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Icositetragonal Number

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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++

// 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;
}

                    

Java

// 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

                    

Python3

# 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

                    

C#

// 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

                    

Javascript

<script>
 
// Javascript program to find nth
// icositetragonal number
 
// Function to find
// icositetragonal number
function Icositetragonal_num(n)
{
     
    // Formula to calculate nth
    // icositetragonal number
    return (22 * n * n - 20 * n) / 2;
}
 
// Driver code
let n = 3;
document.write(Icositetragonal_num(n) + "</br>");
 
n = 10;
document.write(Icositetragonal_num(n));
     
// This code is contributed by Ankita saini
    
</script>

                    

Output: 
69
1000

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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



Last Updated : 06 Apr, 2021
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