Catalan numbers: Difference between revisions

Content added Content deleted

Inline

Revision as of 18:35, 9 September 2015

This page uses content from Wikipedia. The original article was at Catalan numbers. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)

Catalan numbers are a sequence of numbers which can be defined directly:

C_{n}={\frac {1}{n+1}}{2n \choose n}={\frac {(2n)!}{(n+1)!\,n!}}\qquad {\mbox{ for }}n\geq 0.

Or recursively:

C_{0}=1\quad {\mbox{and}}\quad C_{n+1}=\sum _{i=0}^{n}C_{i}\,C_{n-i}\quad {\text{for }}n\geq 0;

Or alternatively (also recursive):

C_{0}=1\quad {\mbox{and}}\quad C_{n}={\frac {2(2n-1)}{n+1}}C_{n-1},

Implement at least one of these algorithms and print out the first 15 Catalan numbers with each. Memoization is not required, but may be worth the effort when using the second method above.

Related tasks:

360 Assembly

Very compact version. <lang 360asm>CATALAN CSECT

        USING  CATALAN,R15
        LA     R7,1               c=1
        LA     R6,1               i=1

LOOPI CH R6,=H'15' do i=1 to 15

        BH     ELOOPI
        XDECO  R6,PG              edit i
        LR     R5,R6              i
        SLA    R5,1               *2
        BCTR   R5,0               -1
        SLA    R5,1               *2
        MR     R4,R7              *c
        LA     R6,1(R6)           i=i+1
        DR     R4,R6              /i
        LR     R7,R5              c=2*(2*i-1)*c/(i+1)
        XDECO  R7,PG+12           edit c
        XPRNT  PG,24              print
        B      LOOPI

ELOOPI BR R14 PG DS CL24

        YREGS 
        END    CATALAN</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

Ada

<lang Ada>with Ada.Text_IO; use Ada.Text_IO;

procedure Test_Catalan is

  function Catalan (N : Natural) return Natural is
     Result : Positive := 1;
  begin
     for I in 1..N loop
        Result := Result * 2 * (2 * I - 1) / (I + 1);
     end loop;
     return Result;
  end Catalan;

begin

  for N in 0..15 loop
     Put_Line (Integer'Image (N) & " =" & Integer'Image (Catalan (N)));
  end loop;

end Test_Catalan;</lang>

Sample output:

 0 = 1
 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

AutoHotkey

As AutoHotkey has no BigInt, the formula had to be tweaked to prevent overflow. It still fails after n=22 <lang AHK>Loop 15

  out .= "`n" Catalan(A_Index)

Msgbox % clipboard := SubStr(out, 2) catalan( n ) {

By [VxE]. Returns ((2n)! / ((n + 1)! * n!)) if 0 <= N <= 22 (higher than 22 results in overflow)

If ( n < 3 ) ; values less than 3 are handled specially

  Return n < 0 ? "" : n = 0 ? 1 : n

i := 1 ; initialize the accumulator to 1

Loop % n - 1 >> 1 ; build the numerator by multiplying odd values between 2N and N+1

  i *= 1 + ( n - A_Index << 1 )

i <<= ( n - 2 >> 1 ) ; multiply the numerator by powers of 2 according to N

Loop % n - 3 >> 1 ; finish up by (integer) dividing by each of the non-cancelling factors

  i //= A_Index + 2

Return i }</lang>

Output:

AWK

<lang AWK># syntax: GAWK -f CATALAN_NUMBERS.AWK BEGIN {

   for (i=0; i<=15; i++) {
     printf("%2d %10d\n",i,catalan(i))
   }
   exit(0)

} function catalan(n, ans) {

   if (n == 0) {
     ans = 1
   }
   else {
     ans = ((2*(2*n-1))/(n+1))*catalan(n-1)
   }
   return(ans)

}</lang>

Output:

 0          1
 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

BASIC

Works with: FreeBASIC

Works with: QuickBASIC version 4.5 (untested)

Use of REDIM PRESERVE means this will not work in QBasic (although that could be worked around if desired).

<lang qbasic>DECLARE FUNCTION catalan (n as INTEGER) AS SINGLE

REDIM SHARED results(0) AS SINGLE

FOR x% = 1 TO 15

   PRINT x%, catalan (x%)

   IF UBOUND(results) < n THEN REDIM PRESERVE results(n)

   IF 0 = n THEN
   	results(0) = 1
   ELSE
   	results(n) = ((2 * ((2 * n) - 1)) / (n + 1)) * catalan(n - 1)
   END IF
   catalan = results(n)

END FUNCTION</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

BBC BASIC

<lang bbcbasic> FOR i% = 1 TO 15

       PRINT FNcatalan(i%)
     NEXT
     END
     
     DEF FNcatalan(n%)
     IF n% = 0 THEN = 1
     = 2 * (2 * n% - 1) * FNcatalan(n% - 1) / (n% + 1)</lang>

Output:

Befunge

Translation of: Ada

Output:

0 = 1
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

Bracmat

<lang bracmat>( out$straight & ( C

 =   
   .   ( F
       =   i prod
         .   !arg:0&1
           |   1:?prod
             & 0:?i
             &   whl
               ' ( 1+!i:~>!arg:?i
                 & !i*!prod:?prod
                 )
             & !prod
       )
     & F$(2*!arg)*(F$(!arg+1)*F$!arg)^-1
 )

& -1:?n & whl

 ' ( 1+!n:~>15:?n
   & out$(str$(C !n " = " C$!n))
   )

& out$"recursive, with memoization, without fractions" & :?seenCs & ( C

 =   i sum
   .   !arg:0&1
     |   ( !seenCs:? (!arg.?sum) ?
         |   0:?sum
           & -1:?i
           &   whl
             ' ( 1+!i:<!arg:?i
               & C$!i*C$(-1+!arg+-1*!i)+!sum:?sum
               )
           & (!arg.!sum) !seenCs:?seenCs
         )
       & !sum
 )

& -1:?n & whl

 ' ( 1+!n:~>15:?n
   & out$(str$(C !n " = " C$!n))
   )

& out$"recursive, without memoization, with fractions" & ( C

 =   
   .   !arg:0&1
     | 2*(2*!arg+-1)*(!arg+1)^-1*C$(!arg+-1)
 )

& -1:?n & whl

 ' ( 1+!n:~>15:?n
   & out$(str$(C !n " = " C$!n))
   )

& out$"Using taylor expansion of sqrt(1-4X). (See http://bababadalgharaghtakamminarronnkonnbro.blogspot.in/2012/10/algebraic-type-systems-combinatorial.html)" & out$(1+(1+-1*tay$((1+-4*X)^1/2,X,16))*(2*X)^-1+-1) & out$ );</lang>

Output:

<lang bracmat>straight C0 = 1 C1 = 1 C2 = 2 C3 = 5 C4 = 14 C5 = 42 C6 = 132 C7 = 429 C8 = 1430 C9 = 4862 C10 = 16796 C11 = 58786 C12 = 208012 C13 = 742900 C14 = 2674440 C15 = 9694845 recursive, with memoization, without fractions C0 = 1 C1 = 1 C2 = 2 C3 = 5 C4 = 14 C5 = 42 C6 = 132 C7 = 429 C8 = 1430 C9 = 4862 C10 = 16796 C11 = 58786 C12 = 208012 C13 = 742900 C14 = 2674440 C15 = 9694845 recursive, without memoization, with fractions C0 = 1 C1 = 1 C2 = 2 C3 = 5 C4 = 14 C5 = 42 C6 = 132 C7 = 429 C8 = 1430 C9 = 4862 C10 = 16796 C11 = 58786 C12 = 208012 C13 = 742900 C14 = 2674440 C15 = 9694845 Using taylor expansion of sqrt(1-4X). (See http://bababadalgharaghtakamminarronnkonnbro.blogspot.in/2012/10/algebraic-type-systems-combinatorial.html)

+ X + 2*X^2 + 5*X^3 + 14*X^4 + 42*X^5 + 132*X^6 + 429*X^7 + 1430*X^8 + 4862*X^9 + 16796*X^10 + 58786*X^11 + 208012*X^12 + 742900*X^13 + 2674440*X^14 + 9694845*X^15 </lang>

Brat

<lang brat>catalan = { n |

 true? n == 0
   { 1 }
   { (2 * ( 2 * n - 1) / ( n + 1 )) * catalan(n - 1) }

}

0.to 15 { n |

 p "#{n} - #{catalan n}"

}</lang>

Output:

0 - 1
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

C

All three methods mentioned in the task: <lang c>#include <stdio.h>

typedef unsigned long long ull;

ull binomial(ull m, ull n) { ull r = 1, d = m - n; if (d > n) { n = d; d = m - n; }

while (m > n) { r *= m--; while (d > 1 && ! (r%d) ) r /= d--; }

return r; }

ull catalan1(int n) { return binomial(2 * n, n) / (1 + n); }

ull catalan2(int n) { int i; ull r = !n;

for (i = 0; i < n; i++) r += catalan2(i) * catalan2(n - 1 - i); return r; }

ull catalan3(int n) { return n ? 2 * (2 * n - 1) * catalan3(n - 1) / (1 + n) : 1; }

int main(void) { int i; puts("\tdirect\tsumming\tfrac"); for (i = 0; i < 16; i++) { printf("%d\t%llu\t%llu\t%llu\n", i, catalan1(i), catalan2(i), catalan3(i)); }

return 0; }</lang>

Output:

       direct  summing frac
0      1       1       1
1      1       1       1
2      2       2       2
3      5       5       5
4      14      14      14
5      42      42      42
6      132     132     132
7      429     429     429
8      1430    1430    1430
9      4862    4862    4862
10     16796   16796   16796
11     58786   58786   58786
12     208012  208012  208012
13     742900  742900  742900
14     2674440 2674440 2674440
15     9694845 9694845 9694845

