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

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

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

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

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 \( a,b \) so that \( \exists i \in \mathbb{N} \left( F_{i,a,b} = n \in \mathbb{N} \right) \) then we can use the properties of general Fibonacci sequences to find relations involving \( n \) and maybe solve a problem.

But which numbers occur in general Fibonacci sequences?

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

\( n = F_{3,1,n-1} \)

But it also occurs in \( F_{3,2,n-1} \), \( F_{3,3,n-3} \) and so on.

\( n = F_{3,a,n-a}, n-1 \geq a \in \mathbb{N} \)

This makes a total of \( n \) sequences that contain \( n \).

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

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

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

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

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

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

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

In general:

\( \begin{align} n & = F_{i,a,b} = aF_{i-2,1,1} + bF_{i-1,1,1} \\ \Leftrightarrow b & = \frac{n - aF_{i-2,1,1}}{F_{i-1,1,1}} \end{align} \)

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

\( \frac n {F_{i-1,1,1} \cdot F_{i-2,1,1}} \)

solutions.

In total, for a given \( n \), there are approximately

\( n \displaystyle \sum_{i=3}^\infty \frac 1 {F_{i-1,1,1} \cdot F_{i-2,1,1}} = n \sum_{i=1}^\infty \frac 1 {F_i \cdot F_{i+1}} \approx n\cdot {\color{red}1.773877583285133} \)

Fibonacci-like sequences that contain \( n \) at some position.

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