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# 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
Recommended Practice

Using the fact that any number n can be expressed as product of prime factors, n = p1k1 x p2k2 x … where p1, p2, … are prime numbers.
All the divisors of n can be expressed as p1a x p2b x …, where 0 <= a <= k1 and 0 <= b <= k2.
Now sum of divisors will be sum of all power of p1 – p10, p11,…., p1k1 multiplied by all power of p2 – p20, p21,…., p2k1
Sum of Divisor of n
= (p10 x p20) + (p11 x p20) +…..+ (p1k1 x p20) +….+ (p10 x p21) + (p11 x p21) +…..+ (p1k1 x p21) +……..+
(p10 x p2k2) + (p11 x p2k2) +……+ (p1k1 x p2k2).
= (p10 + p11 +…+ p1k1) x p20 + (p10 + p11 +…+ p1k1) x p21 +…….+ (p10 + p11 +…+ p1k1) x p2k2
= (p10 + p11 +…+ p1k1) x (p20 + p21 +…+ p2k2).

Now, the divisors of any pa, for p as prime, are p0, p1,……, pa. 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(p1k1)) x (f(p20) + f(p21) +…+ f(p2k2)).

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 ab x ac = ab+c.

Below is the implementation of this approach:

## C++

 // C++ program to find sum of divisors of all// the divisors of a natural number.#includeusing namespace std; // Returns sum of divisors of all the divisors// of nint sumDivisorsOfDivisors(int n){    // Calculating powers of prime factors and    // storing them in a map mp[].    map 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 Programint 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 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 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 ndef 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 Coden = 10print(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 mp = new Dictionary();        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 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



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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