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Pentacontagon number

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Given a number N, the task is to find Nth Pentacontagon number
 

A Pentacontagon number is class of figurate number. It has 50 – sided polygon called pentacontagon. The N-th pentacontagon number count’s the 50 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few pentacontagonol numbers are 1, 50, 147, 292 … 
 


Examples: 
 

Input: N = 2 
Output: 50 
Explanation: 
The second pentacontagonol number is 50. 
Input: N = 3 
Output: 147 
 


 


Approach: The N-th pentacontagon 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 50 sided polygon is
     

Tn =\frac{((50-2)n^2 - (50-4)n)}{2} =\frac{(48n^2 - 46)}{2}


  •  


Below is the implementation of the above approach: 
 

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Finding the nth pentacontagon Number
int pentacontagonNum(int n)
{
    return (48 * n * n - 46 * n) / 2;
}
 
// Driver Code
int main()
{
    int n = 3;
    cout << "3rd pentacontagon Number is = "
         << pentacontagonNum(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 pentacontagon Number
int pentacontagonNum(int n)
{
    return (48 * n * n - 46 * n) / 2;
}
 
// Driver program to test above function
int main()
{
    int n = 3;
    printf("3rd pentacontagon Number is = %d",
           pentacontagonNum(n));
 
    return 0;
}

                    

Java

// Java program for above approach
import java.util.*;
 
class GFG {
 
// Finding the nth pentacontagon number
static int pentacontagonNum(int n)
{
    return (48 * n * n - 46 * n) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
     
    System.out.println("3rd pentacontagon Number is = " +
                                    pentacontagonNum(n));
}
}
 
// This code is contributed by offbeat

                    

Python3

# Python3 program for above approach
 
# Finding the nth pentacontagon Number
def pentacontagonNum(n):
 
    return (48 * n * n - 46 * n) // 2
 
# Driver Code
n = 3
print("3rd pentacontagon Number is = ",
                   pentacontagonNum(n))
 
# This code is contributed by divyamohan123

                    

C#

// C# program for above approach
using System;
 
class GFG {
 
// Finding the nth pentacontagon number
static int pentacontagonNum(int n)
{
    return (48 * n * n - 46 * n) / 2;
}
 
// Driver code
public static void Main(string[] args)
{
    int n = 3;
     
    Console.Write("3rd pentacontagon Number is = " +
                               pentacontagonNum(n));
}
}
 
// This code is contributed by rutvik_56   

                    

Javascript

<script>
 
// javascript program for above approach
 
// Finding the nth pentacontagon Number
function pentacontagonNum( n)
{
    return (48 * n * n - 46 * n) / 2;
}
 
// Driver code
let n = 3;
document.write("3rd pentacontagon Number is " + pentacontagonNum(n));
 
// This code contributed by gauravrajput1
 
</script>

                    

Output: 
3rd pentacontagon Number is = 147

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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


 



Last Updated : 22 Jun, 2021
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