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Program to check if N is a Centered dodecagonal number

  • Last Updated : 08 Apr, 2021

Given an integer N, the task is to check if N is a Centered Dodecagonal Number or not. If the number N is a Centered Dodecagonal Number then print “Yes” else print “No”.

Centered Dodecagonal Number represents a dot in the center and other dots surrounding it in successive Dodecagonal Number(12 sided polygon) layers. The first few Centered Dodecagonal Numbers are 1, 13, 37, 73 … 

Examples:  

Input: N = 13 
Output: Yes 
Explanation: 
Second Centered dodecagonal number is 13.

Input: N = 30 
Output: No 



Approach:  

1 The Kth term of the Centered Dodecagonal Number is given as:
K^{th} Term = 6*K^{2} - 6*K + 1
 

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

=> N = {6*K^{2} - 6*K + 1}
=> K = \frac{6 + \sqrt{24*N + 12}}{12}      

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

4. Else the number N is not a Centered Dodecagonal 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 number N
// is a Centered dodecagonal number
bool isCentereddodecagonal(int N)
{
    float n
        = (6 + sqrt(24 * N + 12))
          / 12;
 
    // Condition to check if N
    // is a Centered Dodecagonal Number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 13;
 
    // Function call
    if (isCentereddodecagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if number N
// is a centered dodecagonal number
static boolean isCentereddodecagonal(int N)
{
    float n = (float) ((6 + Math.sqrt(24 * N +
                                      12)) / 12);
 
    // Condition to check if N is a
    // centered dodecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Number
    int N = 13;
 
    // Function call
    if (isCentereddodecagonal(N))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python3 program for the above approach
import numpy as np
 
# Function to check if the number N
# is a centered dodecagonal number
def isCentereddodecagonal(N):
 
    n = (6 + np.sqrt(24 * N + 12)) / 12
 
    # Condition to check if N
    # is a centered dodecagonal number
    return (n - int(n)) == 0
 
# Driver Code
N = 13
 
# Function call
if (isCentereddodecagonal(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 number N
// is a centered dodecagonal number
static bool isCentereddodecagonal(int N)
{
    float n = (float) ((6 + Math.Sqrt(24 * N +
                                      12)) / 12);
 
    // Condition to check if N is a
    // centered dodecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given Number
    int N = 13;
 
    // Function call
    if (isCentereddodecagonal(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 number N
// is a Centered dodecagonal number
function isCentereddodecagonal(N)
{
    let n
        = (6 + Math.sqrt(24 * N + 12))
          / 12;
 
    // Condition to check if N
    // is a Centered Dodecagonal Number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
// Given Number
let N = 13;
 
// Function call
if (isCentereddodecagonal(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by subham348.
</script>
Output: 
Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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