Catalan numbers/Pascal's triangle: Difference between revisions

[[Pascal's triangle]]

<br><br>

=={{header|11l}}==

{{trans|Python}}

V t = [0] * (n + 2)

t[1] = 1

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

</pre>

=={{header|360 Assembly}}==

For maximum compatibility, this program uses only the basic instruction set.

USING CATALAN,R13,R12

SAVEAREA B STM-SAVEAREA(R15)

02674440

09694845</pre>

=={{header|Action!}}==

{{libheader|Action! Tool Kit}}

DEFINE PTR="CARD"

C(15)=9694845

</pre>

=={{header|Ada}}==

Uses package Pascal from the Pascal triangle solution[[http://rosettacode.org/wiki/Pascal%27s_triangle#Ada]]

procedure Catalan is

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

=={{header|ALGOL 68}}==

{{trans|C++}}

[ 0 : n + 1 ]INT t;

t[0] := 0;

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

</pre>

=={{header|ALGOL W}}==

% print the first 15 Catalan numbers from Pascal's triangle %

integer n;

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

</pre>

=={{header|APL}}==

⍝ Based heavily on the J solution

CATALAN←{¯1↓↑-/1 ¯1↓¨(⊂⎕IO+0 0)⍉¨0 2⌽¨⊂(⎕IO-⍨⍳N){+\⍣⍺⊢⍵}⍤0 1⊢1⍴⍨N←⍵+2}

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

</pre>

=={{header|AutoHotkey}}==

{{works with|AutoHotkey_L}}

//

// smgs: 20th Feb, 2014

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

</pre>

=={{header|AWK}}==

# syntax: GAWK -f CATALAN_NUMBERS_PASCALS_TRIANGLE.AWK

# converted from C

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

</pre>

=={{header|Batch File}}==

setlocal ENABLEDELAYEDEXPANSION

set n=15

2674440

9694845</pre>

=={{header|C}}==

//This code implements the print of 15 first Catalan's Numbers

//Formula used:

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

</pre>

=={{header|C sharp|C#}}==

{{trans|C++}}

int n = 15;

List<int> t = new List<int>() { 0, 1 };

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

</pre>

=={{header|C++}}==

//

// Nigel Galloway: June 9th., 2012

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

</pre>

=={{header|Common Lisp}}==

"Return the n-th Catalan number"

(if (<= n 1) 1

2674440

9694845</pre>

=={{header|D}}==

{{trans|C++}}

import std.stdio;

See [https://rosettacode.org/wiki/Catalan_numbers/Pascal%27s_triangle#Pascal Pascal].

=={{header|EchoLisp}}==

(define dim 100)

(define-syntax-rule (Tidx i j) (+ i (* dim j)))

</syntaxhighlight>

{{out}}

;; 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

</syntaxhighlight>

=={{header|EDSAC order code}}==

Rather than store a triangular array, this solution stores the right-hand half of the current row and updates it in place. It uses the third method in the link, e.g. once we have the half row (70, 56, 28, 8, 1), the Catalan number 42 appears as 70 - 28.

[Catalan numbers from Pascal triangle, Rosetta Code website.

EDSAC program, Initial Orders 2]

15 9694845

</pre>

=={{header|Elixir}}==

def numbers(num) do

{result,_} = Enum.reduce(1..num, {[],{0,1}}, fn i,{list,t0} ->

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

</pre>

=={{header|Erlang}}==

-module(catalin).

-compile(export_all).

Ans=catl(1,2).

</syntaxhighlight>

=={{header|ERRE}}==

PROGRAM CATALAN

15 9694845

</pre>

=={{header|F_Sharp|F#}}==

let mutable nm=uint64(1)

let mutable dm=uint64(1)

printf ", "

</syntaxhighlight>

=={{header|Factor}}==

IN: rosetta-code.catalan-pascal

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

</pre>

=={{header|FreeBASIC}}==

' compile with: fbc -s console

14: 2674440

15: 9694845</pre>

=={{header|Go}}==

{{trans|C++}}

import "fmt"

15 : 9694845

</pre>

=={{header|Groovy}}==

{{trans|C}}

class Catalan

{

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

</pre>

=={{header|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.

-- Pascal's triangle.

