# 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

<lang 11l>V c = 1 L(n) 1..15

  print(c)
c = 2 * (2 * n - 1) * c I/ (n + 1)</lang>

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


## 360 Assembly

Very compact version. <lang 360asm>CATALAN CSECT 08/09/2015

        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


## ABAP

This works for ABAP Version 7.40 and above

<lang ABAP> 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.


</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


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


## ALGOL W

<lang algolw>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_n


end.</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

## Arturo

<lang arturo>catalan: @(n){ if n=0 { return 1 } { return ((4*n-2)*[catalan n-1])/(n+1) } }

loop 0..15 { print [padRight [toString &] 5] + " " + [padLeft [toString [catalan &]] 20] }</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

## 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


### 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).

<lang basic> 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 130


100 PRINT N,FX/(B*A) 110 NEXT N 120 STOP 130 LET FX=1 140 FOR I=1 TO X 150 LET FX=FX*I 160 NEXT I 170 RETURN</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

## BBC BASIC

<lang bbcbasic> 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)</lang>

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

<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>  ## Crystal Translation of: Ruby <lang ruby>require "big" require "benchmark" def factorial(n : BigInt) : BigInt  (1..n).product(1.to_big_i)  end def factorial(n : Int32 | Int64)  factorial n.to_big_i  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).reduce(0) do |sum, i| sum + catalan_rec1(i) * catalan_rec1(n - 1 - i) end  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 Benchmark.bm 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) } }  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> Output: # with --release flag # Using: i7-6700HQ, 3.5GHz user system total real catalan_direct 0.000026 0.000052 0.000078 ( 0.000074) catalan_rec1 0.139766 0.001143 0.140909 ( 0.141418) catalan_rec2 0.000003 0.000000 0.000003 ( 0.000003) 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  ## 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] ## EasyLang <lang>func catalan n . ans .  if n = 0 ans = 1 else call catalan n - 1 h ans = 2 * (2 * n - 1) * h div (1 + n) .  . for i range 15  call catalan i h print h  .</lang> Output: 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, ... <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> ## EDSAC order code The Catalan numbers are here computed by the second method, as a sum of products. The program illustrates a difficulty with multiplying integers on the EDSAC, which was designed to work with fixed-point numbers in the range [-1, 1). When we store a 35-bit integer A, we are really storing A*(2^-34). As long as integer arithmetic is confined to addition and subtraction, this scaling can be ignored. But if we multiply two 35-bit integers A and B, the result is really A*B*(2^-68), and needs to be multiplied by 2^34 to get the same scaling as for A and B. <lang edsac>  [Calculation of Catalan numbers. EDSAC program, Initial Orders 2.]   [Define where to store the list of Catalan numbers.] T 54 K [store address in location 54, so that values are accessed by code letter C (for Catalan)] P 200 F [<------ address here]   [Modification of library subroutine P7. Prints signed integer up to 10 digits, right-justified. 54 storage locations; working position 4D. Must be loaded at an even address. Input: Number is at 0D.] T 56 K GKA3FT42@A47@T31@ADE10@T31@A48@T31@SDTDH44#@NDYFLDT4DS43@TF H17@S17@A43@G23@UFS43@T1FV4DAFG50@SFLDUFXFOFFFSFL4FT4DA49@T31@ A1FA43@G20@XFP1024FP610D@524D!FO46@O26@XFO46@SFL8FT4DE39@   [Main routine] T 120 K [load at 120] G K [set @ (theta) to load address] [Variables] [0] P F [index of Catalan number] [Constants] [1] P 7 D [maximum index required] [2] P D [single-word 1] [3] P 2 F [to change addresses by 2] [4] H #C [these 3 are used to manufacture EDSAC orders] [5] T #C [6] V #C [7] K 4096 F [(1) add to change T order into H order (2) teleprinter null] [8] # F [figures shift] [9] ! F [space] [10] @ F [carriage return] [11] & F [line feed]   [Enter with acc = 0] [12] O 8 @ [set teleprinter to figures] T 4 D [clear 5F and sandwich bit] A 2 @ [load single-word 1] T 4 F [store as double word at 4D; clear acc] [Here with index in acc, Catalan number in 4D] [16] U @ [store index] L 1 F [times 4 by shifting] A 5 @ [make T order to store Catalan number] U 27 @ [plant in code] A 7 @ [make H order with same address] U 45 @ [plant in code] S 47 @ [make A order with same address] T 34 @ [plant in code] A 6 @ [load V order for start of list] T 46 @ [plant in code] A 4 D [Catalan number from temp store] [27] T #C [store in list (manufactured order)] T D [clear 1F and sandwich bit] A @ [load single-word index] T F [store as double word at 0D] [31] A 31 @ [for return from print subroutine] G 56 F [print index] O 9 @ [followed by space] [34] A #C [load Catalan number (manufactured order)] T D [to 0D for printing] [36] A 36 @ [for return from print subroutine] G 56 F [print Catalan number] O 10 @ [followed by new line] O 11 @ T 4 D [clear partial sum] A @ [load index] S 1 @ [reached the maximum?] E 64 @ [if so, jump to exit] [Inner loop to compute sum of products C{i}*C(n-1}] [44] T F [clear acc] [45] H #C [C{n-i} to mult reg (manufactured order)] [46] V #C [acc := C{i}*C{n-i} (manufactiured order)] [Multiply product by 2^34 (see preamble). The 'L F' order is also exploited above to convert an H order into an A order.] [47] L F [shift acc left by 13 (the maximum available)] L F [shift 13 more] L 64 F [shift 8 more, total 34] A 4 D [add partial sum] T 4 D [update partial sum] A 46 @ [inc i in V order] A 3 @ T 46 @ A 45 @ [dec (n - i) in H order] S 3 @ U 45 @ S 4 @ [is (n - i) now negative?] E 44 @ [if not, loop back] [Here with latest Catalan number in temp store 4D] T F [clear acc] A @ [load index] A 2 @ [add 1] E 16 @ [back to start of outer loop] [64] O 7 @ [exit; print null to flush teleprinter buffer] Z F [stop] E 12 Z [define entry point] P F [acc = 0 on entry]  </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  ## Eiffel <lang 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 </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  ## 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 The first method: <lang factor>USING: kernel math math.combinatorics prettyprint ; catalan ( n -- n ) [ 1 + recip ] [ 2 * ] [ nCk * ] tri ; 15 [ catalan . ] each-integer</lang> Output: 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440  The last method, memoized by using arrays. <lang factor>USING: kernel math prettyprint sequences ; next ( seq -- newseq )  [ ] [ last ] [ length ] tri [ 2 * 1 - 2 * ] [ 1 + ] bi / * suffix ;  Catalan ( n -- seq ) V{ 1 } swap 1 - [ next ] times ; 15 Catalan .</lang> Output: Similar to above. ## Fantom  This example is incorrect. Please fix the code and remove this message.Details: series starts 1, 1, 2, ... <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 ===============  ## FreeBASIC <lang 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 sum  End 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 "-", "-----", "------", "-----" Print For i As UInteger = 0 To 15  Print i, catalan1(i), catalan2(i), catalan3(i)  Next Print Print "Press any key to quit" Sleep</lang> 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. <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  ## Groovy <lang 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); } }  } </lang> Output: 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845  ## 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 :: [Integer] cats1 = (\n -> product [n + 2 .. 2 * n] div product [1 .. n]) <$> [0 ..]

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 Works with: Julia version 0.6 <lang julia>catalannum(n::Integer) = binomial(2n, n) ÷ (n + 1) @show catalannum.(1:15) @show catalannum(big(100))</lang> 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 <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 1.1.4 <lang scala>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() }  }</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 ## langur Translation of: Perl <lang langur>val .factorial = f if(.x < 2: 1; .x x self(.x - 1)) val .catalan = f(.n) .factorial(2 x .n) / .factorial(.n+1) / .factorial(.n) for .i in 0..15 {  writeln$"\.i:2;: \(.catalan(.i):10)"


