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This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
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{{task}}
;Task: Print out the first '''15''' Catalan numbers by extracting them from Pascal's triangle.
;See:
- [https://archive.is/0IrNp Catalan Numbers and the Pascal Triangle]. This method enables calculation of Catalan Numbers using only addition and subtraction.
- [http://mathworld.wolfram.com/CatalansTriangle.html Catalan's Triangle] for a Number Triangle that generates Catalan Numbers using only addition.
- Sequence [http://oeis.org/A000108 A000108] on OEIS has a lot of information on Catalan Numbers.
;Related Tasks: [[Pascal's triangle]]
11l
{{trans|Python}}
V n = 15
V t = [0] * (n + 2)
t[1] = 1
L(i) 1 .. n
L(j) (i .< 1).step(-1)
t[j] += t[j - 1]
t[i + 1] = t[i]
L(j) (i + 1 .< 1).step(-1)
t[j] += t[j - 1]
print(t[i + 1] - t[i], end' ‘ ’)
{{out}}
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
360 Assembly
For maximum compatibility, this program uses only the basic instruction set.
CATALAN CSECT
USING CATALAN,R13,R12
SAVEAREA B STM-SAVEAREA(R15)
DC 17F'0'
DC CL8'CATALAN'
STM STM R14,R12,12(R13)
ST R13,4(R15)
ST R15,8(R13)
LR R13,R15
LA R12,4095(R13)
LA R12,1(R12)
* ---- CODE
LA R0,1
ST R0,T t(1)=1
LA R4,0 ix:i=1
LA R6,1 by 1
LH R7,N to n
LOOPI BXH R4,R6,ENDLOOPI loop i
LR R5,R4 ix:j=i+1
LA R5,2(R5) i+2
LA R8,0
BCTR R8,0 by -1
LA R9,1 to 2
LOOP1J BXLE R5,R8,ENLOOP1J loop j
LR R10,R5 j
BCTR R10,0
SLA R10,2
L R2,T(R10) r2=t(j)
LR R1,R10 j
SH R1,=H'4'
L R3,T(R1) r3=t(j-1)
AR R2,R3 r2=r2+r3
ST R2,T(R10) t(j)=t(j)+t(j-1)
B LOOP1J
ENLOOP1J EQU *
LR R1,R4 i
BCTR R1,0
SLA R1,2
L R3,T(R1) t(i)
LA R1,4(R1)
ST R3,T(R1) t(i+1)
LR R5,R4 ix:j=i+2
LA R5,3(R5) i+3
LA R8,0
BCTR R8,0 by -1
LA R9,1 to 2
LOOP2J BXLE R5,R8,ENLOOP2J loop j
LR R10,R5 j
BCTR R10,0
SLA R10,2
L R2,T(R10) r2=t(j)
LR R1,R10 j
SH R1,=H'4'
L R3,T(R1) r3=t(j-1)
AR R2,R3 r2=r2+r3
ST R2,T(R10) t(j)=t(j)+t(j-1)
B LOOP2J
ENLOOP2J EQU *
LR R1,R4 i
BCTR R1,0
SLA R1,2
L R2,T(R1) t(i)
LA R1,4(R1)
L R3,T(R1) t(i+1)
SR R3,R2
CVD R3,P
UNPK Z,P
MVC C,Z
OI C+L'C-1,X'F0'
MVC WTOBUF(8),C+8
WTO MF=(E,WTOMSG)
B LOOPI
ENDLOOPI EQU *
* ---- END CODE
CNOP 0,4
L R13,4(0,R13)
LM R14,R12,12(R13)
XR R15,R15
BR R14
* ---- DATA
N DC H'15'
T DC 17F'0'
P DS PL8
Z DS ZL16
C DS CL16
WTOMSG DS 0F
DC H'80'
DC H'0'
WTOBUF DC CL80' '
YREGS
END
{{out}}
00000001
00000002
00000005
00000014
00000042
00000132
00000429
00001430
00004862
00016796
00058786
00208012
00742900
02674440
09694845
```
## Ada
Uses package Pascal from the Pascal triangle solution[[http://rosettacode.org/wiki/Pascal%27s_triangle#Ada]]
```Ada
with Ada.Text_IO, Pascal;
procedure Catalan is
Last: Positive := 15;
Row: Pascal.Row := Pascal.First_Row(2*Last+1);
begin
for I in 1 .. Last loop
Row := Pascal.Next_Row(Row);
Row := Pascal.Next_Row(Row);
Ada.Text_IO.Put(Integer'Image(Row(I+1)-Row(I+2)));
end loop;
end Catalan;
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## ALGOL 68
{{trans|C++}}
```algol68
INT n = 15;
[ 0 : n + 1 ]INT t;
t[0] := 0;
t[1] := 1;
FOR i TO n DO
FOR j FROM i BY -1 TO 2 DO t[j] := t[j] + t[j-1] OD;
t[i+1] := t[i];
FOR j FROM i+1 BY -1 TO 2 DO t[j] := t[j] + t[j-1] OD;
print( ( whole( t[i+1] - t[i], 0 ), " " ) )
OD
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## ALGOL W
```algolw
begin
% print the first 15 Catalan numbers from Pascal's triangle %
integer n;
n := 15;
begin
integer array pascalLine ( 1 :: n + 1 );
% the Catalan numbers are the differences between the middle and middle - 1 numbers of the odd %
% lines of Pascal's triangle (lines with 3 or more numbers) %
% note - we only need to calculate the left side of the triangle %
pascalLine( 1 ) := 1;
for c := 2 until n + 1 do begin
% even line %
for i := c - 1 step -1 until 2 do pascalLine( i ) := pascalLine( i - 1 ) + pascalLine( i );
pascalLine( c ) := pascalLine( c - 1 );
% odd line %
for i := c step -1 until 2 do pascalLine( i ) := pascalLine( i - 1 ) + pascalLine( i );
writeon( i_w := 1, s_w := 0, " ", pascalLine( c ) - pascalLine( c - 1 ) )
end for_c
end
end.
