# Icosikaioctagon or Icosioctagon Number

Given a number N, the task is to find Nth Icosioctagon number.

An Icosioctagon number is class of figurate number. It has 28 – sided polygon called icosikaioctagon. The N-th icosikaioctagonal number count’s the 28 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few icosikaioctagonol numbers are 1, 28, 81, 160 …

Examples:

Input: N = 2
Output: 28
Explanation:
The second icosikaioctagonol number is 28.

Input: N = 3
Output: 81

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

Approach: The N-th icosikaioctagonal number is given by the formula:

• Nth term of s sided polygon = • Therefore Nth term of 28 sided polygon is Below is the implementation of the above approach:

## C++

 // C++ program for above approach  #include  using namespace std;     // Finding the nth icosikaioctagonal number  int icosikaioctagonalNum(int n)  {      return (26 * n * n - 24 * n) / 2;  }     // Driver code  int main()  {      int n = 3;         cout << "3rd icosikaioctagonal Number is = "           << icosikaioctagonalNum(n);         return 0;  }     // This code is contributed by shubhamsingh10

## C

 // C program for above approach  #include  #include     // Finding the nth icosikaioctagonal Number  int icosikaioctagonalNum(int n)  {      return (26 * n * n - 24 * n) / 2;  }     // Driver program to test above function  int main()  {      int n = 3;      printf("3rd icosikaioctagonal Number is = %d",             icosikaioctagonalNum(n));         return 0;  }

## Java

 // Java program for above approach   class GFG{         // Finding the nth icosikaioctagonal number   public static int icosikaioctagonalNum(int n)   {       return (26 * n * n - 24 * n) / 2;   }      // Driver code  public static void main(String[] args)  {      int n = 3;      System.out.println("3rd icosikaioctagonal Number is = " +                                       icosikaioctagonalNum(n));  }  }     // This code is contributed by divyeshrabadiya07

## Python3

 # Python3 program for above approach      # Finding the nth icosikaioctagonal Number   def icosikaioctagonalNum(n):          return (26 * n * n - 24 * n) // 2    # Driver Code  n = 3 print("3rd icosikaioctagonal Number is = ",                      icosikaioctagonalNum(n))      # This code is contributed by divyamohan123

## C#

 // C# program for above approach   using System;  class GFG{         // Finding the nth icosikaioctagonal number   public static int icosikaioctagonalNum(int n)   {       return (26 * n * n - 24 * n) / 2;   }      // Driver code  public static void Main()  {      int n = 3;      Console.Write("3rd icosikaioctagonal Number is = " +                                  icosikaioctagonalNum(n));  }  }     // This code is contributed by Code_Mech

Output:

3rd icosikaioctagonal Number is = 81


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