Catalan Number Calculator – C(n) Counts

Catalan numbers count recursively nested shapes: balanced bracketings, binary trees, triangulations, monotonic lattice paths. The sequence starts 1, 1, 2, 5, 14, 42, 132, … so C₅ = 14 — the number of triangulations of a convex heptagon (n+2 sides) and the number of ways to fully parenthesize six factors (n+1 of them). The closed form divides a central binomial coefficient by n+1: C(5) = C(10,5)/6 = 252/6 = 14. The calculator prints the factorial substitution, the binomial intermediate, and the recurrence C(n+1) = Σ C(i)C(n−i), verifying 42 = 1·14 + 1·5 + 2·2 + 5·1 + 14·1. Lattice paths that never cross the diagonal are the classic visualization: C(4) = 14 paths from (0,0) to (4,4) staying on one side of y = x.

1. 1

   Enter n

   The Catalan index, e.g. 5

2. 2

   Compute the central binomial

   C(2n, n) = C(10,5) = 252 is shown first

3. 3

   Divide by n+1

   252/6 = 14 = C₅

4. 4

   Browse interpretations

   Trees, brackets, triangulations and paths counted by your C(n)

• 1

  C(n) = C(2n,n)/(n+1) — one division from a binomial coefficient

• 2

  Sequence: 1, 1, 2, 5, 14, 42, 132, 429, …

• 3

  Catalans grow ≈ 4ⁿ, slower than the central binomial itself

• Forgetting the n+1 divisor and reporting C(10,5) = 252 for n = 5

• Indexing off by one: a polygon with n+2 sides uses C(n)

• Confusing Catalan numbers with central binomial coefficients

• Assuming C(0) = 0; it is 1 by convention