Narayana number
= Narayana number
{c}
{title2=$N(n,k)$}
The number $N(n,k)=\frac1n\binom nk\binom n{k-1}$ counts <Dyck paths> of semilength $n$ with exactly $k$ peaks, for $1\le k\le n$. Summing over $k$ gives the $n$th <Catalan number>.
= Narayana number
{c}
{title2=$N(n,k)$}
The number $N(n,k)=\frac1n\binom nk\binom n{k-1}$ counts <Dyck paths> of semilength $n$ with exactly $k$ peaks, for $1\le k\le n$. Summing over $k$ gives the $n$th <Catalan number>.