C#

<lang csharp>namespace CatalanNumbers {

   /// <summary>
   /// Class that holds all options.
   /// </summary>
   public class CatalanNumberGenerator
   {
       private static double Factorial(double n)
       {
           if (n == 0)
               return 1;

           return n * Factorial(n - 1);
       }

       public double FirstOption(double n)
       {
           const double topMultiplier = 2;
           return Factorial(topMultiplier * n) / (Factorial(n + 1) * Factorial(n));
       }

       public double SecondOption(double n)
       {
           if (n == 0)
           {
               return 1;
           }
           double sum = 0;
           double i = 0;
           for (; i <= (n - 1); i++)
           {
               sum += SecondOption(i) * SecondOption((n - 1) - i);
           }
           return sum;
       }

       public double ThirdOption(double n)
       {
           if (n == 0)
           {
               return 1;
           }
           return ((2 * (2 * n - 1)) / (n + 1)) * ThirdOption(n - 1);
       }
   }

}

// Program.cs using System; using System.Configuration;

// Main program // Be sure to add the following to the App.config file and add a reference to System.Configuration: // <?xml version="1.0" encoding="utf-8" ?> // <configuration> // <appSettings> // <clear/> // <add key="MaxCatalanNumber" value="50"/> // </appSettings> // </configuration> namespace CatalanNumbers {

   class Program
   {
       static void Main(string[] args)
       {
           CatalanNumberGenerator generator = new CatalanNumberGenerator();
           int i = 0;
           DateTime initial;
           DateTime final;
           TimeSpan ts;

           try
           {
               initial = DateTime.Now;
               for (; i <= Convert.ToInt32(ConfigurationManager.AppSettings["MaxCatalanNumber"]); i++)
               {
                   Console.WriteLine("CatalanNumber({0}):{1}", i, generator.FirstOption(i));
               }
               final = DateTime.Now;
               ts = final - initial;
               Console.WriteLine("It took {0}.{1} to execute\n", ts.Seconds, ts.Milliseconds);

               i = 0;
               initial = DateTime.Now;
               for (; i <= Convert.ToInt32(ConfigurationManager.AppSettings["MaxCatalanNumber"]); i++)
               {
                   Console.WriteLine("CatalanNumber({0}):{1}", i, generator.SecondOption(i));
               }
               final = DateTime.Now;
               ts = final - initial;
               Console.WriteLine("It took {0}.{1} to execute\n", ts.Seconds, ts.Milliseconds);

               i = 0;
               initial = DateTime.Now;
               for (; i <= Convert.ToInt32(ConfigurationManager.AppSettings["MaxCatalanNumber"]); i++)
               {
                   Console.WriteLine("CatalanNumber({0}):{1}", i, generator.ThirdOption(i));
               }
               final = DateTime.Now;
               ts = final - initial;
               Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds, ts.TotalMilliseconds);
               Console.ReadLine();
           }
           catch (Exception ex)
           {
               Console.WriteLine("Stopped at index {0}:", i);
               Console.WriteLine(ex.Message);
               Console.ReadLine();
           }
       }
   }

}</lang>

Output:

CatalanNumber(0):1
CatalanNumber(1):1
CatalanNumber(2):2
CatalanNumber(3):5
CatalanNumber(4):14
CatalanNumber(5):42
CatalanNumber(6):132
CatalanNumber(7):429
CatalanNumber(8):1430
CatalanNumber(9):4862
CatalanNumber(10):16796
CatalanNumber(11):58786
CatalanNumber(12):208012
CatalanNumber(13):742900
CatalanNumber(14):2674440
CatalanNumber(15):9694845
It took 0.14 to execute

CatalanNumber(0):1
CatalanNumber(1):1
CatalanNumber(2):2
CatalanNumber(3):5
CatalanNumber(4):14
CatalanNumber(5):42
CatalanNumber(6):132
CatalanNumber(7):429
CatalanNumber(8):1430
CatalanNumber(9):4862
CatalanNumber(10):16796
CatalanNumber(11):58786
CatalanNumber(12):208012
CatalanNumber(13):742900
CatalanNumber(14):2674440
CatalanNumber(15):9694845
It took 0.922 to execute

CatalanNumber(0):1
CatalanNumber(1):1
CatalanNumber(2):2
CatalanNumber(3):5
CatalanNumber(4):14
CatalanNumber(5):42
CatalanNumber(6):132
CatalanNumber(7):429
CatalanNumber(8):1430
CatalanNumber(9):4862
CatalanNumber(10):16796
CatalanNumber(11):58786
CatalanNumber(12):208012
CatalanNumber(13):742900
CatalanNumber(14):2674440
CatalanNumber(15):9694845
It took 0.3 to execute

C++

4 Classes

We declare 4 classes representing the four different algorithms for calculating Catalan numbers as given in the description of the task. In addition, we declare two supporting classes for the calculation of factorials and binomial coefficients. Because these two are only internal supporting code they are hidden in namespace 'detail'. Overloading the function call operator to execute the calculation is an obvious decision when using C++. (algorithms.h) <lang cpp>#if !defined __ALGORITHMS_H__

define __ALGORITHMS_H__

namespace rosetta

 {
 namespace catalanNumbers
   {
   namespace detail
     {

     class Factorial
       {
       public:
         unsigned long long operator()(unsigned n)const;
       };

     class BinomialCoefficient
       {
       public:
         unsigned long long operator()(unsigned n, unsigned k)const;
       };

     } //namespace detail

   class CatalanNumbersDirectFactorial
     {
     public:
       CatalanNumbersDirectFactorial();
       unsigned long long operator()(unsigned n)const;
     private:
       detail::Factorial factorial;
     };

   class CatalanNumbersDirectBinomialCoefficient
     {
     public:
       CatalanNumbersDirectBinomialCoefficient();
       unsigned long long operator()(unsigned n)const;
     private:
       detail::BinomialCoefficient binomialCoefficient;
     };

   class CatalanNumbersRecursiveSum
     {
     public:
       CatalanNumbersRecursiveSum();
       unsigned long long operator()(unsigned n)const;
     };

   class CatalanNumbersRecursiveFraction
     {
     public:
       CatalanNumbersRecursiveFraction();
       unsigned long long operator()(unsigned n)const;
     };

   }   //namespace catalanNumbers
 }     //namespace rosetta

endif //!defined __ALGORITHMS_H__</lang>

Here is the implementation of the algorithms. The c'tor of each class tells us the algorithm which will be used. (algorithms.cpp) <lang cpp>#include <iostream> using std::cout; using std::endl;

include <cmath>

using std::floor;

include "algorithms.h"

using namespace rosetta::catalanNumbers;

CatalanNumbersDirectFactorial::CatalanNumbersDirectFactorial()

 {
 cout<<"Direct calculation using the factorial"<<endl;
 }

unsigned long long CatalanNumbersDirectFactorial::operator()(unsigned n)const

 {
 if(n>1)
   {
   unsigned long long nFac = factorial(n);
   return factorial(2 * n) / ((n + 1) * nFac * nFac);
   }
 else
   {
   return 1;
   }
 }

CatalanNumbersDirectBinomialCoefficient::CatalanNumbersDirectBinomialCoefficient()

 {
 cout<<"Direct calculation using a binomial coefficient"<<endl;
 }

unsigned long long CatalanNumbersDirectBinomialCoefficient::operator()(unsigned n)const

 {
 if(n>1)
   return double(1) / (n + 1) * binomialCoefficient(2 * n, n);
 else
   return 1;
 }

CatalanNumbersRecursiveSum::CatalanNumbersRecursiveSum()

 {
 cout<<"Recursive calculation using a sum"<<endl;
 }

unsigned long long CatalanNumbersRecursiveSum::operator()(unsigned n)const

 {
 if(n>1)
   {
   const unsigned n_ = n - 1;
   unsigned long long sum = 0;
   for(unsigned i = 0; i <= n_; i++)
     sum += operator()(i) * operator()(n_ - i);
   return sum;
   }
 else
   {
   return 1;
   }
 }

CatalanNumbersRecursiveFraction::CatalanNumbersRecursiveFraction()

 {
 cout<<"Recursive calculation using a fraction"<<endl;
 }

unsigned long long CatalanNumbersRecursiveFraction::operator()(unsigned n)const

 {
 if(n>1)
   return (double(2 * (2 * n - 1)) / (n + 1)) * operator()(n-1);
 else
   return 1;
 }

unsigned long long detail::Factorial::operator()(unsigned n)const

 {
 if(n>1)
   return n * operator()(n-1);
 else
   return 1;
 }

unsigned long long detail::BinomialCoefficient::operator()(unsigned n, unsigned k)const

 {
 if(k == 0)
   return 1;
 
 if(n == 0)
   return 0;

 double product = 1;
 for(unsigned i = 1; i <= k; i++)
   product *= (double(n - (k - i)) / i);
 return (unsigned long long)(floor(product + 0.5));
 }</lang>

In order to test what we have done, a class Test is created. Using the template parameters N (number of Catalan numbers to be calculated) and A (the kind of algorithm to be used) the compiler will create code for all the test cases we need. What would C++ be without templates ;-) (tester.h) <lang cpp>#if !defined __TESTER_H__

define __TESTER_H__

include <iostream>

namespace rosetta

 {
 namespace catalanNumbers
   {

   template <int N, typename A>
   class Test
     {
     public:
       static void Do()
         {
         A algorithm;
         for(int i = 0; i <= N; i++)
           std::cout<<"C("<<i<<")\t= "<<algorithm(i)<<std::endl;
         }
     };

