The $n=3$'rd Catalan number ([A000108](http://oeis.org/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](http://oeis.org/A000045)) is $1,1,2,3,5,...$ : ***5***.



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> ***Q***. Which other Fibonacci numbers (besides $\{1,2,5\}$) are also Catalan numbers?
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There seems to be no other "small" solutions, at least up to Fibonacci/Catalan numbers
around $10^{60}$.



When does a Catalan number equal a Fibonacci number?
A result of the type you seek follows easily from [Carmichael's theorem](http://en.wikipedia.org/wiki/Carmichael%27s_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 $m \leq 2n$.


