Sum of product of all integers upto N with their count of divisors

Given a positive integer N, the task is to find the sum of the product of all the integers in the range [1, N] with their count of divisors. 

Examples:

Input: N = 3
Output: 11
Explanation:
Number of positive divisors of 1 is 1( i.e. 1 itself). Therefore, f(1) = 1.
Number of positive divisors of 2 is 2( i.e. 1, 2). Therefore, f(2) = 2.
Number of positive divisors of 3 is 2( i.e. 1, 3). Therefore, f(3) = 2.
So the answer is 1*f(1) + 2*f(2) + 3*f(3) = 1*1 + 2*2 + 3*2 = 11.

Input: N = 4
Output: 23
Explanation:
Here f(1) = 1, f(2) = 2, f(3) = 2 and f(4) = 3. So, the answer is 1*1 + 2*2 + 3*2 + 4*3 = 23.

Naive Approach: The naive approach is to traverse from 1 to N and find the sum of all the numbers with their count of divisors.

Time Complexity: O(N*sqrt(N))
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to consider the total contribution each number makes to the answer. Below are the steps: 

1. For any number X in the range [1, N], X contributes K to the sum for each K from 1 to N such that K is a multiple of X.
2. Observe that the list of these K is of form i, 2i, 3i, …, Fi where F is the number of multiples of i between 1 and N.
3. Hence, to find the sum of these numbers generated above is given by 
   

i*(1+2+3 + … F) = i*(F*(F+1))/2.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the sum of the
// product of all the integers and
// their positive divisors up to N
long long sumOfFactors(int N)
{
    long long ans = 0;

    // Iterate for every number
    // between 1 and N
    for (int i = 1; i <= N; i++) {

        // Find the first multiple
        // of i between 1 and N
        long long first = i;

        // Find the last multiple
        // of i between 1 and N
        long long last = (N / i) * i;

        // Find the total count of
        // multiple of in [1, N]
        long long factors
            = (last - first) / i + 1;

        // Compute the contribution of i
        // using the formula
        long long totalContribution
            = (((factors) * (factors + 1)) / 2) * i;

        // Add the contribution
        // of i to the answer
        ans += totalContribution;
    }

    // Return the result
    return ans;
}

// Driver code
int main()
{
    // Given N
    int N = 3;

    // function call
    cout << sumOfFactors(N);
}

Java

// Java program for the above approach
class GFG{

// Function to find the sum of the
// product of all the integers and
// their positive divisors up to N
static int sumOfFactors(int N)
{
    int ans = 0;

    // Iterate for every number
    // between 1 and N
    for (int i = 1; i <= N; i++)
    {

        // Find the first multiple
        // of i between 1 and N
        int first = i;

        // Find the last multiple
        // of i between 1 and N
        int last = (N / i) * i;

        // Find the total count of
        // multiple of in [1, N]
        int factors = (last - first) / i + 1;

        // Compute the contribution of i
        // using the formula
        int totalContribution = (((factors) *
                                  (factors + 1)) / 2) * i;

        // Add the contribution
        // of i to the answer
        ans += totalContribution;
    }

    // Return the result
    return ans;
}

// Driver code
public static void main(String[] args)
{
    // Given N
    int N = 3;

    // function call
    System.out.println(sumOfFactors(N));
}
}

// This code is contributed by Ritik Bansal

Python3

# Python3 implementation of
# the above approach

# Function to find the sum of the
# product of all the integers and
# their positive divisors up to N
def sumOfFactors(N):

    ans = 0

    # Iterate for every number
    # between 1 and N
    for i in range(1, N + 1):

        # Find the first multiple
        # of i between 1 and N
        first = i

        # Find the last multiple
        # of i between 1 and N
        last = (N // i) * i

        # Find the total count of
        # multiple of in [1, N]
        factors = (last - first) // i + 1

        # Compute the contribution of i
        # using the formula
        totalContribution = (((factors *
                              (factors + 1)) // 2) * i)

        # Add the contribution
        # of i to the answer
        ans += totalContribution

    # Return the result
    return ans

# Driver Code

# Given N
N = 3

# Function call
print(sumOfFactors(N))

# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
class GFG{

// Function to find the sum of the
// product of all the integers and
// their positive divisors up to N
static int sumOfFactors(int N)
{
    int ans = 0;

    // Iterate for every number
    // between 1 and N
    for (int i = 1; i <= N; i++)
    {

        // Find the first multiple
        // of i between 1 and N
        int first = i;

        // Find the last multiple
        // of i between 1 and N
        int last = (N / i) * i;

        // Find the total count of
        // multiple of in [1, N]
        int factors = (last - first) / i + 1;

        // Compute the contribution of i
        // using the formula
        int totalContribution = (((factors) *
                                  (factors + 1)) / 2) * i;

        // Add the contribution
        // of i to the answer
        ans += totalContribution;
    }

    // Return the result
    return ans;
}

// Driver code
public static void Main(String[] args)
{
    // Given N
    int N = 3;

    // function call
    Console.WriteLine(sumOfFactors(N));
}
}

// This code is contributed by gauravrajput1

Javascript

<script>
// javascript program for the above approach
// Function to find the sum of the
    // product of all the integers and
    // their positive divisors up to N
    function sumOfFactors(N)
    {
        var ans = 0;

        // Iterate for every number
        // between 1 and N
        for (i = 1; i <= N; i++)
        {

            // Find the first multiple
            // of i between 1 and N
            var first = i;

            // Find the last multiple
            // of i between 1 and N
            var last = parseInt(N / i) * i;

            // Find the total count of
            // multiple of in [1, N]
            var factors = parseInt((last - first) / i )+ 1;

            // Compute the contribution of i
            // using the formula
            var totalContribution = parseInt(((factors) * (factors + 1)) / 2) * i;

            // Add the contribution
            // of i to the answer
            ans += totalContribution;
        }

        // Return the result
        return ans;
    }

    // Driver code

        // Given N
        var N = 3;

        // function call
        document.write(sumOfFactors(N));

// This code is contributed by todaysgaurav.
</script>

Output:

11

Time Complexity: O(N)
Auxiliary Space: O(1)