Catalan's triangle

Summary

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:

  1. .
  2. .
  3. .

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.

General formula edit

The general formula for   is given by[1][4]

 

So

 

When  , 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.

Table of values edit

Some values are given by[5]

 k
n 
0 1 2 3 4 5 6 7 8
0 1
1 1 1
2 1 2 2
3 1 3 5 5
4 1 4 9 14 14
5 1 5 14 28 42 42
6 1 6 20 48 90 132 132
7 1 7 27 75 165 297 429 429
8 1 8 35 110 275 572 1001 1430 1430

Properties edit

  • Formula 3 from the first section can be used to prove both
 

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).

  • If  , then at some stage there must be more Y's than X's, so  .
  • A combinatorial interpretation of the  -th value is the number of non-decreasing partitions with exactly n parts with maximum part k such that each part is less than or equal to its index. So, for example,   counts
 

Generalization edit

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

 k
n 
0 1 2 3 4 5 6 7 8
0 1 1
1 1 2 2
2 1 3 5 5
3 1 4 9 14 14
4 1 5 14 28 42 42
5 1 6 20 48 90 132 132
6 1 7 27 75 165 297 429 429
7 1 8 35 110 275 572 1001 1430 1430

Some values of Catalan's trapezoid of order m = 3 are given by

 k
n 
0 1 2 3 4 5 6 7 8 9
0 1 1 1
1 1 2 3 3
2 1 3 6 9 9
3 1 4 10 19 28 28
4 1 5 15 34 62 90 90
5 1 6 21 55 117 207 297 297
6 1 7 28 83 200 407 704 1001 1001
7 1 8 36 119 319 726 1430 2431 3432 3432

Again, each element is the sum of the one above and the one to the left.

A general formula for   is given by

 

( n = 0, 1, 2, ..., k = 0, 1, 2, ..., m = 1, 2, 3, ...).

Proofs of the general formula edit

Proof 1 edit

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   to the top right   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   to   that do not cross the constraint:

(1) when   the constraint cannot be crossed, so all paths from   to   are valid, i.e.  .

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

(3) when  , then   is the number of 'red' paths   minus the number of 'yellow' paths that cross the constraint, i.e.  .

Therefore the number of paths from   to   that do not cross the constraint   is as indicated in the formula in the previous section "Generalization".

Proof 2 edit

Firstly, we confirm the validity of the recurrence relation   by breaking down   into two parts, the first for XY combinations ending in X and the second for those ending in Y. The first group therefore has   valid combinations and the second has  . Proof 2 is completed by verifying the solution satisfies the recurrence relation and obeys initial conditions for   and  .

References edit

  1. ^ a b Bailey, D. F. (1996). "Counting Arrangements of 1's and -1's". Mathematics Magazine. 69 (2): 128–131. doi:10.1080/0025570X.1996.11996408.
  2. ^ Arbogast, L. F. A. (1800). Du Calcul des Derivations. Levrault. p. 214.
  3. ^ Shapiro, L. W. (1976). "A Catalan Triangle". Discrete Mathematics. 14 (1): 83–90. doi:10.1016/0012-365x(76)90009-1.
  4. ^ Eric W. Weisstein. "Catalan's Triangle". MathWorld − A Wolfram Web Resource. Retrieved March 28, 2012.
  5. ^ Sloane, N. J. A. (ed.). "Sequence A009766 (Catalan's triangle)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved March 28, 2012.
  6. ^ Reuveni, Shlomi (2014). "Catalan's trapezoids". Probability in the Engineering and Informational Sciences. 28 (3): 4391–4396. doi:10.1017/S0269964814000047. S2CID 122765015.