Is there a obvious pattern between a Catalan number and another?

Question

I know that $ C_n = C_0 C_{n-1} + C_1 C_{n-2} + C_2 C_{n-3} + \cdots + C_{n-1} C_0 $., but I was wondering if there was a more obvious pattern between $C_n$ and $C_{n+1}$.

Answer 1

Yes, there is:

$$\begin{align*} \frac{2(2n-1)}{n+1}C_{n-1}&=\frac{2(2n-1)}{n(n+1)}\binom{2n-2}{n-1}\\\\ &=\frac2{n+1}\binom{2n-1}n\\\\ &=\frac1{n+1}\left(\binom{2n-1}n+\binom{2n-1}n\right)\\\\ &=\frac1{n+1}\left(\binom{2n-1}n+\binom{2n-1}{n+1}\right)\\\\ &=\frac1{n+1}\binom{2n}n\\\\ &=C_n\;. \end{align*}$$

Answer 2

Sure, $$C_{n+1} = \frac{2(2n+1)}{n+2}C_n.$$

You can prove this from the formula $$C_n = \frac{1}{n+1}\binom{2n}{n}.$$

More at Wikipedia: https://en.wikipedia.org/wiki/Catalan_number.