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Program to check if N is a Centered Hexadecagonal Number

  • Last Updated : 23 Mar, 2021

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 Kth 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++




// 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




// 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




# 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#




// 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

Javascript




<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>
Output: 
Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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