Catalan numbers/Pascal's triangle

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Task
Catalan numbers/Pascal's triangle
You are encouraged to solve this task according to the task description, using any language you may know.

The task is to print out the first 15 Catalan numbers by extracting them from Pascal's triangle, see Catalan Numbers and the Pascal Triangle.

C++

<lang cpp>// Generate Catalan Numbers // // Nigel Galloway: June 9th., 2012 //

  1. include <iostream>

int main() {

 const int N = 15;
 int t[N+2] = {0,1};
 for(int i = 1; i<=N; i++){
   for(int j = i; j>1; j--) t[j] = t[j] + t[j-1];
   t[i+1] = t[i];
   for(int j = i+1; j>1; j--) t[j] = t[j] + t[j-1];
   std::cout << t[i+1] - t[i] << " ";
 }
 return 0;

}</lang>

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

D

Translation of: C++

<lang d>void main() {

   import std.stdio;
   enum uint N = 15;
   uint[N + 2] t;
   t[1] = 1;
   foreach (immutable i; 1 .. N + 1) {
       foreach_reverse (immutable j; 2 .. i + 1)
           t[j] = t[j] + t[j - 1];
       t[i + 1] = t[i];
       foreach_reverse (immutable j; 2 .. i + 2)
           t[j] = t[j] + t[j - 1];
       write(t[i + 1] - t[i], ' ');
   }

}</lang>

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

MATLAB / Octave

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

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

Perl

Translation of: C++

<lang Perl>use constant N => 15; my @t = (0, 1); for(my $i = 1; $i <= N; $i++) {

   for(my $j = $i; $j > 1; $j--) { $t[$j] += $t[$j-1] }
   $t[$i+1] = $t[$i];
   for(my $j = $i+1; $j>1; $j--) { $t[$j] += $t[$j-1] }
   print $t[$i+1] - $t[$i], " ";

}</lang>

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

Perl 6

<lang perl6>constant @pascal = [1], -> @p { [0, @p Z+ @p, 0] } ... *;

constant @catalan = gather for 2, 4 ... * -> $ix {

   my @row := @pascal[$ix];
   my $mid = +@row div 2;
   take [-] @row[$mid, $mid+1]

}

.say for @catalan[^20];</lang>

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

Python

Translation of: C++

<lang python>>>> n = 15 >>> t = [0] * (n + 2) >>> t[1] = 1 >>> for i in range(1, n + 1): for j in range(i, 1, -1): t[j] += t[j - 1] t[i + 1] = t[i] for j in range(i + 1, 1, -1): t[j] += t[j - 1] print(t[i+1] - t[i], end=' ')


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

Racket

<lang Racket>

  1. lang racket

(define (next-half-row r)

 (define r1 (for/list ([x r] [y (cdr r)]) (+ x y)))
 `(,(* 2 (car r1)) ,@(for/list ([x r1] [y (cdr r1)]) (+ x y)) 1 0))

(let loop ([n 15] [r '(1 0)])

 (cons (- (car r) (cadr r))
       (if (zero? n) '() (loop (sub1 n) (next-half-row r)))))
-> '(1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900
2674440 9694845)

</lang>

REXX

<lang rexx>/*REXX program obtains Catalan numbers from Pascal's triangle. */ numeric digits 200 /*might have large Catalan nums. */ parse arg N .; if N== then N=15 /*Any args? No, then use default*/ @.=0; @.1=1 /*stem array default, 1st value. */

 do i=1  for N;                         ip=i+1
                do j=i  by -1  for N;   jm=j-1;   @.j=@.j+@.jm;  end /*j*
 @.ip=@.i;      do k=ip by -1  for N;   km=k-1;   @.k=@.k+@.km;  end /*k*
 say @.ip-@.i
 end   /*i*
                                      /*stick a fork in it, we're done.*/</lang>

output when using the default input:

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

Run BASIC

<lang runbasic>n = 15 dim t(n+2) t(1) = 1 for i = 1 to n

 for  j = i to 1 step -1  : t(j) = t(j) + t(j-1): next j
 t(i+1) = t(i)
 for  j = i+1 to 1 step -1: t(j) = t(j) + t(j-1 : next j

print t(i+1) - t(i);" "; next i</lang>

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

Ruby

<lang tcl>def catalan(num)

 t = [0, 1] #grows as needed
 1.upto(num).map do |i|
   i.downto(1){|j| t[j] += t[j-1]}
   t[i+1] = t[i]
   (i+1).downto(1) {|j| t[j] += t[j-1]}
   t[i+1] - t[i]
 end

end puts catalan(15).join(", ")</lang>

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

Tcl

<lang tcl>proc catalan n {

   set result {}
   array set t {0 0 1 1}
   for {set i 1} {[set k $i] <= $n} {incr i} {

for {set j $i} {$j > 1} {} {incr t($j) $t([incr j -1])} set t([incr k]) $t($i) for {set j $k} {$j > 1} {} {incr t($j) $t([incr j -1])} lappend result [expr {$t($k) - $t($i)}]

   }
   return $result

}

puts [catalan 15]</lang>

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