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Find all numbers up to N which are both Pentagonal and Hexagonal
  • Last Updated : 05 Apr, 2021

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 ith 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 prvar 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)
 

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