Tetracontagon Number

Given a number N, the task is to find Nth 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 = \frac{((s-2)n^2 - (s-4)n)}{2}
  • Therefore Nth term of 40 sided polygon is

    Tn =\frac{((40-2)n^2 - (40-4)n)}{2} =\frac{(38n^2 - 36)}{2}

Below is the implementation of the above approach:

C++

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// 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 

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C

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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    

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Output:

3rd tetracontagon Number is = 117

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

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