最初の 50 個のカタラン数

The sequence reached print in Europe in 1751, when Leonhard Euler asked how many ways a convex polygon can be cut into triangles by non-crossing diagonals and worked out both the counts and a formula. The name attached to it much later and for an entirely different reason. Eugène Charles Catalan, a Belgian-born mathematician who spent most of his career in France and Liège, found the connection to parenthesised expressions while working on the Towers of Hanoi puzzle. It was the twentieth-century American combinatorialist John Riordan who began calling the numbers Catalan's, and the label stuck.

The dates run further back than Euler, though this only became widely known in 1988. The Mongolian mathematician Mingantu, working in Qing China, had begun writing Ge Yuan Mi Lu Jie Fa — "the quick method for obtaining the precise ratio of division of a circle" — by about 1730, prompted in part by three infinite series that the Jesuit missionary Pierre Jartoux had brought to China early in the century. Mingantu used the sequence as coefficients in series expansions, writing sin 2α and sin 4α in terms of sin α. His student Chen Jixin completed the manuscript in 1774, and it waited roughly another sixty years to be published. Peter Larcombe surveyed this history in 1999.

What makes the sequence remarkable is less its discovery than how often it is rediscovered. Désiré André's reflection argument of 1887 gave a clean way to count Dyck words, and the catalogue of things the numbers count has kept growing: Richard Stanley's Enumerative Combinatorics sets out sixty-six different interpretations as exercises. The pattern is consistent enough that finding a counting problem whose answers are 1, 1, 2, 5, 14, 42 is now taken as a strong hint that a bijection to brackets or trees is waiting to be found.

主な性質

• C(0) = C(1) = 1, and C(n) = C(2n, n) / (n + 1) = (2n)! / (n! · (n+1)!).
• Equivalently C(n) = C(2n, n) − C(2n, n+1), a difference of two binomial coefficients.
• The sequence convolves with itself: C(n+1) = C(0)·C(n) + C(1)·C(n−1) + … + C(n)·C(0).
• C(n+1) = C(n) · 2(2n + 1) / (n + 2), and the result is always a whole number, which is what lets each term be computed from its predecessor in exact integer arithmetic.
• C(n) is odd exactly when n = 2^k − 1; every other Catalan number is even.
• The only prime Catalan numbers are C(2) = 2 and C(3) = 5.
• The ratio C(n+1)/C(n) approaches 4, since C(n) grows like 4ⁿ / (n^(3/2)·√π).
• The n×n Hankel matrix whose (i, j) entry is C(i+j−2) has determinant 1 for every n.