Given a positive integer n, the task is to print the nth Hermite number.
Hermite Number: In mathematics, Hermite numbers are values of Hermite Polynomials at zero arguments.
The Recurrence Relation of Hermite polynomials at x = 0 is given by,
Hn = -2 * (n – 1) * Hn – 2
where H0 = 1 and H1 = 0
First few terms of Hermite number sequence are:
1, 0, -2, 0, 12, 0, -120, 0, 1680, 0, -30240
Examples:
Input: n = 6
Output: -120
Input: n = 8
Output: 1680
Naive Approach: Write a recursive function implementing the above recurrence relation.
Below is the implementation of the above approach:
C++
// C++ program to find nth Hermite number#include <bits/stdc++.h>using namespace std;// Function to return nth Hermite numberint getHermiteNumber(int n){ // Base conditions if (n == 0) return 1; if (n == 1) return 0; else return -2 * (n - 1) * getHermiteNumber(n - 2);}// Driver Codeint main(){ int n = 6; // Print nth Hermite number cout << getHermiteNumber(n); return 0;} |
Java
// Java program to find nth Hermite numberimport java.util.*;class GFG { // Function to return nth Hermite number static int getHermiteNumber(int n) { // Base condition if (n == 0) return 1; else if (n == 1) return 1; else return -2 * (n - 1) * getHermiteNumber(n - 2); } // Driver Code public static void main(String[] args) { int n = 6; // Print nth Hermite number System.out.println(getHermiteNumber(n)); }} |
Python3
# Python3 program to find nth Hermite number# Function to return nth Hermite numberdef getHermiteNumber( n): # Base conditions if n == 0 : return 1 if n == 1 : return 0 else : return (-2 * (n - 1) * getHermiteNumber(n - 2))# Driver Coden = 6# Print nth Hermite numberprint(getHermiteNumber(n));# This code is contributed # by Arnab Kundu |
C#
// C# program to find nth Hermite numberusing System;class GFG { // Function to return nth Hermite number static int getHermiteNumber(int n) { // Base condition if (n == 0) return 1; else if (n == 1) return 1; else return -2 * (n - 1) * getHermiteNumber(n - 2); } // Driver Code public static void Main() { int n = 6; // Print nth Hermite number Console.WriteLine(getHermiteNumber(n)); }} |
PHP
<?php// PHP program to find nth Hermite number// Function to return nth Hermite numberfunction getHermiteNumber($n){ // Base conditions if ($n == 0) return 1; if ($n == 1) return 0; else return -2 * ($n - 1) * getHermiteNumber($n - 2);}// Driver Code$n = 6;// Print nth Hermite numberecho getHermiteNumber($n);// This code is contributed by ajit.?> |
Javascript
<script> // Javascript program to // find nth Hermite number // Function to return nth Hermite number function getHermiteNumber(n) { // Base condition if (n == 0) return 1; else if (n == 1) return 1; else return -2 * (n - 1) * getHermiteNumber(n - 2); } let n = 6; // Print nth Hermite number document.write(getHermiteNumber(n)); </script> |
Output:
-120
Time Complexity: O(n)
Auxiliary Space: O(n)
Efficient Approach: It is clear from the Hermite sequence that if n is odd then nth Hermite number will be 0. Now, nth Hermite number can be found using,
Where (n – 1)!! = 1 * 3 * 5 * (n – 1) i.e. double factorial of (n – 1)
Below is the implementing of the above approach:
C++
// C++ program to find nth Hermite number#include <bits/stdc++.h>using namespace std;// Utility function to calculate// double factorial of a numberint doubleFactorial(int n){ int fact = 1; for (int i = 1; i <= n; i = i + 2) { fact = fact * i; } return fact;}// Function to return nth Hermite numberint hermiteNumber(int n){ // If n is even then return 0 if (n % 2 == 1) return 0; // If n is odd else { // Calculate double factorial of (n-1) // and multiply it with 2^(n/2) int number = (pow(2, n / 2)) * doubleFactorial(n - 1); // If n/2 is odd then // nth Hermite number will be negative if ((n / 2) % 2 == 1) number = number * -1; // Return nth Hermite number return number; }}// Driver Codeint main(){ int n = 6; // Print nth Hermite number cout << hermiteNumber(n); return 0;} |
Java
