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



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

  • Nth term of s sided polygon = \frac{((s-2)n^2 - (s-4)n)}{2}
  • Therefore Nth term of 28 sided polygon is

    Tn =\frac{((28-2)n^2 - (28-4)n)}{2} =\frac{(26n^2 - 24n)}{2}

Below is the implementation of the above approach:

C++

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// C++ program for above approach
#include <iostream>
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

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C

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// C program for above approach
#include <stdio.h>
#include <stdlib.h>
  
// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

3rd icosikaioctagonal Number is = 81

Reference: https://en.wikipedia.org/wiki/Icosioctagon

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