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.

See also

Catalan's Triangle for a Number Triangle that generates Catalan Numbers using only addition.

Related Tasks

Pascal's triangle

360 Assembly

For maximum compatibility, this program uses only the basic instruction set. <lang 360asm>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</lang>

Output:

Ada

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

<lang Ada>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;</lang>

Output:

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

ALGOL 68

Translation of: C++

<lang algol68>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</lang>

Output:

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

APL

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

</lang>

Output:

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

AutoHotkey

Works with: AutoHotkey_L

<lang AutoHotkey>/* 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</lang>

Produces:

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

Batch File

<lang dos>@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</lang>

Output:

C++

<lang cpp>// 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;

}</lang>

Produces:

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

C#

Translation of: C++

<lang csharp> 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]));

} </lang>

Produces:

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

D

Translation of: C++

<lang d>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], ' ');
   }

}</lang>

Output:

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

EchoLisp

<lang scheme> (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)) </lang>

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

</lang>

Elixir

<lang elixir>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)</lang>

Output:

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

ERRE

<lang ERRE> 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 </lang>

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

FreeBASIC

<lang freebasic>' 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</lang>

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

Haskell

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. <lang haskell>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</lang>

Output:

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

Icon and Unicon

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

<lang unicon>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</lang>

Sample run:

J

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

Example use:

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

</lang>

A structured derivation of Catalan follows:

<lang j> 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

}:@:(}.@:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)</lang>

Java

Translation of: C++

<lang java>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]);
       }
   }

}</lang>

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

JavaScript

Translation of: C++

<lang javascript>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]) }</lang>

Output:

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

jq

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. <lang jq>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);</lang>

Example:

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

Output:

<lang sh>$ 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]</lang>

Mathematica / Wolfram Language

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. <lang Mathematica>nextrow[lastrow_] := Module[{output},

 output = ConstantArray[1, Length[lastrow] + 1];
 Do[
  outputi + 1 = lastrowi + lastrowi + 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]</lang>

Output:

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

MATLAB / Octave

<lang MATLAB>p = pascal(17); diag(p(2:end-1,2:end-1))-diag(p,2)</lang>

Output:

Nim

Translation of: Python

<lang nim>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], " "</lang>

Output:

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

Oforth

<lang Oforth>: pascal(n) [ 1 ] #[ dup 0 + 0 rot + zipWith(#+) ] times(n) ;

catalan(n) pascal(n 2 * ) at(n 1+) n 1+ / ;</lang>

Output:

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

PARI/GP

<lang parigp>vector(15,n,binomial(2*n,n)-binomial(2*n,n+1))</lang>

Output:

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

Pascal

<lang pascal>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.</lang>

  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

Translation of: C++

<lang Perl>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], " ";

}</lang>

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

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

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

Perl 6

Works with: Rakudo version 2015.12

<lang perl6>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];</lang>

Output:

Phix

Calculates the minimum pascal triangle in minimum memory. Inspired by the comments in, but not the code of the FreeBasic example <lang Phix>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</lang>

Output:

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

Explanatory comments to accompany the above <lang Phix>-- 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)</lang> The following bigatom version is over ten times faster than the equivalent on Catalan_numbers <lang Phix>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})</lang>

Output:

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

PicoLisp

<lang PicoLisp>(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)</lang>

Python

Translation of: C++

<lang python>>>> 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 >>> </lang>

Works with: Python version 2.7

<lang python>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]</lang>

Racket

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)

</lang>

REXX

explicit subscripts

All of the REXX program examples can handle arbitrary large numbers. <lang rexx>/*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;   @.j=@.j+@.jm;   end /*j*/
 @.ip=@.i
                 do k=ip  by -1  for N;  km=k-1;   @.k=@.k+@.km;   end /*k*/
 say  @.ip - @.i                                /*display the   Ith   Catalan number.  */
 end   /*i*/                                    /*stick a fork in it,  we're all done. */</lang>

output when using the default input:

implicit subscripts

<lang rexx>/*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;  @.j=@.j+@(j-1);   end  /*j*/
            @.ip=@.i;              do k=ip  by -1  for N;  @.k=@.k+@(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)] */</lang> output is the same as the 1^st version.

using binomial coefficients

<lang rexx>/*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)</lang>

output is the same as the 1^st version.

binomial coefficients, memoized

This REXX version uses memoization for the calculation of factorials. <lang rexx>/*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)</lang>

output is the same as the 1^st version.

Ring

<lang ring> 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 </lang> Output:

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

Ruby

<lang tcl>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)</lang>

Output:

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

Run BASIC

<lang runbasic>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</lang>

Output:

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

Scilab

<lang>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</lang>

Output:

Seed7

<lang seed7>$ 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;</lang>

Output:

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

Sidef

Translation of: Ruby

<lang ruby>func catalan(num) {

 var t = [0, 1];
 range(1, num).map { |i|
   range(i, 1, -1).each {|j| t[j] += t[j-1]};
   t[i+1] = t[i];
   range(i+1, 1, -1).each {|j| t[j] += t[j-1]};
   t[i+1] - t[i];
 }

}

say catalan(15).join(' ');</lang>

Output:

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

Tcl

<lang tcl>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]</lang>

Output:

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

TI-83 BASIC

<lang ti83b>"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</lang>

Output:

VBScript

To run in console mode with cscript. <lang vbscript>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</lang>

Visual Basic

Translation of: Rexx

Works with: Visual Basic version VB6 Standard

<lang vb> 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 </lang>

Output:

zkl

Translation of: PARI/GP

using binomial coefficients.

<lang zkl>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) })</lang>

Output:

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

ZX Spectrum Basic

Translation of: C++

<lang zxbasic>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</lang>