}

writeln "10000: ", .catalan(10000)</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
10000: 224537812493385215633593584257360578701103586219365887773293713835854436588700534490998102719114320210209905393799589701149327326500953702713977513001838761306936534407802585494454599941773729984591764542782202886796997833276495496514760245912220654267091568311812071300891219894022165175451441066691435091975969499731921675488934120638046514134965974069039677192984714638704528752769863567952620334847707274529741976558104236293861846622622783294667505268651205024766408784881872997404042356319626323351089169906635603513309014645157443570842822082866699012415455339518777770781742052837799476906230350785959040487158118992753484022865373274100095762968510625236915280143408460651206678398725681703811505423791566261735329550627967717189932855983913468867794806585863794483869239933179341394259456515091026456652770409848702116046445406995085092488210998732255656992243441519938747425554228724734242623566663631968254490897214106655375215196762710825001305055093871863518797311135688370964194817463890187212845332422257193414201244344808864449873736345425670715824582633802476282521798739438044652622163657359012681653473214512797365047989922327391063907061792126264420963262176161781711086630089636828211837643128677915076724947168653050318426339007489738275045346257959685376480042860870398232333705506506342394485443047987642390287346746539674780326188825579548593281319807827279403944008553690033855132088140116099772393778770685018936338194366302053586633406848404622048675525765095697363909787189635178694239275237185046710057476484117945279786897787624602379494797322427251542758312638233073625857897083435831841717971137851874666094337671443717108457737153283641719103639784923520519013700030680553564442331411313831920775983175313709250333784211385811480015293165463406576311626295629412110652218717603537723650144357966952842696678735624157616428716812764985074925414219421312810089785108621126934245959900367104035334200067714905754827856122801987429837706493130435832752072139392743006620396370486473952500144779413596417260472218266529167783118015414918168260722824885550181735638670588682513610805160133611349864194033776132438535863120087679096358696928233598996870302136347936567444208209125300149683552369341937471817860835774359234009557030148123353114950735217736514617017504851011193104728986836180908987352236659629183725016607437110422583156042941955830763092095074443334625318588569114114087985404048889671202396824806275701581378689568449507132793603852731445602923990458926101180821029108808623323378547869169352237448925371763574346501610378415722137519019474474794069155118626291447578558908522430436148987521551911541787974276591708584289036595642180860178815462862735993859177180582760389253540408842580225467216988321950591728369194164290645992782274919561096308372635908842325870580231011459216934235078490764707633348336131667313582584404397290232519769625777374165187949140092779343812345117947306771376053099536367169631889642304360871187460737580808157222861127968703067542270175460553478533349238111434409526724363429611803844595968793121871649699680963646793415774160274520010905236593324062464542927011227158945796188186430711399250096518886617184049325827319276468018789191520522185358895653192882843061349706085770767046601045697944646638311930027354235643643713545212361580694059553720806659066661496416423676930095857438882302891350789287291844752601744462789158506243012088536936184422120232369244564444689340142897415432231452353338115944183447986470689449043710051589958391273681116292415738776171575775695905846247205522469202801517417551374761549677412720803623129527503286287755308576386461385928958587649159872019202866614901547860974883963007792442796064165417207167072370586790722366932349325253877744621251386864069101337572557790214048760202008337611577675840153696735860276810033694744314488435390547908483357054897387317002405793108554524629034558098886977538473481750772616164313845337139245688079995996839933620829828339492800825536599964878893947278408890351634126931068657027524005795713514365098086505030570362785115155293306343520969872400876180105031975302255898787642403303027682634969586730202117121076117629457710028105378124677420093990476071697970354661002217702623344454780740808459286778553016318604430682610618871098652904537323336381304469735192868285840882036271136058499391069436145426450229039329475974178236465920534171895204155964515055983303017823692138977622016292722019365841360360274557488926673754175222061483328914099598663902320310143583379354121664996173733086613692927391384486261610892314450463841637667054196985332620403539011932606618414419229492637564924726411270720189611019154677281846409387514072618176832310721327819277699943226895919915049652045449281057471199978267843961724883768772155477073354744908923995448752333726740642292872107500458349718026322755698226793850983280706045951407323891263270928264657562125955511946782954645656015480418543664557515041692091317941000997342935512311493290722434384401250133402934163457264794261787386862382738330195237770190998115114193014769006071380834085352290585937952429981509893303796306071520571655936820282768086579891336876000368502562579738337809071051261343359121744773055264455701014137255399929760233753812017596045145926791136761130783810840502248142803073720015451941006030172192834375431286154255159659778817089767964922549014569972777126726537787896968876337799235679125368824867754881036161730805613471278633981478858113141202728303435218970292775366288829203013873713349923690394124920402725698544786016048685431525811047414746045227535216327530901827040588505255466803793791888002231571686068617764292584075135236237044383334893874602177596602979234717936820827427229615827657960492946059695301906791494260652411424538532836730097985187522379068364429583532675896349363295120431429006688249818006722311568902288350452581968418068616818268667067741994472455501649753611708445979082338902214467454627107888156489438584617793175431865532382711812960546611287516640


