Sum of divisors of factorial of a number

Given a number n, we need to calculate the sum of divisors of factorial of the number.

Examples:

Input : 4
Output : 60
Factorial of 4 is 24. Divisors of 24 are
1 2 3 4 6 8 12 24, sum of these is 60.

Input : 6
Output : 2418


A Simple Solution is to first compute factorial of given number, then count number divisors of the factorial. This solution is not efficient and may cause overflow due to factorial computation.

Java

// Java program to find sum of proper divisor of
// factorial of a number
import java.io.*;
import java.util.*;
  
public class Division
{
    // function to calculate factorial
    static int fac(int n)
    {
        if (n = = 0)
            return 1;
        return n*fac(n-1);
    }
  
    // function to calculate sum of divisor
    static int div(int x)
    {
        int ans = 0;
        for (int i = 1; i< = x; i++)
            if (x%i == 0)
                ans + = i;
        return ans;
    }
  
    // Returns sum of divisors of n!
    static int sumFactDiv(int n)
    {
        return div(fact(n));
    }
  
    // driver program
    public static void main(String args[])
    {
        int n = 4;
        System.out.println(sumFactDiv(n));
    }
} 

C#

// C# program to find sum of proper 
// divisor of factorial of a number
using System;
class Division {
      
    // function to calculate factorial
    static int fac(int n)
    {
        if (n == 0)
            return 1;
        return n * fac(n - 1);
    }
  
    // function to calculate
    // sum of divisor
    static int div(int x)
    {
        int ans = 0;
        for (int i = 1; i <= x; i++)
            if (x % i == 0)
                ans += i;
        return ans;
    }
  
    // Returns sum of divisors of n!
    static int sumFactDiv(int n)
    {
        return div(fac(n));
    }
  
    // Driver Code
    public static void Main()
    {
        int n = 4;
        Console.Write(sumFactDiv(n));
    }
} 
  
// This code is contributed by Nitin Mittal.


Output : 

60

An efficient solution is based on Legendre’s formula. Below are the steps.

  1. Find all prime numbers less than or equal to n (input number). We can use Sieve Algorithm for this. Let n be 6. All prime numbers less than 6 are {2, 3, 5}.
  2. For each prime number p find the largest power of it that divides n!. We use below Legendre’s formula formula for this purpose.

    3. The value of largest power that divides p is floor value of each term n/p + n/(p2) + n/(p3) + ……
    Let these values be exp1, exp2, exp3, .. Using the above formula, we get below values for n = 6.

    • The largest power of 2 that divides 6!, exp1 = 4.
    • The largest power of 3 that divides 6!, exp2 = 2.
    • The largest power of 5 that divides 6!, exp3 = 1.
  3. The result is based on the Divisor Function.
    // C++ program to find sum of divisors in n!
    #include<bits/stdc++.h>
    #include<math.h>
    using namespace std;
      
    // allPrimes[] stores all prime numbers less
    // than or equal to n.
    vector<int> allPrimes;
      
    // Fills above vector allPrimes[] for a given n
    void sieve(int n)
    {
        // Create a boolean array "prime[0..n]" and
        // initialize all entries it as true. A value
        // in prime[i] will finally be false if i is
        // not a prime, else true.
        vector<bool> prime(n+1, true);
      
        // Loop to update prime[]
        for (int p = 2; p*p <= n; p++)
        {
            // If prime[p] is not changed, then it
            // is a prime
            if (prime[p] == true)
            {
                // Update all multiples of p
                for (int i = p*2; i <= n; i += p)
                    prime[i] = false;
            }
        }
      
        // Store primes in the vector allPrimes
        for (int p = 2; p <= n; p++)
            if (prime[p])
                allPrimes.push_back(p);
    }
      
    // Function to find all result of factorial number
    int factorialDivisors(int n)
    {
        sieve(n);  // create sieve
      
        // Initialize result
        int result = 1;
      
        // find exponents of all primes which divides n
        // and less than n
        for (int i = 0; i < allPrimes.size(); i++)
        {
            // Current divisor
            int p = allPrimes[i];
      
            // Find the highest power (stored in exp)'
            // of allPrimes[i] that divides n using
            // Legendre's formula.
            int exp = 0;
            while (p <= n)
            {
                exp = exp + (n/p);
                p = p*allPrimes[i];
            }
      
            // Using the divisor function to calculate
            // the sum
            result = result*(pow(allPrimes[i], exp+1)-1)/
                                        (allPrimes[i]-1);
        }
      
        // return total divisors
        return result;
    }
      
    // Driver program to run the cases
    int main()
    {
        cout << factorialDivisors(4);
        return 0;
    }

    Output:

    
  15334088

    This article is contributed by Pramod Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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    Improved By : nitin mittal