   } //namespace catalanNumbers
 }   //namespace rosetta

endif //!defined __TESTER_H__</lang>

Finally, we test the four different algorithms. Note that the first one (direct calculation using the factorial) only works up to N = 10 because some intermediate result (namely (2n)! with n = 11) exceeds the boundaries of an unsigned 64 bit integer. (catalanNumbersTest.cpp) <lang cpp>#include "algorithms.h"

include "tester.h"

using namespace rosetta::catalanNumbers;

int main(int argc, char* argv[])

 {
 Test<10, CatalanNumbersDirectFactorial>::Do();
 Test<15, CatalanNumbersDirectBinomialCoefficient>::Do();
 Test<15, CatalanNumbersRecursiveFraction>::Do();
 Test<15, CatalanNumbersRecursiveSum>::Do();
 return 0;
 }</lang>

Output:

(source code is compiled both by MS Visual C++ 10.0 (WinXP 32 bit) and GNU g++ 4.4.3 (Ubuntu 10.04 64 bit) compilers)

Direct calculation using the factorial
C(0)    = 1
C(1)    = 1
C(2)    = 2
C(3)    = 5
C(4)    = 14
C(5)    = 42
C(6)    = 132
C(7)    = 429
C(8)    = 1430
C(9)    = 4862
C(10)   = 16796
Direct calculation using a binomial coefficient
C(0)    = 1
C(1)    = 1
C(2)    = 2
C(3)    = 5
C(4)    = 14
C(5)    = 42
C(6)    = 132
C(7)    = 428
C(8)    = 1430
C(9)    = 4862
C(10)   = 16796
C(11)   = 58786
C(12)   = 208012
C(13)   = 742900
C(14)   = 2674440
C(15)   = 9694845
Recursive calculation using a fraction
C(0)    = 1
C(1)    = 1
C(2)    = 2
C(3)    = 5
C(4)    = 14
C(5)    = 42
C(6)    = 132
C(7)    = 429
C(8)    = 1430
C(9)    = 4862
C(10)   = 16796
C(11)   = 58786
C(12)   = 208012
C(13)   = 742900
C(14)   = 2674440
C(15)   = 9694845
Recursive calculation using a sum
C(0)    = 1
C(1)    = 1
C(2)    = 2
C(3)    = 5
C(4)    = 14
C(5)    = 42
C(6)    = 132
C(7)    = 429
C(8)    = 1430
C(9)    = 4862
C(10)   = 16796
C(11)   = 58786
C(12)   = 208012
C(13)   = 742900
C(14)   = 2674440
C(15)   = 9694845

Clojure

<lang Clojure>(def ! (memoize #(apply * (range 1 (inc %)))))

(defn catalan-numbers-direct []

 (map #(/ (! (* 2 %))

(* (! (inc %)) (! %))) (range)))

(def catalan-numbers-recursive

    #(->> [1 1] ; [c0 n1]

(iterate (fn c n [(* 2 (dec (* 2 n)) (/ (inc n)) c) (inc n)]) ,) (map first ,)))

user> (take 15 (catalan-numbers-direct)) (1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440)

user> (take 15 (catalan-numbers-recursive)) (1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440)</lang>

Common Lisp

With all three methods defined. <lang lisp>(defun catalan1 (n)

 ;; factorial. CLISP actually has "!" defined for this
 (labels ((! (x) (if (zerop x) 1 (* x (! (1- x))))))
   (/ (! (* 2 n)) (! (1+ n)) (! n))))

cache

(defparameter *catalans* (make-array 5 :fill-pointer 0 :adjustable t :element-type 'integer)) (defun catalan2 (n)

   (if (zerop n) 1
   ;; check cache
   (if (< n (length *catalans*)) (aref *catalans* n)
     (loop with c = 0 for i from 0 to (1- n) collect

(incf c (* (catalan2 i) (catalan2 (- n 1 i)))) ;; lower values always get calculated first, so ;; vector-push-extend is safe finally (progn (vector-push-extend c *catalans*) (return c))))))

(defun catalan3 (n)

 (if (zerop n) 1 (/ (* 2 (+ n n -1) (catalan3 (1- n))) (1+ n))))

test all three methods

(loop for f in (list #'catalan1 #'catalan2 #'catalan3)

     for i from 1 to 3 do
     (format t "~%Method ~d:~%" i)
     (dotimes (i 16) (format t "C(~2d) = ~d~%" i (funcall f i))))</lang>

D

<lang d>import std.stdio, std.algorithm, std.bigint, std.functional, std.range;

auto product(R)(R r) { return reduce!q{a * b}(1.BigInt, r); }

const cats1 = sequence!((a, n) => iota(n+2, 2*n+1).product / iota(1, n+1).product)(1);

BigInt cats2a(in uint n) {

   alias mcats2a = memoize!cats2a;
   if (n == 0) return 1.BigInt;
   return n.iota.map!(i => mcats2a(i) * mcats2a(n - 1 - i)).sum;

}

const cats2 = sequence!((a, n) => n.cats2a);

const cats3 = recurrence!q{ (4*n - 2) * a[n - 1] / (n + 1) }(1.BigInt);

void main() {

   foreach (cats; TypeTuple!(cats1, cats2, cats3))
       cats.take(15).writeln;

}</lang>

Output:

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

Eiffel

<lang Eiffel> class APPLICATION

create make

feature {NONE}

make do across 0 |..| 14 as c loop io.put_double (catalan_numbers (c.item)) io.new_line end end

nth_catalan_number (n: INTEGER): DOUBLE --'n'th number in the sequence of Catalan numbers. require n_not_negative: n >= 0 local s, t: DOUBLE do if n = 0 then Result := 1.0 else t := 4 * n.to_double - 2 s := n.to_double + 1 Result := t / s * catalan_numbers (n - 1) end end

end

</lang>

Output:

Erlang

<lang erlang>-module(catalan).

-export([test/0]).

cat(N) ->

  factorial(2 * N) div (factorial(N+1) * factorial(N)).

factorial(N) ->

  fac1(N,1).

fac1(0,Acc) ->

  Acc;

fac1(N,Acc) ->

  fac1(N-1, N * Acc).

cat_r1(0) ->

1;

cat_r1(N) ->

  lists:sum([cat_r1(I)*cat_r1(N-1-I) || I <- lists:seq(0,N-1)]).

cat_r2(0) ->

1;

cat_r2(N) ->

  cat_r2(N - 1) * (2 * ((2 * N) - 1)) div (N + 1).

test() ->

   TestList = lists:seq(0,14),
   io:format("Directly:\n~p\n",| N <- TestList),
   io:format("1st recusive method:\n~p\n",| N <- TestList),
   io:format("2nd recusive method:\n~p\n",| N <- TestList).</lang>

Output:

Directly:
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
1st recusive method:
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
2nd recusive method:
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]

ERRE

<lang>PROGRAM CATALAN

PROCEDURE CATALAN(N->RES)

  RES=1
  FOR I=1 TO N DO
     RES=RES*2*(2*I-1)/(I+1)
  END FOR

END PROCEDURE

BEGIN

  FOR N=0 TO 15 DO
     CATALAN(N->RES)    
     PRINT(N;"=";RES)
  END FOR

END PROGRAM </lang>

Output:

 0 = 1
 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

Euphoria

<lang Euphoria>--Catalan number task from Rosetta Code wiki --User:Lnettnay

--function from factorial task function factorial(integer n) atom f = 1 while n > 1 do

       f *= n
       n -= 1

end while

return f end function

function catalan(integer n) atom numerator = factorial(2 * n) atom denominator = factorial(n+1)*factorial(n) return numerator/denominator end function

for i = 0 to 15 do

       ? catalan(i)

end for</lang>

Output:

Factor

This is the last solution, memoized by using arrays. Run in scratchpad. <lang factor>: next ( seq -- newseq )

 [ ] [ last ] [ length ] tri
 [ 2 * 1 - 2 * ] [ 1 + ] bi /
 * suffix ;

Catalan ( n -- seq ) V{ 1 } swap 1 - [ next ] times ;

15 Catalan . V{

}</lang>

Fantom

<lang fantom>class Main {

 static Int factorial (Int n)
 {
   Int res := 1
   if (n>1)
     (2..n).each |i| { res *= i }
   return res
 }

 static Int catalanA (Int n)
 { 
   return factorial(2*n)/(factorial(n+1) * factorial(n))
 }

 static Int catalanB (Int n)
 {
   if (n == 0)
   {
     return 1
   }
   else
   {
     sum := 0
     n.times |i| { sum += catalanB(i) * catalanB(n-1-i) } 
     return sum
   }
 }

 static Int catalanC (Int n)
 {
   if (n == 0)
   {
     return 1
   }
   else
   {
     return catalanC(n-1)*2*(2*n-1)/(n+1)
   }
 }

 public static Void main ()
 {
   (1..15).each |n|
   {
     echo (n.toStr.padl(4) + 
           catalanA(n).toStr.padl(10) +
           catalanB(n).toStr.padl(10) +
           catalanC(n).toStr.padl(10))
   }
 }

}</lang> 22! exceeds the range of Fantom's Int class, so the first technique fails afer n=10

   1         1         1         1
   2         2         2         2
   3         5         5         5
   4        14        14        14
   5        42        42        42
   6       132       132       132
   7       429       429       429
   8      1430      1430      1430
   9      4862      4862      4862
  10     16796     16796     16796
  11       -65     58786     58786
  12        -2    208012    208012
  13         0    742900    742900
  14        97   2674440   2674440
  15        -2   9694845   9694845

