Natural Numbers

Counting numbers like 1, 2, 3, 4, 5, 6 … Basically, all integers greater than 0 are natural numbers. 

Fact about Natural numbers 

1. They are whole numbers (called integers), and never less than zero (i.e. positive numbers)
2. The next possible natural number can be found by adding 1 to the current natural number
3. The natural numbers are the ordinary numbers, 1, 2, 3, etc., with which we count.
4. The number zero is sometimes considered to be a natural number. Not always because no one counts starting with zero, 0, 1, 2, 3.
5. GCD of all other natural numbers with a prime is always one.
6. The natural numbers can be defined formally by relating them to sets. Then, zero is the number of elements in the empty set; 1 is the number of elements in the set containing one natural number; and so on.

How to print sum of n natural Numbers?

Using Recursion

Given a number n, find sum of first n natural numbers. To calculate the sum, we will use the recursive function recur_sum().

Examples :

Input : 3
Output : 6
Explanation : 1 + 2 + 3 = 6

Input : 5
Output : 15
Explanation : 1 + 2 + 3 + 4 + 5 = 15

C++

// C++ program to find the
// sum of natural numbers up
// to n using recursion
#include <iostream>
using namespace std;

// Returns sum of first
// n natural numbers
int recurSum(int n)
{
    if (n <= 1)
        return n;
    return n + recurSum(n - 1);
}

// Driver code
int main()
{
    int n = 5;
    cout << recurSum(n);
    return 0;
}

Java

// Java program to find the
// sum of natural numbers up
// to n using recursion
import java.util.*;
import java.lang.*;

class GFG
{

    // Returns sum of first
    // n natural numbers
    public static int recurSum(int n)
    {
        if (n <= 1)
            return n;
        return n + recurSum(n - 1);
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(recurSum(n));
    }
}

Python

# Python code to find sum
# of natural numbers upto
# n using recursion

# Returns sum of first
# n natural numbers
def recurSum(n):
    if n <= 1:
        return n
    return n + recurSum(n - 1)

# Driver code
n = 5
print(recurSum(n))

C#

// C# program to find the
// sum of natural numbers
// up to n using recursion
using System;

class GFG
{

    // Returns sum of first
    // n natural numbers
    public static int recurSum(int n)
    {
        if (n <= 1)
            return n;
        return n + recurSum(n - 1);
    }

    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.WriteLine(recurSum(n));
    }
}

PHP

<?php
// PHP program to find the
// sum of natural numbers
// up to n using recursion

// Returns sum of first
// n natural numbers
function recurSum($n)
{
    if ($n <= 1)
        return $n;
    return $n + recurSum($n - 1);
}

// Driver code
$n = 5;
echo(recurSum($n));

?>

Javascript

<script>

// JavaScript program to find the
// sum of natural numbers
// up to n using recursion

// Returns sum of first
// n natural numbers
function recurSum(n)
{
    if (n <= 1)
        return n;
    return n + recurSum(n - 1);
}

// Driver code
n = 5;
document.write(recurSum(n));

</script>

Output :

15 

Time Complexity: O(n)
Auxiliary Space: O(n)

Using Loop

A simple solution is to do the following. 

1) Initialize : sum = 0
2) Run a loop from x = 1 to n and 
   do following in loop.
     sum = sum + x  

C++

// CPP program to find sum of first
// n natural numbers.
#include <iostream>
using namespace std;

// Returns sum of first n natural
// numbers
int findSum(int n)
{
    int sum = 0;
    for (int x = 1; x <= n; x++)
        sum = sum + x;
    return sum;
}

// Driver code
int main()
{
    int n = 5;
    cout << findSum(n);
    return 0;
}

Java

// JAVA program to find sum of first
// n natural numbers.
import java.io.*;

class GFG{

    // Returns sum of first n natural
    // numbers
    static int findSum(int n)
    {
        int sum = 0;
        for (int x = 1; x <= n; x++)
            sum = sum + x;
        return sum;
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(findSum(n));
    }
}

Python

# PYTHON program to find sum of first
# n natural numbers.

# Returns sum of first n natural
# numbers
def findSum(n) :
    sum = 0
    x = 1
    while x <=n :
        sum = sum + x
        x = x + 1
    return sum


# Driver code

n = 5
print findSum(n)

C#

// C# program to find sum of first
// n natural numbers.
using System;

class GFG{

    // Returns sum of first n natural
    // numbers
    static int findSum(int n)
    {
        int sum = 0;
        for (int x = 1; x <= n; x++)
            sum = sum + x;
        return sum;
    }

    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.Write(findSum(n));
    }
}

PHP

<?php
// PHP program to find sum of first
// n natural numbers.

// Returns sum of first n natural
// numbers
function findSum($n)
{
$sum = 0;
for ($x = 1; $x <= $n; $x++)
    $sum = $sum + $x;
return $sum;
}

// Driver code
$n = 5;
echo findSum($n);


?>

Javascript

<script>
// JavaScript program to find the
// sum of natural numbers
// up to n using recursion

// Returns sum of first
// n natural numbers
function findSum(n)
{
let sum = 0;
for (x = 1; x <= n; x++)
    sum = sum + x;
return sum;
}

// Driver code
n = 5;
document.write(findSum(n));

// This code is contributed by sravan kumar

</script>

Output :

15

Time Complexity: O(n)
Auxiliary Space: O(1)

Using Sum of n terms formula

Formula for finding sum of n natural numbers is given by n*(n+1)/2 which implies if the formula is used the program returns output faster than it would take iterating over loop or recursion. Time complexity is O(1). 

Referral Link:  Program to find sum of n natural numbers

More problems related to Natural Number: 

• Count natural numbers whose all permutations are greater than that number
• Sum of squares of first n natural numbers
• Sum of cubes of even and odd natural numbers
• LCM of First n Natural Numbers
• Sum of squares of first n natural numbers