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

</pre>

=={{header|Icon}} and {{header|Unicon}}==

that aren't used.

procedure main(A)

->

</pre>

=={{header|J}}==

{{out|Example use}}

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

</syntaxhighlight>

A structured derivation of Catalan follows:

( PascalTriangle=. i. ((+/\@]^:[)) #&1 ) 5

1 1 1 1 1

Catalan

}:@:(}.@:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)</syntaxhighlight>

=={{header|Java}}==

{{trans|C++}}

public static void main(String[] args) {

int N = 15;

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

=={{header|JavaScript}}==

===ES5===

Iteration

{{trans|C++}}

for (var t = [0, 1], i = 1; i <= n; i++) {

for (var j = i; j > 1; j--) t[j] += t[j - 1];

{{Trans|Haskell}}

'use strict';

{{Out}}

=={{header|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.

if k > n / 2 then binomial(n; n-k)

else reduce range(1; k+1) as $i (1; . * (n - $i + 1) / $i)

(range(0;16), 30, 31, 510, 511) | [., catalan_by_pascal]

{{Out}}

[0,0]

[1,1]

[510,5.491717746183512e+302]

[511,null]</syntaxhighlight>

=={{header|Julia}}==

{{trans|Matlab}}

function pascal(n::Int)

{{out}}

<pre>catalan_num(15) = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]</pre>

=={{header|Kotlin}}==

import java.math.BigInteger

15 : 9694845

</pre>

=={{header|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.

local ret = {}

t[0], t[#t + 1] = 0, 0

{{out}}

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

=={{header|M2000 Interpreter}}==

{{trans|FreeBasic}}

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).

Module CatalanNumbers {

Def Integer count, t_row, size=31

15: 9694845

</pre>

=={{header|Maple}}==

local i,a:=[1],C:=[1];

for i to n do

catalan(10);

# [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796]</syntaxhighlight>

=={{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.

output = ConstantArray[1, Length[lastrow] + 1];

Do[

{{out}}

<pre>{1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845}</pre>

=={{header|MATLAB}} / {{header|Octave}}==

p = pascal(n + 2);

p(n + 4 : n + 3 : end - 1)' - diag(p, 2)</syntaxhighlight>

2674440

9694845</pre>

=={{header|Nim}}==

{{trans|Python}}

var t = newSeq[int](n + 2)

{{Out}}

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

=={{header|OCaml}}==

let catalan : int ref = ref 0 in

Printf.printf "%d ," 1 ;

1 ,2,5,14,42,132,429,1430,4862

</pre>

=={{header|Oforth}}==

: pascal( n -- [] )

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

</pre>

=={{header|PARI/GP}}==

{{out}}

<pre>%1 = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]</pre>

=={{header|Pascal}}==

Program CatalanNumbers

type

</pre>

=={{header|Perl}}==

{{trans|C++}}

my @t = (0, 1);

for(my $i = 1; $i <= N; $i++) {

After the 28th Catalan number, this overflows 64-bit integers. We could add <tt>use bigint;</tt> <tt>use Math::GMP ":constant";</tt> to make it work, albeit not at a fast pace. However we can use a module to do it much faster:

{{libheader|ntheory}}

print join(" ", map { binomial( 2*$_, $_) / ($_+1) } 1 .. 1000), "\n";</syntaxhighlight>

The <tt>Math::Pari</tt> module also has a binomial, but isn't as fast and overflows its stack after 3400.

=={{header|Phix}}==

Calculates the minimum pascal triangle in minimum memory. Inspired by the comments in, but not the code of the FreeBasic example

<span style="color: #008080;">constant</span> <span style="color: #000000;">N</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">15</span> <span style="color: #000080;font-style:italic;">-- accurate to 30, nan/inf for anything over 514 (gmp version is below).</span>

<span style="color: #004080;">sequence</span> <span style="color: #000000;">catalan</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{},</span> <span style="color: #000080;font-style:italic;">-- (&gt;=1 only)</span>

</pre>

Explanatory comments to accompany the above

--' 1 1 1 1 1 1

--' 1 2 3 4 5 6

=== gmp version ===

{{libheader|Phix/mpfr}}

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #008080;">include</span> <span style="color: #000000;">builtins</span><span style="color: #0000FF;">\</span><span style="color: #004080;">mpfr</span><span style="color: #0000FF;">.</span><span style="color: #000000;">e</span>

catalan2m: 0.7s 7.0s 64.9s 644s

</pre>

=={{header|PicoLisp}}==

(let f

'((N)

(bino (* 2 N) (inc N)) ) ) )

