# Catalan numbers

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Catalan numbers
You are encouraged to solve this task according to the task description, using any language you may know.

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

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

Or recursively:

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

Or alternatively (also recursive):

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

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.

## 11l

V c = 1L(n) 1..15   print(c)   c = 2 * (2 * n - 1) * c I/ (n + 1)
Output:
1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440


## 360 Assembly

Very compact version.

CATALAN  CSECT        08/09/2015         USING  CATALAN,R15         LA     R7,1               c=1         LA     R6,1               i=1LOOPI    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              next iELOOPI   BR     R14PG       DS     CL24         YREGS          END    CATALAN
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


## ABAP

This works for ABAP Version 7.40 and above

 report z_catalan_numbers. class catalan_numbers definition.  public section.    class-methods:      get_nth_number        importing          i_n                     type int4        returning          value(r_catalan_number) type int4.endclass. class catalan_numbers implementation.  method get_nth_number.    r_catalan_number = cond int4(      when i_n eq 0      then 1      else reduce int4(        init          result = 1          index = 1        for position = 1 while position <= i_n        next          result = result * 2 * ( 2 * index - 1 ) div ( index + 1 )          index = index + 1 ) ).  endmethod.endclass. start-of-selection.  do 15 times.    write / |C({ sy-index - 1 }) = { catalan_numbers=>get_nth_number( sy-index - 1 ) }|.  enddo.
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


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;
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


## ALGOL 68

# calculate the first few catalan numbers, using LONG INT values        ## (64-bit quantities in Algol 68G which can handle up to C23)           # # returns n!/k!                                                         #PROC factorial over factorial = ( INT n, k )LONG INT:     IF      k > n THEN 0     ELIF    k = n THEN 1     ELSE #  k < n #         LONG INT f := 1;         FOR i FROM k + 1 TO n DO f *:= i OD;         f     FI # factorial over factorial # ; # returns n!                                                             #PROC factorial = ( INT n )LONG INT:     BEGIN         LONG INT f := 1;         FOR i FROM 2 TO n DO f *:= i OD;         f     END # factorial # ; # returnss the nth Catalan number using binomial coefficeients            ## uses the factorial over factorial procedure for a slight optimisation   ## note:     Cn = 1/(n+1)(2n n)                                            ##              = (2n)!/((n+1)!n!)                                         ##              = factorial over factorial( 2n, n+1 )/n!                   #PROC catalan = ( INT n )LONG INT: IF n < 2 THEN 1 ELSE factorial over factorial( n + n, n + 1 ) OVER factorial( n ) FI;  # show the first few catalan numbers                                      #FOR i FROM 0 TO 15 DO    print( ( whole( i, -2 ), ": ", whole( catalan( i ), 0 ), newline ) )OD
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


## ALGOL W

begin    % print the catalan numbers up to C15 %    integer Cprev;    Cprev := 1; % C0 %    write(     s_w := 0, i_w := 3, 0, ": ", i_w := 9, Cprev );    for n := 1 until 15 do begin        Cprev := round( ( ( ( 4 * n ) - 2 ) / ( n + 1 ) ) * Cprev );        write( s_w := 0, i_w := 3, n, ": ", i_w := 9, Cprev );    end for_nend.
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


## APL

      {(!2×⍵)÷(!⍵+1)×!⍵}(⍳15)-1
Output:
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440

## AutoHotkey

As AutoHotkey has no BigInt, the formula had to be tweaked to prevent overflow. It still fails after n=22

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

## AWK

# syntax: GAWK -f CATALAN_NUMBERS.AWKBEGIN {    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)}
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).

DECLARE FUNCTION catalan (n AS INTEGER) AS SINGLE REDIM SHARED results(0) AS SINGLE FOR x% = 1 TO 15    PRINT x%, catalan (x%)NEXT FUNCTION catalan (n AS INTEGER) AS SINGLE    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
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


### Sinclair ZX81 BASIC

Works with 1k of RAM.

The specification asks for the first 15 Catalan numbers. A lot of the other implementations produce either C(0) to C(15), which is 16 numbers, or else C(1) to C(15)—which is 15 numbers, but I'm not convinced they're the first 15. This program produces C(0) to C(14).

