Sum of divisors of factorial of a number

Last Updated : 13 Sep, 2022

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 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 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# 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 :

Time Complexity: O(n!)

Auxiliary Space: O(1)

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# 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:

Time Complexity: O(n*log(log(n)))

Auxiliary Space: O(n)

GFact | 2x + 1(where x > 0) is prime if and only if x is a power of 2

kartik

Improve

Article Tags :

Practice Tags :

Sum of divisors of factorial of a number

Similar Reads

Basics of Number Theory

Modular Arithmetic

Number Theory

Game Theory

Practice Problems

Miscellaneous Practice Problems