The $n^{\text{th}}$ Catalan number is given by the formula $C_n = \frac 1{n+1}\binom{2n}n$.
It also satisfies the recurence \begin{align*}C_n &=\sum_{k=0}^{n-1}C_kC_{n-1-k}\\ &= C_0C_{n-1}+C_1C_{n-2}+C_2C_{n-3}+\cdots+C_{n-1}C_0\end{align*} for all n greater than or equal to 1.
What is the eighth positive integer n for which $C_n$ is odd?
On
The OEIS has a list of odd Catalan Numbers, A038003. The 8th odd Catalan number is 73 digits long. A quick check with Wolfram Alpha tells me that this number is $C_{127}$
In general, it appears that $C_x$ is odd when $x=2^n-1$ for some natural number $n$, and even otherwise.
By considering the greatest power of two that divides $(2n)!$ against the maximum power of two that divides $n!(n+1)!$, we have that $C_n$ is always even unless $n=2^k-1$ for some integer $k$.