(bye)</syntaxhighlight>

=={{header|PureBasic}}==

{{trans|C}}

Declare catalan()

<pre>

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

=={{header|Python}}==

===Procedural===

{{trans|C++}}

>>> t = [0] * (n + 2)

>>> t[1] = 1

{{Works with|Python|2.7}}

nm = dm = 1

for k in range(2, n+1):

{{Trans|Haskell}}

{{Works with|Python|3.7}}

from itertools import (islice)

{{Out}}

<pre>[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]</pre>

=={{header|Quackery}}==

witheach

[ tuck +

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

=={{header|Racket}}==

#lang racket

;; 2674440 9694845)

</syntaxhighlight>

=={{header|Raku}}==

(formerly Perl 6)

{{works with|Rakudo|2015.12}}

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

1767263190

6564120420</pre>

=={{header|REXX}}==

===explicit subscripts===

All of the REXX program examples can handle arbitrary large numbers.

parse arg N . /*Obtain the optional argument from CL.*/

if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/

===implicit subscripts===

parse arg N . /*Obtain the optional argument from CL.*/

if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/

===using binomial coefficients===

parse arg N . /*Obtain the optional argument from CL.*/

if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/

===binomial coefficients, memoized===

This REXX version uses memoization for the calculation of factorials.

parse arg N . /*Obtain the optional argument from CL.*/

if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/

if x-y<y then y=x-y; _=1; do j=x-y+1 to x; _=_*j; end; return _/!(y)</syntaxhighlight>

'''output''' &nbsp; is the same as the 1^st version. <br><br>

=={{header|Ring}}==

n=15

cat = list(n+2)

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

</pre>

=={{header|Ruby}}==

t = [0, 1] #grows as needed

(1..num).map do |i|

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

</pre>

=={{header|Run BASIC}}==

dim t(n+2)

t(1) = 1

{{out}}

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

=={{header|Rust}}==

fn main()

{{out}}

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

=={{header|Scala}}==

if (n <= 1) 1

else (0 until n).map(i => catalan(i) * catalan(n - i - 1)).sum

(1 to 15).map(catalan(_))</syntaxhighlight>

{{Out}}See it in running in your browser by [https://scastie.scala-lang.org/2ybpRZxCTOyrx3mIy8yIDw Scastie (JVM)].

=={{header|Scilab}}==

t=zeros(1,n+2)

t(1)=1

2674440.

9694845. </pre>

=={{header|Seed7}}==

const proc: main is func

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

</pre>

=={{header|Sidef}}==

{{trans|Ruby}}

var t = [0, 1]

(1..num).map { |i|

{{out}}

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

=={{header|smart BASIC}}==

x = 15

DIM t(x+2)

PRINT n,"#######":t(n+1) - t(n)

NEXT n</syntaxhighlight>

=={{header|Tcl}}==

set result {}

array set t {0 0 1 1}

{{out}}

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

=={{header|TI-83 BASIC}}==

15→N

seq(0,I,1,N+2)→L1

9694845

Done </pre>

=={{header|Vala}}==

{{trans|C++}}

const int N = 15;

uint64[] t = {0, 1};

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

</pre>

=={{header|VBScript}}==

To run in console mode with cscript.

if Wscript.arguments.count=1 then

n=Wscript.arguments.item(0)

Wscript.echo t(i+1)-t(i)

next 'i</syntaxhighlight>

=={{header|Visual Basic}}==

{{trans|Rexx}}

{{works with|Visual Basic|VB6 Standard}}

Sub catalan()

Const n = 15

9694845

</pre>

=={{header|Wren}}==

{{trans|C++}}

var t = List.filled(n+2, 0)

t[1] = 1

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

</pre>

=={{header|Zig}}==

Uses comptime functionality to compile Pascal's triangle into the object code, then at runtime, it's simply a table lookup. (uses code from AKS primality task.)

const std = @import("std");

const stdout = std.io.getStdOut().outStream();

15 9694845

</pre>

=={{header|zkl}}==

{{trans|PARI/GP}} using binomial coefficients.

(1).pump(15,List,fcn(n){ binomial(2*n,n)-binomial(2*n,n+1) })</syntaxhighlight>

{{out}}

L(1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845)

</pre>

=={{header|ZX Spectrum Basic}}==

{{trans|C++}}

20 DIM t(N+2)

30 LET t(2)=1