Sum of divisors of factorial of a number

Difficulty Level : Medium
Last Updated : 02 Jul, 2021

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

Examples:

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

Below is the implementation of the above approach:

C++

// C++ program to find sum of proper divisor of
// factorial of a number
#include <bits/stdc++.h>
using namespace std;
 
// function to calculate factorial
int fact(int n)
{
    if (n == 0)
        return 1;
    return n * fact(n - 1);
}
 
// function to calculate sum of divisor
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!
int sumFactDiv(int n)
{
    return div(fact(n));
}
 
// Driver Code
int main()
{
    int n = 4;
    cout << sumFactDiv(n);
}
 
// This code is contributed
// by Akanksha Rai

C

// C program to find sum of proper divisor of
// factorial of a number
#include <stdio.h>
// function to calculate factorial
 
int fact(int n) {
    if (n == 0)
        return 1;
    return n * fact(n - 1);
}
 
// function to calculate sum of divisor
 
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!
 
int sumFactDiv(int n) {
    return div(fact(n));
}
 
// driver program
 
int main() {
    int n = 4;
    printf("%d",sumFactDiv(n));
}

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 fact(int n)
    {
        if (n == 0)
            return 1;
        return n*fact(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));
    }
}

Python3

# Python 3 program to find sum of proper
# divisor of factorial of a number
 
# function to calculate factorial
def fact(n):
     
    if (n == 0):
        return 1
    return n * fact(n - 1)
 
# function to calculate sum
# of divisor
def div(x):
    ans = 0;
    for i in range(1, x + 1):
        if (x % i == 0):
            ans += i
    return ans
 
# Returns sum of divisors of n!
def sumFactDiv(n):
    return div(fact(n))
 
# Driver Code
n = 4
print(sumFactDiv(n))
 
# This code is contributed
# by Rajput-Ji

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.

PHP

<?php
// PHP program to find sum of proper
// divisor of factorial of a number
 
// function to calculate factorial
function fact($n)
{
    if ($n == 0)
        return 1;
    return $n * fact($n - 1);
}
 
// function to calculate sum of divisor
function div($x)
{
    $ans = 0;
    for ($i = 1; $i<= $x; $i++)
        if ($x % $i == 0)
            $ans += $i;
    return $ans;
}
 
// Returns sum of divisors of n!
function sumFactDiv($n)
{
    return div(fact($n));
}
 
// Driver Code
$n = 4;
echo sumFactDiv($n);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>
// JavaScript program to find
// sum of proper divisor of
// factorial of a number
 
// function to calculate factorial
function fact(n)
{
    if (n == 0)
        return 1;
    return n * fact(n - 1);
}
 
// function to calculate sum of divisor
function div(x)
{
    let ans = 0;
    for (let i = 1; i<= x; i++)
        if (x % i == 0)
            ans += i;
    return ans;
}
 
// Returns sum of divisors of n!
function sumFactDiv(n)
{
    return div(fact(n));
}
 
// Driver Code
 
    let n = 4;
    document.write(sumFactDiv(n));
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output :

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

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}.
For each prime number, p find the largest power of it that divides n!. We use Legendre’s formula for this purpose.
- 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.
The result is based on the Divisor Function.

C++

// 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;
}

Java

// Java program to find sum of divisors in n!
import java.util.*;
 
class GFG{
// allPrimes[] stores all prime numbers less
// than or equal to n.
static ArrayList<Integer> allPrimes=new ArrayList<Integer>();
 
// Fills above vector allPrimes[] for a given n
static 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.
    boolean[] prime=new boolean[n+1];
 
    // 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] == false)
        {
            // Update all multiples of p
            for (int i = p*2; i <= n; i += p)
                prime[i] = true;
        }
    }
 
    // Store primes in the vector allPrimes
    for (int p = 2; p <= n; p++)
        if (prime[p]==false)
            allPrimes.add(p);
}
 
// Function to find all result of factorial number
static 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.get(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.get(i);
        }
 
        // Using the divisor function to calculate
        // the sum
        result = result*((int)Math.pow(allPrimes.get(i), exp+1)-1)/
                                    (allPrimes.get(i)-1);
    }
 
    // return total divisors
    return result;
}
 
// Driver program to run the cases
public static void main(String[] args)
{
    System.out.println(factorialDivisors(4));
}
}
// This code is contributed by mits

Python3

# Python3 program to find sum of divisors in n!
 
# allPrimes[] stores all prime numbers
# less than or equal to n.
allPrimes = [];
 
# Fills above vector allPrimes[]
# for a given n
def sieve(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.
    prime = [True] * (n + 1);
 
    # Loop to update prime[]
    p = 2;
    while (p * p <= n):
         
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == True):
             
            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                prime[i] = False;
        p += 1;
 
    # Store primes in the vector allPrimes
    for p in range(2, n + 1):
        if (prime[p]):
            allPrimes.append(p);
 
