Program to check if N is a Icosidigonal Number

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

Icosidigonal number:
The polygon has many gons, depends on their gonal number series. In mathematics, there are a number of gonal numbers and the Icosidigonal Number is one of them and these numbers have 22 -sided polygon(icosidigon). An Icosidigonal Number belong to the class of figurative number. They have one common dots points and other dots pattern is arranged in an n-th nested Icosidigon pattern.
The first few Icosidigonal numbers are 1, 22, 63, 124, 205, 306…

Examples:

Input: N = 22
Output: Yes
Explanation:
Second Icosidigonal number is 22.

Input: 30
Output: No



Approach:

  1. The Kth term of the Icosidigonal number is given as

    K^{th} Term =  \frac{20*K^{2} - 18*K}{2}

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

    => N =  \frac{20*K^{2} - 18*K}{2}
    => K = \frac{18 + \sqrt{160*N + 324}}{40}

  3. If the value of K calculated using the above formula is an integer, then N is a Icosidigonal Number.
  4. Else N is not a Icosidigonal Number.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if the number N
// is a Icosidigonal number
bool isIcosidigonal(int N)
{
    float n
        = (18 + sqrt(160 * N + 324))
          / 40;
  
    // Condition to check if the
    // number is a Icosidigonal number
    return (n - (int)n) == 0;
}
  
// Driver Code
int main()
{
    // Given Number
    int N = 22;
  
    // Function call
    if (isIcosidigonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
class GFG{
  
// Function to check if the number N
// is a icosidigonal number
static boolean isIcosidigonal(int N)
{
    float n = (float) ((18 + Math.sqrt(160 * N + 
                                       324)) / 40);
  
    // Condition to check if the number 
    // is a icosidigonal number
    return (n - (int)n) == 0;
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given number
    int N = 22;
  
    // Function call
    if (isIcosidigonal(N))
    {
        System.out.print("Yes");
    }
    else 
    {
        System.out.print("No");
    }
}
}
  
// This code is contributed by Amit Katiyar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
import numpy as np
  
# Function to check if the number N
# is a icosidigonal number
def isIcosidigonal(N):
  
    n = (18 + np.sqrt(160 * N + 324)) / 40
  
    # Condition to check if N 
    # is a icosidigonal number
    return (n - int(n)) == 0
  
# Driver Code 
N = 22
  
# Function call 
if (isIcosidigonal(N)):
    print ("Yes"
else:
    print ("No")
  
# This code is contributed by PratikBasu

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach 
using System;
  
class GFG{ 
  
// Function to check if the number N 
// is a icosidigonal number 
static bool isIcosidigonal(int N) 
    float n = (float) ((18 + Math.Sqrt(160 * N + 
                                    324)) / 40); 
  
    // Condition to check if the number 
    // is a icosidigonal number 
    return (n - (int)n) == 0; 
  
// Driver Code 
public static void Main(string[] args) 
      
    // Given number 
    int N = 22; 
  
    // Function call 
    if (isIcosidigonal(N)) 
    
        Console.Write("Yes"); 
    
    else
    
        Console.Write("No"); 
    
  
// This code is contributed by rutvik_56

chevron_right


Output:

Yes

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.