Let $F$ be the set of all integers $n>1$ such that in the [Fibonacci sequence](https://en.wikipedia.org/wiki/Fibonacci_sequence) modulo $n$, the value $0$ occurs infinitely often. What is the value of $\lim\sup\_{n\to\infty}\frac{|F\cap\{0,\ldots,n\}|}{n+1}$?



Density of "Fibonacci friends"
The value $0$ always occurs infinitely often regardless of the value $n>1$. To see this, consider the mapping $L\_n:\mathbb{Z}\_n^2\rightarrow\mathbb{Z}\_n^2$ defined by $L\_n(x,y)=(y,x+y)$. Then the mapping $L\_n$ is invertible. Then if we set $F\_{0,n}=[0]\_n,F\_{1,n}=[1]\_n$ and $F\_{i+2,n}=F\_{i+1,n}+F\_{i,n}$, then $F\_{i,n}$ is the $i$-th Fibonacci number modulo $n$. On the other hand,
$L\_n(F\_{i,n},F\_{i+1,n})=(F\_{i+1,n},F\_{i+2,n})$, so $L\_n^m(F\_{i,n},F\_{i+1,n})=(F\_{i+m,n},F\_{i+m+1,n})$ for all $m\geq 0$. Since $L\_n$ is invertible, there must be some $m>0$ where $([0]\_n,[1]\_n)=L^m\_n([0]\_n,[1]\_n)=L^m\_n(F\_{0,n},F\_{1,n})=(F\_{m,n},F\_{m+1,n})$, so $F\_{m,n}=[0]\_n$.


This is in contrast to the Foobonacci sequence where the function $M\_n$ is not injective. Define the Foobonacci sequence $FO\_{1,n}=1,FO\_{2,n}=1,FO\_{i+2,n}=FO\_{i+1,n}\*\_nFO\_{i,n}$ where $\*\_n$ is the self-distributive operation in the $n$-th Laver table. For example, in Hexadecimal ($104\_{\text{hex}}=260\_{\text{dec}}$), we have (according to my artificial intelligence calculations but without proof) $(FO\_{1,104},\dots,FO\_{7,104})=(1,1,2,3, F00...008, \mathbf{F00...00B}, 100...00)$, so we call this sequence the Foobonacci sequence.


The mapping
$M\_n:A\_n^2\rightarrow A\_n^2$ defined by $M\_n(x,y)=(x\*\_ny,x)$ (where $A\_n$ denotes the $n$-th Laver table) is not invertible, and for all $n>1$, there is no $t>2$ where $FO\_{n,t}=1$, but for all $n>1$, we eventually have
$FO\_{n,2t+1}=2^n$ for sufficiently large $t$.