 10 FOR N=0 TO 14 20 LET X=N 30 GOSUB 130 40 LET A=FX 50 LET X=N+1 60 GOSUB 130 70 LET B=FX 80 LET X=2*N 90 GOSUB 130100 PRINT N,FX/(B*A)110 NEXT N120 STOP130 LET FX=1140 FOR I=1 TO X150 LET FX=FX*I160 NEXT I170 RETURN
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

## BBC BASIC

      10 FOR i% = 1 TO 15      20   PRINT FNcatalan(i%)      30 NEXT      40 END      50 DEF FNcatalan(n%)      60   IF n% = 0 THEN = 1      70   = 2 * (2 * n% - 1) * FNcatalan(n% - 1) / (n% + 1)
Output:
         1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440

## Befunge

0>:.:000p1>\:00g-#v_vv 2-1*2p00 :+1g00\< $> **00g1+/^v,*84,"="<_^#<*53:+1>#,.#+5< @ Output: 0 = 1 1 = 1 2 = 2 3 = 5 4 = 14 5 = 42 6 = 132 7 = 429 8 = 1430 9 = 4862 10 = 16796 11 = 58786 12 = 208012 13 = 742900 14 = 2674440 15 = 9694845 ## Bracmat ( out$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$); Output: straightC0 = 1C1 = 1C2 = 2C3 = 5C4 = 14C5 = 42C6 = 132C7 = 429C8 = 1430C9 = 4862C10 = 16796C11 = 58786C12 = 208012C13 = 742900C14 = 2674440C15 = 9694845recursive, with memoization, without fractionsC0 = 1C1 = 1C2 = 2C3 = 5C4 = 14C5 = 42C6 = 132C7 = 429C8 = 1430C9 = 4862C10 = 16796C11 = 58786C12 = 208012C13 = 742900C14 = 2674440C15 = 9694845recursive, without memoization, with fractionsC0 = 1C1 = 1C2 = 2C3 = 5C4 = 14C5 = 42C6 = 132C7 = 429C8 = 1430C9 = 4862C10 = 16796C11 = 58786C12 = 208012C13 = 742900C14 = 2674440C15 = 9694845Using taylor expansion of sqrt(1-4X). (See http://bababadalgharaghtakamminarronnkonnbro.blogspot.in/2012/10/algebraic-type-systems-combinatorial.html) 1+ 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  ## Brat catalan = { n | true? n == 0 { 1 } { (2 * ( 2 * n - 1) / ( n + 1 )) * catalan(n - 1) }} 0.to 15 { n | p "#{n} - #{catalan n}"} 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: #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;} 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# 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.csusing 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(); } } }} 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) #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__ Here is the implementation of the algorithms. The c'tor of each class tells us the algorithm which will be used. (algorithms.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)); } 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) #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__ 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) #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; } 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 (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) ## Common Lisp With all three methods defined. (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)))) ## 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;} 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] ## EchoLisp  This example is incorrect. Please fix the code and remove this message.Details: series starts 1, 1, 2, ...  (lib 'sequences)(lib 'bigint)(lib 'math) ;; function definition(define (C1 n) (/ (factorial (* n 2)) (factorial (1+ n)) (factorial n)))(for ((i [1 .. 16])) (write (C1 i))) → 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 ;; using a recursive procedure with memoization(define (C2 n) ;; ( Σ ...)is the same as (sigma ..) (Σ (lambda(i) (* (C2 i) (C2 (- n i 1)))) 0 (1- n)))(remember 'C2 #(1)) ;; first term defined here (for ((i [1 .. 16])) (write (C2 i))) → 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 ;; using procrastinators = infinite sequence(define (catalan n acc) (/ (* acc 2 (1- (* 2 n))) (1+ n)))(define C3 (scanl catalan 1 [1 ..]))(take C3 15) → (1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845) ;; the same, using infix notation(lib 'match)(load 'infix.glisp) (define (catalan n acc) ((2 * acc * ( 2 * n - 1)) / (n + 1)))(define C3 (scanl catalan 1 [1 ..])) (take C3 15) → (1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845);; or(for ((c C3) (i 15)) (write c)) → 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845  ## Eiffel  class APPLICATION create make feature {NONE} make do across 0 |..