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Program to check if N is a Centered Tridecagonal Number

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  • Last Updated : 08 Apr, 2021

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

Centered tridecagonal number represents a dot at the center and other dots surrounding the center dot in the successive tridecagonal(13 sided polygon) layer. The first few Centered tridecagonal numbers are 1, 14, 40, 79 … 

Examples: 

Input: N = 14 
Output: Yes 
Explanation: 
Second Centered tridecagonal number is 14.

Input: N = 30 
Output: No   

Approach:  

1. The Kth term of the Centered Tridecagonal Number is given as
K^{th} Term = \frac{{13*K^{2} - 13*K + 2}}{2}
 

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

=> N = \frac{{13*K^{2} - 13*K + 2}}{2}
=> K = \frac{13 + \sqrt{104*N + 65}}{26}

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

4. Else the number N is not a Centered Tridecagonal 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 tridecagonal number
bool isCenteredtridecagonal(int N)
{
    float n
        = (13 + sqrt(104 * N + 65))
          / 26;
 
    // Condition to check if the N
    // is a Centered tridecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 14;
 
    // Function call
    if (isCenteredtridecagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
// Function to check if the number N
// is a centered tridecagonal number
static boolean isCenteredtridecagonal(int N)
{
    float n = (float) ((13 + Math.sqrt(104 * N +
                                       65)) / 26);
 
    // Condition to check if the N
    // is a centered tridecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Number
    int N = 14;
 
    // Function call
    if (isCenteredtridecagonal(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 tridecagonal number
def isCenteredtridecagonal(N):
 
    n = (13 + np.sqrt(104 * N + 65)) / 26
 
    # Condition to check if N
    # is centered tridecagonal number
    return (n - int(n)) == 0
 
# Driver Code
N = 14
 
# Function call
if (isCenteredtridecagonal(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 tridecagonal number
static bool isCenteredtridecagonal(int N)
{
    float n = (float) ((13 + Math.Sqrt(104 * N +
                                       65)) / 26);
 
    // Condition to check if the N
    // is a centered tridecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given Number
    int N = 14;
 
    // Function call
    if (isCenteredtridecagonal(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 tridecagonal number
function isCenteredtridecagonal(N)
{
    let n
        = (13 + Math.sqrt(104 * N + 65))
          / 26;
 
    // Condition to check if the N
    // is a Centered tridecagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
// Given Number
let N = 14;
 
// Function call
if (isCenteredtridecagonal(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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