Catalan numbers/Pascal's triangle

Task

Print out the first 15 Catalan numbers by extracting them from Pascal's triangle.

See

Catalan Numbers and the Pascal Triangle. This method enables calculation of Catalan Numbers using only addition and subtraction.
Catalan's Triangle for a Number Triangle that generates Catalan Numbers using only addition.
Sequence A000108 on OEIS has a lot of information on Catalan Numbers.

Related Tasks

360 Assembly[edit]

For maximum compatibility, this program uses only the basic instruction set.

CATALAN  CSECT                         
         USING  CATALAN,R13,R12
SAVEAREA B      STM-SAVEAREA(R15)
         DC     17F'0'
         DC     CL8'CATALAN'
STM      STM    R14,R12,12(R13)
         ST     R13,4(R15)
         ST     R15,8(R13)
         LR     R13,R15
         LA     R12,4095(R13)
         LA     R12,1(R12)
*        ----   CODE
         LA     R0,1
         ST     R0,T            t(1)=1
         LA     R4,0            ix:i=1
         LA     R6,1            by 1
         LH     R7,N            to n 
LOOPI    BXH    R4,R6,ENDLOOPI  loop i
         LR     R5,R4           ix:j=i+1
         LA     R5,2(R5)        i+2
         LA     R8,0
         BCTR   R8,0            by -1 
         LA     R9,1            to 2
LOOP1J   BXLE   R5,R8,ENLOOP1J  loop j
         LR     R10,R5          j
         BCTR   R10,0
         SLA    R10,2
         L      R2,T(R10)       r2=t(j)  
         LR     R1,R10          j
         SH     R1,=H'4'
         L      R3,T(R1)        r3=t(j-1)
         AR     R2,R3           r2=r2+r3 
         ST     R2,T(R10)       t(j)=t(j)+t(j-1)
         B      LOOP1J
ENLOOP1J EQU    *
         LR     R1,R4           i
         BCTR   R1,0
         SLA    R1,2
         L      R3,T(R1)        t(i)
         LA     R1,4(R1)
         ST     R3,T(R1)        t(i+1)
         LR     R5,R4           ix:j=i+2
         LA     R5,3(R5)        i+3
         LA     R8,0
         BCTR   R8,0            by -1 
         LA     R9,1            to 2
LOOP2J   BXLE   R5,R8,ENLOOP2J  loop j
         LR     R10,R5          j
         BCTR   R10,0
         SLA    R10,2
         L      R2,T(R10)       r2=t(j)  
         LR     R1,R10          j
         SH     R1,=H'4'
         L      R3,T(R1)        r3=t(j-1)
         AR     R2,R3           r2=r2+r3 
         ST     R2,T(R10)       t(j)=t(j)+t(j-1)
         B      LOOP2J
ENLOOP2J EQU    *
         LR     R1,R4           i
         BCTR   R1,0
         SLA    R1,2
         L      R2,T(R1)        t(i)
         LA     R1,4(R1)
         L      R3,T(R1)        t(i+1)
         SR     R3,R2
         CVD    R3,P            
         UNPK   Z,P
         MVC    C,Z
         OI     C+L'C-1,X'F0'
         MVC    WTOBUF(8),C+8
         WTO    MF=(E,WTOMSG)		  
         B      LOOPI
ENDLOOPI EQU    *
*        ----   END CODE
         CNOP   0,4
         L      R13,4(0,R13)
         LM     R14,R12,12(R13)
         XR     R15,R15
         BR     R14
*        ----   DATA
N        DC     H'15'
T        DC     17F'0'
P        DS     PL8
Z        DS     ZL16
C        DS     CL16
WTOMSG   DS     0F
         DC     H'80'
         DC     H'0'
WTOBUF   DC     CL80' '
         YREGS  
         END

Output:

Ada[edit]

Uses package Pascal from the Pascal triangle solution[[1]]

with Ada.Text_IO, Pascal;
 
procedure Catalan is
 
   Last: Positive := 15;   
   Row: Pascal.Row := Pascal.First_Row(2*Last+1);
 
begin
   for I in 1 .. Last loop
      Row := Pascal.Next_Row(Row);
      Row := Pascal.Next_Row(Row);
      Ada.Text_IO.Put(Integer'Image(Row(I+1)-Row(I+2)));
   end loop;
end Catalan;

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

ALGOL 68[edit]

Translation of: C++

INT n = 15;
[ 0 : n + 1 ]INT t;
t[0] := 0;
t[1] := 1;
FOR i TO n DO
    FOR j FROM i   BY -1 TO 2 DO t[j] := t[j] + t[j-1] OD;
    t[i+1] := t[i];
    FOR j FROM i+1 BY -1 TO 2 DO t[j] := t[j] + t[j-1] OD;
    print( ( whole( t[i+1] - t[i], 0 ), " " ) )
OD

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

ALGOL W[edit]

begin
    % print the first 15 Catalan numbers from Pascal's triangle %
    integer n;
    n := 15;
    begin
        integer array pascalLine ( 1 :: n + 1 );
        % the Catalan numbers are the differences between the middle and middle - 1 numbers of the odd %
        % lines of Pascal's triangle (lines with 3 or more numbers)                                    %
        % note - we only need to calculate the left side of the triangle                               %
        pascalLine( 1 ) := 1;
        for c := 2 until n + 1 do begin
            % even line %
            for i := c - 1 step -1 until 2 do pascalLine( i ) := pascalLine( i - 1 ) + pascalLine( i );
            pascalLine( c ) := pascalLine( c - 1 );
            % odd line %
            for i := c     step -1 until 2 do pascalLine( i ) := pascalLine( i - 1 ) + pascalLine( i );
            writeon( i_w := 1, s_w := 0, " ", pascalLine( c ) - pascalLine( c - 1 ) )
        end for_c
    end
end.

