# Program to check if N is a Icosihenagonal number

Given an integer N, the task is to check if it is a Icosihenagonal number or not.

Icosihenagonal number is class of figurate number. It has 21 – sided polygon called Icosihenagon. The n-th Icosihenagonal number counts the 21 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icosihenagonal numbers are 1, 21, 60, 118, 195, 291, 406…

Examples:

Input: N = 21
Output: Yes
Explanation:
Second icosihenagonal number is 21.

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 icosihenagonal number is given as

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

=>
=>

3. Finally, check the value of computed using this formulae is an integer, which means that N is a icosihenagonal number.

Below is the implementation of the above approach:

## C++

 // C++ implementation to check that  // a number is icosihenagonal number or not     #include     using namespace std;     // Function to check that the  // number is a icosihenagonal number  bool isicosihenagonal(int N)  {      float n          = (17 + sqrt(152 * N + 289))            / 38;         // Condition to check if the      // number is a icosihenagonal number      return (n - (int)n) == 0;  }     // Driver Code  int main()  {      int i = 21;         // Function call      if (isicosihenagonal(i)) {          cout << "Yes";      }      else {          cout << "No";      }      return 0;  }

## Java

 // Java implementation to check that a   // number is icosihenagonal number or not  class GFG{     // Function to check that the number  // is a icosihenagonal number  static boolean isicosihenagonal(int N)  {      float n = (float) ((17 + Math.sqrt(152 * N +                                          289)) / 38);         // Condition to check if the number      // is a icosihenagonal number      return (n - (int)n) == 0;  }     // Driver Code  public static void main(String[] args)  {      int i = 21;         // Function call      if (isicosihenagonal(i))      {          System.out.print("Yes");      }      else     {          System.out.print("No");      }  }  }     // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation to check that   # a number is icosihenagonal number or not   import math      # Function to check that the number  # is a icosihenagonal number   def isicosihenagonal(N):          n = (17 + math.sqrt(152 * N + 289)) / 38        # Condition to check if the number      # is a icosihenagonal number       return (n - int(n)) == 0    # Driver Code   i = 21    # Function call   if isicosihenagonal(i):       print("Yes")  else :       print("No")         # This code is contributed by divyamohan123

## C#

 // C# implementation to check that a   // number is icosihenagonal number or not  using System;     class GFG{     // Function to check that the number  // is a icosihenagonal number  static bool isicosihenagonal(int N)  {      float n = (float)((17 + Math.Sqrt(152 * N +                                         289)) / 38);         // Condition to check if the number      // is a icosihenagonal number      return (n - (int)n) == 0;  }     // Driver Code  public static void Main()  {      int i = 21;         // Function call      if (isicosihenagonal(i))      {          Console.Write("Yes");      }      else     {          Console.Write("No");      }  }  }     // This code is contributed by Code_Mech

Output:

Yes


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