// Java program to find nth Hermite numberimport java.util.*;class GFG { // Utility function to calculate // double factorial of a number static int doubleFactorial(int n) { int fact = 1; for (int i = 1; i <= n; i = i + 2) { fact = fact * i; } return fact; } // Function to return nth Hermite number static int hermiteNumber(int n) { // If n is even then return 0 if (n % 2 == 1) return 0; // If n is odd else { // Calculate double factorial of (n-1) // and multiply it with 2^(n/2) int number = (int)(Math.pow(2, n / 2)) * doubleFactorial(n - 1); // If n/2 is odd then // nth Hermite number will be negative if ((n / 2) % 2 == 1) number = number * -1; // Return nth Hermite number return number; } } // Driver Code public static void main(String[] args) { int n = 6; // Print nth Hermite number System.out.println(hermiteNumber(n)); }} |
Python3
# Python 3 program to find nth# Hermite numberfrom math import pow# Utility function to calculate# double factorial of a numberdef doubleFactorial(n): fact = 1 for i in range(1, n + 1, 2): fact = fact * i return fact# Function to return nth Hermite numberdef hermiteNumber(n): # If n is even then return 0 if (n % 2 == 1): return 0 # If n is odd else: # Calculate double factorial of (n-1) # and multiply it with 2^(n/2) number = ((pow(2, n / 2)) * doubleFactorial(n - 1)) # If n/2 is odd then nth Hermite # number will be negative if ((n / 2) % 2 == 1): number = number * -1 # Return nth Hermite number return number # Driver Codeif __name__ == '__main__': n = 6 # Print nth Hermite number print(int(hermiteNumber(n)))# This code is contributed by# Surendra_Gangwar |
C#
// C# program to find nth Hermite numberusing System;class GFG { // Utility function to calculate // double factorial of a number static int doubleFactorial(int n) { int fact = 1; for (int i = 1; i <= n; i = i + 2) { fact = fact * i; } return fact; } // Function to return nth Hermite number static int hermiteNumber(int n) { // If n is even then return 0 if (n % 2 == 1) return 0; // If n is odd else { // Calculate double factorial of (n-1) // and multiply it with 2^(n/2) int number = (int)(Math.Pow(2, n / 2)) * doubleFactorial(n - 1); // If n/2 is odd then // nth Hermite number will be negative if ((n / 2) % 2 == 1) number = number * -1; // Return nth Hermite number return number; } } // Driver Code public static void Main() { int n = 6; // Print nth Hermite number Console.WriteLine(hermiteNumber(n)); }} |
PHP
<?php// PHP program to find nth Hermite number// Utility function to calculate double// factorial of a numberfunction doubleFactorial($n){ $fact = 1; for ($i = 1; $i <= $n; $i = $i + 2) { $fact = $fact * $i; } return $fact;}// Function to return nth Hermite numberfunction hermiteNumber($n){ // If n is even then return 0 if ($n % 2 == 1) return 0; // If n is odd else { // Calculate double factorial of (n-1) // and multiply it with 2^(n/2) $number = (pow(2, $n / 2)) * doubleFactorial($n - 1); // If n/2 is odd then nth Hermite // number will be negative if (($n / 2) % 2 == 1) $number = $number * -1; // Return nth Hermite number return $number; }}// Driver Code$n = 6;// Print nth Hermite numberecho hermiteNumber($n); // This code is contributed by akt_mit?> |
Javascript
<script>// Javascript program to find nth Hermite number// Utility function to calculate// double factorial of a numberfunction doubleFactorial(n){ var fact = 1; for (var i = 1; i <= n; i = i + 2) { fact = fact * i; } return fact;}// Function to return nth Hermite numberfunction hermiteNumber(n){ // If n is even then return 0 if (n % 2 == 1) return 0; // If n is odd else { // Calculate double factorial of (n-1) // and multiply it with 2^(n/2) var number = (Math.pow(2, n / 2)) * doubleFactorial(n - 1); // If n/2 is odd then // nth Hermite number will be negative if ((n / 2) % 2 == 1) number = number * -1; // Return nth Hermite number return number; }}// Driver Codevar n = 6;// Print nth Hermite numberdocument.write( hermiteNumber(n));</script> |
Output:
-120
Time Complexity: O(n)
Auxiliary Space: O(1)
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