Output:

 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

APL[edit]

 
      ⍝ Based heavily on the J solution
      CATALAN←{¯1↓↑-/1 ¯1↓¨(⊂⎕IO+0 0)⍉¨0 2⌽¨⊂(⎕IO-⍨⍳N){+\⍣⍺⊢⍵}⍤0 1⊢1⍴⍨N←⍵+2}

Output:

      CATALAN 15
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

AutoHotkey[edit]

Works with: AutoHotkey_L

/* Generate Catalan Numbers
//
// smgs: 20th Feb, 2014
*/
Array := [], Array[2,1] := Array[2,2] := 1 ; Array inititated and 2nd row of pascal's triangle assigned
INI := 3 ; starts with calculating the 3rd row and as such the value
Loop, 31 ; every odd row is taken for calculating catalan number as such to obtain 15 we need 2n+1
{
	if ( A_index > 2 )
	{
		Loop, % A_INDEX
		{
			old := ini-1, 		index := A_index, 		index_1 := A_index + 1
			Array[ini, index_1] := Array[old, index] + Array[old, index_1]
			Array[ini, 1] := Array[ini, ini] := 1
			line .= Array[ini, A_index] " "
		}
	;~ MsgBox % line ; gives rows of pascal's triangle
	; calculating every odd row starting from 1st so as to obtain catalan's numbers
		if ( mod(ini,2) != 0)
		{
			StringSplit, res, line, %A_Space%
			ans := res0//2, ans_1 := ans++
			result := result . res%ans_1% - res%ans% " " 
		}
	line :=
	ini++
	}
}
MsgBox % result

Produces:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

Batch File[edit]

@echo off
setlocal ENABLEDELAYEDEXPANSION
set n=15
set /A nn=n+1
for /L %%i in (0,1,%nn%) do set t.%%i=0
set t.1=1
for /L %%i in (1,1,%n%) do (
    set /A ip=%%i+1
    for /L %%j in (%%i,-1,1) do (
        set /A jm=%%j-1
	    set /A t.%%j=t.%%j+t.!jm!
	)
    set /A t.!ip!=t.%%i	  
    for /L %%j in (!ip!,-1,1) do (
        set /A jm=%%j-1
	    set /A t.%%j=t.%%j+t.!jm!
	)
    set /A ci=t.!ip!-t.%%i
	echo !ci!
  )
)
pause

Output:

C[edit]

 
//This code implements the print of 15 first Catalan's Numbers
//Formula used:
//  __n__
//   | |  (n + k) / k  n>0
//   k=2  
 
#include <stdio.h>
#include <stdlib.h>
 
//the number of Catalan's Numbers to be printed
const int N = 15;
 
int main()
{
    //loop variables (in registers)
    register int k, n;
 
    //necessarily ull for reach big values
    unsigned long long int num, den;
 
    //the nmmber
    int catalan;
 
    //the first is not calculated for the formula
    printf("1 ");
 
    //iterating fro 2 to 15
    for (n=2; n<=N; ++n) {
        //initializaing for products
        num = den = 1;
        //applying the formula
        for (k=2; k<=n; ++k) {
            num *= (n+k);
            den *= k;
            catalan = num /den;
        }
 
        //output
        printf("%d ", catalan);
    }
 
    //the end
    printf("\n");
    return 0;
}

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

C++[edit]

// Generate Catalan Numbers
//
// Nigel Galloway: June 9th., 2012
//
#include <iostream>
int main() {
  const int N = 15;
  int t[N+2] = {0,1};
  for(int i = 1; i<=N; i++){
    for(int j = i; j>1; j--) t[j] = t[j] + t[j-1];
    t[i+1] = t[i];
    for(int j = i+1; j>1; j--) t[j] = t[j] + t[j-1];
    std::cout << t[i+1] - t[i] << " ";
  }
  return 0;
}

Produces:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

C#[edit]

Translation of: C++

 
int n = 15;
List<int> t = new List<int>() { 0, 1 };
for (int i = 1; i <= n; i++)
{
    for (var j = i; j > 1; j--) t[j] += t[j - 1];
    t.Add(t[i]);
    for (var j = i + 1; j > 1; j--) t[j] += t[j - 1];
    Console.Write(((i == 1) ? "" : ", ") + (t[i + 1] - t[i]));
}

Produces:

1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845

D[edit]

Translation of: C++

void main() {
    import std.stdio;
 
    enum uint N = 15;
    uint[N + 2] t;
    t[1] = 1;
 
    foreach (immutable i; 1 .. N + 1) {
        foreach_reverse (immutable j; 2 .. i + 1)
            t[j] += t[j - 1];
        t[i + 1] = t[i];
        foreach_reverse (immutable j; 2 .. i + 2)
            t[j] += t[j - 1];
        write(t[i + 1] - t[i], ' ');
    }
}

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

EchoLisp[edit]

 
(define dim 100)
(define-syntax-rule  (Tidx i j)  (+ i (* dim j)))
 
;; generates Catalan's triangle
;; T (i , j) = T(i-1,j) + T (i, j-1)
 
(define (T n)
	(define i (modulo n dim))
	(define j (quotient n dim))
	(cond 
		((zero? i) 1) ;; left column = 1
		((= i j) (T (Tidx (1- i) j))) ;; diagonal value = left value
		(else (+ (T (Tidx (1- i) j)) (T (Tidx i (1- j)))))))
 
(remember 'T #(1))

Output:

 
;; take elements on diagonal = Catalan numbers
(for ((i (in-range 0 16))) (write (T (Tidx i i))))
 
 → 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

Elixir[edit]

defmodule Catalan do
  def numbers(num) do
    {result,_} = Enum.reduce(1..num, {[],{0,1}}, fn i,{list,t0} ->
      t1 = numbers(i, t0)
      t2 = numbers(i+1, Tuple.insert_at(t1, i+1, elem(t1, i)))
      {[elem(t2, i+1) - elem(t2, i) | list], t2}
    end)
    Enum.reverse(result)
  end
 
  defp numbers(0, t), do: t
  defp numbers(n, t), do: numbers(n-1, put_elem(t, n, elem(t, n-1) + elem(t, n)))
end
 
IO.inspect Catalan.numbers(15)

Output:

[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]

Erlang[edit]

 
-module(catalin).
-compile(export_all).
mul(N,D,S,S)-> 
	N2=N*(S+S),
	D2=D*S,
	K = N2 div D2 ;
mul(N,D,S,L)->
	N2=N*(S+L),
	D2=D*L,
	K = mul(N2,D2,S,L+1).
 
catl(Ans,16) -> Ans;
catl(D,S)->
	C=mul(1,1,S,2),
	catl([D|C],S+1).
main()->
	Ans=catl(1,2).

