Tetracontagon Number

Given a number N, the task is to find N^th Tetracontagon number. 

A Tetracontagon number is class of figurate number. It has 40 – sided polygon called tetracontagon. The N-th tetracontagon number count’s the 40 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few tetracontagonol numbers are 1, 40, 117, 232 …

Examples:

Input: N = 2 
Output: 40 
Explanation:
The second tetracontagonol number is 40. 
Input: N = 3 
Output: 117 

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

• Nth term of s sided polygon = 
  
• Therefore Nth term of 40 sided polygon is
  

• 
  

Below is the implementation of the above approach: 

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Finding the nth tetracontagon Number
int tetracontagonNum(int n)
{
    return (38 * n * n - 36 * n) / 2;
}

// Driver Code
int main()
{
    int n = 3;
    cout << "3rd tetracontagon Number is = "
         << tetracontagonNum(n);

    return 0;
}

// This code is contributed by Akanksha_Rai

C

// C program for above approach

#include <stdio.h>
#include <stdlib.h>

// Finding the nth tetracontagon Number
int tetracontagonNum(int n)
{
    return (38 * n * n - 36 * n) / 2;
}

// Driver program to test above function
int main()
{
    int n = 3;
    printf("3rd tetracontagon Number is = %d",
           tetracontagonNum(n));

    return 0;
}

Java

// Java program for above approach
import java.util.*;

class GFG {

// Finding the nth tetracontagon number
static int tetracontagonNum(int n)
{
    return (38 * n * n - 36 * n) / 2;
}

// Driver code
public static void main(String[] args)
{
    int n = 3;

    System.out.println("3rd tetracontagon Number is = " +
                                    tetracontagonNum(n));
}
}

// This code is contributed by offbeat

Python3

# Python3 program for above approach

# Finding the nth tetracontagon Number
def tetracontagonNum(n):

    return (38 * n * n - 36 * n) // 2

# Driver Code
n = 3
print("3rd tetracontagon Number is = ",
                   tetracontagonNum(n))

# This code is contributed by divyamohan123

C#

// C# program for above approach
using System;

class GFG {

// Finding the nth tetracontagon number
static int tetracontagonNum(int n)
{
    return (38 * n * n - 36 * n) / 2;
}

// Driver code
public static void Main(string[] args)
{
    int n = 3;

    Console.Write("3rd tetracontagon Number is = " +
                               tetracontagonNum(n));
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// javascript program for above approach

// Finding the nth tetracontagon Number
function tetracontagonNum( n)
{
    return (38 * n * n - 36 * n) / 2;
}

// Driver code
let n = 3;
document.write("3rd tetracontagon Number is " + tetracontagonNum(n));

// This code contributed by gauravrajput1

</script>

Output:

3rd tetracontagon Number is = 117

Time Complexity: O(1)

Auxiliary Space: O(1)

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