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## APL
```apl
⍝ Based heavily on the J solution
CATALAN←{¯1↓↑-/1 ¯1↓¨(⊂⎕IO+0 0)⍉¨0 2⌽¨⊂(⎕IO-⍨⍳N){+\⍣⍺⊢⍵}⍤0 1⊢1⍴⍨N←⍵+2}
```
{{out}}
```txt
CATALAN 15
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## AutoHotkey
{{works with|AutoHotkey_L}}
```AutoHotkey
/* Generate Catalan Numbers
//
// smgs: 20th Feb, 2014
*/
Array := [], Array[2,1] := Array[2,2] := 1 ; Array inititated and 2nd row of pascal's triangle assigned
INI := 3 ; starts with calculating the 3rd row and as such the value
Loop, 31 ; every odd row is taken for calculating catalan number as such to obtain 15 we need 2n+1
{
if ( A_index > 2 )
{
Loop, % A_INDEX
{
old := ini-1, index := A_index, index_1 := A_index + 1
Array[ini, index_1] := Array[old, index] + Array[old, index_1]
Array[ini, 1] := Array[ini, ini] := 1
line .= Array[ini, A_index] " "
}
;~ MsgBox % line ; gives rows of pascal's triangle
; calculating every odd row starting from 1st so as to obtain catalan's numbers
if ( mod(ini,2) != 0)
{
StringSplit, res, line, %A_Space%
ans := res0//2, ans_1 := ans++
result := result . res%ans_1% - res%ans% " "
}
line :=
ini++
}
}
MsgBox % result
```
{{out|Produces}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## AWK
```AWK
# syntax: GAWK -f CATALAN_NUMBERS_PASCALS_TRIANGLE.AWK
# converted from C
BEGIN {
printf("1")
for (n=2; n<=15; n++) {
num = den = 1
for (k=2; k<=n; k++) {
num *= (n + k)
den *= k
catalan = num / den
}
printf(" %d",catalan)
}
printf("\n")
exit(0)
}
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## Batch File
```dos
@echo off
setlocal ENABLEDELAYEDEXPANSION
set n=15
set /A nn=n+1
for /L %%i in (0,1,%nn%) do set t.%%i=0
set t.1=1
for /L %%i in (1,1,%n%) do (
set /A ip=%%i+1
for /L %%j in (%%i,-1,1) do (
set /A jm=%%j-1
set /A t.%%j=t.%%j+t.!jm!
)
set /A t.!ip!=t.%%i
for /L %%j in (!ip!,-1,1) do (
set /A jm=%%j-1
set /A t.%%j=t.%%j+t.!jm!
)
set /A ci=t.!ip!-t.%%i
echo !ci!
)
)
pause
```
{{Out}}
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
```
## C
```c
//This code implements the print of 15 first Catalan's Numbers
//Formula used:
// __n__
// | | (n + k) / k n>0
// k=2
#include
#include
//the number of Catalan's Numbers to be printed
const int N = 15;
int main()
{
//loop variables (in registers)
register int k, n;
//necessarily ull for reach big values
unsigned long long int num, den;
//the nmmber
int catalan;
//the first is not calculated for the formula
printf("1 ");
//iterating from 2 to 15
for (n=2; n<=N; ++n) {
//initializaing for products
num = den = 1;
//applying the formula
for (k=2; k<=n; ++k) {
num *= (n+k);
den *= k;
catalan = num /den;
}
//output
printf("%d ", catalan);
}
//the end
printf("\n");
return 0;
}
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## C++
```cpp
// Generate Catalan Numbers
//
// Nigel Galloway: June 9th., 2012
//
#include
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;
}
```
{{out|Produces}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## C#
{{trans|C++}}
```c#
int n = 15;
List t = new List() { 0, 1 };
for (int i = 1; i <= n; i++)
{
for (var j = i; j > 1; j--) t[j] += t[j - 1];
t.Add(t[i]);
for (var j = i + 1; j > 1; j--) t[j] += t[j - 1];
Console.Write(((i == 1) ? "" : ", ") + (t[i + 1] - t[i]));
}
```
{{out|Produces}}
```txt
1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845
```
## Common Lisp
```Lisp
(defun catalan (n)
"Return the n-th Catalan number"
(if (<= n 1) 1
(let ((result 2))
(dotimes (k (- n 2) result)
(setq result (* result (/ (+ n k 2) (+ k 2)))) ))))
(dotimes (n 15)
(print (catalan (1+ n))) )
```
{{out}}
```txt
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
```
## D
{{trans|C++}}
```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 - 1];
t[i + 1] = t[i];
foreach_reverse (immutable j; 2 .. i + 2)
t[j] += t[j - 1];
write(t[i + 1] - t[i], ' ');
}
}
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## EchoLisp
```scheme
(define dim 100)
(define-syntax-rule (Tidx i j) (+ i (* dim j)))
;; generates Catalan's triangle
;; T (i , j) = T(i-1,j) + T (i, j-1)
(define (T n)
(define i (modulo n dim))
(define j (quotient n dim))
(cond
((zero? i) 1) ;; left column = 1
((= i j) (T (Tidx (1- i) j))) ;; diagonal value = left value
(else (+ (T (Tidx (1- i) j)) (T (Tidx i (1- j)))))))
(remember 'T #(1))
```
{{out}}
```scheme
;; take elements on diagonal = Catalan numbers
(for ((i (in-range 0 16))) (write (T (Tidx i i))))
→ 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## Elixir
```elixir
defmodule Catalan do
def numbers(num) do
{result,_} = Enum.reduce(1..num, {[],{0,1}}, fn i,{list,t0} ->
t1 = numbers(i, t0)
t2 = numbers(i+1, Tuple.insert_at(t1, i+1, elem(t1, i)))
{[elem(t2, i+1) - elem(t2, i) | list], t2}
end)
Enum.reverse(result)
end
defp numbers(0, t), do: t
defp numbers(n, t), do: numbers(n-1, put_elem(t, n, elem(t, n-1) + elem(t, n)))
end
IO.inspect Catalan.numbers(15)
```
{{out}}
```txt
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
```
## Erlang
```erlang
-module(catalin).
