Find sum of divisors of all the divisors of a natural number

Given a natural number n, the task is to find sum of divisors of all the divisors of n.

Examples:

Input : n = 54
Output : 232
Divisors of 54 = 1, 2, 3, 6, 9, 18, 27, 54.
Sum of divisors of 1, 2, 3, 6, 9, 18, 27, 54 
are 1, 3, 4, 12, 13, 39, 40, 120 respectively.
Sum of divisors of all the divisors of 54 = 
1 + 3 + 4 + 12 + 13 + 39 + 40 + 120 = 232.

Input : n = 10
Output : 28
Divisors of 10 are 1, 2, 5, 10
Sums of divisors of divisors are 
1, 3, 6, 18.
Overall sum = 1 + 3 + 6 + 18 = 28

Using the fact that any number n can be expressed as product of prime factors, n = p₁^k1 x p₂^k2 x … where p₁, p₂, … are prime numbers. 
All the divisors of n can be expressed as p₁^a x p₂^b x …, where 0 <= a <= k1 and 0 <= b <= k2. 
Now sum of divisors will be sum of all power of p₁ – p₁⁰, p₁¹,…., p₁^k1 multiplied by all power of p₂ – p₂⁰, p₂¹,…., p₂^k1
Sum of Divisor of n 
= (p₁⁰ x p₂⁰) + (p₁¹ x p₂⁰) +…..+ (p₁^k1 x p₂⁰) +….+ (p₁⁰ x p₂¹) + (p₁¹ x p₂¹) +…..+ (p₁^k1 x p₂¹) +……..+ 
   (p₁⁰ x p₂^k2) + (p₁¹ x p₂^k2) +……+ (p₁^k1 x p₂^k2). 
= (p₁⁰ + p₁¹ +…+ p₁^k1) x p₂⁰ + (p₁⁰ + p₁¹ +…+ p₁^k1) x p₂¹ +…….+ (p₁⁰ + p₁¹ +…+ p₁^k1) x p₂^k2. 
= (p₁⁰ + p₁¹ +…+ p₁^k1) x (p₂⁰ + p₂¹ +…+ p₂^k2).

Now, the divisors of any p^a, for p as prime, are p⁰, p¹,……, p^a. And sum of divisors will be (p^(a+1) – 1)/(p -1), let it define by f(p). 
So, sum of divisors of all divisor will be, 
= (f(p₁⁰) + f(p₁¹) +…+ f(p₁^k1)) x (f(p₂⁰) + f(p₂¹) +…+ f(p₂^k2)).

So, given a number n, by prime factorization we can find the sum of divisors of all the divisors. But in this problem we are given that n is product of element of array. So, find prime factorization of each element and by using the fact a^b x a^c = a^b+c.

Below is the implementation of this approach:  

C++

// C++ program to find sum of divisors of all
// the divisors of a natural number.
#include<bits/stdc++.h>
using namespace std;

// Returns sum of divisors of all the divisors
// of n
int sumDivisorsOfDivisors(int n)
{
    // Calculating powers of prime factors and
    // storing them in a map mp[].
    map<int, int> mp;
    for (int j=2; j<=sqrt(n); j++)
    {
        int count = 0;
        while (n%j == 0)
        {
            n /= j;
            count++;
        }

        if (count)
            mp[j] = count;
    }

    // If n is a prime number
    if (n != 1)
        mp[n] = 1;

    // For each prime factor, calculating (p^(a+1)-1)/(p-1)
    // and adding it to answer.
    int ans = 1;
    for (auto it : mp)
    {
        int pw = 1;
        int sum = 0;

        for (int i=it.second+1; i>=1; i--)
        {
            sum += (i*pw);
            pw *= it.first;
        }
        ans *= sum;
    }

    return ans;
}

// Driven Program
int main()
{
    int n = 10;
    cout << sumDivisorsOfDivisors(n);
    return 0;
}

Java

// Java program to find sum of divisors of all
// the divisors of a natural number.
import java.util.HashMap;

class GFG
{

    // Returns sum of divisors of all the divisors
    // of n
    public static int sumDivisorsOfDivisors(int n)
    {

        // Calculating powers of prime factors and
        // storing them in a map mp[].
        HashMap<Integer, Integer> mp = new HashMap<>();
        for (int j = 2; j <= Math.sqrt(n); j++)
        {
            int count = 0;
            while (n % j == 0)
            {
                n /= j;
                count++;
            }
            if (count != 0)
                mp.put(j, count);
        }