Forth

<lang forth>: catalan ( n -- ) 1 swap 1+ 1 do dup cr . i 2* 1- 2* i 1+ */ loop drop ;</lang>

Fortran

Works with: Fortran version 90 and later

<lang fortran>program main

 !=======================================================================================
 implicit none

 !=== Local data
 integer                      :: n

 !=== External procedures
 double precision, external   :: catalan_numbers         
 
 !=== Execution =========================================================================

 write(*,'(1x,a)')'==============='
 write(*,'(5x,a,6x,a)')'n','c(n)'
 write(*,'(1x,a)')'---------------'

 do n = 0, 14
   write(*,'(1x,i5,i10)') n, int(catalan_numbers(n))
 enddo

 write(*,'(1x,a)')'==============='

 !=======================================================================================

end program main !BL !BL !BL double precision recursive function catalan_numbers(n) result(value)

 !=======================================================================================
 implicit none

 !=== Input, ouput data
 integer, intent(in)          :: n

 !=== Execution =========================================================================

 if ( n .eq. 0 ) then
   value = 1
 else 
   value = ( 2.0d0 * dfloat(2 * n - 1) / dfloat( n + 1 ) ) * catalan_numbers(n-1)
 endif

 !=======================================================================================

end function catalan_numbers</lang>

Output:

 ===============
     n      c(n)
 ---------------
     0         1
     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
 ===============

Frink

Frink includes efficient algorithms for calculating arbitrarily-large binomial coefficients and automatically caches factorials. <lang frink>catalan[n] := binomial[2n,n]/(n+1) for n = 0 to 15

  println[catalan[n]]</lang>

FunL

<lang funl>import integers.choose import util.TextTable

def

 catalan( n ) = choose( 2n, n )/(n + 1)

 catalan2( n ) = product( (n + k)/k | k <- 2..n )

 catalan3( 0 ) = 1
 catalan3( n ) = 2*(2n - 1)/(n + 1)*catalan3( n - 1 )

t = TextTable() t.header( 'n', 'definition', 'product', 'recursive' ) t.line()

for i <- 1..4

 t.rightAlignment( i )

for i <- 0..15

 t.row( i, catalan(i), catalan2(i), catalan3(i) )

println( t )</lang>

Output:

+----+------------+---------+-----------+
| n  | definition | product | recursive |
+----+------------+---------+-----------+
|  0 |          1 |       1 |         1 |
|  1 |          1 |       1 |         1 |
|  2 |          2 |       2 |         2 |
|  3 |          5 |       5 |         5 |
|  4 |         14 |      14 |        14 |
|  5 |         42 |      42 |        42 |
|  6 |        132 |     132 |       132 |
|  7 |        429 |     429 |       429 |
|  8 |       1430 |    1430 |      1430 |
|  9 |       4862 |    4862 |      4862 |
| 10 |      16796 |   16796 |     16796 |
| 11 |      58786 |   58786 |     58786 |
| 12 |     208012 |  208012 |    208012 |
| 13 |     742900 |  742900 |    742900 |
| 14 |    2674440 | 2674440 |   2674440 |
| 15 |    9694845 | 9694845 |   9694845 |
+----+------------+---------+-----------+

GAP

<lang gap>Catalan1 := n -> Binomial(2*n, n) - Binomial(2*n, n - 1);

Catalan2 := n -> Binomial(2*n, n)/(n + 1);

Catalan3 := function(n)

   local k, c;
   c := 1;
   k := 0;
   while k < n do
       k := k + 1;
       c := 2*(2*k - 1)*c/(k + 1);
   od;
   return c;

end;

Catalan4_memo := [1]; Catalan4 := function(n)

   if not IsBound(Catalan4_memo[n + 1]) then
       Catalan4_memo[n + 1] := Sum([0 .. n - 1], i -> Catalan4(i)*Catalan4(n - 1 - i));
   fi;
   return Catalan4_memo[n + 1];

end;

The first fifteen: 0 to 14 !

List([0 .. 14], Catalan1); List([0 .. 14], Catalan2); List([0 .. 14], Catalan3); List([0 .. 14], Catalan4);

Same output for all four:
[ 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440 ]</lang>

Go

Direct: <lang go>package main

import (

   "fmt"
   "math/big"

)

func main() {

   var b, c big.Int
   for n := int64(0); n < 15; n++ {
       fmt.Println(c.Div(b.Binomial(n*2, n), c.SetInt64(n+1)))
   }

}</lang>

Output:

Harbour

<lang visualfoxpro> PROCEDURE Main()

  LOCAL i

  FOR i := 0 to 15
     ? PadL( i, 2 ) + ": " + hb_StrFormat("%d", Catalan( i ))
  NEXT

  RETURN

STATIC FUNCTION Catalan( n )

  LOCAL i, nCatalan := 1

  FOR i := 1 TO n 
     nCatalan := nCatalan * 2 * (2 * i - 1) / (i + 1)
  NEXT

  RETURN nCatalan

</lang>

Output:

Haskell

<lang haskell>-- Three infinite lists, corresponding to the three definitions in the problem -- statement.

cats1 = map (\n -> product [n+2..2*n] `div` product [1..n]) [0..]

cats2 = 1 : map (\n -> sum $ zipWith (*) (reverse (take n cats2)) cats2) [1..]

cats3 = scanl (\c n -> c*2*(2*n-1) `div` (n+1)) 1 [1..]

main = mapM_ (print . take 15) [cats1, cats2, cats3]</lang>

Output:

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

Icon and Unicon

<lang Icon>procedure main(arglist) every writes(catalan(i)," ") end

procedure catalan(n) # return catalan(n) or fail static M initial M := table()

if n > 0 then

  return (n = 1) | \M[n] | ( M[n] := (2*(2*n-1)*catalan(n-1))/(n+1))

end</lang>

Output:

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

J

Java

Accuracy may be lost for larger n's due to floating-point arithmetic (seen for C(15) here). This implementation is memoized (factorial and Catalan numbers are stored in the Maps "facts", "catsI", "catsR1", and "catsR2"). <lang java5>import java.util.HashMap; import java.util.Map;

public class Catalan { private static final Map<Long, Double> facts = new HashMap<Long, Double>(); private static final Map<Long, Double> catsI = new HashMap<Long, Double>(); private static final Map<Long, Double> catsR1 = new HashMap<Long, Double>(); private static final Map<Long, Double> catsR2 = new HashMap<Long, Double>();

static{//pre-load the memoization maps with some answers facts.put(0L, 1D); facts.put(1L, 1D); facts.put(2L, 2D);

catsI.put(0L, 1D); catsR1.put(0L, 1D); catsR2.put(0L, 1D); }

private static double fact(long n){ if(facts.containsKey(n)){ return facts.get(n); } double fact = 1; for(long i = 2; i <= n; i++){ fact *= i; //could be further optimized, but it would probably be ugly } facts.put(n, fact); return fact; }

private static double catI(long n){ if(!catsI.containsKey(n)){ catsI.put(n, fact(2 * n)/(fact(n+1)*fact(n))); } return catsI.get(n); }

private static double catR1(long n){ if(catsR1.containsKey(n)){ return catsR1.get(n); } double sum = 0; for(int i = 0; i < n; i++){ sum += catR1(i) * catR1(n - 1 - i); } catsR1.put(n, sum); return sum; }

private static double catR2(long n){ if(!catsR2.containsKey(n)){ catsR2.put(n, ((2.0*(2*(n-1) + 1))/(n + 1)) * catR2(n-1)); } return catsR2.get(n); }

public static void main(String[] args){ for(int i = 0; i <= 15; i++){ System.out.println(catI(i)); System.out.println(catR1(i)); System.out.println(catR2(i)); } } }</lang>

Output:

1.0
1.0
1.0
1.0
1.0
1.0
2.0
2.0
2.0
5.0
5.0
5.0
14.0
14.0
14.0
42.0
42.0
42.0
132.0
132.0
132.0
429.0
429.0
429.0
1430.0
1430.0
1430.0
4862.0
4862.0
4862.0
16796.0
16796.0
16796.0
58786.0
58786.0
58786.0
208012.0
208012.0
208012.0
742900.0
742900.0
742900.0
2674439.9999999995
2674440.0
2674440.0
9694844.999999998
9694845.0
9694845.0

JavaScript

<lang javascript><html><head><title>Catalan</title></head>

<body>

function disp(x) { var e = document.createTextNode(x + '\n'); document.getElementById('x').appendChild(e); }

disp(" meth1 meth2 meth3"); for (var i = 0; i <= 15; i++) disp(i + '\t' + cata1(i) + '\t' + cata2(i) + '\t' + cata3(i));

</script></body></html></lang>

Output:

       meth1   meth2   meth3
0	1	1	1
1	1	1	1
2	2	2	2
3	5	5	5
4	14	14	14
5	42	42	42
6	132	132	132
7	429	429	429
8	1430	1430	1430
9	4862	4862	4862
10	16796	16796	16796
11	58786	58786	58786
12	208012	208012	208012
13	742900	742900	742900
14	2674440	2674440	2674440
15	9694845	9694845	9694845

jq

Works with: jq version 1.4

The recursive formula for C(n) in terms of C(n-1) lends itself directly to efficient implementations in jq so in this section, that formula is used (a) to define a function for computing a single Catalan number; (b) to define a function for generating a sequence of Catalan numbers; and (c) to write a single expression for generating a sequence of Catalan numbers using jq's builtin "recurse/1" filter.

