Hyperfactorial in Java

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

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

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

Output:

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