Hyperfactorial in Java

Last Updated : 17 Mar 2025

Hyperfactorial of a number is the outcome of multiplying a given number of continuous numbers from 1 to that number, where each number is raised to its power.

Mathematically,

Thus,

H(1) = 1 ^ 1 = 1

H(2) = 1 ^ 1 × 2 ^ 2 = 1 × 4 = 4

H(3) = 1 ^ 1 × 2 ^ 2 × 3 ^ 3 = 1 × 4 × 27 = 108

H(4) = 1 ^ 1 × 2 ^ 2 × 3 ^ 3 × 4 ^ 4 = 1 × 4 × 27 × 256 = 27648

H(5) = 1 ^ 1 × 2 ^ 2 × 3 ^ 3 × 4 ^ 4 × 5 ^ 5 = 1 × 4 × 27 × 256 × 3125 = 86400000

Java Program to Find the Hyperfactorial Number

The following program uses the mathematical formula to find the Hyperfactorial numbers.

FileName: HyperFactorialNumber.java

// A program that finds the first 5 hyperfactorial numbers 

public class HyperFactorialNumber 
{
// the method finds the hyperFactorial number using the number no
public int findHyperFactorialNo(int no)
{
int ans = 1;

for(int i = 1; i <= no; i++)
{
int hyperFactorialNo = i;

for(int j = 2; j <= i; j++)
{
hyperFactorialNo = hyperFactorialNo * i;
}

ans = ans * hyperFactorialNo;

}
return ans;
}
// main method
public static void main(String argvs[])
{
// creating an object of the class HyperFactorialNumber
HyperFactorialNumber obj = new HyperFactorialNumber();
System.out.println("The first 5 HyperFactorial numbers are: ");
// loop to iterate till 5 to find
// the first 5 hyperFactorial numbers
for(int i = 1; i <= 5; i++)
{
int hyperFactorialNo = obj.findHyperFactorialNo(i);
System.out.print(hyperFactorialNo + " ");
}
}
}

Output:

The first 5 HyperFactorial numbers are: 
1 4 108 27648 86400000

To find each hyperfactorial number, the above program is taking O(n ^ 2) time. We can do the optimization to reduce the time complexity. Consider the following program.

FileName: HyperFactorialNumber1.java

// A program that finds the first 5 hyperfactorial numbers 

public class HyperFactorialNumber1 
{
// the method finds the value of x ^ y
public int findPow(int x, int y)
{
int ans = 1;
while(true)
{
if( y == 0)
{
break;
}
if(y % 2 == 1)
{
ans = ans * x;
y = y - 1;
}

else
{
x = x * x;
y = y / 2;
}
}

return ans;
}

// the method finds the hyperFactorial number using the number no
public int findHyperFactorialNo(int no)
{
int ans = 1;

for(int i = 1; i <= no; i++)
{
ans = ans * findPow(i , i);
}
return ans;
}
// main method
public static void main(String argvs[])
{
// creating an object of the class HyperFactorialNumber1
HyperFactorialNumber1 obj = new HyperFactorialNumber1();

System.out.println("The first 5 HyperFactorial numbers are: ");
// loop to iterate till 5 to find
// the first 5 hyperFactorial numbers
for(int i = 1; i <= 5; i++)
{
int hyperFactorialNo = obj.findHyperFactorialNo(i);
System.out.print(hyperFactorialNo + " ");
}
}
}

Output:

The first 5 HyperFactorial numbers are: 
1 4 108 27648 86400000

The findPow() method has the time complexity of O(log(n)), and this method has been called by a for-loop that runs with the time complexity of O(n). Thus, the total time complexity is O(nlog(n)), which is better than the previous code.

Finding Hyperfactorial Number Using Recursion

Using recursion also, one can find the HyperFactorial numbers. The recursive formula to find the hyperfactorial number is:

H(1) = 1

H(p) = H(p - 1) × p ^ p, where p >= 2

Thus,

H(2) = H(2 - 1) × 2 ^ 2 = H(1) × 4 = 1 × 4 = 4

H(3) = H(3 - 1) × 3 ^ 3 = H(2) × 27 = 4 × 27 = 108

H(4) = H(4 - 1) × 4 ^ 4 = H(3) × 256 = 108 × 256 = 27648

H(5) = H(5 - 1) × 5 ^ 5 = H(4) × 15625 = 27648 × 15625 = 86400000

Let's translate the above recursive formula into the Java code.

FileName: HyperFactorialNumber2.java

// A program that finds the first 5 hyperfactorial numbers recursively

public class HyperFactorialNumber2
{
    
// the method finds the value of x ^ y
public int findPow(int x, int y)
{
int ans = 1;
while(true)
{
if( y == 0)
{
break;
}
if(y % 2 == 1)
{
ans = ans * x;
y = y - 1;
}

else
{
x = x * x;
y = y / 2;
}
}

return ans;
}


// the method finds the hyperfactorial number recursively
public int findHyperFactorialNo(int no)
{
// handling the base case
if(no == 1)
{
return 1;
}

else
{
// recursively find the hyperfactorial numbers
return findHyperFactorialNo(no - 1) * findPow(no, no);
}

}


// main method
public static void main(String argvs[])
{

// creating an object of the class HyperFactorialNumber2
HyperFactorialNumber2 obj = new HyperFactorialNumber2();

System.out.println("The first 5 HyperFactorial numbers are: ");


// loop to iterate till 5 to find
// the first 5 hyperFactorial numbers
for(int i = 1; i <= 5; i++)
{
int hyperFactorialNo = obj.findHyperFactorialNo(i);

System.out.print(hyperFactorialNo + " ");

}

}
}

Output:

The first 5 HyperFactorial numbers are: 
1 4 108 27648 86400000

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