-compile(export_all).
mul(N,D,S,S)->
N2=N*(S+S),
D2=D*S,
K = N2 div D2 ;
mul(N,D,S,L)->
N2=N*(S+L),
D2=D*L,
K = mul(N2,D2,S,L+1).
catl(Ans,16) -> Ans;
catl(D,S)->
C=mul(1,1,S,2),
catl([D|C],S+1).
main()->
Ans=catl(1,2).
```
## ERRE
```ERRE
PROGRAM CATALAN
!$DOUBLE
DIM CATALAN[50]
FUNCTION ODD(X)
ODD=FRC(X/2)<>0
END FUNCTION
PROCEDURE GETCATALAN(L)
LOCAL J,K,W
LOCAL DIM PASTRI[100]
L=L*2
PASTRI[0]=1
J=0
WHILE J15) then
printf ", "
```
## Factor
```factor
USING: arrays grouping io kernel math prettyprint sequences ;
IN: rosetta-code.catalan-pascal
: next-row ( seq -- seq' )
2 clump [ sum ] map 1 prefix 1 suffix ;
: pascal ( n -- seq )
1 - { { 1 } } swap [ dup last next-row suffix ] times ;
15 2 * pascal [ length odd? ] filter [
dup length 1 = [ 1 ]
[ dup midpoint@ dup 1 + 2array swap nths first2 - ] if
pprint bl
] each drop
```
{{out}}
```txt
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440
```
## FreeBASIC
```freebasic
' version 15-09-2015
' compile with: fbc -s console
#Define size 31 ' (N * 2 + 1)
Sub pascal_triangle(rows As Integer, Pas_tri() As ULongInt)
Dim As Integer x, y
For x = 1 To rows
Pas_tri(1,x) = 1
Pas_tri(x,1) = 1
Next
For x = 2 To rows
For y = 2 To rows + 1 - x
Pas_tri(x, y) = pas_tri(x - 1 , y) + pas_tri(x, y - 1)
Next
Next
End Sub
' ------=< MAIN >=------
Dim As Integer count, row
Dim As ULongInt triangle(1 To size, 1 To size)
pascal_triangle(size, triangle())
' 1 1 1 1 1 1
' 1 2 3 4 5 6
' 1 3 6 10 15 21
' 1 4 10 20 35 56
' 1 5 15 35 70 126
' 1 6 21 56 126 252
' The Pascal triangle is rotated 45 deg.
' to find the Catalan number we need to follow the diagonal
' for top left to bottom right
' take the number on diagonal and subtract the number in de cell
' one up and one to right
' 1 (2 - 1), 2 (6 - 4), 5 (20 - 15) ...
Print "The first 15 Catalan numbers are" : print
count = 1 : row = 2
Do
Print Using "###: #########"; count; triangle(row, row) - triangle(row +1, row -1)
row = row + 1
count = count + 1
Loop Until count > 15
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
```
{{out}}
```txt
The first 15 Catalan numbers are
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
```
## Go
{{trans|C++}}
```go
package main
import "fmt"
func main() {
const n = 15
t := [n + 2]uint64{0, 1}
for i := 1; i <= n; i++ {
for j := i; j > 1; j-- {
t[j] += t[j-1]
}
t[i+1] = t[i]
for j := i + 1; j > 1; j-- {
t[j] += t[j-1]
}
fmt.Printf("%2d : %d\n", i, t[i+1]-t[i])
}
}
```
{{out}}
```txt
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
```
## Groovy
{{trans|C}}
```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;
}
print(" " + catalan);
}
}
}
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## Haskell
As required by the task this implementation extracts the Catalan numbers from Pascal's triangle, rather
than calculating them directly. Also, note that it (correctly) produces [1, 1] as the first two numbers.
```haskell
import System.Environment (getArgs)
-- Pascal's triangle.
pascal :: [[Integer]]
pascal = [1] : map (\row -> 1 : zipWith (+) row (tail row) ++ [1]) pascal
-- The Catalan numbers from Pascal's triangle. This uses a method from
-- http://www.cut-the-knot.org/arithmetic/algebra/CatalanInPascal.shtml
-- (see "Grimaldi").
catalan :: [Integer]
catalan = map (diff . uncurry drop) $ zip [0..] (alt pascal)
where alt (x:_:zs) = x : alt zs -- every other element of an infinite list
diff (x:y:_) = x - y
diff (x:_) = x
main :: IO ()
main = do
ns <- fmap (map read) getArgs :: IO [Int]
mapM_ (print . flip take catalan) ns
```
{{out}}
```txt
./catalan 15
[1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
```
=={{header|Icon}} and {{header|Unicon}}==
The following works in both languages. It avoids computing elements in Pascal's triangle
that aren't used.
