Let $z(n)=\min\{k>0 : n\mid F_k\}$. This function is known as the Fibonacci entry point (for example). A result of Sallé gives the sharpest upper bound for $z(n)$, namely, $z(n)\leq 2n$, for all $n$ (the equality holds for all $n=6\cdot 5^k$). Also, it is well-known that $\lim\sup_{n\to \infty}z(n)/n=0$.

By using the Mathematica software, I am almost convinced that $z(n)$ is not too small, in density. For example, I do believe that the set $$ A:=\{n\geq 1: z(n)>n/4\} $$ has positive upper density, i.e., $$ \lim_{x\to \infty}\sup\displaystyle\frac{|A\cap [1,x]|}{x}>0. $$ However, I was not able to prove that. Someone can give me some suggestion?

Thanks in advance.

Let $z(n)=\min\{k>0 : n\mid F_k\}$. This function is known as the Fibonacci entry point (for example). A result of Sallé gives the sharpest upper bound for $z(n)$, namely, $z(n)\leq 2n$, for all $n$ (the equality holds for all $n=6\cdot 5^k$). Also, it is well-known that $\liminf_{n\to \infty}z(n)/n=0$.

By using the Mathematica software, I am almost convinced that $z(n)$ is not too small, in density. For example, I do believe that the set $$ A:=\{n\geq 1: z(n)>n/4\} $$ has positive upper density, i.e., $$ \lim_{x\to \infty}\sup\displaystyle\frac{|A\cap [1,x]|}{x}>0. $$ However, I was not able to prove that. Someone can give me some suggestion?

Thanks in advance.