Sum of series (n/1) + (n/2) + (n/3) + (n/4) +…….+ (n/n)

Given a value n, find the sum of series, (n/1) + (n/2) + (n/3) + (n/4) +…….+(n/n) where the value of n can be up to 10^12.
Note: Consider only integer division.

Examples:

Input : n = 5
Output : (5/1) + (5/2) + (5/3) + 
        (5/4) + (5/5) = 5 + 2 + 1 + 1 + 1 
                      = 10

Input : 7
Output : (7/1) + (7/2) + (7/3) + (7/4) +
         (7/5) + (7/6) + (7/7) 
         = 7 + 3 + 2 + 1 + 1 + 1 + 1 
         = 16



Below is the program to find the sum of given series:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to find
// sum of given series
#include <bits/stdc++.h>
using namespace std;
  
// function to find sum of series
long long int sum(long long int n)
{
    long long int root = sqrt(n);
    long long int ans = 0;
  
    for (int i = 1; i <= root; i++) 
        ans += n / i;
      
    ans = 2 * ans - (root * root);
    return ans;
}
  
// driver code
int main()
{
    long long int n = 35;
    cout << sum(n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find
// sum of given series
import java.util.*;
  
class GFG {
      
    // function to find sum of series
    static long sum(long n)
    {
        long root = (long)Math.sqrt(n);
        long ans = 0;
       
        for (int i = 1; i <= root; i++) 
            ans += n / i;
           
        ans = 2 * ans - (root * root);
          
        return ans;
    }
      
    /* Driver code */
    public static void main(String[] args) 
    {
        long n = 35;
        System.out.println(sum(n));
    }
}
      
// This code is contributed by Arnav Kr. Mandal.        

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to find
# sum of given series
  
import math
  
# function to find sum of series
def sum(n) :
    root = (int)(math.sqrt(n))
    ans = 0
   
    for i in range(1, root + 1) :
        ans = ans + n // i
       
    ans = 2 * ans - (root * root)
    return ans
  
# driver code
n = 35
print(sum(n))
  
# This code is contributed by Nikita Tiwari.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find 
// sum of given series
using System;
  
class GFG {
      
    // Function to find sum of series
    static long sum(long n)
    {
        long root = (long)Math.Sqrt(n);
        long ans = 0;
      
        for (int i = 1; i <= root; i++) 
            ans += n / i;
          
        ans = 2 * ans - (root * root);
          
        return ans;
    }
      
    // Driver code
    public static void Main() 
    {
        long n = 35;
        Console.Write(sum(n));
    }
}
      
// This code is contributed vt_m. 

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find
// sum of given series
  
// function to find
// sum of series
function sum($n)
{
    $root = intval(sqrt($n));
    $ans = 0;
  
    for ($i = 1; $i <= $root; $i++) 
        $ans += intval($n / $i);
  
    $ans = (2 * $ans) - 
           ($root * $root);
    return $ans;
}
  
// Driver code
$n = 35;
echo (sum($n));
  
// This code is contributed by 
// Manish Shaw(manishshaw1)
?>

chevron_right



Output:

131

Note: If observed closely, we can see that, if we take n common, series turns into an Harmonic Progression.



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : manishshaw1