ERRE[edit]

 
PROGRAM CATALAN
 
!$DOUBLE
 
DIM CATALAN[50]
 
FUNCTION ODD(X)
    ODD=FRC(X/2)<>0
END FUNCTION
 
PROCEDURE GETCATALAN(L)
    LOCAL J,K,W
    LOCAL DIM PASTRI[100]
 
    L=L*2
    PASTRI[0]=1
    J=0
    WHILE J<L DO
       J+=1
       K=INT((J+1)/2)
       PASTRI[K]=PASTRI[K-1]
       FOR W=K TO 1 STEP -1 DO
          PASTRI[W]+=PASTRI[W-1]
       END FOR
       IF NOT(ODD(J)) THEN
          K=INT(J/2)
          CATALAN[K]=PASTRI[K]-PASTRI[K-1]
       END IF
    END WHILE
END PROCEDURE
 
BEGIN
   LL=15
   GETCATALAN(LL)
   FOR I=1 TO LL DO
      WRITE("### ####################";I;CATALAN[I])
   END FOR
END PROGRAM

Output:

  1                    1
  2                    2
  3                    5
  4                   14
  5                   42
  6                  132
  7                  429
  8                 1430
  9                 4862
 10                16796
 11                58786
 12               208012
 13               742900
 14              2674440
 15              9694845

F#[edit]

 
let mutable nm=uint64(1)
let mutable dm=uint64(1)
let mutable a=uint64(1)
 
printf "1, "
for i = 2 to 15 do
    nm<-uint64(1)
    dm<-uint64(1)
    for k = 2 to i do
        nm <-uint64( uint64(nm) * (uint64(i)+uint64(k)))
        dm <-uint64( uint64(dm) * uint64(k))
    let a = uint64(uint64(nm)/uint64(dm))
    printf "%u"a
    if(i<>15) then
        printf ", "

Factor[edit]

USING: arrays grouping io kernel math prettyprint sequences ;
IN: rosetta-code.catalan-pascal
 
: next-row ( seq -- seq' )
    2 clump [ sum ] map 1 prefix 1 suffix ;
 
: pascal ( n -- seq )
    1 - { { 1 } } swap [ dup last next-row suffix ] times ;
 
15 2 * pascal [ length odd? ] filter [
    dup length 1 = [ 1 ]
    [ dup [email protected] dup 1 + 2array swap nths first2 - ] if
    pprint bl
] each drop

Output:

1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440

FreeBASIC[edit]

' version 15-09-2015
' compile with: fbc -s console
 
#Define size 31                 ' (N * 2 + 1)
 
Sub pascal_triangle(rows As Integer, Pas_tri() As ULongInt)
 
    Dim As Integer x, y
 
    For x = 1 To rows
        Pas_tri(1,x) = 1
        Pas_tri(x,1) = 1
    Next
 
    For x = 2 To rows
        For y = 2 To rows + 1 - x
            Pas_tri(x, y) = pas_tri(x - 1 , y) + pas_tri(x, y - 1)
        Next
    Next
 
End Sub
 
' ------=< MAIN >=------
 
Dim As Integer count, row
Dim As ULongInt triangle(1 To size, 1 To size)
 
pascal_triangle(size, triangle())
 
'  1   1   1   1   1   1  
'  1   2   3   4   5   6  
'  1   3   6  10  15  21  
'  1   4  10  20  35  56  
'  1   5  15  35  70 126  
'  1   6  21  56 126 252  
' The Pascal triangle is rotated 45 deg.
' to find the Catalan number we need to follow the diagonal
' for top left to bottom right 
' take the number on diagonal and subtract the number in de cell
' one up and one to right 
' 1 (2 - 1), 2 (6 - 4), 5 (20 - 15) ... 
 
 
Print "The first 15 Catalan numbers are" : print
count = 1 : row = 2
Do
    Print Using "###: #########"; count; triangle(row, row) - triangle(row +1, row -1)
    row = row + 1
    count =  count + 1
Loop Until count > 15
 
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End

Output:

The first 15 Catalan numbers are

  1:         1
  2:         2
  3:         5
  4:        14
  5:        42
  6:       132
  7:       429
  8:      1430
  9:      4862
 10:     16796
 11:     58786
 12:    208012
 13:    742900
 14:   2674440
 15:   9694845

Go[edit]

Translation of: C++

package main
 
import "fmt"
 
func main() {
    const n = 15
    t := [n + 2]uint64{0, 1}
    for i := 1; i <= n; i++ {
        for j := i; j > 1; j-- {
            t[j] += t[j-1]
        }
        t[i+1] = t[i]
        for j := i + 1; j > 1; j-- {
            t[j] += t[j-1]
        }
        fmt.Printf("%2d : %d\n", i, t[i+1]-t[i])
    }
}

Output:

 1 : 1
 2 : 2
 3 : 5
 4 : 14
 5 : 42
 6 : 132
 7 : 429
 8 : 1430
 9 : 4862
10 : 16796
11 : 58786
12 : 208012
13 : 742900
14 : 2674440
15 : 9694845

Groovy[edit]

