# 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):

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

Task

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              next i


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


## ALGOL 68

<lang algol68># calculate the first few catalan numbers, using LONG INT values #

1. (64-bit quantities in Algol 68G which can handle up to C23) #
1. 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 # ;

1. 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 # ;

1. returnss the nth Catalan number using binomial coefficeients #
2. uses the factorial over factorial procedure for a slight optimisation #
3. note: Cn = 1/(n+1)(2n n) #
4. = (2n)!/((n+1)!n!) #
5. = 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;

1. show the first few catalan numbers #

FOR i FROM 0 TO 15 DO

   print( ( whole( i, -2 ), ": ", whole( catalan( i ), 0 ), newline ) )


OD</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


## APL

<lang apl> {(!2×⍵)÷(!⍵+1)×!⍵}(⍳15)-1</lang>

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

## 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%)


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

## Befunge

Translation of: Ada

<lang befunge>0>:.:000p1>\:00g-#v_v v 2-1*2p00 :+1g00\< $> **00g1+/^v,*84,"="< _^#<*53:+1>#,.#+5< @</lang> Output: 0 = 1 1 = 1 2 = 2 3 = 5 4 = 14 5 = 42 6 = 132 7 = 429 8 = 1430 9 = 4862 10 = 16796 11 = 58786 12 = 208012 13 = 742900 14 = 2674440 15 = 9694845 ## Bracmat <lang bracmat>( out$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)  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 </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__ 1. 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  1. 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; 1. include <cmath> using std::floor; 1. 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__ 1. define __TESTER_H__ 1. 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  1. 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" 1. 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] ## EchoLisp <lang scheme> (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  </lang> ## 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: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440  ## Elixir Translation of: Erlang <lang elixir>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))] end  end Catalan.test</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]  ## 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: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845  ## 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{  1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440  }</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; 1. The first fifteen: 0 to 14 ! List([0 .. 14], Catalan1); List([0 .. 14], Catalan2); List([0 .. 14], Catalan3); List([0 .. 14], Catalan4); 1. Same output for all four: 2. [ 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: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440  ## 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: 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 <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

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

## 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>

<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></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

<lang jq>

 [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>

## Kotlin

Works with: Java version 1.7.0
Works with: Kotlin version 1.0.0
Library: Koloboke version 0.6.8

<lang scala>import net.openhft.koloboke.collect.map.hash.HashIntDoubleMaps.*

abstract class Catalan {

   abstract operator fun invoke(n: Int) : Double

   protected val m = newUpdatableMapOf(0 , 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]
var f = n * fact(n -1)
facts[n] = f
return f
}

   private val facts = newUpdatableMapOf(0 , 1.0, 1 , 1.0, 2 , 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()
}


