Find all numbers up to N which are both Pentagonal and Hexagonal

Given an integer N, the task is to find all numbers up to N, which are both Pentagonal as well as Hexagonal.
Example:

Input: N = 1000 
Output: 1
Input: N = 100000 
Output: 1, 40755 

Approach:

• To solve the problem, we are generating all pentagonal numbers up to N and check if those are hexagonal numbers or not.
• Formula to calculate i^th Pentagonal Number: 
  

i * ( 3 * i – 1 ) / 2

• To check if a pentagonal number, say pn, is a hexagonal number or not: 
  

( 1 + sqrt(8 * pn + 1 ) ) / 4 needs to be a Natural number

Below is the implementation of the above approach: 

C++

// C++ Program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to print numbers upto N
// which are both pentagonal as well
// as hexagonal numbers
void pen_hex(long long n)
{
    long long pn = 1;
    for (long long int i = 1;; i++) {

        // Calculate i-th pentagonal number
        pn = i * (3 * i - 1) / 2;

        if (pn > n)
            break;

        // Check if the pentagonal number
        // pn is hexagonal or not
        long double seqNum
            = (1 + sqrt(8 * pn + 1)) / 4;

        if (seqNum == long(seqNum))
            cout << pn << ", ";
    }
}

// Driver Program
int main()
{
    long long int N = 1000000;
    pen_hex(N);
    return 0;
}

Java

// Java program of the above approach
import java.util.*;

class GFG{

// Function to print numbers upto N
// which are both pentagonal as well
// as hexagonal numbers
static void pen_hex(long n)
{
    long pn = 1;
    for(long i = 1; i < n; i++)
    {

        // Calculate i-th pentagonal number
        pn = i * (3 * i - 1) / 2;

        if (pn > n)
            break;

        // Check if the pentagonal number
        // pn is hexagonal or not
        double seqNum = (1 + Math.sqrt(
                        8 * pn + 1)) / 4;

        if (seqNum == (long)seqNum)
            System.out.print(pn + ", ");
    }
}

// Driver code
public static void main(String[] args)
{
    long N = 1000000;
    pen_hex(N);
}
}

// This code is contributed by offbeat

Python3

# Python3 program of the above approach
import math

# Function to print numbers upto N
# which are both pentagonal as well
# as hexagonal numbers
def pen_hex(n):

    pn = 1
    for i in range(1, N):

        # Calculate i-th pentagonal number
        pn = (int)(i * (3 * i - 1) / 2)

        if (pn > n):
            break

        # Check if the pentagonal number
        # pn is hexagonal or not
        seqNum = (1 + math.sqrt(8 * pn + 1)) / 4

        if (seqNum == (int)(seqNum)):
            print(pn, end = ", ")

# Driver Code
N = 1000000

pen_hex(N)

# This code is contributed by divyeshrabadiya07

C#

// C# program of the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to print numbers upto N
// which are both pentagonal as well
// as hexagonal numbers
static void pen_hex(long n)
{
    long pn = 1;
    for(long i = 1;; i++)
    {

        // Calculate i-th pentagonal number
        pn = i * (3 * i - 1) / 2;

        if (pn > n)
            break;

        // Check if the pentagonal number
        // pn is hexagonal or not
        double seqNum = (1 + Math.Sqrt(
                         8 * pn + 1)) / 4;

        if (seqNum == (long)(seqNum))
        {
            Console.Write(pn + ", ");
        }
    }
}

// Driver Code
public static void Main (string[] args)
{
    long N = 1000000;

    pen_hex(N);
}
}

// This code is contributed by rutvik_56

Javascript

<script>
// javascript program of the above approach

// Function to print numbers upto N
// which are both pentagonal as well
// as hexagonal numbers
function pen_hex(n)
{
    var pn = 1;
    for(i = 1; i < n; i++)
    {

        // Calculate i-th pentagonal number
        pn = parseInt(i * (3 * i - 1) / 2);

        if (pn > n)
            break;

        // Check if the pentagonal number
        // pn is hexagonal or not
        var seqNum = (1 + Math.sqrt(
                        8 * pn + 1)) / 4;

        if (seqNum == parseInt(seqNum))
            document.write(pn + ", ");
    }
}

// Driver code
var N = 1000000;
pen_hex(N);

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

Output:

1, 40755,

Time Complexity: O(N) 
Auxiliary space: O(1)