Compute a single Catalan number

<lang jq>def catalan:

 if . == 0 then 1
 elif . < 0 then error("catalan is not defined on \(.)")
 else (2 * (2*. - 1) * ((. - 1) | catalan)) / (. + 1)
 end;</lang>

Example 1 <lang jq>(range(0; 16), 100) as $i | $i | catalan | [$i, .]</lang>

Output:

<lang sh>$ jq -M -n -c -f Catalan_numbers.jq [0,1] [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] [100,8.96519947090131e+56]

</lang>

Generate a sequence of Catalan numbers

<lang jq>def catalan_series(max):

 def _catalan: # state: [n, catalan(n)]
   if .[0] > max then empty 
   else .,
     ((.[0] + 1) as $n | .[1] as $cp
      | [$n,  (2 * (2*$n - 1) * $cp) / ($n + 1) ] | _catalan)
   end;
 [0,1] | _catalan;

</lang> Example 2: <lang jq>catalan_series(15)</lang>

Output:

As above for 0 to 15.

An expression to generate Catalan numbers

 [0,1]
 | recurse( if .[0] == 15 then empty
            else .[1] as $c | (.[0] + 1) | [ ., (2 * (2*. - 1) * $c) / (. + 1) ] 
            end )</lang>

Output:

As above for 0 to 15.

Julia

<lang julia>Catalan(n::Integer) = div(binomial(2n, n), n+1)</lang>

Output:

julia> [Catalan(n) for n=1:15]
15-element Array{Int64,1}:
       1
       2
       5
      14
      42
     132
     429
    1430
    4862
   16796
   58786
  208012
  742900
 2674440
 9694845

julia> Catalan(big(100))
896519947090131496687170070074100632420837521538745909320

(In the second example, we have used arbitrary-precision integers to avoid overflow for large Catalan numbers.)

K

<lang k> catalan: {_{*/(x-i)%1+i:!y-1}[2*x;x+1]%x+1}

 catalan'!:15

1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440</lang>

Liberty BASIC

<lang lb>print "non-recursive version" print catNonRec(5) for i = 0 to 15

   print i;"   =   "; catNonRec(i)

next print

print "recursive version" print catRec(5) for i = 0 to 15

   print i;"   =   "; catRec(i)

next print

print "recursive with memoisation" redim cats(20) 'clear the array print catRecMemo(5) for i = 0 to 15

   print i;"   =   "; catRecMemo(i)

next print

wait

function catNonRec(n) 'non-recursive version

   catNonRec=1
   for i=1 to n
       catNonRec=((2*((2*i)-1))/(i+1))*catNonRec
   next

end function

function catRec(n) 'recursive version

   if n=0 then
       catRec=1
   else
       catRec=((2*((2*n)-1))/(n+1))*catRec(n-1)
   end if

end function

function catRecMemo(n) 'recursive version with memoisation

   if n=0 then
       catRecMemo=1
   else
       if cats(n-1)=0 then    'call it recursively only if not already calculated
           prev = catRecMemo(n-1)
       else
           prev = cats(n-1)
       end if
       catRecMemo=((2*((2*n)-1))/(n+1))*prev
   end if
   cats(n) = catRecMemo    'memoisation for future use

end function</lang>

Output:

non-recursive version
42
0   =   1
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

recursive version
42
0   =   1
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

recursive with memoisation
42
0   =   1
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

<lang Lua>-- recursive with memoization catalan = {[0] = 1} setmetatable(catalan, { __index = function(c, n) c[n] = c[n-1]*2*(2*n-1)/(n+1) return c[n] end } )

for i=0,14 do print(catalan[i]) end</lang>

Output:

Maple

<lang Maple>CatalanNumbers := proc( n::posint )

   return seq( (2*i)!/((i + 1)!*i!), i = 0 .. n - 1 );

end proc: CatalanNumbers(15); </lang> Output:

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

Mathematica

<lang Mathematica>CatalanN[n_Integer /; n >= 0] := (2 n)!/((n + 1)! n!)</lang>

Sample Output:

<lang Mathematica>TableForm[CatalanN/@Range[0,15]] //TableForm= 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845</lang>

MATLAB / Octave

<lang MATLAB>function n = catalanNumbers(n)

   for i = (1:length(n))
       n(i) = (1/(n(i)+1))*nchoosek(2*n(i),n(i));
   end

end</lang> The following version is fully vectorized and does not require a loop <lang MATLAB>function n = catalanNumbers(n)

   n = prod(n+1:2*n)/prod(1:n+1);

end</lang>

Sample Output:

<lang MATLAB>>> catalanNumbers(14)

ans =

>> catalanNumbers((0:17))'

ans =

          1
          1
          2
          5
         14
         42
        132
        429
       1430
       4862
      16796
      58786
     208012
     742900
    2674440
    9694845
   35357670
  129644790</lang>

Maxima

<lang maxima>/* The following is an array function, hence the square brackets. It uses memoization automatically */ cata[n] := sum(cata[i]*cata[n - 1 - i], i, 0, n - 1)$ cata[0]: 1$

cata2(n) := binomial(2*n, n)/(n + 1)$

makelist(cata[n], n, 0, 14);

makelist(cata2(n), n, 0, 14);

/* both return [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440] */</lang>

Nim

<lang nim>import strutils

proc binomial(m, n): auto =

 result = 1
 var
   d = m - n
   n = n
   m = m
 if d > n:
   n = d

 while m > n:
   result *= m
   dec m
   while d > 1 and (result mod d) == 0:
     result = result div d
     dec d

proc catalan1(n): auto =

 binomial(2 * n, n) div (n + 1)

proc catalan2(n): auto =

 if n == 0:
   result = 1
 for i in 0 .. <n:
   result += catalan2(i) * catalan2(n - 1 - i)

proc catalan3(n): int =

 if n > 0: 2 * (2 * n - 1) * catalan3(n - 1) div (1 + n)
 else: 1

for i in 0..15:

 echo align($i, 7), " ", align($catalan1(i), 7), " ", align($catalan2(i), 7), " ", align($catalan3(i), 7)</lang>

Output:

      0       1       1       1
      1       1       1       1
      2       2       2       2
      3       5       5       5
      4      14      14      14
      5      42      42      42
      6     132     132     132
      7     429     429     429
      8    1430    1430    1430
      9    4862    4862    4862
     10   16796   16796   16796
     11   58786   58786   58786
     12  208012  208012  208012
     13  742900  742900  742900
     14 2674440 2674440 2674440
     15 9694845 9694845 9694845

Oforth

<lang Oforth>: catalan(n) { n ifZero: [ 1 ] else: [ catalan(n 1 -) 2 n * 1 - * 2 * n 1 + / ] }</lang>

Output:

seqFrom(0, 15) map(#catalan) println
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]

ooRexx

Three versions of this. <lang ooRexx>loop i = 0 to 15

   say "catI("i") =" .catalan~catI(i)
   say "catR1("i") =" .catalan~catR1(i)
   say "catR2("i") =" .catalan~catR2(i)

end

-- This is implemented as static members on a class object -- so that the code is able to keep state information between calls. This -- memoization will speed up things like factorial calls by remembering previous -- results.

class catalan

-- initialize the class object

method init class

 expose facts catI catR1 catR2
        facts = .table~new
        catI = .table~new
        catR1 = .table~new
        catR2 = .table~new
        -- seed a few items
        facts[0] = 1
        facts[1] = 1
        facts[2] = 2
        catI[0] = 1
        catR1[0] = 1
        catR2[0] = 1

-- private factorial method

method fact private class

 expose facts
 use arg n
 -- see if we've calculated this before
 if facts~hasIndex(n) then return facts[n]
 numeric digits 120

 fact = 1
 loop i = 2 to n
     fact *= i
 end
 -- save this result
 facts[n] = fact
 return fact

method catI class

 expose catI
 use arg n
 numeric digits 20

 res = catI[n]
 if res == .nil then do
     -- dividing by 1 removes insignificant trailing 0s
     res = (self~fact(2 * n)/(self~fact(n + 1) * self~fact(n))) / 1
     catI[n] = res
 end
 return res

method catR1 class

 expose catR1
 use arg n
 numeric digits 20

 if catR1~hasIndex(n) then return catR1[n]
 sum = 0
 loop i = 0 to n - 1
     sum += self~catR1(i) * self~catR1(n - 1 - i)
 end
 -- remove insignificant trailing 0s
 sum = sum / 1
 catR1[n] = sum
 return sum

method catR2 class

 expose catR2
 use arg n
 numeric digits 20

 res = catR2[n]
 if res == .nil then do
    res = ((2 * (2 * n - 1) * self~catR2(n - 1)) /  (n + 1))
    catR2[n] = res
 end
 return res</lang>

Output:

catI(0) = 1
catR1(0) = 1
catR2(0) = 1
catI(1) = 1
catR1(1) = 1
catR2(1) = 1
catI(2) = 2
catR1(2) = 2
catR2(2) = 2
catI(3) = 5
catR1(3) = 5
catR2(3) = 5
catI(4) = 14
catR1(4) = 14
catR2(4) = 14
catI(5) = 42
catR1(5) = 42
catR2(5) = 42
catI(6) = 132
catR1(6) = 132
catR2(6) = 132
catI(7) = 429
catR1(7) = 429
catR2(7) = 429
catI(8) = 1430
catR1(8) = 1430
catR2(8) = 1430
catI(9) = 4862
catR1(9) = 4862
catR2(9) = 4862
catI(10) = 16796
catR1(10) = 16796
catR2(10) = 16796
catI(11) = 58786
catR1(11) = 58786
catR2(11) = 58786
catI(12) = 208012
catR1(12) = 208012
catR2(12) = 208012
catI(13) = 742900
catR1(13) = 742900
catR2(13) = 742900
catI(14) = 2674440
catR1(14) = 2674440
catR2(14) = 2674440
catI(15) = 9694845
catR1(15) = 9694845
catR2(15) = 9694845