| 14 as c loop io.put_double (nth_catalan_number (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 * nth_catalan_number (n - 1) end end end  Output: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440  ## Elixir Translation of: Erlang defmodule Catalan do def cat(n), do: div( factorial(2*n), factorial(n+1) * factorial(n) ) defp factorial(n), do: fac1(n,1) defp fac1(0, acc), do: acc defp fac1(n, acc), do: fac1(n-1, n*acc) def cat_r1(0), do: 1 def cat_r1(n), do: Enum.sum(for i <- 0..n-1, do: cat_r1(i) * cat_r1(n-1-i)) def cat_r2(0), do: 1 def cat_r2(n), do: div(cat_r2(n-1) * 2 * (2*n - 1), n + 1) def test do range = 0..14 :io.format "Directly:~n~p~n", [(for n <- range, do: cat(n))] :io.format "1st recusive method:~n~p~n", [(for n <- range, do: cat_r1(n))] :io.format "2nd recusive method:~n~p~n", [(for n <- range, do: cat_r2(n))] endend Catalan.test 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]  ## 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",[[cat(N) || N <- TestList]]), io:format("1st recusive method:\n~p\n",[[cat_r1(N) || N <- TestList]]), io:format("2nd recusive method:\n~p\n",[[cat_r2(N) || N <- TestList]]). 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 PROGRAM CATALAN PROCEDURE CATALAN(N->RES) RES=1 FOR I=1 TO N DO RES=RES*2*(2*I-1)/(I+1) END FOREND PROCEDURE BEGIN FOR N=0 TO 15 DO CATALAN(N->RES) PRINT(N;"=";RES) END FOREND PROGRAM  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 --Catalan number task from Rosetta Code wiki--User:Lnettnay --function from factorial taskfunction factorial(integer n)atom f = 1while n > 1 do f *= n n -= 1end while return fend function function catalan(integer n) atom numerator = factorial(2 * n)atom denominator = factorial(n+1)*factorial(n)return numerator/denominatorend function for i = 0 to 15 do ? catalan(i)end for Output: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845  ## F# In the REPL (with 3rd equation): > Seq.unfold(fun (c,n) -> let cc = 2*(2*n-1)*c/(n+1) in Some(c,(cc,n+1))) (1,1) |> Seq.take 15 |> Seq.iter (printf "%i, ");; 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, val it : unit = ()  ## Factor This is the last solution, memoized by using arrays. Run in scratchpad. : next ( seq -- newseq ) [ ] [ last ] [ length ] tri [ 2 * 1 - 2 * ] [ 1 + ] bi / * suffix ;: Catalan ( n -- seq ) V{ 1 } swap 1 - [ next ] times ;15 Catalan .V{ 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440} ## Fantom  This example is incorrect. Please fix the code and remove this message.Details: series starts 1, 1, 2, ... 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)) } }} 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 : catalan ( n -- ) 1 swap 1+ 1 do dup cr . i 2* 1- 2* i 1+ */ loop drop ; ## Fortran Works with: Fortran version 90 and later 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!BLdouble 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 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 ===============  ## FreeBASIC ' FB 1.05.0 Win64 Function factorial(n As UInteger) As UInteger If n = 0 Then Return 1 Return n * factorial(n - 1)End Function Function catalan1(n As UInteger) As UInteger Dim prod As UInteger = 1 For i As UInteger = n + 2 To 2 * n prod *= i Next Return prod / factorial(n)End Function Function catalan2(n As UInteger) As UInteger If n = 0 Then Return 1 Dim sum As UInteger = 0 For i As UInteger = 0 To n - 1 sum += catalan2(i) * catalan2(n - 1 - i) Next Return sumEnd Function Function catalan3(n As UInteger) As UInteger If n = 0 Then Return 1 Return catalan3(n - 1) * 2 * (2 * n - 1) \ (n + 1)End Function Print "n", "First", "Second", "Third"Print "-", "-----", "------", "-----"PrintFor i As UInteger = 0 To 15 Print i, catalan1(i), catalan2(i), catalan3(i)NextPrintPrint "Press any key to quit"Sleep Output: n First Second Third - ----- ------ ----- 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  ## Frink Frink includes efficient algorithms for calculating arbitrarily-large binomial coefficients and automatically caches factorials. catalan[n] := binomial[2n,n]/(n+1)for n = 0 to 15 println[catalan[n]] ## FunL import integers.chooseimport 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 ) 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 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 ] ## Go Direct: 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))) }} Output: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440  ## Groovy  class Catalan{ public static void main(String[] args) { BigInteger N = 15; BigInteger k,n,num,den; BigInteger catalan; print(1); for(n=2;n<=N;n++) { num = 1; den = 1; for(k=2;k<=n;k++) { num = num*(n+k); den = den*k; catalan = num/den; } println(catalan); } }}  Output: 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845  ## Harbour  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  Output: 0: 1 1: 1 2: 2 3: 5 4: 14 5: 42 6: 132 7: 429 8: 1430 9: 4862 0: 16796 1: 58786 2: 208012 3: 742900 4: 2674440 5: 9694845  ## 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]
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

