# Program to check if N is a Icositrigonal number

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

Icositrigonal number is a class of figurate number. It has 23 – sided polygon called Icositrigon. The N-th Icositrigonal number count’s the 23 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icositrigonol numbers are 1, 23, 66, 130, 215, 321, 448 …

Examples:

Input: N = 23
Output: Yes
Explanation:
Second icositrigonal number is 23.

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 icositrigonal number is given as 2. As we have to check that the given number can be expressed as a icositrigonal 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 icositrigonal number.

Below is the implementation of the above approach:

## C++

 // C++ implementation to check that  // a number is a icositrigonal number or not     #include     using namespace std;     // Function to check that the  // number is a icositrigonal number  bool isicositrigonal(int N)  {      float n          = (19 + sqrt(168 * N + 361))            / 42;         // Condition to check if the      // number is a icositrigonal number      return (n - (int)n) == 0;  }     // Driver Code  int main()  {      int i = 23;         // Function call      if (isicositrigonal(i)) {          cout << "Yes";      }      else {          cout << "No";      }      return 0;  }

## Java

 // Java implementation to check that  // a number is a icositrigonal number or not  import java.util.*;  class GFG{     // Function to check that the  // number is a icositrigonal number  static boolean isicositrigonal(int N)  {      float n = (float)(19 + Math.sqrt(168 * N + 361)) / 42;         // Condition to check if the      // number is a icositrigonal number      return (n - (int)n) == 0;  }     // Driver Code  public static void main(String args[])  {      int i = 23;         // Function call      if (isicositrigonal(i))       {          System.out.print("Yes");      }      else     {          System.out.print("No");      }  }  }     // This code is contributed by Akanksha_Rai

## Python3

 # Python3 implementation to check that a    # number is a icositrigonal number or not   import math     # Function to check that the number   # is a icositrigonal number   def isicositrigonal(N):          n = (19 + math.sqrt(168 * N + 361)) / 42        # Condition to check if the number        # is a icositrigonal number       return (n - int(n)) == 0    # Driver Code   i = 23    # Function call   if (isicositrigonal(i)):       print("Yes")   else:       print("No")      # This code is contributed by divyamohan123

## C#

 // C# implementation to check that  // a number is a icositrigonal number or not  using System;  class GFG{     // Function to check that the  // number is a icositrigonal number  static bool isicositrigonal(int N)  {      float n = (float)(19 + Math.Sqrt(168 * N + 361)) / 42;         // Condition to check if the      // number is a icositrigonal number      return (n - (int)n) == 0;  }     // Driver Code  public static void Main()  {      int i = 23;         // Function call      if (isicositrigonal(i))       {          Console.Write("Yes");      }      else     {          Console.Write("No");      }  }  }     // This code is contributed by Nidhi_Biet

Output:

Yes


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