n'th Pentagonal Number

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Given an integer n, find the nth Pentagonal number. The first three pentagonal numbers are 1, 5, and 12 (Please see the below diagram). 
The n'th pentagonal number P_n is the number of distinct dots in a pattern of dots consisting of the outlines of regular pentagons with sides up to n dots when the pentagons are overlaid so that they share one vertex [Source Wiki]
Examples : 

Input: n = 1
Output: 1

Input: n = 2
Output: 5

Input: n = 3
Output: 12

In general, a polygonal number (triangular number, square number, etc) is a number represented as dots or pebbles arranged in the shape of a regular polygon. The first few pentagonal numbers are: 1, 5, 12, etc. 
If s is the number of sides in a polygon, the formula for the nth s-gonal number P (s, n) is 
 

nth s-gonal number P(s, n) = (s - 2)n(n-1)/2 + n

If we put s = 5, we get

n'th Pentagonal number Pn = 3*n*(n-1)/2 + n

Examples: 

Pentagonal Number

Below are the implementations of the above idea in different programming languages.
 

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

// Finding the nth pentagonal number
int pentagonalNum(int n)
{
    return (3 * n * n - n) / 2;
}

// Driver code
int main()
{
    int n = 10;
    
    cout << "10th Pentagonal Number is = " 
         << pentagonalNum(n);

    return 0;
}

// This code is contributed by Code_Mech

// C program for above approach
#include <stdio.h>
#include <stdlib.h>

// Finding the nth Pentagonal Number
int pentagonalNum(int n)
{
    return (3*n*n - n)/2;
}

// Driver program to test above function
int main()
{
    int n = 10;
    printf("10th Pentagonal Number is = %d \n \n", 
                             pentagonalNum(n));

    return 0;
}

// Java program for above approach
class Pentagonal
{
    int pentagonalNum(int n)
    {
        return (3*n*n - n)/2;
    }
}

public class GeeksCode
{
    public static void main(String[] args)
    {
        Pentagonal obj = new Pentagonal();
        int n = 10;    
        System.out.printf("10th petagonal number is = "
                          + obj.pentagonalNum(n));
    }
}

# Python program for finding pentagonal numbers
def pentagonalNum( n ):
    return (3*n*n - n)/2
#Script Begins

n = 10
print ("10th Pentagonal Number is = ", pentagonalNum(n))
 
#Scripts Ends

// C# program for above approach
using System;

class GFG {
    
    static int pentagonalNum(int n)
    {
        return (3 * n * n - n) / 2;
    }

    public static void Main()
    {
        int n = 10; 
        
        Console.WriteLine("10th petagonal"
        + " number is = " + pentagonalNum(n));
    }
}

// This code is contributed by vt_m.

<?php
// PHP program for above approach

// Finding the nth Pentagonal Number
function pentagonalNum($n)
{
    return (3 * $n * $n - $n) / 2;
}

// Driver Code
$n = 10;
echo "10th Pentagonal Number is = ", 
                  pentagonalNum($n);

// This code is contributed by ajit
?>

<script>

// Javascript program for above approach

    function pentagonalNum(n)
    {
        return (3 * n * n - n) / 2;
    }

// Driver code to test above methods

        let n = 10; 
          
        document.write("10th petagonal"
        + " number is = " + pentagonalNum(n));
         
         // This code is contributed by avijitmondal1998.
</script>

10th Pentagonal Number is = 145

Time Complexity: O(1) // since no loop or recursion is used the algorithm takes up constant time to perform the operations
Auxiliary Space: O(1) // since no extra array or data structure is used so the space taken by the algorithm is constant

Another Approach:

The formula indicates that the n-th pentagonal number depends quadratically on n. Therefore, try to find the positive integral root of N = P(n) equation. 
P(n) = nth pentagonal number 
N = Given Number
Solve for n: 
P(n) = N 
or (3*n*n - n)/2 = N 
or 3*n*n - n - 2*N = 0 ... (i)
The positive root of equation (i) 
n = (1 + sqrt(24N+1))/6
After obtaining n, check if it is an integer or not. n is an integer if n - floor(n) is 0.

// C++ Program to check a
// pentagonal number
#include <bits/stdc++.h>
using namespace std;

// Function to determine if
// N is pentagonal or not.
bool isPentagonal(int N)
{    
    // Get positive root of
    // equation P(n) = N.
    float n = (1 + sqrt(24*N + 1))/6;
    
    // Check if n is an integral
    // value of not. To get the
    // floor of n, type cast to int.
    return (n - (int) n) == 0;
}

// Driver Code
int main()
{
    int N = 145;    
    if (isPentagonal(N)) 
        cout << N << " is pentagonal " << endl;    
    else
        cout << N << " is not pentagonal" << endl;    
    return 0;
}

// Java Program to check a
// pentagonal number
import java.util.*;

public class Main {

    // Function to determine if
    // N is pentagonal or not.
    public static boolean isPentagonal(int N)
    {
        // Get positive root of
        // equation P(n) = N.
        float n = (1 + (float)Math.sqrt(24 * N + 1)) / 6;

        // Check if n is an integral
        // value of not. To get the
        // floor of n, type cast to int.
        return (n - (int)n) == 0;
    }

    // Driver Code
    public static void main(String[] args)
    {

        int N = 145;
        if (isPentagonal(N))
            System.out.println(N + " is pentagonal ");
        else
            System.out.println(N + " is not pentagonal");
    }
}

import math

# Function to determine if N is pentagonal or not.
def isPentagonal(N):
    # Get positive root of equation P(n) = N
    n = (1 + math.sqrt(24*N + 1))/6
    
    # Check if n is an integral value or not.
    # To get the floor of n, use the int() function
    return (n - int(n)) == 0

# Driver code
if __name__ == "__main__":
    N = 145
    
    if isPentagonal(N):
        print(N, "is pentagonal")
    else:
        print(N, "is not pentagonal")

using System;

public class Program
{

  // Function to determine if
  // N is pentagonal or not.
  public static bool IsPentagonal(int n)
  {
    // Get positive root of equation P(n) = N.
    float x = (1 + MathF.Sqrt(24 * n + 1)) / 6;

    // Check if x is an integral value or not
    return MathF.Floor(x) == x;
  }

  public static void Main()
  {
    int n = 145;
    if (IsPentagonal(n))
      Console.WriteLine(n + " is pentagonal");
    else
      Console.WriteLine(n + " is not pentagonal");

    // Pause the console so that we can see the output
    Console.ReadLine();
  }
}
// This code is contributed by divyansh2212

// js equivalent 

// import math functions
function isPentagonal(N) {
    // Get positive root of equation P(n) = N
    let n = (1 + Math.sqrt(24 * N + 1)) / 6;
    
    // Check if n is an integral value or not.
    // To get the floor of n, use the Math.floor() function
    return (n - Math.floor(n)) === 0;
}

// Driver code 
let N = 145;

if (isPentagonal(N)) {
    console.log(`${N} is pentagonal`);
} else {
    console.log(`${N} is not pentagonal`);
}

145 is pentagonal 

Time Complexity: O(log n) //the inbuilt sqrt function takes logarithmic time to execute
Auxiliary Space: O(1) // since no extra array or data structure is used so the space taken by the algorithm is constant

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