Sum of all divisors from 1 to N | Set 2

0 0 5 minutes read

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 occurrence
    while(true)
    {
          
        // Sum of product of each
        // number and its occurrence
        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 occurrence
    while(true)
    {
           
        // Sum of product of each
        // number and its occurrence
        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 occurrence
    while True:
         
        # Sum of product of each
        # number and its occurrence
        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 occurrence
    while(true)
    {
           
        // Sum of product of each
        // number and its occurrence
        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

Javascript

<script>
 
// JavaScript program for
// the above approach
let mod = 1000000007;
 
// Functions returns sum
// of numbers from 1 to n
function linearSum(n)
{
    return (n * (n + 1) / 2) % mod;
}
   
// Functions returns sum
// of numbers from a+1 to b
function rangeSum(b, a)
{
    return (linearSum(b) - 
            linearSum(a)) % mod;
}
  
// Function returns total
// sum of divisors
function totalSum(n)
{
      
    // Stores total sum
    let result = 0;
    let i = 1;
   
    // Finding numbers and
    //its occurrence
    while(true)
    {
           
        // Sum of product of each
        // number and its occurrence
        result += rangeSum(Math.floor(n / i),
                           Math.floor(n / (i + 1))) * 
                                     (i % mod) % mod;
           
        result %= mod;
          
        if (i == n)
            break;
              
        i = Math.floor(n / (n / (i + 1)));
    } 
    return result;
} 
   
// Driver Code
let N = 4;
document.write(totalSum(N) + "<br/>");
 
N = 12;
document.write(totalSum(N));
   
// This code is contributed by susmitakundugoaldanga
 
</script>

Output:

15
127

Time complexity: O(?n)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!