Translation of: C

 
class Catalan
{
 public static void main(String[] args)
  {
    BigInteger N = 15;
    BigInteger k,n,num,den;
    BigInteger  catalan;
      print(1);
       for(n=2;n<=N;n++)
          {
            num = 1;
            den = 1;
              for(k=2;k<=n;k++)
                 {
                    num = num*(n+k);
                    den = den*k;
                    catalan = num/den; 
                 }
            print(" " + catalan);
          }
 
  }
}

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

As required by the task this implementation extracts the Catalan numbers from Pascal's triangle, rather than calculating them directly. Also, note that it (correctly) produces [1, 1] as the first two numbers.

import System.Environment (getArgs)
 
-- Pascal's triangle.
pascal :: [[Integer]]
pascal = [1] : map (\row -> 1 : zipWith (+) row (tail row) ++ [1]) pascal
 
-- The Catalan numbers from Pascal's triangle.  This uses a method from
-- http://www.cut-the-knot.org/arithmetic/algebra/CatalanInPascal.shtml
-- (see "Grimaldi").
catalan :: [Integer]
catalan = map (diff . uncurry drop) $ zip [0..] (alt pascal)
  where alt (x:_:zs) = x : alt zs -- every other element of an infinite list
        diff (x:y:_) = x - y
        diff (x:_)   = x
 
main :: IO ()
main = do
  ns <- fmap (map read) getArgs :: IO [Int]
  mapM_ (print . flip take catalan) ns

Output:

./catalan 15
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]

Icon and Unicon[edit]

The following works in both languages. It avoids computing elements in Pascal's triangle that aren't used.

link math
 
procedure main(A)
    limit := (integer(A[1])|15)+1
    every write(right(binocoef(i := 2*seq(0)\limit,i/2)-binocoef(i,i/2+1),30))
end

Sample run:

J[edit]

   Catalan=. }:@:(}[email protected]:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)

Example use:

   Catalan 15
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

A structured derivation of Catalan follows:

   o=. @: NB. Composition of verbs (functions) 
   ( PascalTriangle=. i. ((+/\@]^:[)) #&1 ) 5
1 1  1  1  1
1 2  3  4  5
1 3  6 10 15
1 4 10 20 35
1 5 15 35 70
   ( MiddleDiagonal=. (<0 1)&|: )               o PascalTriangle 5
1 2 6 20 70
   ( AdjacentLeft=.   MiddleDiagonal o (2&|.) ) o PascalTriangle 5
1 4 15 1 5
 
   ( Catalan=. }: o (}. o MiddleDiagonal - }: o AdjacentLeft) o PascalTriangle o (2&+) f. ) 5
1 2 5 14 42
 
   Catalan
}:@:(}[email protected]:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)

Java[edit]

Translation of: C++

public class Test {
    public static void main(String[] args) {
        int N = 15;
        int[] t = new int[N + 2];
        t[1] = 1;
 
        for (int i = 1; i <= N; i++) {
 
            for (int j = i; j > 1; j--)
                t[j] = t[j] + t[j - 1];
 
            t[i + 1] = t[i];
 
            for (int j = i + 1; j > 1; j--)
                t[j] = t[j] + t[j - 1];
 
            System.out.printf("%d ", t[i + 1] - t[i]);
        }
    }
}

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

JavaScript[edit]

ES5[edit]

Iteration

Translation of: C++

var n = 15;
for (var t = [0, 1], i = 1; i <= n; i++) {
    for (var j = i; j > 1; j--) t[j] += t[j - 1];
    t[i + 1] = t[i];
    for (var j = i + 1; j > 1; j--) t[j] += t[j - 1];
    document.write(i == 1 ? '' : ', ', t[i + 1] - t[i]);
}

Output:

1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845

ES6[edit]

Functional composition

Translation of: Haskell

(() => {
    'use strict';
 
    // CATALAN
 
    // catalanSeries :: Int -> [Int]
    let catalanSeries = n => {
        let alternate = xs => xs.reduce(
                (a, x, i) => i % 2 === 0 ? a.concat([x]) : a, []
            ),
            diff = xs => xs.length > 1 ? xs[0] - xs[1] : xs[0];
 
        return alternate(pascal(n * 2))
            .map((xs, i) => diff(drop(i, xs)));
    }
 
    // PASCAL
 
    // pascal :: Int -> [[Int]]
    let pascal = n => until(
            m => m.level <= 1,
            m => {
                let nxt = zipWith(
                    (a, b) => a + b, [0].concat(m.row), m.row.concat(0)
                );
                return {
                    row: nxt,
                    triangle: m.triangle.concat([nxt]),
                    level: m.level - 1
                }
            }, {
                level: n,
                row: [1],
                triangle: [
                    [1]
                ]
            }
        )
        .triangle;
 
 
    // GENERIC FUNCTIONS
 
    // zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
    let zipWith = (f, xs, ys) =>
        xs.length === ys.length ? (
            xs.map((x, i) => f(x, ys[i]))
        ) : undefined;
 
    // until :: (a -> Bool) -> (a -> a) -> a -> a
    let until = (p, f, x) => {
        let v = x;
        while (!p(v)) v = f(v);
        return v;
    }
 
    // drop :: Int -> [a] -> [a]
    let drop = (n, xs) => xs.slice(n);
 
    // tail :: [a] -> [a]
    let tail = xs => xs.length ? xs.slice(1) : undefined;
 
    return tail(catalanSeries(16));
})();

Output:

[1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845]

jq[edit]

The first identity (C(2n,n) - C(2n, n-1)) given in the reference is used in accordance with the task description, but it would of course be more efficient to factor out C(2n,n) and use the expression C(2n,n)/(n+1). See also Catalan_numbers#jq for other alternatives.

