Program to check if N is a Pentagonal Number

Last Updated : 06 Dec, 2023

Given a number (N), check if it is pentagonal or not.

Examples :

Input: 12
Output: Yes
Explanation: 12 is the third pentagonal number

Input: 19
Output: No
Explanation:
The third pentagonal number is 12 while the fourth pentagonal number is 22.
Hence 19 is not a pentagonal number.

Pentagonal numbers are numbers which can be arranged to form a pentagon. If N is a pentagonal number then we can use N dots or points to generate a regular pentagon (Please see figure below).
The first few pentagonal numbers are 1, 5, 12, 22, 35, 51, 70, ...

Method I (Iterative):
We begin by noting that the nth Pentagonal Number is given by
P_n = \frac{3*n^2-n}{2}
Follow an iterative process. Consecutively substitute n = 1, 2, 3 ... into the formula and store the result in some variable M. Stop, if M >= N. After iteration if M equals N then N must be a pentagonal number. Else if M exceeds N then N cannot be a pentagonal number.
Algorithm

function isPentagonal(N) 
    Set i = 1
    do 
        M = (3*i*i - i)/2
        i += 1
    while M < N
    
    if M == N
        print Yes
    else
        print No

Below is the implementation of the algorithm

C++

// C++ program to check
// pentagonal numbers.
#include <iostream>
using namespace std;

// Function to determine
// if N is pentagonal or not.
bool isPentagonal(int N) 
{
    int i = 1, M;
    
    do {

        // Substitute values of i
        // in the formula.
        M = (3*i*i - i)/2;
        i += 1;
    }
    while ( M < N );
    
    return (M == N);
}

// Driver Code
int main()
{
    int N = 12;
    
    if (isPentagonal(N)) 
        cout << N << " is pentagonal " << endl;    
    else
        cout << N << " is not pentagonal" << endl;
    
    return 0;
}

Java

// Java program to check
// pentagonal numbers.
import java.io.*;

class GFG {
    
// Function to determine
// if N is pentagonal or not.
static Boolean isPentagonal(int N) 
{
    int i = 1, M;
     
    do {
 
        // Substitute values of
        // i in the formula.
        M = (3*i*i - i)/2;
        i += 1;
    }
    while ( M < N );
     
    return (M == N);
}
    public static void main (String[] args) {
    int N = 12;
     
    if (isPentagonal(N)) 
        System.out.println( N + " is pentagonal " );    
    else
        System.out.println( N + " is not pentagonal");

    }
}

// This code is contributed by Gitanjali.

Python3

# python3 program to check
# pentagonal numbers.
import math 

# Function to determine if
# N is pentagonal or not.
def isPentagonal( N ) :

    i = 1
    while True:

        # Substitute values of i
        # in the formula.
        M = (3 * i * i - i) / 2
        i += 1
    
        if ( M >= N ):
            break
    
    return (M == N)
    
# Driver method
N = 12
if (isPentagonal(N)):
    print(N , end = ' ')
    print ("is pentagonal " ) 
else:
    print (N , end = ' ')
    print ("is not pentagonal")

# This code is contributed by Gitanjali.

// C# program to check pentagonal numbers.
using System;

class GFG {
    
// Function to determine
// if N is pentagonal or not.
static bool isPentagonal(int N) 
{
    int i = 1, M;
    
    do {

        // Substitute values of
        // i in the formula.
        M = (3 * i * i - i) / 2;
        i += 1;
    }
    while ( M < N );
    
    return (M == N);
}

// Driver Code
public static void Main () 
{
    int N = 12;
    
    if (isPentagonal(N)) 
    Console.Write( N + " is pentagonal " ); 
    else
    Console.Write( N + " is not pentagonal");

}
}

// This code is contributed by vt_m.

JavaScript

<script>
// javascript program to check
// pentagonal numbers.
   
// Function to determine
// if N is pentagonal or not.
function isPentagonal(N) 
{
    var i = 1, M;  
    do 
    {
 
        // Substitute values of
        // i in the formula.
        M = (3 * i * i - i)/2;
        i += 1;
    }
    while ( M < N ); 
        return (M == N);
}

var N = 12;

if (isPentagonal(N)) 
    document.write( N + " is pentagonal " );    
else
    document.write( N + " is not pentagonal");

// This code is contributed by Amit Katiyar 
</script>

PHP

<?php
// PHP program to check
// pentagonal numbers.

