Exercise · ZW-CATALAN-NUMBER-006 · ★★☆ Intermediate

Compute \(C_4\) using the recurrence \(C_{n+1}=\sum_{k=0}^{n} C_k C_{n-k}\) with \(C_0=1\), \(C_1=1\), \(C_2=2\), \(C_3=5\).

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  • A. 14

Explanation

For \(n=3\) the sum is \(C_0C_3+C_1C_2+C_2C_1+C_3C_0=5+2+2+5=14\). The value 42 is \(C_5\), a common index shift.

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Review the concept: Catalan number

The Catalan number \(C_n\) counts, among many other structures, the triangulations of a convex polygon with \(n+2\) sides and the Dyck paths of length \(2n\).

\[C_n=\frac{1}{n+1}\binom{2n}{n}=\frac{(2n)!}{n!\,(n+1)!}\]

More practice problems: catalan number

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