Sum of all divisors from 1 to n

Given a positive integer n. Find the value of $\sum_{i=1}^{i=n} F(i)$ where function F(i) for number i be defined as sum of all divisors of ‘i‘.
Examples :

Input: 4
Output: 15
Explanation
F(1) = 1
F(2) = 1 + 2 = 3
F(3) = 1 + 3 = 4
F(4) = 1 + 2 + 4 = 7
ans = F(1) + F(2) + F(3) + F(4)
    = 1 + 3 + 4 + 7
    = 15

Input: 5
Output: 21

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

Naive approach is to traverse for every number(1 to n), find all divisors and keep updating the sum with that divisor. See this to understand more.

C++

// CPP program to find sum of all 
// divisor of number up to 'n' 
#include <bits/stdc++.h> 
  
// Utility function to find sum of 
// all divisor of number up to 'n' 
int divisorSum(int n) i 
{ 
    int sum = 0; 
  
    for (int i = 1; i <= n; ++i) { 
  
        // Find all divisors of i and add them 
        for (int j = 1; j * j <= i; ++j) { 
            if (i % j == 0) { 
                if (i / j == j) 
                    sum += j; 
                else
                    sum += j + i / j; 
            } 
        } 
    } 
    return sum; 
} 
  
// Driver code 
int main() 
{ 
    int n = 4; 
    printf("%d\n", divisorSum(n)); 
    n = 5; 
    printf("%d", divisorSum(n)); 
    return 0; 
} 

Java

// JAVA program to find sum of all 
// divisor of number up to 'n' 
import java.io.*; 
  
class GFG { 
  
    // Utility function to find sum of 
    // all divisor of number up to 'n' 
    static int divisorSum(int n) 
    { 
        int sum = 0; 
  
        for (int i = 1; i <= n; ++i) { 
  
            // Find all divisors of i 
            // and add them 
            for (int j = 1; j * j <= i; ++j) { 
                if (i % j == 0) { 
                    if (i / j == j) 
                        sum += j; 
                    else
                        sum += j + i / j; 
                } 
            } 
        } 
        return sum; 
    } 
  
    // Driver code 
    public static void main(String args[]) 
    { 
        int n = 4; 
        System.out.println(divisorSum(n)); 
        n = 5; 
        System.out.println(divisorSum(n)); 
    } 
} 
  
/*This code is contributed by Nikita tiwari.*/

Python3

# Python3 code to find sum of all 
# divisor of number up to 'n' 
  
# Utility function to find sum of 
# all divisor of number up to 'n' 
def divisorSum( n ): 
    sum = 0
      
    for i in range(1, n + 1): 
          
        # Find all divisors of i 
        # and add them 
        j = 1
        while j * j <= i: 
            if i % j == 0: 
                if i / j == j: 
                    sum += j 
                else: 
                    sum += j + i / j 
            j = j + 1
    return int(sum) 
  
# Driver code 
n = 4
print( divisorSum(n)) 
n = 5
print( divisorSum(n)) 
  
# This code is contributed by "Sharad_Bhardwaj". 

C#

// C# program to find sum of all 
// divisor of number up to 'n' 
using System; 
  
class GFG { 
  
    // Utility function to find sum of 
    // all divisor of number up to 'n' 
    static int divisorSum(int n) 
    { 
        int sum = 0; 
  
        for (int i = 1; i <= n; ++i) { 
  
            // Find all divisors of i 
            // and add them 
            for (int j = 1; j * j <= i; ++j) { 
                if (i % j == 0) { 
                    if (i / j == j) 
                        sum += j; 
                    else
                        sum += j + i / j; 
                } 
            } 
        } 
        return sum; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int n = 4; 
        Console.WriteLine(divisorSum(n)); 
        n = 5; 
        Console.WriteLine(divisorSum(n)); 
    } 
} 
  
/*This code is contributed by vt_m.*/

PHP

<?php 
// PHP program to find sum of all 
// divisor of number up to 'n' 
  
// Utility function to find sum of 
// all divisor of number up to 'n' 
  
function divisorSum($n)  
{ 
    $sum = 0; 
  
    for ($i = 1; $i <= $n; ++$i)  
    { 
  
        // Find all divisors of i 
        // and add them  
        for ($j = 1; $j * $j <= $i; ++$j)  
        { 
            if ($i % $j == 0)  
            { 
                if ($i / $j == $j) 
                    $sum += $j; 
                else
                    $sum += $j + $i / $j; 
            } 
        } 
    } 
    return $sum; 
} 
  