```unicon
link math
procedure main(A)
limit := (integer(A[1])|15)+1
every write(right(binocoef(i := 2*seq(0)\limit,i/2)-binocoef(i,i/2+1),30))
end
```
Sample run:
```txt
->cn
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
->
```
## J
```j
Catalan=. }:@:(}.@:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)
```
{{out|Example use}}
```j
Catalan 15
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
A structured derivation of Catalan follows:
```j
o=. @: NB. Composition of verbs (functions)
( PascalTriangle=. i. ((+/\@]^:[)) #&1 ) 5
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 35
1 5 15 35 70
( MiddleDiagonal=. (<0 1)&|: ) o PascalTriangle 5
1 2 6 20 70
( AdjacentLeft=. MiddleDiagonal o (2&|.) ) o PascalTriangle 5
1 4 15 1 5
( Catalan=. }: o (}. o MiddleDiagonal - }: o AdjacentLeft) o PascalTriangle o (2&+) f. ) 5
1 2 5 14 42
Catalan
}:@:(}.@:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)
```
## Java
{{trans|C++}}
```java
public class Test {
public static void main(String[] args) {
int N = 15;
int[] t = new int[N + 2];
t[1] = 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];
System.out.printf("%d ", t[i + 1] - t[i]);
}
}
}
```
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## JavaScript
### ES5
Iteration
{{trans|C++}}
```javascript
var n = 15;
for (var t = [0, 1], i = 1; i <= n; i++) {
for (var j = i; j > 1; j--) t[j] += t[j - 1];
t[i + 1] = t[i];
for (var j = i + 1; j > 1; j--) t[j] += t[j - 1];
document.write(i == 1 ? '' : ', ', t[i + 1] - t[i]);
}
```
{{out}}
```txt
1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845
```
### ES6
Functional composition
{{Trans|Haskell}}
```JavaScript
(() => {
'use strict';
// CATALAN
// catalanSeries :: Int -> [Int]
let catalanSeries = n => {
let alternate = xs => xs.reduce(
(a, x, i) => i % 2 === 0 ? a.concat([x]) : a, []
),
diff = xs => xs.length > 1 ? xs[0] - xs[1] : xs[0];
return alternate(pascal(n * 2))
.map((xs, i) => diff(drop(i, xs)));
}
// PASCAL
// pascal :: Int -> [[Int]]
let pascal = n => until(
m => m.level <= 1,
m => {
let nxt = zipWith(
(a, b) => a + b, [0].concat(m.row), m.row.concat(0)
);
return {
row: nxt,
triangle: m.triangle.concat([nxt]),
level: m.level - 1
}
}, {
level: n,
row: [1],
triangle: [
[1]
]
}
)
.triangle;
// GENERIC FUNCTIONS
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
let zipWith = (f, xs, ys) =>
xs.length === ys.length ? (
xs.map((x, i) => f(x, ys[i]))
) : undefined;
// until :: (a -> Bool) -> (a -> a) -> a -> a
let until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
}
// drop :: Int -> [a] -> [a]
let drop = (n, xs) => xs.slice(n);
// tail :: [a] -> [a]
let tail = xs => xs.length ? xs.slice(1) : undefined;
return tail(catalanSeries(16));
})();
```
{{Out}}
```JavaScript
[1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845]
```
## jq
The first identity (C(2n,n) - C(2n, n-1)) given in the reference is used in accordance with the task description, but it would of course be more efficient to factor out C(2n,n) and use the expression C(2n,n)/(n+1). See also [[Catalan_numbers#jq]] for other alternatives.
''Warning'': jq uses IEEE 754 64-bit arithmetic,
so the algorithm used here for Catalan numbers loses precision for n > 30 and fails completely for n > 510.
```jq
def binomial(n; k):
if k > n / 2 then binomial(n; n-k)
else reduce range(1; k+1) as $i (1; . * (n - $i + 1) / $i)
end;
# Direct (naive) computation using two numbers in Pascal's triangle:
def catalan_by_pascal: . as $n | binomial(2*$n; $n) - binomial(2*$n; $n-1);
```
'''Example''':
(range(0;16), 30, 31, 510, 511) | [., catalan_by_pascal]
{{Out}}
```sh
$ jq -n -c -f Catalan_numbers_Pascal.jq
[0,0]
[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]
[30,3814986502092304]
[31,14544636039226880]
[510,5.491717746183512e+302]
[511,null]
```
## Julia
{{trans|Matlab}}
```julia
# v0.6
function pascal(n::Int)
r = ones(Int, n, n)
for i in 2:n, j in 2:n
r[i, j] = r[i-1, j] + r[i, j-1]
end
return r
end
function catalan_num(n::Int)
p = pascal(n + 2)
p[n+4:n+3:end-1] - diag(p, 2)
end
@show catalan_num(15)
```
{{out}}
```txt
catalan_num(15) = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
```
## Kotlin
```scala
// version 1.1.2
import java.math.BigInteger
val ONE = BigInteger.ONE
fun pascal(n: Int, k: Int): BigInteger {
if (n == 0 || k == 0) return ONE
val num = (k + 1..n).fold(ONE) { acc, i -> acc * BigInteger.valueOf(i.toLong()) }
val den = (2..n - k).fold(ONE) { acc, i -> acc * BigInteger.valueOf(i.toLong()) }
return num / den
}
fun catalanFromPascal(n: Int) {
for (i in 1 until n step 2) {
val mi = i / 2 + 1
val catalan = pascal(i, mi) - pascal(i, mi - 2)
println("${"%2d".format(mi)} : $catalan")
}
}
fun main(args: Array) {
val n = 15
catalanFromPascal(n * 2)
}
```
{{out}}
```txt
1 : 1
2 : 2
3 : 5
4 : 14
5 : 42
6 : 132
7 : 429
8 : 1430
9 : 4862
10 : 16796
11 : 58786
12 : 208012
13 : 742900
14 : 2674440
15 : 9694845
```
## Lua
For each line of odd-numbered length from Pascal's triangle, print the middle number minus the one immediately to its right.
