Program to check if N is a Hexadecagonal Number

Given an integer N, the task is to check if N is a Hexadecagonal Number or not. If the number N is an Hexadecagonal Number then print "Yes" else print "No".

Hexadecagonal Number is class of figurate number and a perfect squares. It has 16-sided polygon called Hexadecagon or Hexakaidecagon. The nth Hexadecagonal Number counts the sixteen number of dots and all others dots are surrounding to its successive layer.The first few Hexadecagonal Numbers are 1, 16, 45, 88, 145, 216... 

Examples:  

Input: N = 16 
Output: Yes 
Explanation: 
Second hexadecagonal number is 16.

Input: N = 30 
Output: No 

Approach:  

1. The K^th term of the hexadecagonal number is given as
K^{th} Term = \frac{14*K^{2} - 12*K}{2}  

2. As we have to check that the given number can be expressed as a Hexadecagonal Number or not. This can be checked as: 

=> N = \frac{14*K^{2} - 12*K}{2}      
=> K = \frac{12 + \sqrt{112*N + 144}}{28}      
 

3. If the value of K calculated using the above formula is an integer, then N is a Hexadecagonal Number.

4. Else N is not a Hexadecagonal Number.

Below is the implementation of the above approach: 

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if N is a
// hexadecagonal number
bool ishexadecagonal(int N)
{
    float n
        = (12 + sqrt(112 * N + 144))
          / 28;

    // Condition to check if the
    // number is a hexadecagonal number
    return (n - (int)n) == 0;
}

// Driver Code
int main()
{
    // Given Number
    int N = 16;

    // Function call
    if (ishexadecagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

// Java program for the above approach 
import java.lang.Math;

class GFG{
    
// Function to check if N is a 
// hexadecagonal number 
public static boolean ishexadecagonal(int N) 
{ 
    double n = (12 + Math.sqrt(112 * N +
                               144)) / 28; 
    
    // Condition to check if the 
    // number is a hexadecagonal number 
    return (n - (int)n) == 0; 
} 
    
// Driver code    
public static void main(String[] args)
{
        
    // Given number 
    int N = 16; 
    
    // Function call 
    if (ishexadecagonal(N))
    { 
        System.out.println("Yes");
    } 
    else
    { 
        System.out.println("No");
    } 
}
}

// This code is contributed by divyeshrabadiya07    

# Python3 program for the above approach
from math import sqrt

# Function to check if N is 
# a hexadecagonal number
def ishexadecagonal(N):

    n = (12 + sqrt(112 * N + 144)) / 28;

    # Condition to check if the number 
    # is a hexadecagonal number
    return (n - int(n)) == 0;

# Driver code
if __name__ == "__main__":

    # Given number
    N = 16;

    # Function call
    if (ishexadecagonal(N)):
        print("Yes");
    else:
        print("No");
    
# This code is contributed by AnkitRai01

// C# program for the above approach 
using System;

class GFG{
    
// Function to check if N is a 
// hexadecagonal number 
public static bool ishexadecagonal(int N) 
{ 
    double n = (12 + Math.Sqrt(112 * N +
                             144)) / 28; 
    
    // Condition to check if the 
    // number is a hexadecagonal number 
    return (n - (int)n) == 0; 
} 
    
// Driver code 
public static void Main(string[] args)
{
        
    // Given number 
    int N = 16; 
    
    // Function call 
    if (ishexadecagonal(N))
    { 
        Console.Write("Yes");
    } 
    else
    { 
        Console.Write("No");
    } 
}
}

// This code is contributed by rutvik_56

<script>

// javascript program for the above approach


// Function to check if N is a
// hexadecagonal number
function ishexadecagonal( N)
{
    let n
        = (12 + Math.sqrt(112 * N + 144))
          / 28;

    // Condition to check if the
    // number is a hexadecagonal number
    return (n - parseInt(n)) == 0;
}

// Driver Code

    // Given Number
    let N = 16;

    // Function Call
    if (ishexadecagonal(N)) {
        document.write( "Yes");
    }
    else {
        document.write( "No");
    }

// This code contributed by Rajput-Ji 

</script>

Yes

Time Complexity: O(logN) because sqrt() function is being used
Auxiliary Space: O(1)