Program to check if N is a Tetradecagonal Number

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

Tetradecagonal Number is 14-sided polygon called Tetrakaidecagon or Tetradecagon and belongs to the figurative number. The nth tetradecagonal number dotted with some dots and create a series of the pattern. They have a common sharing corner point and dotted with their spaces to each other. The dots continue with nth nested loop.The first few Tetradecagonal Numbers are 1, 14, 39, 76, 125, 186, …

Examples:

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

Input: N = 40 
Output: No  

Approach:

• The K^th term of the tetradecagonal number is given as
  
• As we have to check whether the given number can be expressed as a Tetradecagonal Number or not. This can be checked as: 

=> 
=> 

• If the value of K calculated using the above formula is an integer, then N is a Tetradecagonal Number.
• Else N is not a Tetradecagonal 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 N is a
// Tetradecagonal Number
bool istetradecagonal(int N)
{
    float n
        = (10 + sqrt(96 * N + 100))
          / 24;

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

// Driver Code
int main()
{
    // Given Number
    int N = 11;

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

Java

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

class GFG{

// Function to check if N is a
// tetradecagonal number
public static boolean istetradecagonal(int N)
{
    double n = (10 + Math.sqrt(96 * N +
                               100)) / 24;

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

// Driver Code
public static void main(String[] args)
{

    // Given number
    int N = 11;

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

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
import math

# Function to check if N is a
# Tetradecagonal Number
def istetradecagonal(N):

    n = (10 + math.sqrt(96 * N + 100)) / 24

    # Condition to check if the
    # number is a tetradecagonal number
    if (n - int(n)) == 0:
        return True

    return False

# Driver Code

# Given Number
N = 11

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

# This code is contributed by shubhamsingh10

C#

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

class GFG{

// Function to check if N is a
// tetradecagonal number
public static bool istetradecagonal(int N)
{
    double n = (10 + Math.Sqrt(96 * N +
                               100)) / 24;

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

// Driver Code
static public void Main ()
{

    // Given number
    int N = 11;

    // Function call
    if (istetradecagonal(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}

// This code is contributed by shubhamsingh10

Javascript

<script>

// Javascript program for the above approach

// Function to check if N is a
// Tetradecagonal Number
function istetradecagonal(N)
{
    n = (10 + Math.sqrt(96 * N + 100))
          / 24;

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

// Driver Code
// Given Number
N = 11;
// Function call
if (istetradecagonal(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}

</script>

Output:

No

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