PARI/GP

Memoization is not worthwhile; PARI has fast built-in facilities for calculating binomial coefficients and factorials. <lang parigp>catalan(n)=binomial(2*n,n+1)/n</lang> A second version: <lang parigp>catalan(n)=(2*n)!/(n+1)!/n!</lang> Naive version with binary splitting: <lang parigp>catalan(n)=prod(k=n+2,2*n,k)/prod(k=2,n,k)</lang> Naive version: <lang parigp>catalan(n)={

 my(t=1);
 for(k=n+2,2*n,t*=k);
 for(k=2,n,t/=k);
 t

};</lang> The first version takes about 1.5 seconds to compute the millionth Catalan number, while the second takes 3.9 seconds. The naive implementations, for comparison, take 21 and 45 minutes. In any case, printing the first 15 is simple: <lang parigp>vector(15,n,catalan(n))</lang>

Pascal

<lang pascal>Program CatalanNumbers(output);

function catalanNumber1(n: integer): double;

 begin
   if n = 0 then
     catalanNumber1 := 1.0
   else 
     catalanNumber1 := double(4 * n - 2) / double(n + 1) * catalanNumber1(n-1);
 end;

var

 number: integer;

begin

 writeln('Catalan Numbers');
 writeln('Recursion with a fraction:');
 for number := 0 to 14 do
   writeln (number:3, round(catalanNumber1(number)):9);

end.</lang>

Output:

:> ./CatalanNumbers
Catalan Numbers
Recursion with a fraction:
  0        1
  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

Perl

<lang perl>sub factorial { my $f = 1; $f *= $_ for 2 .. $_[0]; $f; } sub catalan {

 my $n = shift;
 factorial(2*$n) / factorial($n+1) / factorial($n);

}

print "$_\t@{[ catalan($_) ]}\n" for 0 .. 20;</lang> For computing up to 20 ish, memoization is not needed. For much bigger numbers, this is faster: <lang perl>my @c = (1); sub catalan {

       use bigint;
       $c[$_[0]] //= catalan($_[0]-1) * (4 * $_[0]-2) / ($_[0]+1)

}

most of the time is spent displaying the long numbers, actually

print "$_\t", catalan($_), "\n" for 0 .. 10000;</lang>

That has two downsides: high memory use and slow access to an isolated large value. Using a fast binomial function can solve both these issues. The downside here is if the platform doesn't have the GMP library then binomials won't be fast.

Library: ntheory

<lang perl>use ntheory qw/binomial/; sub catalan {

 my $n = shift;
 binomial(2*$n,$n)/($n+1);

} print "$_\t", catalan($_), "\n" for 0 .. 10000;</lang>

Perl 6

<lang perl6>constant Catalan = 1, [\*] (2, 6 ... *) Z/ 2 .. *;

.say for Catalan[^15];</lang>

Output:

Phix

<lang Phix> -- returns inf/-nan for n>85, and needs the rounding for n>=14, accurate to n=29 function catalan1(integer n)

   return floor(factorial(2*n)/(factorial(n+1)*factorial(n))+0.5)

end function

-- returns inf for n>519, accurate to n=30: function catalan2(integer n) -- NB: very slow! atom res = not n

   n -= 1
   for i=0 to n do
       res += catalan2(i)*catalan2(n-i)
   end for
   return res

end function

-- returns inf for n>514, accurate to n=30: function catalan3(integer n)

   if n=0 then return 1 end if
   return 2*(2*n-1)/(1+n)*catalan3(n-1)

end function

for i=0 to 15 do

   printf(1,"%2d: %10d %10d %10d\n",{i,catalan1(i),catalan2(i),catalan3(i)})

end for

-- An explicitly memoized version of what seems to be the best, and the one that really needed it: -- (and in fact it turned out to be faster than similarly memoized versions of 1 and 3, when atom) -- I also converted this to use bigatoms.

include builtins\bigatom.e

sequence c2cache = {}

function catalan2bc(integer n) -- very fast! object r -- result (a bigatom)

   if n<=0 then return BA_ONE end if
   if n<=length(c2cache) then
       r = c2cache[n]
       if r!=0 then return r end if
   else
       c2cache &= repeat(0,n-length(c2cache))
   end if
   r = BA_ZERO
   for i=0 to n-1 do
       r = ba_add(r,ba_multiply(catalan2bc(i),catalan2bc(n-1-i)))
   end for
   c2cache[n] = r
   return r

end function

atom t0 = time() -- (this last call only) string sc100 = ba_sprint(catalan2bc(100)) printf(1,"100: %s (%3.2fs)\n",{sc100,time()-t0})</lang>

Output:

 0:          1          1          1
 1:          1          1          1
 2:          2          2          2
 3:          5          5          5
 4:         14         14         14
 5:         42         42         42
 6:        132        132        132
 7:        429        429        429
 8:       1430       1430       1430
 9:       4862       4862       4862
10:      16796      16796      16796
11:      58786      58786      58786
12:     208012     208012     208012
13:     742900     742900     742900
14:    2674440    2674440    2674440
15:    9694845    9694845    9694845
100: 896519947090131496687170070074100632420837521538745909320 (0.42s)

PHP

<lang php><?php

class CatalanNumbersSerie {

 private static $cache = array(0 => 1);
  
 private function fill_cache($i)
 {
   $accum = 0;
   $n = $i-1;
   for($k = 0; $k <= $n; $k++)
   {
     $accum += $this->item($k)*$this->item($n-$k);
   } 
   self::$cache[$i] = $accum;
 }
 function item($i)
 {
   if (!isset(self::$cache[$i]))
   {
     $this->fill_cache($i);
   }
   return self::$cache[$i];
 }

}

$cn = new CatalanNumbersSerie(); for($i = 0; $i <= 15;$i++) {

 $r = $cn->item($i);
 echo "$i = $r\r\n";

} ?></lang>

Output:

0 = 1
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

PicoLisp

<lang PicoLisp># Factorial (de fact (N)

  (if (=0 N)
     1
     (* N (fact (dec N))) ) )

Directly

(de catalanDir (N)

  (/ (fact (* 2 N)) (fact (inc N)) (fact N)) )

Recursively

(de catalanRec (N)

  (if (=0 N)
     1
     (cache '(NIL) N  # Memoize
        (sum
           '((I) (* (catalanRec I) (catalanRec (- N I 1))))
           (range 0 (dec N)) ) ) ) )

Alternatively

(de catalanAlt (N)

  (if (=0 N)
     1
     (*/ 2 (dec (* 2 N)) (catalanAlt (dec N)) (inc N)) ) )

Test

(for (N 0 (> 15 N) (inc N))

  (tab (2 4 8 8 8)
     N
     " => "
     (catalanDir N)
     (catalanRec N)
     (catalanAlt N) ) )</lang>

Output:

 0 =>        1       1       1
 1 =>        1       1       1
 2 =>        2       2       2
 3 =>        5       5       5
 4 =>       14      14      14
 5 =>       42      42      42
 6 =>      132     132     132
 7 =>      429     429     429
 8 =>     1430    1430    1430
 9 =>     4862    4862    4862
10 =>    16796   16796   16796
11 =>    58786   58786   58786
12 =>   208012  208012  208012
13 =>   742900  742900  742900
14 =>  2674440 2674440 2674440

PL/I

<lang PL/I>catalan: procedure options (main); /* 23 February 2012 */

  declare (i, n) fixed;

  put skip list ('How many catalan numbers do you want?');
  get list (n);

  do i = 0 to n;
     put skip list (c(i));
  end;

c: procedure (n) recursive returns (fixed decimal (15));

  declare n fixed;

  if n <= 1 then return (1);

  return ( 2*(2*n-1) * c(n-1) / (n + 1) );

end c;

end catalan;</lang>

Output:

How many catalan numbers do you want? 

                 1 
                 1 
                 2 
                 5 
                14 
                42 
               132 
               429 
              1430 
              4862 
             16796 
             58786 
            208012 
            742900 
           2674440 
           9694845 
          35357670 
         129644790 
         477638700 
        1767263190 
        6564120420

Plain $T e X$

<lang tex>\newcount\n \newcount\r \newcount\x \newcount\ii

\def\catalan#1{% \n#1\advance\n by1\ii1\r1% \loop{% \x\ii% \multiply\x by 2 \advance\x by -1 \multiply\x by 2% \global\multiply\r by\x% \global\advance\ii by1% \global\divide\r by\ii% } \ifnum\number\ii<\n\repeat% \the\r }

\rightskip=0pt plus1fil\parindent=0pt \loop{${\rm Catalan}(\the\x) = \catalan{\the\x}$\hfil\break}% \advance\x by 1\ifnum\x<15\repeat

\bye</lang>

Prolog

Works with: SWI-Prolog

<lang Prolog>catalan(N) :- length(L1, N), L = [1 | L1], init(1,1,L1), numlist(0, N, NL), maplist(my_write, NL, L).

init(_, _, []).

init(V, N, [H | T]) :- N1 is N+1, H is 2 * (2 * N - 1) * V / N1, init(H, N1, T).

my_write(N, V) :- format('~w : ~w~n', [N, V]).</lang>

Output:

 ?- catalan(15).
0 : 1
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
true .