## 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

## Logo

<lang logo>to factorial :n

 output ifelse [less? :n 1] 1 [product :n factorial difference :n 1]


end to choose :n :r

 output quotient factorial :n product factorial :r factorial difference :n :r


end to catalan :n

 output product (quotient sum :n 1) choose product 2 :n :n


end

repeat 15 [print catalan repcount]</lang>

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

## 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 = catalanNumber(n)

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


end</lang> The following version computes at the same time the n first Catalan numbers (including C0). <lang MATLAB>function n = catalanNumbers(n)

   n = [1 cumprod((2:4:4*n-6) ./ (2:n))];


end</lang>

Sample Output:

<lang MATLAB>>> 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</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> ## Modula-2 <lang modula2>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 r  END 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 END  END 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 END  END 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; ReadChar  END CatalanNumbers.</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 ## OCaml <lang 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  </lang> 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

<lang Oforth>: catalan( n -- m )

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

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. <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>

## Phix

See also Catalan_numbers/Pascal's_triangle#Phix which may be faster. <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

sequence res = repeat(repeat(0,4),16),

        times = repeat(0,3)


for t=1 to 4 do

   atom t0 = time()
for i=0 to 15 do
switch t do
case 1: res[i+1][2] = catalan1(i)
case 2: res[i+1][3] = catalan2(i)
case 3: res[i+1][4] = catalan3(i)
case 4: res[i+1][1] = i; printf(1,"%2d: %10d %10d %10d\n",res[i+1])
end switch
end for
if t=4 then exit end if
times[t] = elapsed(time()-t0)


end for printf(1,"times:%8s %10s %10s\n",times)</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
times:      0s       1.6s         0s


As expected, catalan2() is by far the slowest, so let's memoise that one!

### memoized recursive gmp version

Library: Phix/mpfr

<lang Phix>include builtins\mpfr.e

sequence c2cache = {}

function catalan2m(integer n) -- very fast! object r -- result (a [cached/shared] mpz)

        -- (nb: modifying result will mess up cache)
if n<=0 then return mpz_init(1) 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 = mpz_init(0)
mpz t = mpz_init()
for i=0 to n-1 do
mpz_mul(t,catalan2m(i),catalan2m(n-1-i))
end for
t = mpz_free(t)
c2cache[n] = r
return r


end function

sequence s = {} for i=0 to 15 do s = append(s,mpz_get_str(catalan2m(i))) end for printf(1,"0..15: %s\n",join(s,",")) printf(1,"100: %s\n",{mpz_get_str(catalan2m(100))})</lang>

Output:
0..15: 1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845
100: 896519947090131496687170070074100632420837521538745909320


## PHP

<lang php><?php

 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


<lang php> <?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";


} </lang>

Output:
1	2	5	14	42	132	429	1430	4862	16796	58786	208012	742900	2674440	9694845	35357670


## 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


### An Alternate Version

This version could easily be modified to work with big integers. <lang PowerShell> 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
}
}
}


} </lang> Get the first fifteen Catalan numbers as a PSCustomObject: <lang PowerShell> 0..14 | Get-CatalanNumber </lang>

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: <lang PowerShell> (0..14 | Get-CatalanNumber).CatalanNumber </lang>

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


## 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.

Works with: Python version 3

<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(0:15)