# Function to find all result of factorial number
def factorialDivisors(n):
 
    sieve(n); # create sieve
 
    # Initialize result
    result = 1;
 
    # find exponents of all primes which
    # divides n and less than n
    for i in range(len(allPrimes)):
         
        # Current divisor
        p = allPrimes[i];
 
        # Find the highest power (stored in exp)'
        # of allPrimes[i] that divides n using
        # Legendre's formula.
        exp = 0;
        while (p <= n):
            exp = exp + int(n / p);
            p = p * allPrimes[i];
 
        # Using the divisor function to 
        # calculate the sum
        result = int(result * (pow(allPrimes[i], exp + 1) - 1) /
                                           (allPrimes[i] - 1));
 
    # return total divisors
    return result;
 
# Driver Code
print(factorialDivisors(4));
 
# This code is contributed by mits

C#

// C# program to find sum of divisors in n!
using System;
using System.Collections;
 
class GFG{
// allPrimes[] stores all prime numbers less
// than or equal to n.
static ArrayList allPrimes=new ArrayList();
 
// Fills above vector allPrimes[] for a given n
static 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.
    bool[] prime=new bool[n+1];
 
    // 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] == false)
        {
            // Update all multiples of p
            for (int i = p*2; i <= n; i += p)
                prime[i] = true;
        }
    }
 
    // Store primes in the vector allPrimes
    for (int p = 2; p <= n; p++)
        if (prime[p]==false)
            allPrimes.Add(p);
}
 
// Function to find all result of factorial number
static 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.Count; i++)
    {
        // Current divisor
        int p = (int)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*(int)allPrimes[i];
        }
 
        // Using the divisor function to calculate
        // the sum
        result = result*((int)Math.Pow((int)allPrimes[i], exp+1)-1)/
                                    ((int)allPrimes[i]-1);
    }
 
    // return total divisors
    return result;
}
 
// Driver program to run the cases
static void Main()
{
    Console.WriteLine(factorialDivisors(4));
}
}
// This code is contributed by mits

PHP

<?php
// PHP program to find sum of divisors in n!
 
// allPrimes[] stores all prime numbers less
// than or equal to n.
$allPrimes=array();
 
// Fills above vector allPrimes[] for a given n
function sieve($n)
{
    global $allPrimes;
    // 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.
    $prime=array_fill(0,$n+1,true);
 
    // Loop to update prime[]
    for ($p = 2; $p*$p <= $n; $p++)
    {
        // If prime[p] is not changed, then it
        // is a prime
        if ($prime[$p] == true)
        {
            // Update all multiples of p
            for ($i = $p*2; $i <= $n; $i += $p)
                $prime[$i] = false;
        }
    }
 
    // Store primes in the vector allPrimes
    for ($p = 2; $p <= $n; $p++)
        if ($prime[$p])
            array_push($allPrimes,$p);
}
 
// Function to find all result of factorial number
function factorialDivisors($n)
{
    global $allPrimes;
    sieve($n); // create sieve
 
    // Initialize result
    $result = 1;
 
    // find exponents of all primes which divides n
    // and less than n
    for ($i = 0; $i < count($allPrimes); $i++)
    {
        // Current divisor
        $p = $allPrimes[$i];
 
        // Find the highest power (stored in exp)'
        // of allPrimes[i] that divides n using
        // Legendre's formula.
        $exp = 0;
        while ($p <= $n)
        {
            $exp = $exp + (int)($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
 
    print(factorialDivisors(4));
 
 
// This code is contributed by mits
?>

Javascript

<script>
 
// Javascript program to find sum of divisors in n!
 
// allPrimes[] stores all prime numbers less
// than or equal to n.
let allPrimes=[];
 
// Fills above vector allPrimes[] for a given n
function sieve(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.
    let prime = new Array(n + 1);
    for(let i = 0; i < n + 1; i++)
        prime[i] = true;
  
    // Loop to update prime[]
    for(let 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(let i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
  
    // Store primes in the vector allPrimes
    for(let p = 2; p <= n; p++)
        if (prime[p])
            allPrimes.push(p);
}
 
// Function to find all result of
// factorial number
function factorialDivisors(    n)
{
     
    // Create sieve
    sieve(n);
  
    // Initialize result
    let result = 1;
  
    // Find exponents of all primes which
    // divides n and less than n
    for(let i = 0; i < allPrimes.length; i++)
    {
         
        // Current divisor
        let p = allPrimes[i];
  
        // Find the highest power (stored in exp)'
        // of allPrimes[i] that divides n using
        // Legendre's formula.
        let exp = 0;
         
        while (p <= n)
        {
            exp = exp + Math.floor(n/p);
            p = p*allPrimes[i];
        }
  
        // Using the divisor function to calculate
        // the sum
        result = Math.floor(result * (Math.pow(
                 allPrimes[i], exp + 1) - 1) /
                     (allPrimes[i] - 1));
    }
  
    // Return total divisors
    return result;
}
 
// Driver code
document.write(factorialDivisors(4));
 
// This code is contributed by rag2127
 
</script>

Output:

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