PureBasic

Using the third formula... <lang PureBasic>; saving the division for last ensures we divide the largest

numerator by the smallest denominator

Procedure.q CatalanNumber(n.q) If n<0:ProcedureReturn 0:EndIf If n=0:ProcedureReturn 1:EndIf ProcedureReturn (2*(2*n-1))*CatalanNumber(n-1)/(n+1) EndProcedure

ls=25 rs=12

a.s="" a.s+LSet(RSet("n",rs),ls)+"CatalanNumber(n)"

cw(a.s)

Debug a.s

For n=0 to 33 ;33 largest correct quad for n a.s="" a.s+LSet(RSet(Str(n),rs),ls)+Str(CatalanNumber(n))

cw(a.s)

Debug a.s Next</lang>

Sample Output:

           n             CatalanNumber(n)
           0             1
           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
          16             35357670
          17             129644790
          18             477638700
          19             1767263190
          20             6564120420
          21             24466267020
          22             91482563640
          23             343059613650
          24             1289904147324
          25             4861946401452
          26             18367353072152
          27             69533550916004
          28             263747951750360
          29             1002242216651368
          30             3814986502092304
          31             14544636039226909
          32             55534064877048198
          33             212336130412243110

Python

Three algorithms including explicit memoization. (Pythons factorial built-in function is not memoized internally).

Python will transparently switch to bignum-type integer arithmetic, so the code below works unchanged on computing larger catalan numbers such as cat(50) and beyond. <lang python>from math import factorial import functools

def memoize(func):

   cache = {}
   def memoized(key):
       # Returned, new, memoized version of decorated function
       if key not in cache:
           cache[key] = func(key)
       return cache[key]
   return functools.update_wrapper(memoized, func)

@memoize def fact(n):

   return factorial(n)

def cat_direct(n):

   return fact(2*n) // fact(n + 1) // fact(n)

@memoize def catR1(n):

   return ( 1 if n == 0
            else sum( catR1(i) * catR1(n - 1 - i)
                      for i in range(n) ) )

@memoize def catR2(n):

   return ( 1 if n == 0
            else ( ( 4 * n - 2 ) * catR2( n - 1) ) // ( n + 1 ) )

if __name__ == '__main__':

   def pr(results):
       fmt = '%-10s %-10s %-10s'
       print ((fmt % tuple(c.__name__ for c in defs)).upper())
       print (fmt % (('='*10,)*3))
       for r in zip(*results):
           print (fmt % r)

   defs = (cat_direct, catR1, catR2)
   results = [ tuple(c(i) for i in range(15)) for c in defs ]
   pr(results)</lang>

Sample Output:

CAT_DIRECT CATR1      CATR2     
========== ========== ==========
1          1          1         
1          1          1         
2          2          2         
5          5          5         
14         14         14        
42         42         42        
132        132        132       
429        429        429       
1430       1430       1430      
4862       4862       4862      
16796      16796      16796     
58786      58786      58786     
208012     208012     208012    
742900     742900     742900    
2674440    2674440    2674440

R

<lang r>catalan <- function(n) choose(2*n, n)/(n + 1) catalan(1:15)

[1] 1 2 5 14 42 132 429 1430 4862
[10] 16796 58786 208012 742900 2674440 9694845</lang>

Racket

<lang racket>#lang racket (require planet2)

(install "this-and-that") ; uncomment to install

(require memoize/memo)

(define/memo* (catalan m)

 (if (= m 0) 
     1
     (for/sum ([i m]) 
       (* (catalan i) (catalan (- m i 1))))))

(map catalan (range 1 15))</lang>

Output:

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

REXX

version 1

All methods use (independent) memoization for the computation of factorials. <lang rexx>/*REXX program calculates Catalan numbers using four different methods. */ parse arg bot top . /*get optional arguments from the C.L. */ if bot== then do; top=15; bot=0; end /*No args? Use a range of 0 ───► 15. */ if top== then top=bot /*No top? Use the bottom for default. */ numeric digits max(20, 5*top) /*this allows gihugic Catalan numbers. */ @cat=' Catalan' /*a nice literal to have for the SAY. */ w=length(top) /*width of the largest number for SAY. */ call hdr 1A; do j=bot to top; say @cat right(j,w)": " Catalan1A(j); end call hdr 1B; do j=bot to top; say @cat right(j,w)": " Catalan1B(j); end call hdr 2 ; do j=bot to top; say @cat right(j,w)": " Catalan2(j); end call hdr 3 ; do j=bot to top; say @cat right(j,w)": " Catalan3(j); end exit /*stick a fork in it, we're all done. */ /*────────────────────────────────────────────────────────────────────────────*/ Catalan1A: procedure expose !.; parse arg n; return comb(n+n, n) % (n+1) Catalan1B: procedure expose !.; parse arg n; return !(n+n) % ((n+1) * !(n)**2) comb: procedure; parse arg x,y; return pFact(x-y+1,x) % pFact(2,y) pFact: procedure; !=1; do k=arg(1) to arg(2); !=!*k; end; return ! /*────────────────────────────────────────────────────────────────────────────*/ hdr: !.=.; c.=.; c.0=1; say say center(' Catalan numbers, method' left(arg(1),3), 79, '─'); return /*────────────────────────────────────────────────────────────────────────────*/ !: procedure expose !.; parse arg x; !=1; if !.x\==. then return !.x

                          do k=1  for x;  !=!*k;  end  /*k*/
          !.x=!; return !

/*──────────────────────────────────Catalan method 2──────────────────────────*/ Catalan2: procedure expose c.; parse arg n; $=0; if c.n\==. then return c.n

                          do k=0  to n-1; $=$+catalan2(k)*catalan2(n-k-1);  end
          c.n=$; return $             /*use a REXX memoization technique.    */

/*──────────────────────────────────Catalan method 3──────────────────────────*/ Catalan3: procedure expose c.; parse arg n

          if c.n==.  then c.n=(4*n-2) * catalan3(n-1) % (n+1);       return c.n</lang>

output when using the input of: 0 16

───────────────────────── Catalan numbers, method 1A ──────────────────────────
     Catalan  0:  1
     Catalan  1:  1
     Catalan  2:  2
     Catalan  3:  5
     Catalan  4:  14
     Catalan  5:  42
     Catalan  6:  132
     Catalan  7:  429
     Catalan  8:  1430
     Catalan  9:  4862
     Catalan 10:  16796
     Catalan 11:  58786
     Catalan 12:  208012
     Catalan 13:  742900
     Catalan 14:  2674440
     Catalan 15:  9694845 

───────────────────────── Catalan numbers, method 1B ──────────────────────────
···  (elided, same as first method) ··· 

───────────────────────── Catalan numbers, method 2  ──────────────────────────
···  (elided, same as first method) ···

───────────────────────── Catalan numbers, method 3  ──────────────────────────
···  (elided, same as first method) ···

Timing notes of the four methods:

For Catalan numbers 1 ──► 200:

method 1A is about 50 times slower than method 3
method 1B is about 100 times slower than method 3
method 2 is about 85 times slower than method 3

For Catalan numbers 1 ──► 300:

method 1A is about 100 times slower than method 3
method 1B is about 200 times slower than method 3
method 2 is about 200 times slower than method 3

Method 3 is really quite fast; even in the thousands range, computation time is still quite reasonable.

version 2

Implements the 3 methods shown in the task description <lang rexx>/* REXX ---------------------------------------------------------------

01.07.2014 Walter Pachl
--------------------------------------------------------------------*/

Numeric Digits 1000 Parse Arg m . If m= Then m=20 Do i=0 To m

 c1.i=c1(i)
 End

c2.=1 Do i=1 To m

 c2.i=c2(i)
 End

c3.=1 Do i=1 To m

 im1=i-1
 c3.i=2*(2*i-1)*c3.im1/(i+1)
 End

l=length(c3.m) hdr=' n' right('c1.n',l),

        right('c2.n',l),
        right('c3.n',l)

Say hdr Do i=0 To m

 Say right(i,2) format(c1.i,l),
                format(c2.i,l),
                format(c3.i,l)
 End

Say hdr Exit

c1: Procedure Parse Arg n return fact(2*n)/(fact(n)*fact(n+1))

c2: Procedure Expose c2. Parse Arg n res=0 Do i=0 To n-1

 nmi=n-i-1
 res=res+c2.i*c2.nmi
 End

Return res

fact: Procedure Parse Arg n f=1 Do i=1 To n

 f=f*i
 End

Return f</lang>

Output:

 n       c1.n       c2.n       c3.n
 0          1          1          1
 1          1          1          1
 2          2          2          2
 3          5          5          5
 4         14         14         14
 5         42         42         42
 6        132        132        132
 7        429        429        429
 8       1430       1430       1430
 9       4862       4862       4862
10      16796      16796      16796
11      58786      58786      58786
12     208012     208012     208012
13     742900     742900     742900
14    2674440    2674440    2674440
15    9694845    9694845    9694845
16   35357670   35357670   35357670
17  129644790  129644790  129644790
18  477638700  477638700  477638700
19 1767263190 1767263190 1767263190
20 6564120420 6564120420 6564120420
 n       c1.n       c2.n       c3.n

Ruby

Library: RubyGems

<lang ruby>def factorial(n)

 (1..n).reduce(1, :*)

end

direct

def catalan_direct(n)

 factorial(2*n) / (factorial(n+1) * factorial(n))

end

recursive

def catalan_rec1(n)

 return 1 if n == 0
 (0...n).inject(0) {|sum, i| sum + catalan_rec1(i) * catalan_rec1(n-1-i)}

end

def catalan_rec2(n)

 return 1 if n == 0
 2*(2*n - 1) * catalan_rec2(n-1) / (n+1)

end

performance and results

require 'benchmark' require 'memoize' include Memoize

Benchmark.bm(17) do |b|

 b.report('catalan_direct')    {16.times {|n| catalan_direct(n)} }
 b.report('catalan_rec1')      {16.times {|n| catalan_rec1(n)} }
 b.report('catalan_rec2')      {16.times {|n| catalan_rec2(n)} }
 
 memoize :catalan_rec1
 b.report('catalan_rec1(memo)'){16.times {|n| catalan_rec1(n)} }

end

puts "\n direct rec1 rec2" 16.times {|n| puts "%2d :%9d%9d%9d" % [n, catalan_direct(n), catalan_rec1(n), catalan_rec2(n)]}</lang> The output shows the dramatic difference memoizing makes.

