TITLE: When does a Catalan number equal a Fibonacci number?
QUESTION [14 upvotes]: The $n=3$'rd Catalan number (A000108) is $1,1,2,5$ : $\frac{\binom{2n}{n}}{n+1}=\frac{\binom{6}{3}}{4}=\frac{20}{4}=$ 5.
The $n=4$'th Fibonacci number (A000045) is $1,1,2,3,5,...$ : 5.

Q. Which other Fibonacci numbers (besides $\{1,2,5\}$) are also Catalan numbers?

There seems to be no other "small" solutions, at least up to Fibonacci/Catalan numbers
around $10^{60}$.

REPLY [38 votes]: A result of the type you seek follows easily from Carmichael's theorem, that if $m > 12$, then there is a prime $p$ that divides $F_{m}$, but does not divide $F_{k}$ for $k < m$. 
Suppose $C_{n} = \binom{2n}{n}/(n+1)$ and we assume that $C_{n} = F_{m}$. All the prime factors of the left hand side are $\leq 2n$, while by Carmichael's theorem there is a prime $p | F_{m}$ that does not divide any earlier Fibonacci number. However, since for $p > 5$, $p$ divides either $F_{p-1}$ or $F_{p+1}$ we have that $m \leq p+1 \leq 2n$. However, from the asymptotic size of $C_{n}$ and $F_{m}$, we must have that $m \approx \frac{\log(4)}{\log(\phi)} n$, and this contradicts the inequality $m \leq 2n$.