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 = 4Output:1 8 17 32Explanation:

1 + 8 = 9 = 3^{2}

8 + 17 = 25 = 5^{2}

17 + 32 = 49 = 7^{2}

Input:N = 2Output: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:

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 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); ` `} ` |

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## 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 code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` `int` `n = ` `6` `; ` ` ` ` ` `// Function call ` ` ` `findNumbers(n); ` `} ` `} ` ` ` `// This code is contributed by offbeat` |

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**Output:**

1 8 17 32 49 72 97

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

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