# Sum of all proper divisors from 1 to N

• Last Updated : 06 Jun, 2021

Given a positive integer N, the task is to find the value of where function F(x) can be defined as sum of all proper divisors of ‘x‘.
Examples:

Input: N = 4
Output:
Explanation:
Sum of all proper divisors of numbers:
F(1) = 0
F(2) = 1
F(3) = 1
F(4) = 1 + 2 = 3
Total Sum = F(1) + F(2) + F(3) + F(4) = 0 + 1 + 1 + 3 = 5
Input: N = 5
Output:
Explanation:
Sum of all proper divisors of numbers:
F(1) = 0
F(2) = 1
F(3) = 1
F(4) = 1 + 2 = 3
F(5) = 1
Total Sum = F(1) + F(2) + F(3) + F(4) + F(5) = 0 + 1 + 1 + 3 + 1 = 6

Naive approach: The idea is to find the sum of proper divisors of each number in the range [1, N] individually, and then add them to find the required sum.
Below is the implementation of the above approach:

## C++

 // C++ implementation to find sum of all// proper divisor of number up to N#include using namespace std; // Utility function to find sum of// all proper divisor of number up to Nint properDivisorSum(int n){    int sum = 0;     // Loop to iterate over all the    // numbers from 1 to N    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;            }        }         // Subtracting 'i' so that the        // number itself is not included        sum = sum - i;    }    return sum;} // Driver Codeint main(){    int n = 4;    cout << properDivisorSum(n) << endl;     n = 5;    cout << properDivisorSum(n) << endl;     return 0;}

## Java

 // Java implementation to find sum of all// proper divisor of number up to Nclass GFG {         // Utility function to find sum of    // all proper divisor of number up to N    static int properDivisorSum(int n)    {        int sum = 0;             // Loop to iterate over all the        // numbers from 1 to N        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;                }            }                 // Subtracting 'i' so that the            // number itself is not included            sum = sum - i;        }        return sum;    }         // Driver Code    public static void main (String[] args)    {        int n = 4;        System.out.println(properDivisorSum(n));             n = 5;        System.out.println(properDivisorSum(n)) ;         }} // This code is contributed by Yash_R

## Python3

 # Python3 implementation to find sum of all# proper divisor of number up to N # Utility function to find sum of# all proper divisor of number up to Ndef properDivisorSum(n):     sum = 0     # Loop to iterate over all the    # numbers from 1 to N    for i in range(n+1):         # Find all divisors of        # i and add them        for j in range(1, i + 1):            if j * j > i:                break            if (i % j == 0):                if (i // j == j):                    sum += j                else:                    sum += j + i // j         # Subtracting 'i' so that the        # number itself is not included        sum = sum - i     return sum # Driver Codeif __name__ == '__main__':     n = 4    print(properDivisorSum(n))     n = 5    print(properDivisorSum(n)) # This code is contributed by mohit kumar 29

## C#

 // C# implementation to find sum of all// proper divisor of number up to Nusing System; class GFG {         // Utility function to find sum of    // all proper divisor of number up to N    static int properDivisorSum(int n)    {        int sum = 0;             // Loop to iterate over all the        // numbers from 1 to N        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;                }            }                 // Subtracting 'i' so that the            // number itself is not included            sum = sum - i;        }        return sum;    }         // Driver Code    public static void Main (string[] args)    {        int n = 4;        Console.WriteLine(properDivisorSum(n));             n = 5;        Console.WriteLine(properDivisorSum(n)) ;       }} // This code is contributed by Yash_R

## Javascript

 

Output:

5
6

Time complexity: O(N * √N)
Auxiliary space: O(1)
Efficient approach: Upon observing the pattern in the function, it can be seen that “For a given number N, every number ‘x’ in the range [1, N] occurs (N/x) number of times”.
For example:

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

From above observation, it can easily be observed that number x occurs only in its multiple less than or equal to N. Therefore, we just need to find the count of such multiples, for each value of x in [1, N], and then multiply it with x. This value is then added to the final sum.
Below is the implementation of the above approach:

## C++

 // C++ implementation to find sum of all// proper divisor of numbers up to N #include using namespace std; // Utility function to find sum of// all proper divisor of number up to Nint properDivisorSum(int n){    int sum = 0;     // Loop to find the proper    // divisor of every number    // from 1 to N    for (int i = 1; i <= n; ++i)        sum += (n / i) * i;     return sum - n * (n + 1) / 2;} // Driver Codeint main(){    int n = 4;    cout << properDivisorSum(n) << endl;     n = 5;    cout << properDivisorSum(n) << endl;    return 0;}

## Java

 // Java implementation to find sum of all// proper divisor of numbers up to N  // Utility function to find sum of// all proper divisor of number up to N class GFG{    static int properDivisorSum(int n)    {        int sum = 0;        int i;        // Loop to find the proper        // divisor of every number        // from 1 to N        for (i = 1; i <= n; ++i)            sum += (n / i) * i;              return sum - n * (n + 1) / 2;    }          // Driver Code    public static void main(String []args)    {        int n = 4;        System.out.println(properDivisorSum(n));              n = 5;        System.out.println(properDivisorSum(n));             }}

## Python3

 # Python3 implementation to find sum of all# proper divisor of numbers up to N # Utility function to find sum of# all proper divisor of number up to Ndef properDivisorSum(n):         sum = 0         # Loop to find the proper    # divisor of every number    # from 1 to N    for i in range(1, n + 1):        sum += (n // i) * i             return sum - n * (n + 1) // 2  # Driver Coden = 4print(properDivisorSum(n)) n = 5print(properDivisorSum(n)) # This code is contributed by shubhamsingh10

## C#

 // C# implementation to find sum of all// proper divisor of numbers up to N   // Utility function to find sum of// all proper divisor of number up to Nusing System; class GFG{    static int properDivisorSum(int n)    {        int sum = 0;        int i;        // Loop to find the proper        // divisor of every number        // from 1 to N        for (i = 1; i <= n; ++i)            sum += (n / i) * i;               return sum - n * (n + 1) / 2;    }           // Driver Code    public static void Main(String []args)    {        int n = 4;        Console.WriteLine(properDivisorSum(n));               n = 5;        Console.WriteLine(properDivisorSum(n));              }} // This code is contributed by 29AjayKumar

## Javascript

 

Output:

5
6

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

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