Catalan's triangle

In combinatorial mathematics, Catalan's triangle is a number triangle whose entries

give the number of strings consisting of n X's and k Y's such that no initial segment of the string has more Y's than X's.

It is a generalization of the Catalan numbers, and is named after Eugène Charles Catalan.

Bailey[1] shows that

satisfy the following properties: Formula 3 shows that the entry in the triangle is obtained recursively by adding numbers to the left and above in the triangle.

The earliest appearance of the Catalan triangle along with the recursion formula is in page 214 of the treatise on Calculus published in 1800[2] by Louis François Antoine Arbogast.

Shapiro[3] introduces another triangle which he calls the Catalan triangle that is distinct from the triangle being discussed here.

, the diagonal C(n, n) is the n-th Catalan number.

The row sum of the n-th row is the (n + 1)-th Catalan number, using the hockey-stick identity and an alternative expression for Catalan numbers.

Some values are given by[5] That is, an entry is the partial sum of the above row and also the partial sum of the column to the left (except for the entry on the diagonal).

Catalan's trapezoids are a countable set of number trapezoids which generalize Catalan’s triangle.

Catalan's trapezoid of order m = 1, 2, 3, ... is a number trapezoid whose entries

give the number of strings consisting of n X-s and k Y-s such that in every initial segment of the string the number of Y-s does not exceed the number of X-s by m or more.

[6] By definition, Catalan's trapezoid of order m = 1 is Catalan's triangle, i.e.,

Some values of Catalan's trapezoid of order m = 2 are given by Some values of Catalan's trapezoid of order m = 3 are given by Again, each element is the sum of the one above and the one to the left.

This proof involves an extension of Andre's reflection method as used in the second proof for the Catalan number to different diagonals.

The following shows how every path from the bottom left

of the diagram that crosses the constraint

can also be reflected to the end point

We consider three cases to determine the number of paths from

that do not cross the constraint: (1) when

the constraint cannot be crossed, so all paths from

it is impossible to form a path that does not cross the constraint, i.e.

minus the number of 'yellow' paths that cross the constraint, i.e.

that do not cross the constraint

is as indicated in the formula in the previous section "Generalization".

Firstly, we confirm the validity of the recurrence relation

into two parts, the first for XY combinations ending in X and the second for those ending in Y.

valid combinations and the second has

Proof 2 is completed by verifying the solution satisfies the recurrence relation and obeys initial conditions for