// Driver code 
$n = 4; 
echo "\n", divisorSum($n), "\n"; 
$n = 5; 
echo divisorSum($n), "\n"; 
  
// This code is contributed by aj_36 
?> 

Output :

15
21

Time complexity: O( $n\sqrt{n}$ )
Auxiliary space: O(1)

Efficient approach is to observe the function and co-relate the pattern. For a given number n, every number from 1 to n contribute it’s presence up to the highest multiple less than n. For instance,

Let n = 6,
=> F(1) + F(2) + F(3) + F(4) + F(5) + F(6)
=> 1 will occurs 6 times in F(1), F(2),
   F(3), F(4), F(5) and F(6)
=> 2 will occurs 3 times in F(2), F(4) and
   F(6)
=> 3 will occur 2 times in F(3) and F(6)
=> 4 will occur 1 times in F(4)
=> 5 will occur 1 times in F(5)
=> 6 will occur 1 times in F(6)

From above observation, it can easily be observed that number i is occurring only in there multiples less than or equal to n. Thus, we just need to find the count of multiples and then multiply it with i for full contribution in the final sum. It can easily be done in O(1) time by taking floor of (n / i) and then multiply it with i for the sum.

C++

// CPP program to find sum of all 
// divisor of number up to 'n' 
#include <stdio.h> 
  
// Utility function to find sum of 
// all divisor of number up to 'n' 
int divisorSum(int n) 
{ 
    int sum = 0; 
    for (int i = 1; i <= n; ++i) 
        sum += (n / i) * i; 
    return sum; 
} 
  
// Driver code 
int main() 
{ 
    int n = 4; 
    printf("%d\n", divisorSum(n)); 
    n = 5; 
    printf("%d", divisorSum(n)); 
    return 0; 
} 

Java

// Java program to find sum of all 
// divisor of number up to 'n' 
import java.io.*; 
  
class GFG { 
  
    // Utility function to find sum of 
    // all divisor of number up to 'n' 
    static int divisorSum(int n) 
    { 
        int sum = 0; 
        for (int i = 1; i <= n; ++i) 
            sum += (n / i) * i; 
        return sum; 
    } 
  
    // Driver code 
    public static void main(String args[]) 
    { 
        int n = 4; 
        System.out.println(divisorSum(n)); 
        n = 5; 
        System.out.println(divisorSum(n)); 
    } 
} 
  
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python3 code to find sum of all 
# divisor of number up to 'n' 
  
# Utility function to find sum of 
# all divisor of number up to 'n' 
def divisorSum( n ): 
    sum = 0
    for i in range(1, n + 1): 
        sum += int(n / i) * i 
    return int(sum) 
      
# Driver code 
n = 4
print( divisorSum(n)) 
n = 5
print( divisorSum(n)) 
  
# This code is contributed by "Sharad_Bhardwaj". 

C#

// C# program to find sum of all 
// divisor of number up to 'n' 
using System; 
  
class GFG { 
  
    // Utility function to find sum of 
    // all divisor of number up to 'n' 
    static int divisorSum(int n) 
    { 
        int sum = 0; 
        for (int i = 1; i <= n; ++i) 
            sum += (n / i) * i; 
        return sum; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int n = 4; 
        Console.WriteLine(divisorSum(n)); 
        n = 5; 
        Console.WriteLine(divisorSum(n)); 
    } 
} 
  
/*This code is contributed by vt_m.*/

PHP

<?php 
// PHP program to find sum of all 
// divisor of number up to 'n' 
  
// Utility function to find sum of 
// all divisor of number up to 'n' 
function divisorSum( $n) 
{ 
    $sum = 0; 
    for ( $i = 1; $i <= $n; ++$i) 
        $sum += floor($n / $i) * $i; 
    return $sum; 
} 
  
// Driver code 
$n = 4; 
echo divisorSum($n),"\n"; 
$n = 5; 
echo divisorSum($n),"\n"; 
  
// This code is contributed by anuj_67. 
?> 

Output :

15
21

Time complexity: O(n)
Auxiliary space: O(1)

My Personal Notes

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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