// Function to determine
// if N is pentagonal or not.
function isPentagonal(int $N) 
{
    $i = 1; 
    $M;
    
    do {

        // Substitute values of i
        // in the formula.
        $M = (3 * $i * $i - $i) / 2;
        $i += 1;
    }
    while ($M < $N);
    
    return ($M == $N);
}

    // Driver Code
    $N = 12;
    
    if (isPentagonal($N)) 
        echo $N , " is pentagonal " ; 
    else
        echo $N ," is not pentagonal" ;
    
// This code is contributed by anuj_67.
?>

Output

12 is pentagonal

Time Complexity: O(n), since we need to compute successive values of pentagonal numbers up to N.
Auxiliary Space: O(1) because it is using constant space for variables

Method 2 (Efficient):

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.

Implementation of the method is given below :

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 = 19;    
    if (isPentagonal(N)) 
        cout << N << " is pentagonal " << endl;    
    else
        cout << N << " is not pentagonal" << endl;    
    return 0;
}

Java

// Java program to check
// pentagonal numbers.
import java.io.*;

class GFG {
    
// Function to determine if
// N is pentagonal or not.
static Boolean isPentagonal(int N) 
{
        // Get positive root of
    // equation P(n) = N.
    double n = (1 + 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;
}
    public static void main (String[] args) {
    int N = 19;
     
    if (isPentagonal(N)) 
        System.out.println( N + " is pentagonal " );    
    else
        System.out.println( N + " is not pentagonal");

    }
}

// This code is contributed by Gitanjali.

Python3

# Python3 code Program to  
# check a pentagonal number

# Import math library
import math as m

# Function to determine if
# N is pentagonal or not
def isPentagonal( n ):
    
    # Get positive root of
    # equation P(n) = N.
    n = (1 + m.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
N = 19

if (isPentagonal(N)):
    print ( N, " is pentagonal " ) 
else:
    print ( N, " is not pentagonal" ) 

# This code is contributed by 'saloni1297'

// C# program to check pentagonal numbers.
using System;

class GFG {

    // Function to determine if
    // N is pentagonal or not.
    static bool isPentagonal(int N)
    {
        // Get positive root of
        // equation P(n) = N.
        double n = (1 + 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()
    {
        int N = 19;

        if (isPentagonal(N))
            Console.Write(N + " is pentagonal ");
        else
            Console.Write(N + " is not pentagonal");
    }
}

// This code is contributed by vt_m.

JavaScript

<script>
// javascript program to check
// pentagonal numbers.
   
// Function to determine if
// N is pentagonal or not.
function isPentagonal(N) 
{
     // Get positive root of
    // equation P(n) = N.
    var n = (1 + 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 - parseInt( n) == 0);
}

var N = 19;

if (isPentagonal(N)) 
    document.write( N + " is pentagonal " );    
else
    document.write( N + " is not pentagonal");

// This code is contributed by Amit Katiyar 
</script>

PHP

<?php
// PHP Program to check
// a pentagonal number

// Function to determine if
// N is pentagonal or not.
function isPentagonal($N)
{ 
    // Get positive root of
    // equation P(n) = N.
    $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
$N = 19; 
if (isPentagonal($N)) 
    echo $N . " is pentagonal "; 
else
    echo $N . " is not pentagonal";

// This code is contributed by mits.
?>

Output

19 is not pentagonal

Time complexity: O(log N) for given n, as it is using inbuilt sqrt function
Auxiliary Space: O(1)

References :
1) Wikipedia - Pentagonal Numbers
2) Wolfram Alpha - Pentagonal Numbers

Approach#3: Using binary search

This approach checks if a given number is a pentagonal number using binary search. It calculates the maximum value of n for the given number, creates a list of pentagonal numbers up to that limit, and searches for the given number in the list using binary search. If the number is found, it returns "Yes"; otherwise, it returns "No"

Algorithm

1. Use the formula to calculate the maximum possible value of n for a given number.
2. Use binary search to find the position of the given number in the list of pentagonal numbers from 1 to n.
3. If the value at the calculated position is equal to the given number, return "Yes". Otherwise, return "No".

C++

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

// Function to check if a number is a pentagonal number
string isPentagonal(int num) {
    // Calculate the maximum 'n' for pentagonal numbers
    int maxN = static_cast<int>((sqrt(24 * num + 1) + 1) / 6);

    // Generate a list of pentagonal numbers up to maxN
    vector<int> pentagonalList;
    for (int n = 1; n <= maxN; n++) {
        // Calculate the nth pentagonal number using the formula
        int pentagonal = n * (3 * n - 1) / 2;
        pentagonalList.push_back(pentagonal);
    }

    // Binary search to check if 'num' is a pentagonal number
    int left = 0;
    int right = pentagonalList.size() - 1;
    while (left <= right) {
        int mid = (left + right) / 2;
        if (pentagonalList[mid] == num) {
            // 'num' is found in the list, so it's a pentagonal number
            return "Yes, it is a pentagonal number";
        } else if (pentagonalList[mid] < num) {
            // 'num' is larger, so search the right half of the list
            left = mid + 1;
        } else {
            // 'num' is smaller, so search the left half of the list
            right = mid - 1;
        }
    }