        // If n is a prime number
        if (n != 1)
            mp.put(n, 1);

        // For each prime factor, calculating (p^(a+1)-1)/(p-1)
        // and adding it to answer.
        int ans = 1;

        for (HashMap.Entry<Integer, Integer> entry : mp.entrySet())
        {
            int pw = 1;
            int sum = 0;
            for (int i = entry.getValue() + 1; i >= 1; i--)
            {
                sum += (i * pw);
                pw *= entry.getKey();
            }
            ans *= sum;
        }

        return ans;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 10;
        System.out.println(sumDivisorsOfDivisors(n));
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python3 program to find sum of divisors
# of all the divisors of a natural number.
import math as mt

# Returns sum of divisors of all
# the divisors of n
def sumDivisorsOfDivisors(n):

    # Calculating powers of prime factors
    # and storing them in a map mp[].
    mp = dict()
    for j in range(2, mt.ceil(mt.sqrt(n))):

        count = 0
        while (n % j == 0):
            n //= j
            count += 1

        if (count):
            mp[j] = count

    # If n is a prime number
    if (n != 1):
        mp[n] = 1

    # For each prime factor, calculating
    # (p^(a+1)-1)/(p-1) and adding it to answer.
    ans = 1
    for it in mp:
        pw = 1
        summ = 0

        for i in range(mp[it] + 1, 0, -1):
            summ += (i * pw)
            pw *= it

        ans *= summ

    return ans

# Driver Code
n = 10
print(sumDivisorsOfDivisors(n))

# This code is contributed
# by mohit kumar 29

C#

// C# program to find sum of divisors of all
// the divisors of a natural number.
using System;
using System.Collections.Generic;

class GFG
{

    // Returns sum of divisors of
    // all the divisors of n
    public static int sumDivisorsOfDivisors(int n)
    {

        // Calculating powers of prime factors and
        // storing them in a map mp[].
        Dictionary<int,
                   int> mp = new Dictionary<int,
                                            int>();
        for (int j = 2; j <= Math.Sqrt(n); j++)
        {
            int count = 0;
            while (n % j == 0)
            {
                n /= j;
                count++;
            }
            if (count != 0)
                mp.Add(j, count);
        }

        // If n is a prime number
        if (n != 1)
            mp.Add(n, 1);

        // For each prime factor,
        // calculating (p^(a+1)-1)/(p-1)
        // and adding it to answer.
        int ans = 1;

        foreach(KeyValuePair<int, int> entry in mp)
        {
            int pw = 1;
            int sum = 0;
            for (int i = entry.Value + 1;
                     i >= 1; i--)
            {
                sum += (i * pw);
                pw = entry.Key;
            }
            ans *= sum;
        }

        return ans;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int n = 10;
        Console.WriteLine(sumDivisorsOfDivisors(n));
    }
}

// This code is contributed
// by Princi Singh

Javascript

<script>
// Javascript program to find sum of divisors of all
// the divisors of a natural number.

     // Returns sum of divisors of all the divisors
    // of n
    function sumDivisorsOfDivisors(n)
    {
        // Calculating powers of prime factors and
        // storing them in a map mp[].
        let mp = new Map();
        for (let j = 2; j <= Math.sqrt(n); j++)
        {
            let count = 0;
            while (n % j == 0)
            {
                n = Math.floor(n/j);
                count++;
            }
            if (count != 0)
                mp.set(j, count);
        }

        // If n is a prime number
        if (n != 1)
            mp.set(n, 1);

        // For each prime factor, calculating (p^(a+1)-1)/(p-1)
        // and adding it to answer.
        let ans = 1;

        for (let [key, value] of mp.entries())
        {
            let pw = 1;
            let sum = 0;
            for (let i = value + 1; i >= 1; i--)
            {
                sum += (i * pw);
                pw = key;
            }
            ans *= sum;
        }

        return ans;
    }

    // Driver code
    let n = 10;
    document.write(sumDivisorsOfDivisors(n));



// This code is contributed by patel2127
</script>

Output:

28

Time Complexity: O(√n log n) 
Auxiliary Space: O(n)

Optimizations : 
For the cases when there are multiple inputs for which we need find the value, we can use Sieve of Eratosthenes as discussed in this post.

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