 This example is incorrect. Please fix the code and remove this message.Details: series starts 1, 1, 2, ...
procedure main(arglist)every writes(catalan(i)," ")end procedure catalan(n) # return catalan(n) or failstatic Minitial M := table() if n > 0 then    return (n = 1) | \M[n] | ( M[n] := (2*(2*n-1)*catalan(n-1))/(n+1))end
Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845

## J

   ((! +:) % >:) i.15x1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440

## 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").

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));		}	}}
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

<html><head><title>Catalan</title></head><body><pre id='x'></pre><script type="application/javascript">function disp(x) {	var e = document.createTextNode(x + '\n');	document.getElementById('x').appendChild(e);} var fc = [], c2 = [], c3 = [];function fact(n) { return fc[n] ? fc[n] : fc[n] = (n ? n * fact(n - 1) : 1); }function cata1(n) { return Math.floor(fact(2 * n) / fact(n + 1) / fact(n) + .5); }function cata2(n) {	if (n == 0) return 1;	if (!c2[n]) {		var s = 0;		for (var i = 0; i < n; i++) s += cata2(i) * cata2(n - i - 1);		c2[n] = s;	}	return c2[n];}function cata3(n) {	if (n == 0) return 1;	return c3[n] ? c3[n] : c3[n] = (4 * n - 2) * cata3(n - 1) / (n + 1);} 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>
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

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

Example 1

(range(0; 16), 100) as $i |$i | catalan | [$i, .] Output: $ 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]

#### Generate a sequence of Catalan numbers

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;

Example 2:

catalan_series(15)
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 )
Output:
As above for 0 to 15.


## Julia

Works with: Julia version 0.6
catalannum(n::Integer) = binomial(2n, n) ÷ (n + 1) @show catalannum.(1:15)@show catalannum(big(100))
Output:
catalannum.(1:15) = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
catalannum(big(100)) = 896519947090131496687170070074100632420837521538745909320

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

## K

  catalan: {_{*/(x-i)%1+i:!y-1}[2*x;x+1]%x+1}  catalan'!:151 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440

## Kotlin

Works with: Java version 1.7.0
Works with: Kotlin version 1.1.4
abstract class Catalan {    abstract operator fun invoke(n: Int) : Double     protected val m = mutableMapOf(0 to 1.0)} object CatalanI : Catalan() {    override fun invoke(n: Int): Double {        if (n !in m)            m[n] = Math.round(fact(2 * n) / (fact(n + 1) * fact(n))).toDouble()        return m[n]!!    }     private fun fact(n: Int): Double {        if (n in facts)            return facts[n]!!        val f = n * fact(n -1)        facts[n] = f        return f    }     private val facts = mutableMapOf(0 to 1.0, 1 to 1.0, 2 to 2.0)} object CatalanR1 : Catalan() {    override fun invoke(n: Int): Double {        if (n in m)            return m[n]!!         var sum = 0.0        for (i in 0..n - 1)            sum += invoke(i) * invoke(n - 1 - i)        sum = Math.round(sum).toDouble()        m[n] = sum        return sum    }} object CatalanR2 : Catalan() {    override fun invoke(n: Int): Double {        if (n !in m)            m[n] = Math.round(2.0 * (2 * (n - 1) + 1) / (n + 1) * invoke(n - 1)).toDouble()        return m[n]!!    }} fun main(args: Array<String>) {    val c = arrayOf(CatalanI, CatalanR1, CatalanR2)    for(i in 0..15) {        c.forEach { print("%9d".format(it(i).toLong())) }        println()    }}
Output:
        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
9694845  9694845  9694845

## Liberty BASIC

print "non-recursive version"print catNonRec(5)for i = 0 to 15    print i;"   =   "; catNonRec(i)nextprint print "recursive version"print catRec(5)for i = 0 to 15    print i;"   =   "; catRec(i)nextprint print "recursive with memoisation"redim cats(20)  'clear the arrayprint catRecMemo(5)for i = 0 to 15    print i;"   =   "; catRecMemo(i)nextprint  wait function catNonRec(n)   'non-recursive version    catNonRec=1    for i=1 to n        catNonRec=((2*((2*i)-1))/(i+1))*catNonRec    nextend function function catRec(n)  'recursive version    if n=0 then        catRec=1    else        catRec=((2*((2*n)-1))/(n+1))*catRec(n-1)    end ifend 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 useend function
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

-- recursive with memoizationcatalan = {[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
Output:
1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440

## Logo

to factorial :n  output ifelse [less? :n 1] 1 [product :n factorial difference :n 1]endto choose :n :r  output quotient factorial :n product factorial :r factorial difference :n :rendto catalan :n  output product (quotient sum :n 1) choose product 2 :n :nend repeat 15 [print catalan repcount]
Output:
1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440

## Maple

CatalanNumbers := proc( n::posint )    return seq( (2*i)!/((i + 1)!*i!), i = 0 .. n - 1 );end proc:CatalanNumbers(15);

Output:

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


## Mathematica / Wolfram Language

CatalanN[n_Integer /; n >= 0] := (2 n)!/((n + 1)! n!)
Sample Output:
TableForm[CatalanN/@Range[0,15]]//TableForm= 1125144213242914304862167965878620801274290026744409694845

