n’th Pentagonal Number

Last Updated : 08 May, 2023

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 P_n = 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) // 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++

   // 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) //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

Centered pentagonal number

Mazhar Imam Khan

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