                        user     system      total        real
catalan_direct      0.000000   0.000000   0.000000 (  0.000124)
catalan_rec1        6.178000   0.000000   6.178000 (  6.195141)
catalan_rec2        0.000000   0.000000   0.000000 (  0.000023)
catalan_rec1(memo)  0.000000   0.000000   0.000000 (  0.000641)

       direct     rec1     rec2
 0 :        1        1        1
 1 :        1        1        1
 2 :        2        2        2
 3 :        5        5        5
 4 :       14       14       14
 5 :       42       42       42
 6 :      132      132      132
 7 :      429      429      429
 8 :     1430     1430     1430
 9 :     4862     4862     4862
10 :    16796    16796    16796
11 :    58786    58786    58786
12 :   208012   208012   208012
13 :   742900   742900   742900
14 :  2674440  2674440  2674440
15 :  9694845  9694845  9694845

Run BASIC

<lang Runbasic>FOR i = 1 TO 15

   PRINT i;" ";catalan(i)

catalan = 1
if n <> 0 then catalan = ((2 * ((2 * n) - 1)) / (n + 1)) * catalan(n - 1)

END FUNCTION</lang>

Rust

<lang rust>fn c_n(n: u64) -> u64 { match n { 0 => 1, _ => c_n(n-1)*2*(2*n-1)/(n+1) } }

fn main() {

   for i in 1..16 {
       println!("c_n({}) = {}", i, c_n(i))
   }

}</lang>

Output:

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

Scala

Simple and straightforward. Noticeably out of steam without memoizing at about 5000. <lang scala>object Catalan {

 def factorial(n: BigInt) = BigInt(1).to(n).foldLeft(BigInt(1))(_ * _)
 def catalan(n: BigInt) = factorial(2 * n) / (factorial(n + 1) * factorial(n))

 def main(args: Array[String]) {
   for (n <- 1 to 15) {
     println("catalan(" + n + ") = " + catalan(n))
   }
 }

}</lang>

Output:

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

Scheme

Tail recursive implementation. <lang scheme>(define (catalan m)

   (let loop ((c 1)(n 0))
       (if (not (eqv? n m))
           (begin
               (display n)(display ": ")(display c)(newline)
               (loop (* (/ (* 2 (- (* 2 (+ n 1)) 1)) (+ (+ n 1) 1)) c) (+ n 1) )))))

(catalan 15)</lang>

Output:

0: 1
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

Seed7

<lang seed7>$ include "seed7_05.s7i";

 include "bigint.s7i";

const proc: main is func

 local
   var bigInteger: n is 0_;
 begin
   for n range 0_ to 15_ do
     writeln((2_ * n) ! n div succ(n));
   end for;
 end func;</lang>

Output:

Sidef

<lang ruby>func f(i) { i.isZero ? 1 : (i * f(i-1)) }; func c(n) { f(2*n) / f(n) / f(n+1) };</lang> Recursive (with memoization): <lang ruby>func c(n) {

   static mem = [1];
   mem[n-1] := (c(n-1) * (4 * n - 2) / (n + 1));

}</lang>

Calling the function: <lang ruby>15.times { |i|

   say "#{i-1}\t#{c(i-1)}";

}</lang>

Output:

Standard ML

<lang sml>(*

* val catalan : int -> int
* Returns the nth Catalan number.
*)

fun catalan 0 = 1 | catalan n = ((4 * n - 2) * catalan(n - 1)) div (n + 1);

(*

* val print_catalans : int -> unit
* Prints out Catalan numbers 0 through 15.
*)

fun print_catalans(n) =

   if n > 15 then ()
   else (print (Int.toString(catalan n) ^ "\n"); print_catalans(n + 1)); print_catalans(0);

(*

* 1
* 1
* 2
* 5
* 14
* 42
* 132
* 429
* 1430
* 4862
* 16796
* 58786
* 208012
* 742900
* 2674440
* 9694845
*)</lang>

Tcl

<lang tcl>package require Tcl 8.5

Memoization wrapper

proc memoize {function value generator} {

   variable memoize
   set key $function,$value
   if {![info exists memoize($key)]} {

set memoize($key) [uplevel 1 $generator]

   }
   return $memoize($key)

}

The simplest recursive definition

proc tcl::mathfunc::catalan n {

   if {[incr n 0] < 0} {error "must not be negative"}
   memoize catalan $n {expr {

$n == 0 ? 1 : 2 * (2*$n - 1) * catalan($n - 1) / ($n + 1)

}}

}</lang> Demonstration: <lang tcl>for {set i 0} {$i < 15} {incr i} {

   puts "C_$i = [expr {catalan($i)}]"

}</lang>

Output:

C_0 = 1
C_1 = 1
C_2 = 2
C_3 = 5
C_4 = 14
C_5 = 42
C_6 = 132
C_7 = 429
C_8 = 1430
C_9 = 4862
C_10 = 16796
C_11 = 58786
C_12 = 208012
C_13 = 742900
C_14 = 2674440

Of course, this code also works unchanged (apart from extending the loop) for producing higher Catalan numbers. For example, here is the end of the output when asked to produce the first fifty:

C_45 = 2257117854077248073253720
C_46 = 8740328711533173390046320
C_47 = 33868773757191046886429490
C_48 = 131327898242169365477991900
C_49 = 509552245179617138054608572

TI-83 BASIC

This problem is perfectly suited for a TI calculator. <lang TI-83 BASIC>:For(I,1,15

Disp (2I)!/((I+1)!I!

End</lang>

Output:

Ursala

<lang ursala>#import std

import nat

catalan = quotient^\successor choose^/double ~&

cast %nL

t = catalan* iota 16</lang>

Output:

VBA

<lang vb>Public Sub Catalan1(n As Integer) 'Computes the first n Catalan numbers according to the first recursion given Dim Cat() As Long Dim sum As Long

ReDim Cat(n) Cat(0) = 1 For i = 0 To n - 1

 sum = 0
 For j = 0 To i
   sum = sum + Cat(j) * Cat(i - j)
 Next j
 Cat(i + 1) = sum

Next i Debug.Print For i = 0 To n

 Debug.Print i, Cat(i)

Next End Sub

Public Sub Catalan2(n As Integer) 'Computes the first n Catalan numbers according to the second recursion given Dim Cat() As Long

ReDim Cat(n) Cat(0) = 1 For i = 1 To n

 Cat(i) = 2 * Cat(i - 1) * (2 * i - 1) / (i + 1)

Next i Debug.Print For i = 0 To n

 Debug.Print i, Cat(i)

Next End Sub</lang>

Result:

Catalan1 15

 0             1 
 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

(Expect same result with "Catalan2 15")

VBScript

<lang vb> Function catalan(n) catalan = factorial(2*n)/(factorial(n+1)*factorial(n)) End Function

Function factorial(n) If n = 0 Then Factorial = 1 Else For i = n To 1 Step -1 If i = n Then factorial = n Else factorial = factorial * i End If Next End If End Function

'Find the first 15 Catalan numbers. For j = 1 To 15 WScript.StdOut.Write j & " = " & catalan(j) WScript.StdOut.WriteLine Next </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

Wortel

<lang wortel>; the following number expression calculcates the nth Catalan number

~ddiFSFmSoFSn

which stands for: dup dup inc fac swap fac mult swap double fac swap divide
to get the first 15 Catalan numbers we map this function over a list from 0 to 15

!*#~ddiFSFmSoFSn @til 15

returns [1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674439.9999999995]</lang>

XPL0

<lang XPL0>code CrLf=9, IntOut=11; int C, N; [C:= 1; IntOut(0, C); CrLf(0); for N:= 1 to 14 do

   [C:= C*2*(2*N-1)/(N+1);
   IntOut(0, C);  CrLf(0);
   ];

]</lang>

Output:

zkl

Uses GMP to calculate big factorials. <lang zkl>var BN=Import("zklBigNum"); fcn catalan(n){

  BN(2*n).factorial() / BN(n+1).factorial() / BN(n).factorial();

}

foreach n in (16){

  println("%2d --> %,d".fmt(n, catalan(n)));

} println("%2d --> %,d".fmt(100, catalan(100)));</lang> And an iterative solution at works up to the limit of 64 bit ints (n=33). Would be 35 but need to avoid factional intermediate results. <lang zkl>fcn catalan(n){ (1).reduce(n,fcn(p,n){ 2*(2*n-1)*p/(n+1) },1) }</lang>

Output:

 0 --> 1
 1 --> 1
 2 --> 2
 3 --> 5
 4 --> 14
 5 --> 42
 6 --> 132
 7 --> 429
 8 --> 1,430
 9 --> 4,862
10 --> 16,796
11 --> 58,786
12 --> 208,012
13 --> 742,900
14 --> 2,674,440
15 --> 9,694,845
100 --> 896,519,947,090,131,496,687,170,070,074,100,632,420,837,521,538,745,909,320