Warning: jq uses IEEE 754 64-bit arithmetic, so the algorithm used here for Catalan numbers loses precision for n > 30 and fails completely for n > 510.

def binomial(n; k): 
  if k > n / 2 then binomial(n; n-k)
  else reduce range(1; k+1) as $i (1; . * (n - $i + 1) / $i)
  end;
 
# Direct (naive) computation using two numbers in Pascal's triangle:
def catalan_by_pascal: . as $n | binomial(2*$n; $n) - binomial(2*$n; $n-1);

Example:

(range(0;16), 30, 31, 510, 511) | [., catalan_by_pascal]

Output:

$ jq -n -c -f Catalan_numbers_Pascal.jq
[0,0]
[1,1]
[2,2]
[3,5]
[4,14]
[5,42]
[6,132]
[7,429]
[8,1430]
[9,4862]
[10,16796]
[11,58786]
[12,208012]
[13,742900]
[14,2674440]
[15,9694845]
[30,3814986502092304]
[31,14544636039226880]
[510,5.491717746183512e+302]
[511,null]

Julia[edit]

Translation of: Matlab

# v0.6
 
function pascal(n::Int)
    r = ones(Int, n, n)
    for i in 2:n, j in 2:n
        r[i, j] = r[i-1, j] + r[i, j-1]
    end
    return r
end
 
function catalan_num(n::Int)
    p = pascal(n + 2)
    p[n+4:n+3:end-1] - diag(p, 2)
end
 
@show catalan_num(15)

Output:

catalan_num(15) = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]

Kotlin[edit]

// version 1.1.2
 
import java.math.BigInteger
 
val ONE = BigInteger.ONE
 
fun pascal(n: Int, k: Int): BigInteger {
    if (n == 0 || k == 0) return ONE
    val num = (k + 1..n).fold(ONE) { acc, i -> acc * BigInteger.valueOf(i.toLong()) }
    val den = (2..n - k).fold(ONE) { acc, i -> acc * BigInteger.valueOf(i.toLong()) }
    return num / den
}
 
fun catalanFromPascal(n: Int) {
    for (i in 1 until n step 2) {
        val mi = i / 2 + 1
        val catalan = pascal(i, mi) - pascal(i, mi - 2) 
        println("${"%2d".format(mi)} : $catalan")
    }
}
 
fun main(args: Array<String>) {
    val n = 15
    catalanFromPascal(n * 2)
}

Output:

 1 : 1
 2 : 2
 3 : 5
 4 : 14
 5 : 42
 6 : 132
 7 : 429
 8 : 1430
 9 : 4862
10 : 16796
11 : 58786
12 : 208012
13 : 742900
14 : 2674440
15 : 9694845

Lua[edit]

For each line of odd-numbered length from Pascal's triangle, print the middle number minus the one immediately to its right. This solution is heavily based on the Lua code to generate Pascal's triangle from the page for that task.

function nextrow (t)
    local ret = {}
    t[0], t[#t + 1] = 0, 0
    for i = 1, #t do ret[i] = t[i - 1] + t[i] end
    return ret
end
 
function catalans (n)
    local t, middle = {1}
    for i = 1, n do
        middle = math.ceil(#t / 2)
        io.write(t[middle] - (t[middle + 1] or 0) .. " ")
        t = nextrow(nextrow(t))
    end
end
 
catalans(15)

Output:

1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440

Mathematica / Wolfram Language[edit]

This builds the entire Pascal triangle that's needed and holds it in memory. Very inefficienct, but seems to be what is asked in the problem.

nextrow[lastrow_] := Module[{output},
  output = ConstantArray[1, Length[lastrow] + 1];
  Do[
   output[[i + 1]] = lastrow[[i]] + lastrow[[i + 1]];
   , {i, 1, Length[lastrow] - 1}];
  output
  ]
pascaltriangle[size_] := NestList[nextrow, {1}, size]  
catalannumbers[length_] := Module[{output, basetriangle},
  basetriangle = pascaltriangle[2 length];
  list1 = basetriangle[[# *2 + 1, # + 1]] & /@ Range[length];
  list2 = basetriangle[[# *2 + 1, # + 2]] & /@ Range[length];
  list1 - list2
  ]
(* testing *)
catalannumbers[15]

Output:

{1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845}

MATLAB / Octave[edit]

n = 15;
p = pascal(n + 2);
p(n + 4 : n + 3 : end - 1)' - diag(p, 2)

Output:

Nim[edit]

Translation of: Python

const n = 15
var t = newSeq[int](n + 2)
 
t[1] = 1
for i in 1..n:
  for j in countdown(i, 1): t[j] += t[j-1]
  t[i+1] = t[i]
  for j in countdown(i+1, 1): t[j] += t[j-1]
  stdout.write t[i+1] - t[i], " "

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

OCaml[edit]

 
let catalan : int ref = ref 0 in
Printf.printf "%d ," 1 ;
for i = 2 to 9  do
let nm : int ref = ref 1 in
let den : int ref = ref 1 in
for k = 2 to i do
nm := (!nm)*(i+k);
den := (!den)*k;
catalan := (!nm)/(!den) ;
done;
print_int (!catalan); print_string "," ;
done;;

Output:

OUTPUT:
1 ,2,5,14,42,132,429,1430,4862

Oforth[edit]

import: mapping
 
: pascal( n -- [] )  
   [ 1 ] n #[ dup [ 0 ] + [ 0 ] rot + zipWith( #+ ) ] times ;
 
: catalan( n -- m )
   n 2 * pascal at( n 1+ ) n 1+ / ;

Output:

>#catalan 15 seq map .
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]

PARI/GP[edit]

vector(15,n,binomial(2*n,n)-binomial(2*n,n+1))

Output:

%1 = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]

Pascal[edit]

type
  tElement = Uint64;
var
  Catalan : array[0..50] of tElement;
procedure GetCatalan(L:longint);
var
  PasTri : array[0..100] of tElement;
  j,k: longInt;
begin
  l := l*2;
  PasTri[0] := 1;
  j    := 0;
  while (j<L) do
  begin
    inc(j);
    k := (j+1) div 2;
    PasTri[k] :=PasTri[k-1];
    For k := k downto 1 do
      inc(PasTri[k],PasTri[k-1]);
    IF NOT(Odd(j)) then
    begin
      k := j div 2;
      Catalan[k] :=PasTri[k]-PasTri[k-1];
    end;
  end;
end;
 
var
  i,l: longint;
Begin
  l := 15;
  GetCatalan(L);
  For i := 1 to L do
    Writeln(i:3,Catalan[i]:20);
end.

