# Program to check if N is a Icosagonal Number

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

Icosagonal Number is a twenty-sided polygon. The number derived from the figurative class. There are different patterns observed in this series. The dots are countable, arrange in a specific way of position, and create a diagram. All the dots have common dots points, all other dots are connected to these points and except this common point the dots connected to their ith dots with their respective successive layer… The first few Icosagonal numbers are 1, 20, 57, 112, 185, 276…

Examples:

Input: N = 20
Output: Yes
Explanation:
Second Icosagonal Number is 20.

Input: N = 30
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. The Kth term of the Icosagonal Number is given as 2. As we have to check that the given number can be expressed as a icosagonal number or not. This can be checked as follows –

=> => 3. If the value of K calculated using the above formula is an integer, then N is an Icosagonal Number.
4. Else the number N is not an Icosagonal Number.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach  #include  using namespace std;     // Function to check if the number  // N is a icosagonal number  bool iicosagonal(int N)  {      float n          = (16 + sqrt(144 * N + 256))            / 36;         // Condition to check if the      // N is a icosagonal number      return (n - (int)n) == 0;  }     // Driver Code  int main()  {      // Given Number      int N = 20;         // Function call      if (iicosagonal(N)) {          cout << "Yes";      }      else {          cout << "No";      }      return 0;  }

## Java

 // Java program for the above approach  import java.util.*;     class GFG{     // Function to check if the number  // N is a icosagonal number  static boolean iicosagonal(int N)  {      float n = (float)((16 + Math.sqrt(144 * N +                                         256)) / 36);         // Condition to check if the      // N is a icosagonal number      return (n - (int)n) == 0;  }     // Driver Code  public static void main(String[] args)  {             // Given Number      int N = 20;         // Function call      if (iicosagonal(N))      {          System.out.print("Yes");      }      else      {          System.out.print("No");      }  }  }     // This code is contributed by Rohit_ranjan

## Python3

 # Python3 program for the above approach  import numpy as np     # Function to check if the number   # N is a icosagonal number   def iicosagonal(N):         n = (16 + np.sqrt(144 * N + 256)) / 36        # Condition to check if the       # N is a icosagonal number       return (n - int(n)) == 0    # Driver Code   N = 20    # Function call   if (iicosagonal(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 icosagonal number   static bool iicosagonal(int N)   {       float n = (float)((16 + Math.Sqrt(144 * N +                                         256)) / 36);                                                // Condition to check if the       // N is a icosagonal number       return (n - (int)n) == 0;   }      // Driver Code   public static void Main(string[] args)   {              // Given Number       int N = 20;          // Function call       if (iicosagonal(N))       {           Console.Write("Yes");       }       else     {           Console.Write("No");       }   }   }      // This code is contributed by rutvik_56

Output:

Yes


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