Program to check if N is a Centered Hexadecagonal Number

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

Centered Hexadecagonal Number represents a dot in the centre and other dots around it in successive Hexadecagonal(16 sided polygon) layers... The first few Centered Hexadecagonal Numbers are 1, 17, 49, 97, 161, 241 ... 
 

Examples:  

Input: N = 17 
Output: Yes 
Explanation: 
Second Centered hexadecagonal number is 17.
Input: N = 20 
Output: No 

Approach:  

1. The K^th term of the Centered Hexadecagonal Number is given as
K^{th} Term = 8*K^{2} - 8*K + 1       
 

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

=> N = 8*K^{2} - 8*K + 1       
=> K = \frac{8 + \sqrt{32*N + 32}}{16}  

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

4. Else the number N is not a Centered 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 the number N
// is a Centered hexadecagonal number
bool isCenteredhexadecagonal(int N)
{
    float n
        = (8 + sqrt(32 * N + 32))
          / 16;

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

// Driver Code
int main()
{
    // Given Number
    int N = 17;

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

// Java program for the above approach
import java.io.*; 
import java.util.*; 

class GFG { 
    
// Function to check if the number N
// is a centered hexadecagonal number
static boolean isCenteredhexadecagonal(int N)
{
    double n = (8 + Math.sqrt(32 * N + 32)) / 16;

    // Condition to check if the N is a
    // centered hexadecagonal number
    return (n - (int)n) == 0;
}
    
// Driver code 
public static void main(String[] args) 
{ 
    
    // Given Number
    int N = 17;

    // Function call
    if (isCenteredhexadecagonal(N)) 
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
} 
} 

// This code is contributed by coder001

# Python3 program for the above approach 
import numpy as np

# Function to check if the number N 
# is a Centered hexadecagonal number 
def isCenteredhexadecagonal(N): 

    n = (8 + np.sqrt(32 * N + 32)) / 16

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

# Driver Code 
N = 17

# Function call 
if (isCenteredhexadecagonal(N)):
    print ("Yes")
else:
    print ("No")

# This code is contributed by PratikBasu

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

class GFG { 
    
// Function to check if the number N
// is a centered hexadecagonal number
static bool isCenteredhexadecagonal(int N)
{
    double n = (8 + Math.Sqrt(32 * N + 32)) / 16;

    // Condition to check if the N is a
    // centered hexadecagonal number
    return (n - (int)n) == 0;
}
    
// Driver code 
public static void Main(string[] args) 
{ 
    
    // Given Number
    int N = 17;

    // Function call
    if (isCenteredhexadecagonal(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 the number N
// is a Centered hexadecagonal number
function isCenteredhexadecagonal( N)
{
    let n
        = (8 + Math.sqrt(32 * N + 32))
          / 16;

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

// Driver Code

    // Given Number
    let N = 17;

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

// This code contributed by Rajput-Ji 

</script>

Yes

Time Complexity: O(logN), for using inbuilt sqrt function.
Auxiliary Space: O(1)

spp____

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Article Tags :

• Mathematical
• DSA

Practice Tags :

• Mathematical