# Program to check if N is a Icosihexagonal Number

• Last Updated : 23 Jun, 2021

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

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 –

=> => 1.
2. 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

## Javascript

 

Output:

Yes

Time Complexity: O(sqrt(n))

Auxiliary Space: O(1)

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