Program to check if N is a Hendecagonal Number

Last Updated : 15 Dec, 2022

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

Hendecagonal Number is a figurate number that extends the concept of triangular and square numbers to the decagon(11-sided polygon). The nth hendecagonal number counts the number of dots in a pattern of n nested decagons, all sharing a common corner, where the ith hendecagon in the pattern has sides made of i dots spaced one unit apart from each other. The first few hendecagonal numbers are 1, 11, 30, 58, 95, 141...

Examples:

Input: N = 11
Output: Yes
Explanation:
Second hendecagonal number is 11.

Input: N = 40
Output: No

Approach:

1. The K^th term of the Hendecagonal Number is given as
K^{th} Term = \frac{9*K^{2} - 7*K}{2}
2. As we have to check whether the given number can be expressed as a Hendecagonal Number or not. This can be checked as:

=> N = \frac{9*K^{2} - 7*K}{2}
=> K = \frac{7 + \sqrt{72*N + 49}}{18}

3. If the value of K calculated using the above formula is an integer, then N is a Hendecagonal Number.
4. Else N is not a Hendecagonal 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
// Hendecagonal Number
bool ishendecagonal(int N)
{
    float n
        = (7 + sqrt(72 * N + 49))
          / 18;

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

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

    // Function call
    if (ishendecagonal(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 
// hendecagonal number 
public static boolean ishendecagonal(int N) 
{ 
    double n = (7 + Math.sqrt(72 * N + 49)) / 18; 
    
    // Condition to check if the 
    // number is a hendecagonal number 
    return (n - (int)n) == 0; 
} 

// Driver code    
public static void main(String[] args)
{
        
    // Given number 
    int N = 11; 
    
    // Function call 
    if (ishendecagonal(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
# Hendecagonal Number
def ishendecagonal(N):

    n = (7 + math.sqrt(72 * N + 49))// 18;

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

# Driver Code

# Given Number
N = 11;

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

# This code is contributed by Nidhi_biet

// C# program for the above approach 
using System;
class GFG{
    
// Function to check if N is a 
// hendecagonal number 
public static bool ishendecagonal(int N) 
{ 
    double n = (7 + Math.Sqrt(72 * N + 49)) / 18; 
    
    // Condition to check if the 
    // number is a hendecagonal number 
    return (n - (int)n) == 0; 
} 

// Driver code 
public static void Main(string[] args)
{
        
    // Given number 
    int N = 11; 
    
    // Function call 
    if (ishendecagonal(N))
    { 
        Console.Write("Yes");
    } 
    else
    { 
        Console.Write("No\n");
    } 
}
}

// This code is contributed by rutvik_56

JavaScript

<script>

// javascript program for the above approach


// Function to check if N is a
// Hendecagonal Number
function ishendecagonal( N)
{
    let n
        = (7 + Math.sqrt(72 * N + 49))
          / 18;

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


// Driver Code

    // Given Number
    let N = 11;

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

// This code contributed by gauravrajput1 

</script>