This solution is heavily based on the Lua code to generate Pascal's triangle from the page for that task.
```Lua
function nextrow (t)
local ret = {}
t[0], t[#t + 1] = 0, 0
for i = 1, #t do ret[i] = t[i - 1] + t[i] end
return ret
end
function catalans (n)
local t, middle = {1}
for i = 1, n do
middle = math.ceil(#t / 2)
io.write(t[middle] - (t[middle + 1] or 0) .. " ")
t = nextrow(nextrow(t))
end
end
catalans(15)
```
{{out}}
```txt
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440
```
## M2000 Interpreter
{{trans|FreeBasic}}
We have to add -1 in For x=2 to rows, because in FreeBasic when x=rows then inner loop never happen because end value for y is 1, so lower than start value 2. In M2000 this should run from 2 to 1, so we have to exclude this situation from outer loop, adding -1, and before loop we have to include en exit from sub if rows are less than 2.
We can define integer variables (16 bit), and we can use integer literals numbers using % as last char.
Inside triangle array we use decimal numbers, using @ for first literals, so all additions next produce decimals too.
We use & to pass by reference, here anarray, to sub, but because a sub can see anything in module we can change array name inside sub to same as triangle and we can remove arguments (including size).
```M2000 Interpreter
Module CatalanNumbers {
Def Integer count, t_row, size=31
Dim triangle(1 to size, 1 to size)
\\ call sub
pascal_triangle(size, &triangle())
Print "The first 15 Catalan numbers are"
count = 1% : t_row = 2%
Do {
Print Format$("{0:0:-3}:{1:0:-15}", count, triangle(t_row, t_row) - triangle(t_row +1, t_row -1))
t_row++
count++
} Until count > 15
End
Sub pascal_triangle(rows As Integer, &Pas_tri())
Local x=0%, y=0%
For x = 1 To rows
Pas_tri( 1%, x ) = 1@
Pas_tri( x, 1% ) = 1@
Next x
if rows<2 then exit sub
For x = 2 To rows-1
For y = 2 To rows + 1 - x
Pas_tri(x, y) = pas_tri(x - 1 , y) + pas_tri(x, y - 1)
Next y
Next x
End Sub
}
CatalanNumbers
```
{{out}}
```txt
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
```
=={{header|Mathematica}} / {{header|Wolfram Language}}==
This builds the entire Pascal triangle that's needed and holds it in memory. Very inefficienct, but seems to be what is asked in the problem.
```Mathematica
nextrow[lastrow_] := Module[{output},
output = ConstantArray[1, Length[lastrow] + 1];
Do[
output[[i + 1]] = lastrow[[i]] + lastrow[[i + 1]];
, {i, 1, Length[lastrow] - 1}];
output
]
pascaltriangle[size_] := NestList[nextrow, {1}, size]
catalannumbers[length_] := Module[{output, basetriangle},
basetriangle = pascaltriangle[2 length];
list1 = basetriangle[[# *2 + 1, # + 1]] & /@ Range[length];
list2 = basetriangle[[# *2 + 1, # + 2]] & /@ Range[length];
list1 - list2
]
(* testing *)
catalannumbers[15]
```
{{out}}
```txt
{1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845}
```
=={{header|MATLAB}} / {{header|Octave}}==
```MATLAB
n = 15;
p = pascal(n + 2);
p(n + 4 : n + 3 : end - 1)' - diag(p, 2)
```
{{Out}}
```txt
ans =
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
```
## Nim
{{trans|Python}}
```nim
const n = 15
var t = newSeq[int](n + 2)
t[1] = 1
for i in 1..n:
for j in countdown(i, 1): t[j] += t[j-1]
t[i+1] = t[i]
for j in countdown(i+1, 1): t[j] += t[j-1]
stdout.write t[i+1] - t[i], " "
```
{{Out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## OCaml
```ocaml
let catalan : int ref = ref 0 in
Printf.printf "%d ," 1 ;
for i = 2 to 9 do
let nm : int ref = ref 1 in
let den : int ref = ref 1 in
for k = 2 to i do
nm := (!nm)*(i+k);
den := (!den)*k;
catalan := (!nm)/(!den) ;
done;
print_int (!catalan); print_string "," ;
done;;
```
{{out}}
```txt
OUTPUT:
1 ,2,5,14,42,132,429,1430,4862
```
## Oforth
```Oforth
import: mapping
: pascal( n -- [] )
[ 1 ] n #[ dup [ 0 ] + [ 0 ] rot + zipWith( #+ ) ] times ;
: catalan( n -- m )
n 2 * pascal at( n 1+ ) n 1+ / ;
```
{{out}}
```txt
>#catalan 15 seq map .
