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n’th Pentagonal Number

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  • Difficulty Level : Medium
  • Last Updated : 15 Mar, 2023
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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 Pn 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
Recommended Practice

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

 

Pentagonal Number

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

C++




// 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




// 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




// 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));
    }
}

Python3




# 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#




// 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
// 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
?>

Javascript




<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>

Output

10th Pentagonal Number is = 145

Time Complexity: O(1)
Auxiliary Space: O(1)

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++




// 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




// 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");
    }
}

Python3




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")

C#




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

Javascript




// 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`);
}

Output

145 is pentagonal 

Time Complexity: O(log n)
Auxiliary Space: O(1)

Reference: 
https://en.wikipedia.org/wiki/Polygonal_number
This article is contributed by Mazhar Imam Khan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
 


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