}</lang>

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

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

## 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 / Wolfram Language

<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 =

    2674440


>> 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) . [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) / ($_[0]+1)  } 1. most of the time is spent displaying the long numbers, actually print "$_\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 Works with: Rakudo version 2015.12 The recursive formulas are easily written into a constant array, either: <lang perl6>constant Catalan = 1, { [+] @_ Z* @_.reverse } ... *;</lang> or <lang perl6>constant Catalan = 1, |[\*] (2, 6 ... *) Z/ 2 .. *;</lang> In both cases, the sixteen first values can be seen with: <lang perl6>.say for Catalan[^15];</lang> Output: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 ## 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 = $r\r\n";  } ?></lang> Output: 0 = 1 1 = 1 2 = 2 3 = 5 4 = 14 5 = 42 6 = 132 7 = 429 8 = 1430 9 = 4862 10 = 16796 11 = 58786 12 = 208012 13 = 742900 14 = 2674440 15 = 9694845  ## PicoLisp <lang PicoLisp># Factorial (de fact (N)  (if (=0 N) 1 (* N (fact (dec N))) ) )  1. Directly (de catalanDir (N)  (/ (fact (* 2 N)) (fact (inc N)) (fact N)) )  1. Recursively (de catalanRec (N)  (if (=0 N) 1 (cache '(NIL) N # Memoize (sum '((I) (* (catalanRec I) (catalanRec (- N I 1)))) (range 0 (dec N)) ) ) ) )  1. Alternatively (de catalanAlt (N)  (if (=0 N) 1 (*/ 2 (dec (* 2 N)) (catalanAlt (dec N)) (inc N)) ) )  1. Test (for (N 0 (> 15 N) (inc N))  (tab (2 4 8 8 8) N " => " (catalanDir N) (catalanRec N) (catalanAlt N) ) )</lang>  Output:  0 => 1 1 1 1 => 1 1 1 2 => 2 2 2 3 => 5 5 5 4 => 14 14 14 5 => 42 42 42 6 => 132 132 132 7 => 429 429 429 8 => 1430 1430 1430 9 => 4862 4862 4862 10 => 16796 16796 16796 11 => 58786 58786 58786 12 => 208012 208012 208012 13 => 742900 742900 742900 14 => 2674440 2674440 2674440 ## PL/I <lang PL/I>catalan: procedure options (main); /* 23 February 2012 */  declare (i, n) fixed;   put skip list ('How many catalan numbers do you want?'); get list (n);   do i = 0 to n; put skip list (c(i)); end;  c: procedure (n) recursive returns (fixed decimal (15));  declare n fixed;   if n <= 1 then return (1);   return ( 2*(2*n-1) * c(n-1) / (n + 1) );  end c; end catalan;</lang> Output: How many catalan numbers do you want? 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 35357670 129644790 477638700 1767263190 6564120420  ## Plain TeX <lang tex>\newcount\n \newcount\r \newcount\x \newcount\ii \def\catalan#1{% \n#1\advance\n by1\ii1\r1% \loop{% \x\ii% \multiply\x by 2 \advance\x by -1 \multiply\x by 2% \global\multiply\r by\x% \global\advance\ii by1% \global\divide\r by\ii% } \ifnum\number\ii<\n\repeat% \the\r } \rightskip=0pt plus1fil\parindent=0pt \loop{${\rm Catalan}(\the\x) = \catalan{\the\x}$\hfil\break}% \advance\x by 1\ifnum\x<15\repeat \bye</lang> ## PowerShell <lang PowerShell> function Catalan([uint64]$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 _)"} </lang> 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  ## 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] 1 2 5 14 42 132 429 1430 4862 2. [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 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); end call hdr 1B; do j=LO to HI; say ' Catalan' right(j, w)": " Cat1B(j); end call hdr 2 ; do j=LO to HI; say ' Catalan' right(j, w)": " Cat2(j) ; end call hdr 3 ; do j=LO to HI; say ' Catalan' right(j, w)": " Cat3(j) ; end exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ !: arg z; if !.z\==. then return !.z; !=1; do k=2 for 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.n comb: 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,'─'); return pFact: 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  to n-1;   $=$ + cat2(k) * cat2(n-k-1);  end
c.n=$; return$    /*use a memoization technique.*/</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

1. direct

def catalan_direct(n)

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


end

1. 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

1. 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)


NEXT

FUNCTION catalan(n)

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


END FUNCTION</lang>

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

## 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

<lang seed7>$include "seed7_05.s7i";  include "bigint.s7i";  const proc: main is func  local var bigInteger: n is 0_; begin for n range 0_ to 15_ do writeln((2_ * n) ! n div succ(n)); end for; end func;</lang>  Output: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845  ## Sidef <lang ruby>func f(i) { i==0 ? 1 : (i * f(i-1)) } func c(n) { f(2*n) / f(n) / f(n+1) }</lang> With memoization: <lang ruby>func c(n) is cached {  n == 0 ? 1 : (c(n-1) * (4 * n - 2) / (n + 1));  }</lang> Calling the function: <lang ruby>15.times { |i|  say "#{i-1}\t#{c(i-1)}";  }</lang> Output: 0 1 1 1 2 2 3 5 4 14 5 42 6 132 7 429 8 1430 9 4862 10 16796 11 58786 12 208012 13 742900 14 2674440  ## Standard ML <lang sml>(* * val catalan : int -> int * Returns the nth Catalan number. *)  fun catalan 0 = 1 | catalan n = ((4 * n - 2) * catalan(n - 1)) div (n + 1); (* * val print_catalans : int -> unit * Prints out Catalan numbers 0 through 15. *)  fun print_catalans(n) =  if n > 15 then () else (print (Int.toString(catalan n) ^ "\n"); print_catalans(n + 1)); print_catalans(0);  (* * 1 * 1 * 2 * 5 * 14 * 42 * 132 * 429 * 1430 * 4862 * 16796 * 58786 * 208012 * 742900 * 2674440 * 9694845 *)</lang>  ## Tcl <lang tcl>package require Tcl 8.5 1. Memoization wrapper proc memoize {function value generator} {  variable memoize set key$function,$value if {![info exists memoize($key)]} {


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

   }
return $memoize($key)


}

1. 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:
               1
2
4
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
Done

## Ursala

<lang ursala>#import std

1. import nat

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

1. cast %nL

t = catalan* iota 16</lang>

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

## 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

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


## 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