  1                   1
  2                   2
  3                   5
  4                  14
  5                  42
  6                 132
  7                 429
  8                1430
  9                4862
 10               16796
 11               58786
 12              208012
 13              742900
 14             2674440
 15             9694845

Perl[edit]

Translation of: C++

use constant N => 15;
my @t = (0, 1);
for(my $i = 1; $i <= N; $i++) {
    for(my $j = $i; $j > 1; $j--) { $t[$j] += $t[$j-1] }
    $t[$i+1] = $t[$i];
    for(my $j = $i+1; $j>1; $j--) { $t[$j] += $t[$j-1] }
    print $t[$i+1] - $t[$i], " ";
}

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

After the 28th Catalan number, this overflows 64-bit integers. We could add use bigint; use Math::GMP ":constant"; to make it work, albeit not at a fast pace. However we can use a module to do it much faster:

Library: ntheory

use ntheory qw/binomial/;
print join(" ", map { binomial( 2*$_, $_) / ($_+1) } 1 .. 1000), "\n";

The Math::Pari module also has a binomial, but isn't as fast and overflows its stack after 3400.

Perl 6[edit]

Works with: Rakudo version 2015.12

constant @pascal = [1], -> @p { [0, |@p Z+ |@p, 0] } ... *;
 
constant @catalan = gather for 2, 4 ... * -> $ix {
    my @row := @pascal[$ix];
    my $mid = +@row div 2;
    take [-] @row[$mid, $mid+1]
}
 
.say for @catalan[^20];

Output:

Phix[edit]

Calculates the minimum pascal triangle in minimum memory. Inspired by the comments in, but not the code of the FreeBasic example

constant N = 15 -- accurate to 30, nan/inf for anything over 514 (bigatom version is below).
sequence catalan = {},      -- (>=1 only)
         p = repeat(1,N+1)
atom p1
for i=1 to N do
    p1 = p[1]*2
    catalan = append(catalan,p1-p[2])
    for j=1 to N-i+1 do
        p1 += p[j+1]
        p[j] = p1
    end for
--  ?p[1..N-i+1]
end for
?catalan

Output:

{1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845}

Explanatory comments to accompany the above

-- FreeBASIC said:
--'  1   1   1   1   1   1  
--'  1   2   3   4   5   6  
--'  1   3   6  10  15  21  
--'  1   4  10  20  35  56  
--'  1   5  15  35  70 126  
--'  1   6  21  56 126 252  
--' The Pascal triangle is rotated 45 deg.
--' to find the Catalan number we need to follow the diagonal
--' for top left to bottom right 
--' take the number on diagonal and subtract the number in de cell
--' one up and one to right 
--' 1 (2 - 1), 2 (6 - 4), 5 (20 - 15) ... 
--
-- The first thing that struck me was it is twice as big as it needs to be, 
--  something like this would do...
--    1   1   1   1   1   1  
--        2   3   4   5   6  
--            6  10  15  21  
--               20  35  56  
--                   70 126  
--                      252  
-- It is more obvious from the upper square that the diagonal on that, which is 
--  that same as column 1 on this, is twice the previous, which on the second 
--  diagram is in column 2. Further, once we have calculated the value for column 
--  one above, we can use it immediately to calculate the next catalan number and 
--  do not need to store it. Lastly we can overwrite row 1 with row 2 etc in situ, 
--  and the following shows what we need for subsequent rounds:
--    1   1   1   1   1
--    3   4   5   6  
--   10  15  21  
--   35  56  
--  126  (unused)

The following bigatom version is over ten times faster than the equivalent on Catalan_numbers

include builtins\bigatom.e
 
function catalanB(integer n)    -- very very fast!
sequence catalan = {},
         p = repeat(1,n+1)
bigatom p1
    if n=0 then return 1 end if
    for i=1 to n do
        p1 = ba_multiply(p[1],2)
        catalan = append(catalan,ba_sub(p1,p[2]))
        for j=1 to n-i+1 do
            p1 = ba_add(p1,p[j+1])
            p[j] = p1
        end for
    end for
    return catalan[n]
end function
 
atom t0 = time()
string sc100 = ba_sprint(catalanB(100))
printf(1,"%d: %s (%3.2fs)\n",{100,sc100,time()-t0})
atom t0 = time()
string sc250 = ba_sprint(catalanB(250))
printf(1,"%d: %s (%3.2fs)\n",{250,sc250,time()-t0})

Output:

100: 896519947090131496687170070074100632420837521538745909320 (0.08s)
250: 465116795969233796497747947259667807407291160080922096111953326525143875193659257831340309862635877995262413955019878805418475969029457769094808256 (1.01s)

PicoLisp[edit]

(de bino (N K)
   (let f
      '((N)
         (if (=0 N) 1 (apply * (range 1 N))) )
      (/
         (f N)
         (* (f (- N K)) (f K)) ) ) )
 
(for N 15
  (println
     (-
        (bino (* 2 N) N)
        (bino (* 2 N) (inc N)) ) ) )
(bye)

PureBasic[edit]

Translation of: C

#MAXNUM = 15
Declare catalan()
 
If OpenConsole("Catalan numbers")
  catalan()
  Input()
  End 0
Else
  End -1
EndIf
 
Procedure catalan()
  Define k.i, n.i, num.d, den.d, cat.d
 
  Print("1 ")
 
  For n=2 To #MAXNUM
    num=1 : den =1
    For k=2 To n
      num * (n+k)
      den * k
      cat = num / den
    Next
    Print(Str(cat)+" ")
  Next
EndProcedure

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

Python[edit]