    // If 'num' is not found in the list, it's not a pentagonal number
    return "Not a pentagonal number";
}

int main() {
    int num = 19;
    // Call the isPentagonal function and print the result
    cout << isPentagonal(num) << endl;
    return 0;
}

Java

import java.util.ArrayList;

public class PentagonalChecker {

    // Function to check if a number is a pentagonal number
    static String isPentagonal(int num)
    {
        // Calculate the maximum 'n' for pentagonal numbers
        int maxN = (int)((Math.sqrt(24 * num + 1) + 1) / 6);

        // Generate a list of pentagonal numbers up to maxN
        ArrayList<Integer> pentagonalList
            = new ArrayList<>();
        for (int n = 1; n <= maxN; n++) {
            // Calculate the nth pentagonal number using the
            // formula
            int pentagonal = n * (3 * n - 1) / 2;
            pentagonalList.add(pentagonal);
        }

        // Binary search to check if 'num' is a pentagonal
        // number
        int left = 0;
        int right = pentagonalList.size() - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (pentagonalList.get(mid) == num) {
                // 'num' is found in the list, so it's a
                // pentagonal number
                return "Yes, it is a pentagonal number";
            }
            else if (pentagonalList.get(mid) < num) {
                // 'num' is larger, so search the right half
                // of the list
                left = mid + 1;
            }
            else {
                // 'num' is smaller, so search the left half
                // of the list
                right = mid - 1;
            }
        }

        // If 'num' is not found in the list, it's not a
        // pentagonal number
        return "Not a pentagonal number";
    }

    public static void main(String[] args)
    {
        int num = 19;
        // Call the isPentagonal function and print the
        // result
        System.out.println(isPentagonal(num));
    }
}

Python3

import math
def is_pentagonal(num):
    max_n = int((math.sqrt(24*num + 1) + 1) / 6)
    pentagonal_list = [(n * (3*n - 1) // 2) for n in range(1, max_n+1)]
    left = 0
    right = len(pentagonal_list) - 1
    while left <= right:
        mid = (left + right) // 2
        if pentagonal_list[mid] == num:
            return "Yes, it is pentagonal number"
        elif pentagonal_list[mid] < num:
            left = mid + 1
        else:
            right = mid - 1
    return "Not a pentagonal number"
num=19
print(is_pentagonal(num))

using System;
using System.Collections.Generic;

class Program
{
    // Function to check if a number is a pentagonal number
    static string IsPentagonal(int num)
    {
        // Calculate the maximum 'n' for pentagonal numbers
        int maxN = (int)((Math.Sqrt(24 * num + 1) + 1) / 6);

        // Generate a list of pentagonal numbers up to maxN
        List<int> pentagonalList = new List<int>();
        for (int n = 1; n <= maxN; n++)
        {
            // Calculate the nth pentagonal number using the formula
            int pentagonal = n * (3 * n - 1) / 2;
            pentagonalList.Add(pentagonal);
        }

        // Binary search to check if 'num' is a pentagonal number
        int left = 0;
        int right = pentagonalList.Count - 1;
        while (left <= right)
        {
            int mid = (left + right) / 2;
            if (pentagonalList[mid] == num)
            {
                // 'num' is found in the list, so it's a pentagonal number
                return "Yes, it is a pentagonal number";
            }
            else if (pentagonalList[mid] < num)
            {
                // 'num' is larger, so search the right half of the list
                left = mid + 1;
            }
            else
            {
                // 'num' is smaller, so search the left half of the list
                right = mid - 1;
            }
        }

        // If 'num' is not found in the list, it's not a pentagonal number
        return "Not a pentagonal number";
    }

    static void Main()
    {
        int num = 19;
        // Call the IsPentagonal function and print the result
        Console.WriteLine(IsPentagonal(num));
    }
}

JavaScript

// Function to check if a number is a pentagonal number
function isPentagonal(num) {
    // Calculate the maximum 'n' for pentagonal numbers
    const maxN = Math.floor((Math.sqrt(24 * num + 1) + 1) / 6);

    // Check if the number is pentagonal using the inverse formula
    const checkPentagonal = (Math.sqrt(1 + 24 * num) + 1) / 6;
    
    // If the result is an integer and within the valid range, it's pentagonal
    if (Number.isInteger(checkPentagonal) && checkPentagonal <= maxN) {
        return "Yes, it is a pentagonal number";
    } else {
        return "Not a pentagonal number";
    }
}

// Driver code
const num = 19;
// Call the isPentagonal function and print the result
console.log(isPentagonal(num));

Output

Not a pentagonal number

Time complexity: O(log n)
Space complexity: O(sqrt(n))

Sayan Mahapatra

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Approach#3: Using binary search

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