## MATLAB / Octave

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

The following version computes at the same time the n first Catalan numbers (including C0).

function n = catalanNumbers(n)    n = [1 cumprod((2:4:4*n-6) ./ (2:n))];end
Sample Output:
>> catalanNumber(14) ans =      2674440 >> catalanNumbers(18)' ans =            1           1           2           5          14          42         132         429        1430        4862       16796       58786      208012      742900     2674440     9694845    35357670   129644790

/* 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] */ ## Modula-2 MODULE CatalanNumbers;FROM FormatString IMPORT FormatString;FROM Terminal IMPORT WriteString,WriteLn,ReadChar; PROCEDURE binomial(m,n : LONGCARD) : LONGCARD;VAR r,d : LONGCARD;BEGIN r := 1; d := m - n; IF d>n THEN n := d; d := m - n; END; WHILE m>n DO r := r * m; DEC(m); WHILE (d>1) AND NOT (r MOD d # 0) DO r := r DIV d; DEC(d) END END; RETURN rEND binomial; PROCEDURE catalan1(n : LONGCARD) : LONGCARD;BEGIN RETURN binomial(2*n,n) DIV (1+n)END catalan1; PROCEDURE catalan2(n : LONGCARD) : LONGCARD;VAR i,sum : LONGCARD;BEGIN IF n>1 THEN sum := 0; FOR i:=0 TO n-1 DO sum := sum + catalan2(i) * catalan2(n - 1 - i) END; RETURN sum ELSE RETURN 1 ENDEND catalan2; PROCEDURE catalan3(n : LONGCARD) : LONGCARD;BEGIN IF n#0 THEN RETURN 2 *(2 * n - 1) * catalan3(n - 1) DIV (1 + n) ELSE RETURN 1 ENDEND catalan3; VAR blah : LONGCARD = 123; buf : ARRAY[0..63] OF CHAR; i : LONGCARD;BEGIN FormatString("\tdirect\tsumming\tfrac\n", buf); WriteString(buf); FOR i:=0 TO 15 DO FormatString("%u\t%u\t%u\t%u\n", buf, i, catalan1(i), catalan2(i), catalan3(i)); WriteString(buf) END; ReadCharEND CatalanNumbers. ## 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) 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 ## OCaml let imp_catalan n = let return = ref 1 in for i = 1 to n do return := !return * 2 * (2 * i - 1) / (i + 1) done; !return let rec catalan = function | 0 -> 1 | n -> catalan (n - 1) * 2 * (2 * n - 1) / (n + 1) let memoize f = let cache = Hashtbl.create 20 in fun n -> match Hashtbl.find_opt cache n with | None -> let x = f n in Hashtbl.replace cache n x; x | Some x -> x let catalan_cache = Hashtbl.create 20 let rec memo_catalan n = if n = 0 then 1 else match Hashtbl.find_opt catalan_cache n with | None -> let x = memo_catalan (n - 1) * 2 * (2 * n - 1) / (n + 1) in Hashtbl.replace catalan_cache n x; x | Some x -> x let () = if not !Sys.interactive then let bench label f n times = let start = Unix.gettimeofday () in begin for i = 1 to times do f n done; let stop = Unix.gettimeofday () in Printf.printf "%s (%d runs) : %.3f\n" label times (stop -. start) end in let show f g h f' n = for i = 0 to n do Printf.printf "%2d %7d %7d %7d %7d\n" i (f i) (g i) (h i) (f' i) done in List.iter (fun (l, f) -> bench l f 15 10_000_000) ["imperative", imp_catalan; "recursive", catalan; "hand-memoized", memo_catalan; "memoized", (memoize catalan)]; show imp_catalan catalan memo_catalan (memoize catalan) 15  Output: $ ocaml unix.cma catalan.ml
imperative (10000000 runs) : 3.420
recursive (10000000 runs) : 3.821
hand-memoized (10000000 runs) : 0.531
memoized (10000000 runs) : 0.515
0       1       1       1       1
1       1       1       1       1
2       2       2       2       2
3       5       5       5       5
4      14      14      14      14
5      42      42      42      42
6     132     132     132     132
7     429     429     429     429
8    1430    1430    1430    1430
9    4862    4862    4862    4862
10   16796   16796   16796   16796
11   58786   58786   58786   58786
12  208012  208012  208012  208012
13  742900  742900  742900  742900
14 2674440 2674440 2674440 2674440
15 9694845 9694845 9694845 9694845

$ocamlopt -O2 unix.cmxa catalan.ml -o catalan$ ./catalan
imperative (10000000 runs) : 2.020
recursive (10000000 runs) : 2.283
hand-memoized (10000000 runs) : 0.159
memoized (10000000 runs) : 0.167
...

