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 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++

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// 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;
}

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Java

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// 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

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Output:

1, 40755,

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

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Improved By : offbeat