[1]       1       1       2       5      14      42     132     429    1430


[10] 4862 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)


## Raku

(formerly 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 .. *;

1. In both cases, the sixteen first values can be seen with:

.say for Catalan[^16];</lang>

Output:
1
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]

<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 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.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; $=0; if c.n\==. then return c.n  do k=0 for n;$=$+ Cat2(k) * Cat2(n-k-1); end c.n=$;           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 ## Ring <lang ring> for n = 1 to 15  see catalan(n) + nl  next func catalan n  if n = 0 return 1 ok cat = 2 * (2 * n - 1) * catalan(n - 1) / (n + 1) return cat  </lang> Output: 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845  ## 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 CatalanNumbers {  def main(args: Array[String]): Unit = { for (n <- 0 to 15) { println("catalan(" + n + ") = " + catalan(n)) } }   def catalan(n: BigInt): BigInt = factorial(2 * n) / (factorial(n + 1) * factorial(n))   def factorial(n: BigInt): BigInt = BigInt(1).to(n).foldLeft(BigInt(1))(_ * _)  } </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  ## Scheme Tail recursive implementation. <lang scheme>(define (catalan m)  (let loop ((c 1)(n 0)) (if (not (eqv? n m)) (begin (display n)(display ": ")(display c)(newline) (loop (* (/ (* 2 (- (* 2 (+ n 1)) 1)) (+ (+ n 1) 1)) c) (+ n 1) )))))  (catalan 15)</lang> Output: 0: 1 1: 1 2: 2 3: 5 4: 14 5: 42 6: 132 7: 429 8: 1430 9: 4862 10: 16796 11: 58786 12: 208012 13: 742900 14: 2674440 ## Seed7 <lang seed7>$ include "seed7_05.s7i";

 include "bigint.s7i";


const proc: main is func

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

Output:
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}\t#{c(i)}"


}</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


## smart BASIC

<lang qbasic>PRINT "Recursive:"!PRINT FOR n = 0 TO 15

   PRINT n,"#######":catrec(n)


NEXT n PRINT!PRINT

PRINT "Non-recursive:"!PRINT FOR 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 = temp


END DEF

DEF catnonrec(x)

   temp = 1
FOR n = 1 TO x
temp = (2*((2*n)-1))/(n+1)*temp
NEXT n
catnonrec = temp


END DEF</lang>

## 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>


## Stata

<lang stata>clear set obs 15 gen catalan=1 in 1 replace catalan=catalan[_n-1]*2*(2*_n-3)/_n in 2/l list, noobs noh</lang>

Output

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

## Swift

Translation of: Rust

<lang swift>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))")


} </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


## 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>

## Vala

<lang vala>namespace CatalanNumbers {

 public class CatalanNumberGenerator {
private static double factorial(double n) {
if (n == 0)
return 1;
return n * factorial(n - 1);
}

public double first_method(double n) {
const double top_multiplier = 2;
return factorial(top_multiplier * n) / (factorial(n + 1) * factorial(n));
}

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

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

 void main() {
CatalanNumberGenerator generator = new CatalanNumberGenerator();
DateTime initial_time;
DateTime final_time;
TimeSpan ts;

   stdout.printf("Direct Method\n");
stdout.printf(" n%9s\n", "C_n");
stdout.printf("............\n");
initial_time = new DateTime.now();
for (double i = 0; i <= 15; i++) {
stdout.printf("%2s %8s\n", i.to_string(), Math.ceil(generator.first_method(i)).to_string());
}
final_time = new DateTime.now();
ts = final_time.difference(initial_time);
stdout.printf("............\n");
stdout.printf("Time Elapsed: %s μs\n", ts.to_string());

   stdout.printf("\nRecursive Method 1\n");
stdout.printf(" n%9s\n", "C_n");
stdout.printf("............\n");
initial_time = new DateTime.now();
for (double i = 0; i <= 15; i++) {
stdout.printf("%2s %8s\n", i.to_string(), Math.ceil(generator.second_method(i)).to_string());
}
final_time = new DateTime.now();
ts = final_time.difference(initial_time);
stdout.printf("............\n");
stdout.printf("Time Elapsed: %s μs\n", ts.to_string());

   stdout.printf("\nRecursive Method 2\n");
stdout.printf(" n%9s\n", "C_n");
stdout.printf("............\n");
initial_time = new DateTime.now();
for (double i = 0; i <= 15; i++) {
stdout.printf("%2s %8s\n", i.to_string(), Math.ceil(generator.third_method(i)).to_string());
}
final_time = new DateTime.now();
ts = final_time.difference(initial_time);
stdout.printf("............\n");
stdout.printf("Time Elapsed: %s μs\n", ts.to_string());