## Oforth

: catalan( n -- m )     n ifZero: [ 1 ] else: [ catalan( n 1- ) 2 n * 1- * 2 * n 1+ / ] ;
Output:
import: mapping
seqFrom(0, 15) map( #catalan ) .
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]


## ooRexx

Three versions of this.

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
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.

catalan(n)=binomial(2*n,n+1)/n

A second version:

catalan(n)=(2*n)!/(n+1)!/n!

Naive version with binary splitting:

catalan(n)=prod(k=n+2,2*n,k)/prod(k=2,n,k)

Naive version:

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

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:

vector(15,n,catalan(n))

## 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.
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

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($_) ]}\n" for 0 .. 20; For computing up to 20 ish, memoization is not needed. For much bigger numbers, this is faster: my @c = (1);sub catalan { use bigint;$c[$_[0]] //= catalan($_[0]-1) * (4 * $_[0]-2) / ($_[0]+1)} # most of the time is spent displaying the long numbers, actuallyprint "$_\t", catalan($_), "\n" for 0 .. 10000;

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
use ntheory qw/binomial/;sub catalan {  my $n = shift; binomial(2*$n,$n)/($n+1);}print "$_\t", catalan($_), "\n" for 0 .. 10000;

## Perl 6

Works with: Rakudo version 2015.12

The recursive formulas are easily written into a constant array, either:

constant Catalan = 1, { [+] @_ Z* @_.reverse } ... *;

or

constant Catalan = 1, |[\*] (2, 6 ... *) Z/ 2 .. *;  # In both cases, the sixteen first values can be seen with:.say for Catalan[^16];
Output:
1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440

## Phix

Library: bigatom

-- returns inf/-nan for n>85, and needs the rounding for n>=14, accurate to n=29function 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 resend 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 rend function atom t0 = time() -- (this last call only)string sc100 = ba_sprint(catalan2bc(100))printf(1,"100: %s (%3.2fs)\n",{sc100,time()-t0})
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

<?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";}?>
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

 <?php $n = 15;$t[1] = 1;foreach (range(1, $n+1) as$i) {    foreach (range($i, 1-1) as$j) {        $t[$j] += $t[$j - 1];    }    $t[$i +1] = $t[$i];    foreach (range($i+1, 1-1) as$j) {        $t[$j] += $t[$j -1];    }    print ($t[$i+1]-$t[$i])."\t";}
Output:
1	2	5	14	42	132	429	1430	4862	16796	58786	208012	742900	2674440	9694845	35357670


## 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) ) )
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

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;
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 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

## PowerShell

 function Catalan([uint64]$m) { function fact([bigint]$n) {        if($n -lt 2) {[bigint]::one} else{2..$n | foreach -Begin {$prod = [bigint]::one} -Process {$prod = [bigint]::Multiply($prod,$_)} -End {$prod}} }$fact = fact $m$fact1 = [bigint]::Multiply($m+1,$fact)    [bigint]::divide((fact (2*$m)), [bigint]::Multiply($fact,$fact1))}0..15 | foreach {"catalan($_): $(catalan$_)"}

Output:

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


### An Alternate Version

This version could easily be modified to work with big integers.

 function Get-CatalanNumber{    [CmdletBinding()]    [OutputType([PSCustomObject])]    Param    (        [Parameter(Mandatory=$true, ValueFromPipeline=$true,                   ValueFromPipelineByPropertyName=$true, Position=0)] [uint32[]]$InputObject    )     Begin    {        function Get-Factorial ([int]$Number) { if ($Number -eq 0)            {                return 1            }             $factorial = 1 1..$Number | ForEach-Object {$factorial *=$_}             $factorial } function Get-Catalan ([int]$Number)        {            if ($Number -eq 0) { return 1 } (Get-Factorial (2 *$Number)) / ((Get-Factorial (1 + $Number)) * (Get-Factorial$Number))        }    }    Process    {        foreach ($number in$InputObject)        {            [PSCustomObject]@{                Number        = $number CatalanNumber = Get-Catalan$number            }        }    }}

Get the first fifteen Catalan numbers as a PSCustomObject:

 0..14 | Get-CatalanNumber
Output:
Number CatalanNumber
------ -------------
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


To return only the array of Catalan numbers:

 (0..14 | Get-CatalanNumber).CatalanNumber
Output:
1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440


## Prolog

Works with: SWI-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]).
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...

