Generate an alternate odd-even sequence having sum of all consecutive pairs as a perfect square

Given an integer N, the task is to print a sequence of length N consisting of alternate odd and even numbers in increasing order such that the sum of any two consecutive terms is a perfect square.

Examples:

Input: N = 4
Output: 1 8 17 32
Explanation:
1 + 8 = 9 = 3²
8 + 17 = 25 = 5²
17 + 32 = 49 = 7²

Input: N = 2
Output: 1 8

Approach: The given problem can be solved based on the observation from the above examples, that for an integer N, sequence will be of the form 1, 8, 17, 32, 49 and so on. Therefore, the N^th term can be calculated by the following equation:

2(n+1)^2 + n\%2   

Therefore, to solve the problem, traverse the range [1, N] to calculate and print every term of the sequence using the above formula.

Below is the implementation of the above approach:

// C++ Program to implement
// the above approach

#include <iostream>
using namespace std;

// Function to print the
// required sequence
void findNumbers(int n)
{
    int i = 0;
    while (i <= n) {

        // Print ith odd number
        cout << 2 * i * i + 4 * i
                    + 1 + i % 2
             << " ";
        i++;
    }
}

// Driver Code
int main()
{
    int n = 6;
    findNumbers(n);
}

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

class GFG{
    
// Function to print the 
// required sequence 
static void findNumbers(int n) 
{ 
    int i = 0; 
    while (i <= n)
    { 

        // Print ith odd number 
        System.out.print(2 * i * i + 4 * i + 
                         1 + i % 2 + " "); 
        i++; 
    } 
} 

// Driver code
public static void main (String[] args)
{
    int n = 6;
    
    // Function call
    findNumbers(n); 
}
}

// This code is contributed by offbeat

# Python3 program to implement
# the above approach
 
# Function to print the
# required sequence
def findNumbers(n):
    
    i = 0
    while (i <= n):
 
        # Print ith odd number
        print(2 * i * i + 4 * i + 
              1 + i % 2, end = " ")
              
        i += 1
    
# Driver Code
n = 6

findNumbers(n)

# This code is contributed by sanjoy_62

// C# program to implement 
// the above approach 
using System;
 
class GFG{
     
// Function to print the 
// required sequence 
static void findNumbers(int n) 
{ 
    int i = 0; 
    while (i <= n)
    { 
        
        // Print ith odd number 
        Console.Write(2 * i * i + 4 * i + 
                      1 + i % 2 + " "); 

        i++; 
    } 
} 
 
// Driver code
public static void Main ()
{
    int n = 6;
     
    // Function call
    findNumbers(n); 
}
}
 
// This code is contributed by sanjoy_62

<script>


// Javascript Program to implement
// the above approach

// Function to print the
// required sequence
function findNumbers(n)
{
    var i = 0;
    while (i <= n) {

        // Print ith odd number
        document.write(2 * i * i + 4 * i + 1 + i % 2+" ");
        i++;
    }
}

// Driver Code
var n = 6;
findNumbers(n);


</script>

1 8 17 32 49 72 97

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