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 = 72Input: 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 Nthterm can be calculated by the following equation:
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> |
Output:
1 8 17 32 49 72 97
Time Complexity: O(N)
Auxiliary Space: O(1)