; 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:EndIfIf n=0:ProcedureReturn 1:EndIfProcedureReturn (2*(2*n-1))*CatalanNumber(n-1)/(n+1)EndProcedure ls=25rs=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.sNext
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.

from math import factorialimport 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)  @memoizedef 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)
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

catalan <- function(n) choose(2*n, n)/(n + 1)catalan(0:15) [1]       1       1       2       5      14      42     132     429    1430[10]    4862   16796   58786  208012  742900 2674440 9694845

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


## REXX

### version 1

All four methods of calculate the Catalan numbers use independent memoization for the computation of factorials.

In the 1st equation, the 2nd version's denominator:

(n+1)! n!

has been rearranged to:

(n+1) * [fact(n) **2]
/*REXX program calculates and displays  Catalan numbers  using  four different methods. */parse arg LO HI .                                /*obtain optional arguments from the CL*/if LO=='' | LO==","  then do;  HI=15; LO=0;  end /*No args? Then use a range of 0 ──► 15*/if HI=='' | HI==","  then      HI=LO             /*No HI?   Then use  LO for the default*/numeric digits max(20, 5*HI)                     /*this allows gihugic Catalan numbers. */w=length(HI)                                     /*W:  is used for aligning the output. */call hdr 1A;  do j=LO  to HI;  say '     Catalan'     right(j, w)": "      Cat1A(j);   endcall hdr 1B;  do j=LO  to HI;  say '     Catalan'     right(j, w)": "      Cat1B(j);   endcall hdr 2 ;  do j=LO  to HI;  say '     Catalan'     right(j, w)": "      Cat2(j) ;   endcall hdr 3 ;  do j=LO  to HI;  say '     Catalan'     right(j, w)": "      Cat3(j) ;   endexit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/!:     arg z; if !.z\==. then return !.z; !=1;  do k=2  to z; !=!*k; end;  !.z=!; return !Cat1A: procedure expose !.;  parse arg n;     return comb(n+n, n)    %  (n+1)Cat1B: procedure expose !.;  parse arg n;     return !(n+n) % ((n+1) * !(n)**2)Cat3:  procedure expose c.;  arg n; if c.n==. then c.n=(4*n-2)*cat3(n-1)%(n+1); return c.ncomb:  procedure;            parse arg x,y;   return pFact(x-y+1, x) % pFact(2, y)hdr:   !.=.; c.=.; c.0=1; say; say center('Catalan numbers, method' arg(1),79,'─'); returnpFact: procedure;            !=1;      do k=arg(1)  to arg(2);  !=!*k;  end;    return !/*──────────────────────────────────────────────────────────────────────────────────────*/Cat2:  procedure expose c.;  parse arg n;  $=0; if c.n\==. then return c.n do k=0 for n;$=$+ Cat2(k) * Cat2(n-k-1); end c.n=$;           return $/*use a memoization technique.*/ 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 /* REXX ---------------------------------------------------------------* 01.07.2014 Walter Pachl*--------------------------------------------------------------------*/Numeric Digits 1000Parse Arg m .If m='' Then m=20Do i=0 To m c1.i=c1(i) Endc2.=1Do i=1 To m c2.i=c2(i) Endc3.=1Do i=1 To m im1=i-1 c3.i=2*(2*i-1)*c3.im1/(i+1) Endl=length(c3.m)hdr=' n' right('c1.n',l), right('c2.n',l), right('c3.n',l)Say hdrDo i=0 To m Say right(i,2) format(c1.i,l), format(c2.i,l), format(c3.i,l) EndSay hdrExit c1: ProcedureParse Arg nreturn fact(2*n)/(fact(n)*fact(n+1)) c2: Procedure Expose c2.Parse Arg nres=0Do i=0 To n-1 nmi=n-i-1 res=res+c2.i*c2.nmi EndReturn res fact: ProcedureParse Arg nf=1Do i=1 To n f=f*i EndReturn f 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 ## Ring  for n = 1 to 15 see catalan(n) + nlnext func catalan n if n = 0 return 1 ok cat = 2 * (2 * n - 1) * catalan(n - 1) / (n + 1) return cat  Output: 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845  ## Ruby Library: RubyGems 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)]} 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 FOR i = 1 TO 15 PRINT i;" ";catalan(i)NEXT FUNCTION catalan(n) catalan = 1 if n <> 0 then catalan = ((2 * ((2 * n) - 1)) / (n + 1)) * catalan(n - 1)END FUNCTION 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 ## 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)); }} 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. 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 <- 0 to 15) { println("catalan(" + n + ") = " + catalan(n)) } }} Output: 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 ## Scheme Tail recursive implementation. (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) 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 $ 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;
Output:
1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845


## Sidef

func f(i) { i==0 ? 1 : (i * f(i-1)) }func c(n) { f(2*n) / f(n) / f(n+1) }

With memoization:

func c(n) is cached {    n == 0 ? 1 : (c(n-1) * (4 * n - 2) / (n + 1))}

Calling the function:

