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 Nth term 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++
// C++ Program to implement// the above approach#include <iostream>using namespace std;// Function to print the// required sequencevoid findNumbers(int n){ int i = 0; while (i <= n) { // Print ith odd number cout << 2 * i * i + 4 * i + 1 + i % 2 << " "; i++; }}// Driver Codeint main(){ int n = 6; findNumbers(n);} |
Java
// 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 codepublic static void main (String[] args){ int n = 6; // Function call findNumbers(n); }}// This code is contributed by offbeat |
Python3
# Python3 program to implement# the above approach # Function to print the# required sequencedef findNumbers(n): i = 0 while (i <= n): # Print ith odd number print(2 * i * i + 4 * i + 1 + i % 2, end = " ") i += 1 # Driver Coden = 6findNumbers(n)# This code is contributed by sanjoy_62 |
C#
// 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 codepublic static void Main (){ int n = 6; // Function call findNumbers(n); }} // This code is contributed by sanjoy_62 |
Javascript
<script>// Javascript Program to implement// the above approach// Function to print the// required sequencefunction findNumbers(n){ var i = 0; while (i <= n) { // Print ith odd number document.write(2 * i * i + 4 * i + 1 + i % 2+" "); i++; }}// Driver Codevar n = 6;findNumbers(n);</script> |
1 8 17 32 49 72 97
Time Complexity: O(N)
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
