Which numbers occur in the Fibonacci and similar sequences?

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

Fi={1,i=11,i=2Fi1+Fi2,i>2 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}}

solutions.

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
1 year ago

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