Gcd of Fibonacci and Catalan Let $F_n=1,1,2,3,5,\ldots$ (starting with $n=1$) be the Fibonacci sequence and let $C_n=\frac{1}{n+1}\binom{2n}{n}$ be the Catalan sequence.
Define $B_z$ to be the cardinality of $$B_z := \#\bigl\{ n \leq z | \gcd(F_n,C_n)=1 \bigr\}.$$ 
It seems $$\lim_{z\to\infty} \frac{B_z}{z}=\frac14.$$ 
Posted a related question here: 
https://math.stackexchange.com/questions/2131648/gcd-of-catalan-and-fibonacci-numbers, but it got no answer and, thinking a bit about it, the question might be more research related than elementary (but I have no real training in such questions).
So question is: Does $B_z/z$ converge, and if so, to what?
Here some values of $g_z := B_z/z$,
$$g_{1000}=\frac{71}{250}=0.284, 
\quad g_{2000}=\frac{13}{50}=0.260, 
\quad g_{3000}= \frac{187}{750}\approx0.249, 
\quad g_{4000}=\frac{979}{4000}\approx0.245.$$
The sequence of integers such that the gcd is one starts with $$1,2,3,4,5,8,10,11,13,14,17,22,23,25,\ldots$$ and is probably not known in the sequence database.
 A: This is a bit overlong for a comment, and not really an answer, but includes some information about a related problem that might offer approaches to this one. In particular, the fact that the Fibonacci sequence is a divisibility sequence means that gcd properties involving $F_n$ are tied into properties for $F_m$ for $m\mid n$. So you might try looking at the proportion of primes in your set. 
The Fibonacci sequence is a linear recursion and a divisibility sequence. For a similar question in which one takes two Fibonacci-type sequences, there are conjectures, but not even an inkling of a proof. For example, Ailon and Rudnick conjectured that $$\{n\ge1 : \gcd(2^n-1,3^n-1)=1\}$$ is infinite, and I published a somewhat dubious heuristic argument that $$\text{Density}\Bigl(\{p~\text{prime}:\gcd(2^p-1,3^p-1)=1\}\Bigr)=1.$$ (The density over all $n\in\mathbb{N}$ seems harder to guess.) More generally, due to the divisibility property of these sequences, for $a,b\ge2$ multiplicatively independent, it is natural to look at $$\{n\ge1 : \gcd(a^n-1,b^n-1)=\gcd(a-1,b-1)\}.$$ I'll also mention that Ailon and Rudnick proved a stronger version of their conjecture when one replaces the integers $a$ and $b$ with polynomials $a(T),b(T)\in\mathbb{C}[T]$.
