Program to check if N is a Hexagonal Number or not

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Given a number N, check if it is Hexagonal or not. If it is then print “Yes” otherwise output “No”.

Examples:

Input: N = 6
Output: Yes
Explanation:
6 is the second hexagonal number.

Input: N = 14
Output: No
Explanation:
The second hexagonal number is 6 while the third hexagonal number is 15. Hence 14 is not a hexagonal number.

Approach: To solve the problem mentioned above we have to note that the nth hexagonal Number is given by the formula: H(n) = n * (2n – 1). The formula indicates that the n-th hexagonal number depends quadratically on n. Therefore, find the positive integral root of N = H(n) equation.

Therefore, H(n) = nth hexagonal number and N is the given Number. Hence solve the following equation:

Here, H(n) = N
=> n * (2n – 1) = N
=> 2 * n * n – n – N = 0
The positive root of this equation is:
n = (1 + sqrt(8 N + 1)) / 4

After obtaining the value for n, check if it is an integer or not where n is an integer if n – floor(n) is 0.

Below is the implementation of the above approach:

C++

// C++ Program to check
// if N is a Hexagonal Number
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check
// if number is hexagonal
bool isHexagonal(int N)
{
    float val = 8 * N + 1;
 
    float x = 1 + sqrt(val);
 
    // Calculate the value for n
    float n = (x) / 4;
 
    // Check if n - floor(n)
    // is equal to 0
    if ((n - (int)n) == 0)
        return true;
 
    else
        return false;
}
 
// Driver code
int main()
{
    int N = 14;
 
    if (isHexagonal(N) == true)
        cout << "Yes" << endl;
 
    else
        cout << "No" << endl;
 
    return 0;
}

Java

// Java program to check
// if N is a hexagonal number
import java.util.*;
 
class GFG {
 
// Function to check
// if number is hexagonal
static boolean isHexagonal(int N)
{
    float val = 8 * N + 1;
 
    float x = 1 + (float)Math.sqrt(val);
 
    // Calculate the value for n
    float n = (x) / 4;
 
    // Check if n - floor(n)
    // is equal to 0
    if ((n - (int)n) == 0)
        return true;
 
    else
        return false;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 14;
 
    if (isHexagonal(N) == true)
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to check
# if N is a hexagonal number
from math import sqrt
 
# Function to check
# if number is hexagonal
def isHexagonal(N):
     
    val = 8 * N + 1
    x = 1 + sqrt(val)
 
    # Calculate the value for n
    n = x / 4
 
    # Check if n - floor(n)
    # is equal to 0
    if ((n - int(n)) == 0):
        return True
    else:
        return False
 
# Driver code
if __name__ == '__main__':
     
    N = 14
     
    if (isHexagonal(N) == True):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by BhupendraSingh

C#

// C# program to check if 
// N is a hexagonal number
using System;
 
class GFG{
 
// Function to check
// if number is hexagonal
static bool isHexagonal(int N)
{
    float val = 8 * N + 1;
    float x = 1 + (float)Math.Sqrt(val);
 
    // Calculate the value for n
    float n = (x) / 4;
 
    // Check if n - floor(n)
    // is equal to 0
    if ((n - (int)n) == 0)
    {
        return true;
    }
    else
    {
        return false;
    }
}
 
// Driver code
public static void Main(string[] args)
{
    int N = 14;
 
    if (isHexagonal(N) == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
 
// Javascript program to check
// if N is a hexagonal number
 
// Function to check
// if number is hexagonal
function isHexagonal(N)
{
    let val = 8 * N + 1;
    let x = 1 + Math.sqrt(val);
     
    // Calculate the value for n
    let n = (x) / 4;
     
    // Check if n - floor(n)
    // is equal to 0
    if ((n - parseInt(n)) == 0)
    {
        return true;
    }
    else
    {
        return false;
    }
}
 
// Driver code
let N = 14;
 
if (isHexagonal(N) == true)
{
    document.write("Yes" + "</br>");
}
else
{
    document.write("No");
}
 
// This code is contributed by Ankita saini
    
</script>

Output:

No

Time Complexity: O(log(N)), for calculation square root.
Auxiliary Space: O(1)

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