# Program to check if N is a Octagonal Number

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

Octagonal Number is the figure number that represent octagonal. Octagonal Numbers can be formed by placing triangular numbers on the four sides of a square. The first few Octagonal Numbers are 1, 8, 21, 40, 65, 96 …

Examples:

Input: N = 8
Output: Yes
Explanation:
Second octagonal number is 8.

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 octagonal number is given as 2. As we have to check that the given number can be expressed as a Octagonal Number or not. This can be checked as:

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

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach  #include  using namespace std;     // Function to check if N is a  // Octagonal Number  bool isoctagonal(int N)  {      float n          = (2 + sqrt(12 * N + 4))            / 6;         // Condition to check if the      // number is a octagonal number      return (n - (int)n) == 0;  }     // Driver Code  int main()  {      // Given Number      int N = 8;         // Function call      if (isoctagonal(N)) {          cout << "Yes";      }      else {          cout << "No";      }      return 0;  }

## Java

 // Java program for the above approach  import java.io.*;   import java.util.*;      class GFG {          // Function to check if N is a  // octagonal number  public static boolean isoctagonal(int N)  {      double n = (2 + Math.sqrt(12 * N + 4)) / 6;             // Condition to check if the      // number is a octagonal number      return (n - (int)n) == 0;  }         // Driver code   public static void main(String[] args)   {              // Given Number      int N = 8;             // Function call      if (isoctagonal(N))      {          System.out.println("Yes");      }      else     {          System.out.println("No");      }  }   }      // This code is contributed by coder001

## Python3

 # Python3 program for the above approach   from math import sqrt     # Function to check if N is a   # octagonal number   def isoctagonal(N):         n = (2 + sqrt(12 * N + 4)) / 6;          # Condition to check if the       # number is a octagonal number       return (n - int(n)) == 0;          # Driver Code   if __name__ == "__main__":          # Given number       N = 8;          # Function call       if (isoctagonal(N)):          print("Yes");              else:          print("No");      # This code is contributed by AnkitRai01

## C#

 // C# program for the above approach  using System;     class GFG {          // Function to check if N is a  // octagonal number  public static bool isoctagonal(int N)  {      double n = (2 + Math.Sqrt(12 * N +                                4)) / 6;             // Condition to check if the      // number is a octagonal number      return (n - (int)n) == 0;  }         // Driver code   public static void Main(String[] args)   {              // Given number      int N = 8;             // Function call      if (isoctagonal(N))      {          Console.WriteLine("Yes");      }      else     {          Console.WriteLine("No");      }  }   }      // This code is contributed by Rohit_ranjan

Output:

Yes


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