Calculate a triangle number

From CodeCodex

A triangular number is the number of dots in an equilateral triangle evenly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangle number. The nth triangle number is the number of dots in a triangle with n dots on a side.

Contents

[edit] Implementations

[edit] Java

This code will print the first 1000 triangle numbers to the system console. 

	public static void main(String[] args) {
		
		for (int i = 1 ; i < 1000 ; i++) {
			
			int n = i*(i+1)/2;
			System.out.println( "n="+i+" triangle number="+n );
				
		}
		
	}

[edit] OCaml

# let triangle_numbers n = Array.init n (fun i -> i*(i+1)/2);;
val triangle_numbers : int -> int array

For example: 

# triangle_numbers 10;;
- : int array = [|0; 1; 3; 6; 10; 15; 21; 28; 36; 45|]

[edit] Perl


print for map {"n=$_ triangle number=".$_*($_+1)/2, "\n"} 1..1000;

[edit] PHP

<?php
foreach ( range(1, 1000) as $i ) {
        $triangle_numbers[] = $i * ( $i + 1 ) / 2;
}
?>

[edit] Python


def triangle_numbers(n):
    return [i*(i+1)/2 for i in xrange(n)]

For example: 

>>> triangle_numbers(10)
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45]

[edit] Seed7

This code will print the first 1000 triangle numbers to the system console. 

$ include "seed7_05.s7i";

const proc: main is func
  local
    var integer: num is 0;
  begin
    for num range 1 to 1000 do
      writeln("n=" <& num <&
          " triangle number=" <& num * succ(num) div 2);
    end for;
  end func;

[edit] Ruby

def triangle_numbers(n)
    Array.new(n) {|i| i*(i+1)/2 }
end

[edit] Common Lisp

 (loop for i from 0 below 1000 collecting (* i (1+ i) 1/2))

[edit] Zsh

for i in {0..9}; print $((i * (i+1) / 2))

[edit] C


#include <stdio.h>

unsigned triangularNumber(unsigned n);

const char *suffix[] = {" ","st","nd","rd"};

int main(int argc, char *argv[]) {
        int n = 0,m;
        unsigned t;

        if ((argc < 2) ||
        (sscanf(argv[1],"%d",&n) != 1) ||
        (n <= 0)) {
                printf("\nusage %s <n>\n",argv[0]);
                printf(" where <n> is a positive integer\n");
                return -1;
        }
        t = (int)(n * (((float)n + 1.0) / 2.0));
        if (t == triangularNumber(n)) {
                m = n % 10;
                printf("\nThe %d%s triangular number is %d\n",
                n,
                ((m > 0) && (m < 4)) ? suffix[m] : "th",
                t);
        } else {
                printf("\nerror\n");
        }
        return 0;
}

unsigned triangularNumber(unsigned n) {
        if (n == 1) return 1;
        return n + triangularNumber(n-1);
}
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