Icositetragonal Number

Given a number N, the task is to find the N^th 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 N^th icositetragonal number is given by the formula: 

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