • $\begingroup$ I know, :-P, that's why I quoted that coprime thing. $\endgroup$ – Vidyanshu Mishra Jun 9 '17 at 16:32
  • $\begingroup$ Well for the first one of your divisibility identity, expanding the binet formula will work right? $\endgroup$ – Vidyanshu Mishra Jun 9 '17 at 16:33
  • $\begingroup$ Yes, but... where's the fun in that? It can be proven using induction, which is a lot more beautiful than yucky algebraic expansion. $\endgroup$ – Franklin Pezzuti Dyer Jun 9 '17 at 16:34

The error when approximating the golden ratio from above as a ratio of consecutive Fibonacci numbers is an Egyptian fraction of Fibonacci numbers.

$$\frac{F(2n+1)}{F(2n)}-\varphi = \sum_{k=0}^\infty \frac{1}{F(n2^{k+2})}$$



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