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
```
## PARI/GP
```parigp
vector(15,n,binomial(2*n,n)-binomial(2*n,n+1))
```
{{out}}
```txt
%1 = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
```
## Pascal
```pascal
type
tElement = Uint64;
var
Catalan : array[0..50] of tElement;
procedure GetCatalan(L:longint);
var
PasTri : array[0..100] of tElement;
j,k: longInt;
begin
l := l*2;
PasTri[0] := 1;
j := 0;
while (juse 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], " ";
}
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
After the 28th Catalan number, this overflows 64-bit integers. We could add use bigint; use Math::GMP ":constant"; to make it work, albeit not at a fast pace. However we can use a module to do it much faster:
{{libheader|ntheory}}
```Perl
use ntheory qw/binomial/;
print join(" ", map { binomial( 2*$_, $_) / ($_+1) } 1 .. 1000), "\n";
```
The Math::Pari module also has a binomial, but isn't as fast and overflows its stack after 3400.
## Perl 6
{{works with|Rakudo|2015.12}}
```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];
```
{{out}}
```txt
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
35357670
129644790
477638700
1767263190
6564120420
```
## Phix
Calculates the minimum pascal triangle in minimum memory. Inspired by the comments in, but not the code of the FreeBasic example
```Phix
constant N = 15 -- accurate to 30, nan/inf for anything over 514 (bigatom version is below).
sequence catalan = {}, -- (>=1 only)
p = repeat(1,N+1)
atom p1
for i=1 to N do
p1 = p[1]*2
catalan = append(catalan,p1-p[2])
for j=1 to N-i+1 do
p1 += p[j+1]
p[j] = p1
end for
-- ?p[1..N-i+1]
end for
?catalan
```
{{out}}
```txt
{1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845}
```
Explanatory comments to accompany the above
```Phix
-- FreeBASIC said:
--' 1 1 1 1 1 1
--' 1 2 3 4 5 6
--' 1 3 6 10 15 21
--' 1 4 10 20 35 56
--' 1 5 15 35 70 126
--' 1 6 21 56 126 252
--' The Pascal triangle is rotated 45 deg.
--' to find the Catalan number we need to follow the diagonal
--' for top left to bottom right
--' take the number on diagonal and subtract the number in de cell
--' one up and one to right
--' 1 (2 - 1), 2 (6 - 4), 5 (20 - 15) ...
--
-- The first thing that struck me was it is twice as big as it needs to be,
-- something like this would do...
-- 1 1 1 1 1 1
-- 2 3 4 5 6
-- 6 10 15 21
-- 20 35 56
-- 70 126
-- 252
-- It is more obvious from the upper square that the diagonal on that, which is
-- that same as column 1 on this, is twice the previous, which on the second
-- diagram is in column 2. Further, once we have calculated the value for column
-- one above, we can use it immediately to calculate the next catalan number and
-- do not need to store it. Lastly we can overwrite row 1 with row 2 etc in situ,
-- and the following shows what we need for subsequent rounds:
-- 1 1 1 1 1
-- 3 4 5 6
-- 10 15 21
-- 35 56
-- 126 (unused)
```
### gmp version
{{libheader|mpfr}}
```Phix
include builtins\mpfr.e
function catalanB(integer n) -- very very fast!
sequence catalan = mpz_inits(n),
p = mpz_inits(n+1,1)
mpz p1 = mpz_init(1)
if n=0 then return p1 end if
for i=1 to n do
mpz_mul_si(p1,p[1],2)
mpz_sub(catalan[i],p1,p[2])
for j=1 to n-i+1 do
mpz_add(p1,p1,p[j+1])
mpz_set(p[j],p1)
end for
end for
return catalan[n]
end function
printf(1,"%d: %s (%s)\n",{100,mpz_get_str(catalanB(100))})
printf(1,"%d: %s (%s)\n",{250,mpz_get_str(catalanB(250))})
```
{{out}}
```txt
100: 896519947090131496687170070074100632420837521538745909320
250: 465116795969233796497747947259667807407291160080922096111953326525143875193659257831340309862635877995262413955019878805418475969029457769094808256
```
The above is significantly faster than the equivalent(s) on [[Catalan_numbers#Phix|Catalan_numbers]],
a quick comparison showing the latter getting exponentially worse (then again I memoised the slowest recursive version):
```txt
800 2000 4000 8000
catalanB: 0.6s 3.5s 14.5s 64s
catalan2m: 0.7s 7.0s 64.9s 644s
```
## PicoLisp
```PicoLisp
(de bino (N K)
(let f
'((N)
(if (=0 N) 1 (apply * (range 1 N))) )
(/
(f N)
(* (f (- N K)) (f K)) ) ) )
(for N 15
(println
(-
(bino (* 2 N) N)
(bino (* 2 N) (inc N)) ) ) )
(bye)
```
## PureBasic
{{trans|C}}
```PureBasic
#MAXNUM = 15
Declare catalan()
If OpenConsole("Catalan numbers")
catalan()
Input()
End 0
Else
End -1
EndIf
Procedure catalan()
Define k.i, n.i, num.d, den.d, cat.d
Print("1 ")
For n=2 To #MAXNUM
num=1 : den =1
For k=2 To n
num * (n+k)
den * k
cat = num / den
Next
Print(Str(cat)+" ")
Next
EndProcedure
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## Python
### Procedural
{{trans|C++}}
```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
>>>
```
{{Works with|Python|2.7}}
```python
def catalan_number(n):
nm = dm = 1
for k in range(2, n+1):
nm, dm = ( nm*(n+k), dm*k )
return nm/dm
print [catalan_number(n) for n in range(1, 16)]
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
```
### Composition of pure functions
Note that sequence [http://oeis.org/A000108 A000108] on OEIS (referenced in the task description) confirms that the first four Catalan numbers are indeed 1, 1, 2, 5 ...
(Several scripts on this page appear to lose the first 1).