 }


}</lang>

Output:
Direct Method
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
15  9694845
............
Time Elapsed: 132 μs

Recursive Method 1
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
15  9694845
............
Time Elapsed: 130430 μs

Recursive Method 2
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
15  9694845
............
Time Elapsed: 76 μs


## 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


## Visual Basic .NET

Translation of: C#

<lang vbnet>Module Module1

   Function Factorial(n As Double) As Double
If n < 1 Then
Return 1
End If

       Dim result = 1.0
For i = 1 To n
result = result * i
Next

       Return result
End Function

   Function FirstOption(n As Double) As Double
Return Factorial(2 * n) / (Factorial(n + 1) * Factorial(n))
End Function

   Function SecondOption(n As Double) As Double
If n = 0 Then
Return 1
End If

       Dim sum = 0
For i = 0 To n - 1
sum = sum + SecondOption(i) * SecondOption((n - 1) - i)
Next
Return sum
End Function

   Function ThirdOption(n As Double) As Double
If n = 0 Then
Return 1
End If

       Return ((2 * (2 * n - 1)) / (n + 1)) * ThirdOption(n - 1)
End Function

   Sub Main()
Const MaxCatalanNumber = 15

       Dim initial As DateTime
Dim final As DateTime
Dim ts As TimeSpan

       initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, FirstOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
Console.WriteLine()

       initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, SecondOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
Console.WriteLine()

       initial = DateTime.Now
For i = 0 To MaxCatalanNumber
Console.WriteLine("CatalanNumber({0}:{1})", i, ThirdOption(i))
Next
final = DateTime.Now
ts = final - initial
Console.WriteLine("It took {0}.{1} to execute", ts.Seconds, ts.Milliseconds)
End Sub


End Module</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 0.19 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 0.831 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 0.8 to execute

## 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>

## Wren

Library: Wren-fmt
Library: Wren-math

<lang ecmascript>import "/fmt" for Fmt import "/math" for Int

var catalan = Fn.new { |n|

   if (n < 0) Fiber.abort("Argument must be a non-negative integer")
var prod = 1
var i = n + 2
while (i <= n * 2) {
prod = prod * i
i = i + 1
}
return prod / Int.factorial(n)


}

var catalanRec catalanRec = Fn.new { |n| (n != 0) ? 2 * (2 * n - 1) * catalanRec.call(n - 1) / (n + 1) : 1 }

System.print(" n Catalan number") System.print("------------------") for (i in 0..15) System.print("%(Fmt.d(2, i))  %(catalan.call(i))") System.print("\nand again using a recursive function:\n") for (i in 0..15) System.print("%(Fmt.d(2, i))  %(catalanRec.call(i))")</lang>

Output:
 n  Catalan number
------------------
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

and again using a recursive function:

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


## XLISP

<lang lisp>(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)))</lang>

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

## 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


## ZX Spectrum Basic

Translation of: C

<lang zxbasic>10 FOR i=0 TO 15 20 LET n=i: LET m=2*n 30 LET r=1: LET d=m-n 40 IF d>n THEN LET n=d: LET d=m-n 50 IF m<=n THEN GO TO 90 60 LET r=r*m: LET m=m-1 70 IF (d>1) AND NOT FN m(r,d) THEN LET r=r/d: LET d=d-1: GO TO 70 80 GO TO 50 90 PRINT i;TAB 4;r/(1+n) 100 NEXT i 110 STOP 120 DEF FN m(a,b)=a-INT (a/b)*b: REM Modulus function </lang>