Program to check if N is a Octadecagon number

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

Octadecagon Number is a 18-sided polygon. The first few Octadecagon Numbers are 1, 18, 51, 100, 165, 246, 343, …

Examples:

Input: N = 18 
Output: Yes 
Explanation:
Second Octadecagon number is 18.

Input: N = 30 
Output: No 

Approach:

1. The K^th term of the Octadecagon number is given as

2. As we have to check whether the given number can be expressed as an Octadecagon Number or not. This can be checked as: 

=> 
=> 

3. If the value of K calculated using the above formula is an integer, then N is an Octadecagon Number.
4. Else N is not an Octadecagon 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
// Octadecagon Number
bool isOctadecagon(int N)
{
    float n
        = (14 + sqrt(128 * N + 196))
          / 32;

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

// Driver Code
int main()
{
    // Given Number
    int N = 18;

    // Function call
    if (isOctadecagon(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
// octadecagon Number
public static boolean isOctadecagon(int N)
{
    double n = (14 + Math.sqrt(128 * N +
                               196)) / 32;

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

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

    // Given Number
    int N = 18;

    // Function call
    if (isOctadecagon(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
# octadecagon number
def isOctadecagon(N):

    n = (14 + math.sqrt(128 * N + 196)) // 32

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

# Driver code
if __name__=='__main__':

    # Given number
    N = 18

    # Function Call
    if isOctadecagon(N):
        print('Yes')
    else:
        print('No')

# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
class GFG{

// Function to check if N is a
// octadecagon Number
public static bool isOctadecagon(int N)
{
    double n = (14 + Math.Sqrt(128 * N +
                               196)) / 32;

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

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

    // Given Number
    int N = 18;

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

// This code is contributed by 29AjayKumar

Javascript

<script>

// javascript program for the above approach


/// Function to check if N is a
// Octadecagon Number
function isOctadecagon( N)
{
    let n
        = (14 + Math.sqrt(128 * N + 196))
          / 32;

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


// Driver Code

    // Given Number
    let N = 18;

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

// This code contributed by gauravrajput1

</script>

Output

Yes

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