Sum of all divisors from 1 to N | Set 2

Difficulty Level : Hard
Last Updated : 12 Oct, 2020

Given a positive integer N, the task is to find the sum of divisors of first N natural numbers.

Examples:

Input: N = 4
Output: 15
Explanation:
Sum of divisors of 1 = (1)
Sum of divisors of 2 = (1+2)
Sum of divisors of 3 = (1+3)
Sum of divisors of 4 = (1+2+4)
Hence, total sum = 1 + (1+2) + (1+3) + (1+2+4) = 15

Input: N = 5
Output: 21
Explanation:
Sum of divisors of 1 = (1)
Sum of divisors of 2 = (1+2)
Sum of divisors of 3 = (1+3)
Sum of divisors of 4 = (1+2+4)
Sum of divisors of 5 = (1+5)
Hence, total sum = (1) + (1+2) + (1+3) + (1+2+4) + (1+5) = 21

For linear time approach, refer to Sum of all divisors from 1 to N

Approach:
To optimize the approach in the post mentioned above, we need to look for a solution with logarithmic complexity. A number D is added multiple times in the final answer. Let us try to observe a pattern of repetitive addition.
Considering N = 12:

D	Number of times added
1	12
2	6
3	4
5, 6	2
7, 8, 9, 10, 11, 12	1

From the above pattern, observe that every number D is added (N / D) times. Also, there are multiple D that have same (N / D). Hence, we can conclude that for a given N, and a particular i, numbers from (N / (i + 1)) + 1 to (N / i) will be added i times.

Illustration:

N = 12, i = 1
(N/(i+1))+1 = 6+1 = 7 and (N/i) = 12
All numbers will be 7, 8, 9, 10, 11, 12 and will be added 1 time only.

N = 12, i = 2
(N/(i+1))+1 = 4+1 = 5 and (N/i) = 6
All numbers will be 5, 6 and will be added 2 times.

Now, assume A = (N / (i + 1)), B = (N / i)
Sum of numbers from A + 1 to B = Sum of numbers from 1 to B – Sum of numbers from 1 to A
Also, instead of just incrementing i each time by 1, find next i like this, i = N/(N/(i+1))

Below is the implementation of the above approach:

C++

// C++ program for
// the above approach
#include<bits/stdc++.h>
using namespace std;
 
int mod = 1000000007;
  
// Functions returns sum
// of numbers from 1 to n
int linearSum(int n)
{
    return (n * (n + 1) / 2) % mod;
}
  
// Functions returns sum
// of numbers from a+1 to b
int rangeSum(int b, int a)
{
    return (linearSum(b) - 
            linearSum(a)) % mod;
}
 
// Function returns total
// sum of divisors
int totalSum(int n)
{
     
    // Stores total sum
    int result = 0;
    int i = 1;
  
    // Finding numbers and
    //its occurence
    while(true)
    {
          
        // Sum of product of each
        // number and its occurence
        result += rangeSum(n / i, n / (i + 1)) * 
                          (i % mod) % mod;
          
        result %= mod;
         
        if (i == n)
            break;
             
        i = n / (n / (i + 1));
    } 
    return result;
}        
  
// Driver code
int main()
{
    int N = 4;
    cout << totalSum(N) << endl;
  
    N = 12;
    cout << totalSum(N) << endl;
    return 0;
}
 
// This code is contributed by rutvik_56

Java

// Java program for
// the above approach
class GFG{
     
static final int mod = 1000000007;
 
// Functions returns sum
// of numbers from 1 to n
public static int linearSum(int n)
{
    return (n * (n + 1) / 2) % mod;
}
   
// Functions returns sum
// of numbers from a+1 to b
public static int rangeSum(int b, int a)
{
    return (linearSum(b) - 
            linearSum(a)) % mod;
}
  
// Function returns total
// sum of divisors
public static int totalSum(int n)
{
      
    // Stores total sum
    int result = 0;
    int i = 1;
   
    // Finding numbers and
    //its occurence
    while(true)
    {
           
        // Sum of product of each
        // number and its occurence
        result += rangeSum(n / i,
                           n / (i + 1)) * 
                          (i % mod) % mod;
           
        result %= mod;
          
        if (i == n)
            break;
              
        i = n / (n / (i + 1));
    } 
    return result;
} 
 
// Driver code
public static void main(String[] args)
{
    int N = 4;
    System.out.println(totalSum(N));
   
    N = 12;
    System.out.println(totalSum(N));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for
# the above approach
 
mod = 1000000007
 
# Functions returns sum
# of numbers from 1 to n
def linearSum(n):
    return n*(n + 1)//2 % mod
 
# Functions returns sum
# of numbers from a+1 to b
def rangeSum(b, a):
    return (linearSum(b) - (
          linearSum(a))) % mod
 
# Function returns total
# sum of divisors
def totalSum(n):
 
    # Stores total sum
    result = 0
    i = 1
 
    # Finding numbers and
    # its occurence
    while True:
         
        # Sum of product of each
        # number and its occurence
        result += rangeSum(
            n//i, n//(i + 1)) * (
                  i % mod) % mod;
         
        result %= mod;
        if i == n:
            break
        i = n//(n//(i + 1))
         
    return result        
 
# Driver code
 
N= 4
print(totalSum(N))
 
N= 12
print(totalSum(N))

C#

// C# program for
// the above approach
using System;
 
class GFG{
     
static readonly int mod = 1000000007;
 
// Functions returns sum
// of numbers from 1 to n
public static int linearSum(int n)
{
    return (n * (n + 1) / 2) % mod;
}
   
// Functions returns sum
// of numbers from a+1 to b
public static int rangeSum(int b, int a)
{
    return (linearSum(b) - 
            linearSum(a)) % mod;
}
  
// Function returns total
// sum of divisors
public static int totalSum(int n)
{
     
    // Stores total sum
    int result = 0;
    int i = 1;
   
    // Finding numbers and
    //its occurence
    while(true)
    {
           
        // Sum of product of each
        // number and its occurence
        result += rangeSum(n / i,
                           n / (i + 1)) * 
                          (i % mod) % mod;
           
        result %= mod;
          
        if (i == n)
            break;
              
        i = n / (n / (i + 1));
    } 
    return result;
} 
 
// Driver code
public static void Main(String[] args)
{
    int N = 4;
    Console.WriteLine(totalSum(N));
   
    N = 12;
    Console.WriteLine(totalSum(N));
}
}
 
// This code is contributed by Amit Katiyar

Output:

15
127

Time complexity: O(log N)

My Personal Notes

Related Articles

C++

Java

Python3

C#