# Program to check if N is a Icosihexagonal Number

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

Icosihexagonal number is class of figurate number. It has 26 – sided polygon called Icosihexagon. The N-th Icosihexagonal number count’s the 26 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few Icosihexagonol numbers are 1, 26, 75, 148 …

Examples:

Input: N = 26
Output: Yes
Explanation:
Second icosihexagonal number is 26.

Input: 30
Output: No

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

Approach:

1. The Kth term of the icosihexagonal number is given as 2. As we have to check that the given number can be expressed as a icosihexagonal 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 icosihexagonal number.

Below is the implementation of the above approach:

## C++

 // C++ program to check whether  // a number is a icosihexagonal number   // or not     #include     using namespace std;     // Function to check whether the  // number is a icosihexagonal number  bool isicosihexagonal(int N)  {      float n          = (22 + sqrt(192 * N + 484))            / 48;         // Condition to check if the      // number is a icosihexagonal number      return (n - (int)n) == 0;  }     // Driver code  int main()  {      int i = 26;         if (isicosihexagonal(i)) {          cout << "Yes";      }      else {          cout << "No";      }      return 0;  }

## Java

 // Java program to check whether the  // number is a icosihexagonal number  // or not  class GFG{      // Function to check whether the  // number is a icosihexagonal number  static boolean isicosihexagonal(int N)   {       float n = (float) ((22 + Math.sqrt(192 * N +                                         484)) / 48);             // Condition to check if the number       // is a icosihexagonal number       return (n - (int)n) == 0;   }      // Driver Code   public static void main(String[] args)   {              // Given number       int N = 26;              // Function call       if (isicosihexagonal(N))       {           System.out.print("Yes");       }       else     {           System.out.print("No");       }   }   }      // This code is contributed by shubham

## Python3

 # Python3 program to check whether   # a number is a icosihexagonal number   # or not   import numpy as np     # Function to check whether the   # number is a icosihexagonal number   def isicosihexagonal(N):         n = (22 + np.sqrt(192 * N + 484)) / 48        # Condition to check if the       # number is a icosihexagonal number       return (n - (int(n))) == 0    # Driver code   i = 26    if (isicosihexagonal(i)):       print ("Yes")  else:       print ("No")     # This code is contributed by PratikBasu

## C#

 // C# program to check whether the  // number is a icosihexagonal number  // or not  using System;  class GFG{      // Function to check whether the  // number is a icosihexagonal number  static bool isicosihexagonal(int N)  {       float n = (float)((22 + Math.Sqrt(192 * N +                                        484)) / 48);             // Condition to check if the number       // is a icosihexagonal number       return (n - (int)n) == 0;   }      // Driver Code   public static void Main()   {              // Given number       int N = 26;              // Function call       if (isicosihexagonal(N))       {           Console.Write("Yes");       }       else     {           Console.Write("No");       }   }   }      // This code is contributed by Code_Mech

Output:

Yes


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