{{Trans|Haskell}}
{{Works with|Python|3.7}}
```python
'''Catalan numbers from Pascal's triangle'''
from itertools import (islice)
from operator import (add)
# nCatalans :: Int -> [Int]
def nCatalans(n):
'''The first n Catalan numbers,
derived from Pascal's triangle.'''
# diff :: [Int] -> Int
def diff(xs):
'''Difference between the first two items in the list,
if its length is more than one.
Otherwise, the first (only) item in the list.'''
return (
xs[0] - (xs[1] if 1 < len(xs) else 0)
) if xs else None
return list(map(
compose(diff)(uncurry(drop)),
enumerate(map(fst, take(n)(
everyOther(
pascalTriangle()
)
)))
))
# pascalTriangle :: Gen [[Int]]
def pascalTriangle():
'''A non-finite stream of
Pascal's triangle rows.'''
return iterate(nextPascal)([1])
# nextPascal :: [Int] -> [Int]
def nextPascal(xs):
'''A row of Pascal's triangle
derived from a preceding row.'''
return zipWith(add)([0] + xs)(xs + [0])
# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''First 16 Catalan numbers.'''
print(
nCatalans(16)
)
# GENERIC -------------------------------------------------
# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
'''Right to left function composition.'''
return lambda f: lambda x: g(f(x))
# drop :: Int -> [a] -> [a]
# drop :: Int -> String -> String
def drop(n):
'''The sublist of xs beginning at
(zero-based) index n.'''
def go(xs):
if isinstance(xs, list):
return xs[n:]
else:
take(n)(xs)
return xs
return lambda xs: go(xs)
# everyOther :: Gen [a] -> Gen [a]
def everyOther(g):
'''Every other item of a generator stream.'''
while True:
yield take(1)(g)
take(1)(g) # Consumed, not yielded.
# fst :: (a, b) -> a
def fst(tpl):
'''First component of a pair.'''
return tpl[0]
# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
'''An infinite list of repeated applications of f to x.'''
def go(x):
v = x
while True:
yield v
v = f(v)
return lambda x: go(x)
# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.'''
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
# uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
'''A function over a tuple
derived from a curried function.'''
return lambda xy: f(xy[0])(
xy[1]
)
# zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
def zipWith(f):
'''A list constructed by zipping with a
custom function, rather than with the
default tuple constructor.'''
return lambda xs: lambda ys: (
list(map(f, xs, ys))
)
# MAIN ---
if __name__ == '__main__':
main()
```
{{Out}}
```txt
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
```
## Racket
```Racket
#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)
```
## REXX
### explicit subscripts
All of the REXX program examples can handle arbitrary large numbers.
```rexx
/*REXX program obtains and displays Catalan numbers from a Pascal's triangle. */
parse arg N . /*Obtain the optional argument from CL.*/
if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/
numeric digits max(9, N%2 + N%8) /*so we can handle huge Catalan numbers*/
@.=0; @.1=1 /*stem array default; define 1st value.*/
do i=1 for N; ip=i+1
do j=i by -1 for N; jm=j-1; @[email protected][email protected]; end /*j*/
@[email protected]; do k=ip by -1 for N; km=k-1; @[email protected][email protected]; end /*k*/
say @.ip - @.i /*display the Ith Catalan number. */
end /*i*/ /*stick a fork in it, we're all done. */
```
'''output''' when using the default input:
```txt
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
```
### implicit subscripts
```rexx
/*REXX program obtains and displays Catalan numbers from a Pascal's triangle. */
parse arg N . /*Obtain the optional argument from CL.*/
if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/
numeric digits max(9, N%2 + N%8) /*so we can handle huge Catalan numbers*/
@.=0; @.1=1 /*stem array default; define 1st value.*/
do i=1 for N; ip=i+1
do j=i by -1 for N; @[email protected]+@(j-1); end /*j*/
@[email protected]; do k=ip by -1 for N; @[email protected]+@(k-1); end /*k*/
say @.ip - @.i /*display the Ith Catalan number. */
end /*i*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
@: parse arg !; return @.! /*return the value of @.[arg(1)] */
```
'''output''' is the same as the 1st version.
### using binomial coefficients
```rexx
/*REXX program obtains and displays Catalan numbers from a Pascal's triangle. */
parse arg N . /*Obtain the optional argument from CL.*/
if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/
numeric digits max(9, N%2 + N%8) /*so we can handle huge Catalan numbers*/
do j=1 for N /* [↓] display N Catalan numbers. */
say comb(j+j, j) % (j+1) /*display the Jth Catalan number. */
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!: procedure; parse arg z; _=1; do j=1 for arg(1); _=_*j; end; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb: procedure; parse arg x,y; if x=y then return 1; if y>x then return 0
if x-yst version.
===binomial coefficients, memoized===
This REXX version uses memoization for the calculation of factorials.
```rexx
/*REXX program obtains and displays Catalan numbers from a Pascal's triangle. */
parse arg N . /*Obtain the optional argument from CL.*/
if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/
numeric digits max(9, N%2 + N%8) /*so we can handle huge Catalan numbers*/
!.=.
do j=1 for N /* [↓] display N Catalan numbers. */
say comb(j+j, j) % (j+1) /*display the Jth Catalan number. */
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!: procedure expose !.; parse arg z; if !.z\==. then return !.z; _=1
do j=1 for arg(1); _=_*j; end; !.z=_; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb: procedure expose !.; parse arg x,y; if x=y then return 1; if y>x then return 0
if x-yst version.
