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Pentadecagonal Number

Given a number N, the task is to find the Nth Pentadecagonal number. 
 

A Pentadecagonal number is a figurate number that extends the concept of triangular and square numbers to the pentadecagon(a 15-sided polygon). The Nth pentadecagonal number counts the number of dots in a pattern of N nested pentadecagons, all sharing a common corner, where the ith tridecagon in the pattern has sides made of ‘i’ dots spaced one unit apart from each other. The first few Pentadecagonal numbers are 1, 15, 42, 82, 135, 201, 280 … 
 




Examples: 
 

Input: N = 2 
Output: 15 
Explanation: 
The second Pentadecagonal number is 15. 
Input: N = 6 
Output: 201 
 




 


 


Approach: The Nth Pentadecagonal number is given by the formula: 
 


Below is the implementation of the above approach:
 

// C++ program to find Nth
// Pentadecagonal number
#include <bits/stdc++.h>
using namespace std;
 
// Function to find N-th
// Pentadecagonal number
int Pentadecagonal_num(int n)
{
    // Formula to calculate nth
    // Pentadecagonal number
    return (13 * n * n - 11 * n) / 2;
}
 
// Driver code
int main()
{
    int n = 3;
    cout << Pentadecagonal_num(n) << endl;
    
    n = 10;
    cout << Pentadecagonal_num(n) << endl;
 
    return 0;
}

                    
// Java program to find Nth
// pentadecagonal number
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to find N-th
// pentadecagonal number
static int Pentadecagonal_num(int n)
{
     
    // Formula to calculate nth
    // Pentadecagonal number
    return (13 * n * n - 11 * n) / 2;
}
     
// Driver code
public static void main(String[] args)
{
    int n = 3;
    System.out.println(Pentadecagonal_num(n));
     
    n = 10;
    System.out.println(Pentadecagonal_num(n));
}
}
 
// This code is contributed by coder001

                    
# Python3 program to find Nth
# pentadecagonal number
 
# Function to find N-th
# pentadecagonal number
def Pentadecagonal_num(n):
     
    # Formula to calculate nth
    # pentadecagonal number
    return (13 * n * n - 11 * n) / 2
 
# Driver code   
n = 3
print(int(Pentadecagonal_num(n)))
 
n = 10
print(int(Pentadecagonal_num(n)))
 
# This code is contributed by divyeshrabadiya07

                    
// C# program to find Nth
// pentadecagonal number
using System;
class GFG{
     
// Function to find N-th
// pentadecagonal number
static int Pentadecagonal_num(int n)
{
     
    // Formula to calculate nth
    // Pentadecagonal number
    return (13 * n * n - 11 * n) / 2;
}
     
// Driver code
public static void Main(string[] args)
{
    int n = 3;
    Console.Write(Pentadecagonal_num(n) + "\n");
     
    n = 10;
    Console.Write(Pentadecagonal_num(n) + "\n");
}
}
 
// This code is contributed by rutvik_56

                    
<script>
 
    // Javascript program to find Nth
    // Pentadecagonal number
     
    // Function to find N-th
    // Pentadecagonal number
    function Pentadecagonal_num(n)
    {
        // Formula to calculate nth
        // Pentadecagonal number
        return (13 * n * n - 11 * n) / 2;
    }
 
    let n = 3;
    document.write(Pentadecagonal_num(n) + "</br>");
      
    n = 10;
    document.write(Pentadecagonal_num(n));
     
</script>

                    

Output: 
42
595

 

Time complexity: O(1) as it is doing constant operations

Auxiliary space: O(1) as it is using constant space for variables

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


 


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