Icositetragonal Number

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

An Icositetragonal number is a class of figurate number. It has a 24-sided polygon called Icositetragon. The N-th Icositetragonal number count’s the number of dots and all others dots are surrounding with a common sharing corner and make a pattern.

Examples:

Input: N = 2
Output: 24

Input: N = 6
Output: 336



Approach: The Nth icositetragonal number is given by the formula:

Tn = (22n^2 - 20n)/2

Below is the implementation of the above approach:

C++

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// C++ program to find nth
// Icositetragonal number
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find
// Icositetragonal number
int Icositetragonal_num(int n)
{
    // Formula to calculate nth
    // Icositetragonal number
    return (22 * n * n - 20 * n) / 2;
}
  
// Driver Code
int main()
{
    int n = 3;
  
    cout << Icositetragonal_num(n) << endl;
  
    n = 10;
  
    cout << Icositetragonal_num(n);
  
    return 0;
}

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Java

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// Java program to find nth
// icositetragonal number
import java.util.*;
  
class GFG {
      
// Function to find
// icositetragonal number
static int Icositetragonal_num(int n)
{
      
    // Formula to calculate nth
    // icositetragonal number
    return (22 * n * n - 20 * n) / 2;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 3;
    System.out.println(Icositetragonal_num(n));
      
    n = 10;
    System.out.println(Icositetragonal_num(n));
}
}
  
// This code is contributed by offbeat

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Python3

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# Python3 program to find nth 
# Icositetragonal number 
  
# Function to find
# Icositetragonal number
def Icositetragonal_num(n):
      
    # Formula to calculate nth 
    # Icositetragonal number 
    return (22 * n * n - 20 * n) / 2
  
# Driver Code
n = 3
print(int(Icositetragonal_num(n))) 
  
n = 10
print(int(Icositetragonal_num(n))) 
  
# This code is contributed by divyeshrabadiya07

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C#

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// C# program to find nth
// icositetragonal number
using System;
  
class GFG{
      
// Function to find
// icositetragonal number
static int Icositetragonal_num(int n)
{
      
    // Formula to calculate nth
    // icositetragonal number
    return (22 * n * n - 20 * n) / 2;
}
  
// Driver code
public static void Main(string[] args)
{
    int n = 3;
    Console.Write(Icositetragonal_num(n) + "\n");
      
    n = 10;
    Console.Write(Icositetragonal_num(n) + "\n");
}
}
  
// This code is contributed by rutvik_56

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Output:

69
1000

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

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