## Ring
```ring
n=15
cat = list(n+2)
cat[1]=1
for i=1 to n
for j=i+1 to 2 step -1
cat[j]=cat[j]+cat[j-1]
next
cat[i+1]=cat[i]
for j=i+2 to 2 step -1
cat[j]=cat[j]+cat[j-1]
next
see "" + (cat[i+1]-cat[i]) + " "
next
```
Output:
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## Ruby
```tcl
def catalan(num)
t = [0, 1] #grows as needed
(1..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
p catalan(15)
```
{{out}}
```txt
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
```
## Run BASIC
```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
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## Rust
```rust
fn main()
{let n=15usize;
let mut t= [0; 17];
t[1]=1;
let mut j:usize;
for i in 1..n+1
{
j=i;
loop{
if j==1{
break;
}
t[j]=t[j] + t[j-1];
j=j-1;
}
t[i+1]= t[i];
j=i+1;
loop{
if j==1{
break;
}
t[j]=t[j] + t[j-1];
j=j-1;
}
print!("{} ", t[i+1]-t[i]);
}
}
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## Scala
```Scala
def catalan(n: Int): Int =
if (n <= 1) 1
else (0 until n).map(i => catalan(i) * catalan(n - i - 1)).sum
(1 to 15).map(catalan(_))
```
{{Out}}See it in running in your browser by [https://scastie.scala-lang.org/2ybpRZxCTOyrx3mIy8yIDw Scastie (JVM)].
## Scilab
n=15
t=zeros(1,n+2)
t(1)=1
for i=1:n
for j=i+1:-1:2
t(j)=t(j)+t(j-1)
end
t(i+1)=t(i)
for j=i+2:-1:2
t(j)=t(j)+t(j-1)
end
disp(t(i+1)-t(i))
end
```
{{out}}
1.
2.
5.
14.
42.
132.
429.
1430.
4862.
16796.
58786.
208012.
742900.
2674440.
9694845.
```
## Seed7
```seed7
$ include "seed7_05.s7i";
const proc: main is func
local
const integer: N is 15;
var array integer: t is [] (1) & N times 0;
var integer: i is 0;
var integer: j is 0;
begin
for i range 1 to N do
for j range i downto 2 do
t[j] +:= t[j - 1];
end for;
t[i + 1] := t[i];
for j range i + 1 downto 2 do
t[j] +:= t[j - 1];
end for;
write(t[i + 1] - t[i] <& " ");
end for;
writeln;
end func;
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## Sidef
{{trans|Ruby}}
```ruby
func catalan(num) {
var t = [0, 1]
(1..num).map { |i|
flip(^i ).each {|j| t[j+1] += t[j] }
t[i+1] = t[i]
flip(^i.inc).each {|j| t[j+1] += t[j] }
t[i+1] - t[i]
}
}
say catalan(15).join(' ')
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
## smart BASIC
```qbasic
PRINT "Catalan Numbers from Pascal's Triangle"!PRINT
x = 15
DIM t(x+2)
t(1) = 1
FOR n = 1 TO x
FOR m = n TO 1 STEP -1
t(m) = t(m) + t(m-1)
NEXT m
t(n+1) = t(n)
FOR m = n+1 TO 1 STEP -1
t(m) = t(m) + t(m-1)
NEXT m
PRINT n,"#######":t(n+1) - t(n)
NEXT n
```
## Tcl
```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]
```
{{out}}
```txt
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
```
=={{header|TI-83 BASIC}}==
```ti83b
"CATALAN
15→N
seq(0,I,1,N+2)→L1
1→L1(1)
For(I,1,N)
For(J,I+1,2,-1)
L1(J)+L1(J-1)→L1(J)
End
L1(I)→L1(I+1)
For(J,I+2,2,-1)
L1(J)+L1(J-1)→L1(J)
End
Disp L1(I+1)-L1(I)
End
```
{{out}}
```txt
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
Done
```
## VBScript
To run in console mode with cscript.
```vbscript
dim t()
if Wscript.arguments.count=1 then
n=Wscript.arguments.item(0)
else
n=15
end if
redim t(n+1)
't(*)=0
t(1)=1
for i=1 to n
ip=i+1
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
Wscript.echo t(i+1)-t(i)
next 'i
```
## Visual Basic
{{trans|Rexx}}
{{works with|Visual Basic|VB6 Standard}}
```vb
Sub catalan()
Const n = 15
Dim t(n + 2) As Long
Dim i As Integer, j As Integer
t(1) = 1
For i = 1 To n
For j = i + 1 To 2 Step -1
t(j) = t(j) + t(j - 1)
Next j
t(i + 1) = t(i)
For j = i + 2 To 2 Step -1
t(j) = t(j) + t(j - 1)
Next j
Debug.Print i, t(i + 1) - t(i)
Next i
End Sub 'catalan
```
{{Out}}
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
```
## zkl
{{trans|PARI/GP}} using binomial coefficients.
```zkl
fcn binomial(n,k){ (1).reduce(k,fcn(p,i,n){ p*(n-i+1)/i },1,n) }
(1).pump(15,List,fcn(n){ binomial(2*n,n)-binomial(2*n,n+1) })
```
{{out}}
```txt
L(1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845)
```
## ZX Spectrum Basic
{{trans|C++}}
```zxbasic
10 LET N=15
20 DIM t(N+2)
30 LET t(2)=1
40 FOR i=2 TO N+1
50 FOR j=i TO 2 STEP -1: LET t(j)=t(j)+t(j-1): NEXT j
60 LET t(i+1)=t(i)
70 FOR j=i+1 TO 2 STEP -1: LET t(j)=t(j)+t(j-1): NEXT j
80 PRINT t(i+1)-t(i);" ";
90 NEXT i
```