# Print first N terms of Lower Wythoff sequence

Given an integer N, the task is to print the first N terms of the Lower Wythoff sequence.

Examples:

Input: N = 5
Output: 1, 3, 4, 6, 8

Input: N = 10
Output: 1, 3, 4, 6, 8, 9, 11, 12, 14, 16

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Lower Wythoff sequence is a sequence whose nth term is a(n) = floor(n * phi) where phi = (1 + sqrt(5)) / 2. So, we run a loop and find the first n terms of the sequence.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print the first n terms ` `// of the lower Wythoff sequence ` `void` `lowerWythoff(``int` `n) ` `{ ` ` `  `    ``// Calculate value of phi ` `    ``double` `phi = (1 + ``sqrt``(5)) / 2.0; ` ` `  `    ``// Find the numbers ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` ` `  `        ``// a(n) = floor(n * phi) ` `        ``double` `ans = ``floor``(i * phi); ` ` `  `        ``// Print the nth numbers ` `        ``cout << ans; ` ` `  `        ``if` `(i != n) ` `            ``cout << ``", "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 5; ` ` `  `    ``lowerWythoff(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Function to print the first n terms ` `// of the lower Wythoff sequence ` `static` `void` `lowerWythoff(``int` `n) ` `{ ` ` `  `    ``// Calculate value of phi ` `    ``double` `phi = (``1` `+ Math.sqrt(``5``)) / ``2.0``; ` ` `  `    ``// Find the numbers ` `    ``for` `(``int` `i = ``1``; i <= n; i++)  ` `    ``{ ` ` `  `        ``// a(n) = floor(n * phi) ` `        ``double` `ans = Math.floor(i * phi); ` ` `  `        ``// Print the nth numbers ` `        ``System.out.print((``int``)ans); ` ` `  `        ``if` `(i != n) ` `            ``System.out.print(``" , "``); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `n = ``5``; ` ` `  `    ``lowerWythoff(n); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# from math import sqrt,floor ` `from` `math ``import` `sqrt, floor ` ` `  `# Function to print the first n terms  ` `# of the lower Wythoff sequence  ` `def` `lowerWythoff(n) :  ` ` `  `    ``# Calculate value of phi  ` `    ``phi ``=` `(``1` `+` `sqrt(``5``)) ``/` `2``;  ` ` `  `    ``# Find the numbers  ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``) : ` ` `  `        ``# a(n) = floor(n * phi)  ` `        ``ans ``=` `floor(i ``*` `phi);  ` ` `  `        ``# Print the nth numbers  ` `        ``print``(ans,end``=``"");  ` ` `  `        ``if` `(i !``=` `n) : ` `            ``print``( ``", "``,end ``=` `"");  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``n ``=` `5``;  ` `    ``lowerWythoff(n); ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to print the first n terms ` `// of the lower Wythoff sequence ` `static` `void` `lowerWythoff(``int` `n) ` `{ ` ` `  `    ``// Calculate value of phi ` `    ``double` `phi = (1 + Math.Sqrt(5)) / 2.0; ` ` `  `    ``// Find the numbers ` `    ``for` `(``int` `i = 1; i <= n; i++)  ` `    ``{ ` ` `  `        ``// a(n) = floor(n * phi) ` `        ``double` `ans = Math.Floor(i * phi); ` ` `  `        ``// Print the nth numbers ` `        ``Console.Write((``int``)ans); ` ` `  `        ``if` `(i != n) ` `            ``Console.Write(``" , "``); ` `    ``} ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` `    ``int` `n = 5; ` ` `  `    ``lowerWythoff(n); ` `} ` `} ` ` `  `// This code is contributed by ajit. `

Output:

```1, 3, 4, 6, 8
```

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