Translation of: C++

>>> n = 15
>>> t = [0] * (n + 2)
>>> t[1] = 1
>>> for i in range(1, n + 1):
	for j in range(i, 1, -1): t[j] += t[j - 1]
	t[i + 1] = t[i]
	for j in range(i + 1, 1, -1): t[j] += t[j - 1]
	print(t[i+1] - t[i], end=' ')
 
 
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 
>>>

Works with: Python version 2.7

def catalan_number(n):
    nm = dm = 1
    for k in range(2, n+1):
      nm, dm = ( nm*(n+k), dm*k )
    return nm/dm
 
print [catalan_number(n) for n in range(1, 16)]
 
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]

Racket[edit]

 
#lang racket
 
(define (next-half-row r)
  (define r1 (for/list ([x r] [y (cdr r)]) (+ x y)))
  `(,(* 2 (car r1)) ,@(for/list ([x r1] [y (cdr r1)]) (+ x y)) 1 0))
 
(let loop ([n 15] [r '(1 0)])
  (cons (- (car r) (cadr r))
        (if (zero? n) '() (loop (sub1 n) (next-half-row r)))))
;; -> '(1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900
;;      2674440 9694845)

REXX[edit]

explicit subscripts[edit]

All of the REXX program examples can handle arbitrary large numbers.

/*REXX program  obtains and displays  Catalan numbers  from  a  Pascal's triangle.      */
parse arg N .                                    /*Obtain the optional argument from CL.*/
if N=='' | N==","  then N=15                     /*Not specified?  Then use the default.*/
numeric digits max(9, N%2 + N%8)                 /*so we can handle huge Catalan numbers*/
@.=0;   @.1=1                                    /*stem array default; define 1st value.*/
 
  do i=1  for N;                               ip=i+1
                      do j=i   by -1  for N;   jm=j-1;   @.[email protected].[email protected].jm;    end /*j*/
  @.[email protected].i;           do k=ip  by -1  for N;   km=k-1;   @.[email protected].[email protected].km;    end /*k*/
  say  @.ip - @.i                                /*display the   Ith   Catalan number.  */
  end   /*i*/                                    /*stick a fork in it,  we're all done. */

output when using the default input:

implicit subscripts[edit]

/*REXX program  obtains and displays  Catalan numbers  from  a  Pascal's triangle.      */
parse arg N .                                    /*Obtain the optional argument from CL.*/
if N=='' | N==","  then N=15                     /*Not specified?  Then use the default.*/
numeric digits max(9, N%2 + N%8)                 /*so we can handle huge Catalan numbers*/
@.=0;  @.1=1                                     /*stem array default; define 1st value.*/
               do i=1  for N;  ip=i+1
                                      do j=i   by -1  for N;  @.[email protected].[email protected](j-1);   end  /*j*/
               @.[email protected].i;              do k=ip  by -1  for N;  @.[email protected].[email protected](k-1);   end  /*k*/
               say  @.ip - @.i                   /*display the   Ith   Catalan number.  */
               end   /*i*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
@:  parse arg !;   return @.!                    /*return the value of   @.[arg(1)]     */

output is the same as the 1^st version.

using binomial coefficients[edit]

/*REXX program  obtains and displays  Catalan numbers  from  a  Pascal's triangle.      */
parse arg N .                                    /*Obtain the optional argument from CL.*/
if N=='' | N==","  then N=15                     /*Not specified?  Then use the default.*/
numeric digits max(9, N%2 + N%8)                 /*so we can handle huge Catalan numbers*/
                      do j=1  for N              /* [↓]  display   N   Catalan numbers. */
                      say  comb(j+j, j) % (j+1)  /*display the   Jth   Catalan number.  */
                      end   /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!:    procedure; parse arg z;   _=1;     do j=1  for arg(1);  _=_*j;  end;        return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb: procedure; parse arg x,y;        if x=y  then return 1;   if y>x  then return 0
      if x-y<y  then y=x-y;     _=1;   do j=x-y+1  to x;  _=_*j;  end;       return _/!(y)

output is the same as the 1^st version.

binomial coefficients, memoized[edit]

This REXX version uses memoization for the calculation of factorials.

/*REXX program  obtains and displays  Catalan numbers  from  a  Pascal's triangle.      */
parse arg N .                                    /*Obtain the optional argument from CL.*/
if N=='' | N==","  then N=15                     /*Not specified?  Then use the default.*/
numeric digits max(9, N%2 + N%8)                 /*so we can handle huge Catalan numbers*/
!.=.
                      do j=1  for N              /* [↓]  display   N   Catalan numbers. */
                      say  comb(j+j, j) % (j+1)  /*display the   Jth   Catalan number.  */
                      end   /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!:    procedure expose !.;  parse arg z;     if !.z\==. then return !.z;  _=1
                         do j=1  for arg(1);   _=_*j;   end;        !.z=_;   return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb: procedure expose !.;  parse arg x,y;   if x=y  then return 1;  if y>x  then return 0
      if x-y<y  then y=x-y;     _=1;   do j=x-y+1  to x;  _=_*j;  end;       return _/!(y)

output is the same as the 1^st version.

Ring[edit]

 
n=15
cat = list(n+2)
cat[1]=1                        
for i=1 to n
    for j=i+1 to 2 step -1 
        cat[j]=cat[j]+cat[j-1]
    next
    cat[i+1]=cat[i]
    for j=i+2 to 2 step -1
        cat[j]=cat[j]+cat[j-1]
    next
    see "" + (cat[i+1]-cat[i]) + " "
next

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

Ruby[edit]

def catalan(num)
  t = [0, 1] #grows as needed
  (1..num).map do |i|
    i.downto(1){|j| t[j] += t[j-1]}
    t[i+1] = t[i]
    (i+1).downto(1) {|j| t[j] += t[j-1]}
    t[i+1] - t[i]
  end
end
 
p catalan(15)

Output:

[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]

