Which numbers occur in the Fibonacci and similar sequences?

The well-known Fibonacci Sequence is generated by the simple recursive formula

\( F_i = \begin{cases} 1 , & \text{i=1} \\ 1 , & \text{i=2} \\ F_{i-1}+F_{i-2} , & \text{i>2} \end{cases} \).

The first few terms are

1,1,2,3,5,8,13,21, 1,1,2,3,5,8,13,21,\ldots .

Of course, choosing 1 1 and 1 1 as the first two terms is rather arbitrary, so let's replace them by general natural numbers a a and b b . Then, we get the recursive definition

Fi,a,b={a,i=1b,i=2Fi1,a,b+Fi2,a,b,i>2 F_{i,a,b} = \begin{cases} a , & \text{i=1} \\ b , & \text{i=2} \\ F_{i-1,a,b}+F_{i-2,a,b} , & \text{i>2} \end{cases} .

The Fibinacci numbers have many special properties (which have been called and used many times), but why should these properties be limited to Fibonacci numbers? If we can find other starting values a,b a,b so that iN(Fi,a,b=nN) \exists i \in \mathbb{N} \left( F_{i,a,b} = n \in \mathbb{N} \right) then we can use the properties of general Fibonacci sequences to find relations involving n n and maybe solve a problem.

But which numbers occur in general Fibonacci sequences?

Obviously, every number n n occurs in the sequence starting with 1 1 and n1 n-1 at i=3 i=3 .

n=F3,1,n1 n = F_{3,1,n-1}

But it also occurs in F3,2,n1 F_{3,2,n-1} , F3,3,n3 F_{3,3,n-3} and so on.

n=F3,a,na,n1aN n = F_{3,a,n-a}, n-1 \geq a \in \mathbb{N}

This makes a total of n n sequences that contain n n .

Are there more? Is there a formula or a asymptote to the number of sequences that contain n n ? Which n n have the most occurences?

For searching Fibonacci sequences that contain n n , let's sort them by index.

How many starting values a,b a,b are there so that n=F3,a,b n = F_{3,a,b} ? We've already solved this as the number of natural numbers a a so that na n-a is also a natural number. There are n n possible values for a a .

Now, n=F4,a,b=a+2b n = F_{4,a,b} = a+2b . This is equivalent to b=na2 b = \frac {n-a}2 which has n2 \left\lceil \frac n2 \right\rceil solutions.

n=F5,a,b=2a+3bb=n2a3 n = F_{5,a,b} = 2a+3b \Leftrightarrow b = \frac {n-2a}3 . This has n61 \left\lceil \frac n6 \right\rceil - 1 solutions, unless n5(mod6) n \equiv 5 \pmod 6 , then it has n6 \left\lceil \frac n6 \right\rceil solutions.

From now on, I won't write exact solutions any more but rather the asymptotic number of solutions. So, n2=O(n2) \left\lceil \frac n2 \right\rceil = O \left( \frac n2 \right) and n61=O(n6) \left\lceil \frac n6 \right\rceil -1 = O \left( \frac n6 \right) , using Big O Notation.

n=F6,a,b=3a+5bb=n3a5 n = F_{6,a,b} = 3a+5b \Leftrightarrow b = \frac {n-3a}5 . This has asymptotically O(n35)=O(n15) O \left( \frac n{3 \cdot 5} \right) = O \left( \frac n{15} \right) solutions since there are about n5 \frac n5 naturals less than n n , but only 13 \frac 13 of them can be reached by taking steps of size 3 3 .

In general:

n=Fi,a,b=aFi2,1,1+bFi1,1,1b=naFi2,1,1Fi1,1,1 \begin{aligned} n & = F_{i,a,b} = aF_{i-2,1,1} + bF_{i-1,1,1} \\ \Leftrightarrow b & = \frac{n - aF_{i-2,1,1}}{F_{i-1,1,1}} \end{aligned}

There are approximately nFi1,1,1 \frac n{F_{i-1,1,1}} naturals less than n n that are divisible by Fi1,1,1 F_{i-1,1,1} , but only 1Fi2,1,1 \frac 1{F^{i-2,1,1}} of them can be reached with steps of size Fi3,1,1 F_{i-3,1,1} , so the equation has about

nFi1,1,1Fi2,1,1 \frac n {F_{i-1,1,1} \cdot F_{i-2,1,1}}


In total, for a given n n , there are approximately

ni=31Fi1,1,1Fi2,1,1=ni=11FiFi+1n1.773877583285133 \boxed{n \displaystyle \sum_{i=3}^\infty \frac 1 {F_{i-1,1,1} \cdot F_{i-2,1,1}} = n \sum_{i=1}^\infty \frac 1 {F_i \cdot F_{i+1}} \approx n\cdot {\color{#D61F06}1.773877583285133}}

Fibonacci-like sequences that contain n n at some position.

Note by Henry U
2 years, 7 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

Hello, can we make friends? My e-mail:552141604@qq.com. Wish to have your answers!

Shenghui Lin - 2 years, 6 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...