15.times { |i|    say "#{i}\t#{c(i)}"}
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


## smart BASIC

PRINT "Recursive:"!PRINTFOR n = 0 TO 15    PRINT n,"#######":catrec(n)NEXT nPRINT!PRINT PRINT "Non-recursive:"!PRINTFOR n = 0 TO 15    PRINT n,"#######":catnonrec(n)NEXT n END DEF catrec(x)    IF x = 0 THEN        temp = 1    ELSE         n = x        temp = ((2*((2*n)-1))/(n+1))*catrec(n-1)    END IF    catrec = tempEND DEF DEF catnonrec(x)    temp = 1    FOR n = 1 TO x         temp = (2*((2*n)-1))/(n+1)*temp    NEXT n    catnonrec = tempEND DEF

## Standard ML

(* * 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 *)

## Stata

clearset obs 15gen catalan=1 in 1replace catalan=catalan[_n-1]*2*(2*_n-3)/_n in 2/llist, noobs noh

Output

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

## Swift

Translation of: Rust
func catalan(_ n: Int) -> Int {  switch n {  case 0:    return 1  case _:    return catalan(n - 1) * 2 * (2 * n - 1) / (n + 1)  }} for i in 1..<16 {  print("catalan(\(i)) => \(catalan(i))")}
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


## Tcl

package require Tcl 8.5 # Memoization wrapperproc memoize {function value generator} {    variable memoize    set key $function,$value    if {![info exists memoize($key)]} { set memoize($key) [uplevel 1 $generator] } return$memoize($key)} # The simplest recursive definitionproc tcl::mathfunc::catalan n { if {[incr n 0] < 0} {error "must not be negative"} memoize catalan$n {expr {	$n == 0 ? 1 : 2 * (2*$n - 1) * catalan($n - 1) / ($n + 1)    }}}

Demonstration:

for {set i 0} {$i < 15} {incr i} { puts "C_$i = [expr {catalan(\$i)}]"}
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.

:For(I,1,15:Disp (2I)!/((I+1)!I!:End
Output:
               1
2
4
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
Done

## Ursala

#import std#import nat catalan = quotient^\successor choose^/double ~& #cast %nL t = catalan* iota 16
Output:
<
1,
1,
2,
5,
14,
42,
132,
429,
1430,
4862,
16796,
58786,
208012,
742900,
2674440,
9694845>

## VBA

Public Sub Catalan1(n As Integer)'Computes the first n Catalan numbers according to the first recursion givenDim Cat() As LongDim sum As Long ReDim Cat(n)Cat(0) = 1For i = 0 To n - 1  sum = 0  For j = 0 To i    sum = sum + Cat(j) * Cat(i - j)  Next j  Cat(i + 1) = sumNext iDebug.PrintFor i = 0 To n  Debug.Print i, Cat(i)NextEnd Sub Public Sub Catalan2(n As Integer)'Computes the first n Catalan numbers according to the second recursion givenDim Cat() As Long ReDim Cat(n)Cat(0) = 1For i = 1 To n  Cat(i) = 2 * Cat(i - 1) * (2 * i - 1) / (i + 1)Next iDebug.PrintFor i = 0 To n  Debug.Print i, Cat(i)NextEnd Sub
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

 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 IfEnd Function 'Find the first 15 Catalan numbers.For j = 1 To 15	WScript.StdOut.Write j & " = " & catalan(j)	WScript.StdOut.WriteLineNext
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

; 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]

## XLISP

(defun catalan (n)	(if (= n 0)		1		(* (/ (* 2 (- (* 2 n) 1)) (+ n 1)) (catalan (- n 1))) ) ) (defun range (x y)	(cons x		(if (< x y)			(range (+ x 1) y) ) ) ) (print (mapcar catalan (range 0 14)))
Output:
(1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440)

## 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);    ];]
Output:
1
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440


## zkl

Uses GMP to calculate big factorials.

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)));

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.

fcn catalan(n){ (1).reduce(n,fcn(p,n){ 2*(2*n-1)*p/(n+1) },1) }
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


## ZX Spectrum Basic

Translation of: C
10 FOR i=0 TO 1520 LET n=i: LET m=2*n30 LET r=1: LET d=m-n40 IF d>n THEN LET n=d: LET d=m-n50 IF m<=n THEN GO TO 9060 LET r=r*m: LET m=m-170 IF (d>1) AND NOT FN m(r,d) THEN LET r=r/d: LET d=d-1: GO TO 7080 GO TO 5090 PRINT i;TAB 4;r/(1+n)100 NEXT i110 STOP 120 DEF FN m(a,b)=a-INT (a/b)*b: REM Modulus function