Thank you Gerry Myerson for bringing this problem to [WCNT 2021: Problems in Number Theory][1]

The answer to this question is **No** as we can prove that the density of $A$ is zero. Simon Rubinstein-Salzedo outlined a solution with one step to be confirmed. Then I filled in the proof of the one step. 

****Theorem**** **$A$ has density 0. (Simon Rubinstein-Salzedo)**

The argument relies on the following known facts:

**Fact** The Fibonacci numbers form a strong divisibility sequence, i.e. $\gcd(F_m,F_n)=F_{\gcd(m,n)}$. It follows that $z(\text{lcm}(m,n))=\text{lcm}(z(m),z(n))$.

**Fact** If $p$ is a prime, then $z(p)\mid p-\left(\frac{p}{5}\right)$.

Thus if $p_1.\ldots,p_k$ are distinct primes different from 2 and 5, then $\frac{z(p_1\cdots p_k)}{p_1\cdots p_k}\le\frac{1}{2^{k-1}}\prod_{i=1}^k \left(1+\frac{1}{p_i}\right)$, and this tends to 0 as $k\to\infty$. Let's write

$$
m(k)=\frac{1}{2^{k-1}} \prod_{i=1}^k \left(1+\frac{1}{p_i}\right),
$$

where the product is over the first $k$ odd primes other than 5.

For an integer $n$, let

$$
r(n)=\prod_{\substack{p\text{ prime} \\ p\neq 2,5 \\ p\mid n \\ p^2\nmid n}} p
$$

and $s(n)=\frac{n}{r(n)}$. I believe (to be confirmed) that for any $k$,

$$
\lim_{x\to\infty} \frac{\#\{n:1\le n\le x, \omega(r(n))\ge k\}}{x}=1.
$$

For any $n$ with $\omega(r(n))\ge k$, we have

$$
\frac{z(n)}{n}\le \frac{z(r(n))}{r(n)}\cdot\frac{z(s(n))}{s(n)}\le m(k)\cdot 2,
$$

which tends to 0 as $k\to\infty$. Thus the upper density of $A$ is 0.

However, this says nothing about what happens if we restrict to Fibonacci entry points of primes.

**Confirming the Density 1 result (Sungjin Kim)**

The constant $C>0$ may appear several times, not necessarily the same everytime.

Let $P_k$ be the set of positive integers with $< k$ distinct prime factors. Let $A_k(t)=\sum_{n\le t, n\in P_k} 1$ be the counting function of $P_k$. By Hardy-Ramanujan, we have an estimate

$$
A_k(t) \le C \Psi_k(t):=C\frac{t(\log\log (t+C)+C)^{k-2}}{\log (t+C)}.
$$

The numbers satisfying $\omega(r(n))<k$ can be decomposed as

$$
n=my
$$

with $m=2^{\nu_2(n)}5^{\nu_5(n)}r(n)$ so that $\omega(m)<k+2$ and $y$ is power-full, that is, $p|y \Rightarrow p^2|y$.

Let $\mathcal{F}$ be the set of power-full numbers. The estimate of the number of power-full numbers is obtained by Bateman and Grossward in 1958 (as a stronger form than below),

$$
\sum_{y\le x, y\in\mathcal{F}}1 \le C \sqrt x.
$$

Combining these to estimate the numbers $n\leq x$ with $\omega(r(n))<k$,

$$
\leq C\sum_{m\leq x, m\in P_{k+2}} \sum_{y\leq \frac xm, y\in\mathcal{F}} 1 \leq C\sum_{m\leq x, m\in P_{k+2}} \sqrt{\frac xm}.
$$

Applying the partial summations to the last sum,

$$
\sum_{m\leq x, m\in P_{k+2}} \sqrt{\frac xm} \leq C\Psi_{k+2}(x) +C\sqrt x
\int_{1}^x \frac{\Psi_{k+2}(t)}{t\sqrt t} dt\leq C\Psi_{k+2}(x).
$$

Hence,

$$
\sum_{n\leq x, \omega(r(n))<k}1\leq  C\Psi_{k+2}(x).
$$




  [1]: https://westcoastnumbertheory.org/2021-schedule-of-talks/