Run BASIC[edit]

n = 15
dim t(n+2)
t(1) = 1
for i = 1 to n
  for  j = i to 1 step -1  : t(j) = t(j) + t(j-1): next j
  t(i+1) = t(i)
  for  j = i+1 to 1 step -1: t(j) = t(j) + t(j-1 : next j
print t(i+1) - t(i);" ";
next i

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

Rust[edit]

 
 
fn main()
{let n=15usize;
 let mut t= [0; 17];
 t[1]=1;
 let mut j:usize;
 for i in 1..n+1
 {
	j=i;
	loop{
	    if j==1{
		      break; 
		}
		t[j]=t[j] + t[j-1];
		j=j-1;
	}
	t[i+1]= t[i];
	j=i+1;
	loop{
		if j==1{
		break;
		}
		t[j]=t[j] + t[j-1];
		j=j-1;
	}
	print!("{} ", t[i+1]-t[i]);
 }
}

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

Scala[edit]

def catalan(n: Int): Int =
  if (n <= 1) 1
  else (0 until n).map(i => catalan(i) * catalan(n - i - 1)).sum
 
(1 to 15).map(catalan(_))

Output:

See it in running in your browser by Scastie (JVM).

Scilab[edit]

n=15
t=zeros(1,n+2)
t(1)=1                        
for i=1:n
  for j=i+1:-1:2 
    t(j)=t(j)+t(j-1)
  end
  t(i+1)=t(i)
  for j=i+2:-1:2
    t(j)=t(j)+t(j-1)
  end
  disp(t(i+1)-t(i))
end

Output:

Seed7[edit]

$ include "seed7_05.s7i";
 
const proc: main is func
  local
    const integer: N is 15;
    var array integer: t is [] (1) & N times 0;
    var integer: i is 0;
    var integer: j is 0;
  begin
    for i range 1 to N do
      for j range i downto 2 do
        t[j] +:= t[j - 1];
      end for;
      t[i + 1] := t[i];
      for j range i + 1 downto 2 do
        t[j] +:= t[j - 1];
      end for;
      write(t[i + 1] - t[i] <& " ");
    end for;
    writeln;
  end func;

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

Sidef[edit]

Translation of: Ruby

func catalan(num) {
  var t = [0, 1]
  (1..num).map { |i|
    flip(^i    ).each {|j| t[j+1] += t[j] }
    t[i+1] = t[i]
    flip(^i.inc).each {|j| t[j+1] += t[j] }
    t[i+1] - t[i]
  }
}
 
say catalan(15).join(' ')

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

smart BASIC[edit]

PRINT "Catalan Numbers from Pascal's Triangle"!PRINT
x = 15
DIM t(x+2)
t(1) = 1
FOR n = 1 TO x
  FOR  m = n TO 1 STEP -1
    t(m) = t(m) + t(m-1)
  NEXT m
    t(n+1) = t(n)
  FOR  m = n+1 TO 1 STEP -1
    t(m) = t(m) + t(m-1)
  NEXT m
PRINT n,"#######":t(n+1) - t(n)
NEXT n

Tcl[edit]

proc catalan n {
    set result {}
    array set t {0 0 1 1}
    for {set i 1} {[set k $i] <= $n} {incr i} {
	for {set j $i} {$j > 1} {} {incr t($j) $t([incr j -1])}
	set t([incr k]) $t($i)
	for {set j $k} {$j > 1} {} {incr t($j) $t([incr j -1])}
	lappend result [expr {$t($k) - $t($i)}]
    }
    return $result
}
 
puts [catalan 15]

Output:

1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

TI-83 BASIC[edit]

"CATALAN
15→N
seq(0,I,1,N+2)→L1
1→L1(1)
For(I,1,N)
For(J,I+1,2,-1)
L1(J)+L1(J-1)→L1(J)
End
L1(I)→L1(I+1)
For(J,I+2,2,-1)
L1(J)+L1(J-1)→L1(J)
End
Disp L1(I+1)-L1(I)
End

Output:

VBScript[edit]

To run in console mode with cscript.

dim t()
if Wscript.arguments.count=1 then
  n=Wscript.arguments.item(0)
else
  n=15
end if
redim t(n+1)   
't(*)=0
t(1)=1                          
for i=1 to n
  ip=i+1
  for j = i to 1 step -1 
    t(j)=t(j)+t(j-1)
  next 'j
  t(i+1)=t(i)
  for j = i+1 to 1 step -1 
    t(j)=t(j)+t(j-1)
  next 'j
  Wscript.echo t(i+1)-t(i)
next 'i

Visual Basic[edit]

Translation of: Rexx

Works with: Visual Basic version VB6 Standard

 
Sub catalan()
    Const n = 15
    Dim t(n + 2) As Long
    Dim i  As Integer, j As Integer
    t(1) = 1
    For i = 1 To n
        For j = i + 1 To 2 Step -1
            t(j) = t(j) + t(j - 1)
        Next j
        t(i + 1) = t(i)
        For j = i + 2 To 2 Step -1
            t(j) = t(j) + t(j - 1)
        Next j
        Debug.Print i, t(i + 1) - t(i)
    Next i
End Sub 'catalan

Output:

zkl[edit]

Translation of: PARI/GP

using binomial coefficients.

fcn binomial(n,k){ (1).reduce(k,fcn(p,i,n){ p*(n-i+1)/i },1,n) }
(1).pump(15,List,fcn(n){ binomial(2*n,n)-binomial(2*n,n+1) })

Output:

L(1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845)

ZX Spectrum Basic[edit]

Translation of: C++

10 LET N=15
20 DIM t(N+2)
30 LET t(2)=1
40 FOR i=2 TO N+1
50 FOR j=i TO 2 STEP -1: LET t(j)=t(j)+t(j-1): NEXT j
60 LET t(i+1)=t(i)
70 FOR j=i+1 TO 2 STEP -1: LET t(j)=t(j)+t(j-1): NEXT j
80 PRINT t(i+1)-t(i);" ";
90 NEXT i

Catalan numbers/Pascal's triangle

Contents

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zkl[edit]

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